...but programmers are just asked in interviews if they can google/lift code, this crazy problem is for the cooks. You know... to make sure they get portion sizes correct =)
I came up with an alternative solution: To find the side length of the small circle I used the following equations: x^2 + y^2 = 1 (formula for a unit circle) and y = 2x + root(2)/2 because the circle should bisect the square sitting on the line y = root(2)/2 (the height of the large square). Using this method I arrived at the same result.
Most of your puzzles are WAY over my head. A few of the Pythagorean type puzzles are a delight because they boost my ego. Sunday is 'Daisy Day' the one day I can devote to my wife, my daughter and Little Daisy, my grandaughter. Your video comes around that 'my cup runneth over' moment. Thank you. You are a true friend.
A fun follow-up question on this is to look at the subsequent, even smaller squares you get between the circle and the small (and following) square(s). It's actually possible to obtain a recursive formula that describes the side length (or area) of the i-th square constructed this way. d_0 = r*√2 is our starting value, the biggest inner square. I'll leave the way to solve this open as a challenge. If you want to check your final result, here's what I got: d_i = 2/5 * ( √(4*r^2 + 1/4 * d_{i-1}^2) - √(4*r^2 - d_{i-1}^2) ) Or the whole thing squared for the area. Taking the ratio to the previous term if you want to compare ratios. As you'd expect intuitively, the incremental ratios get bigger and bigger (because the circle segment approaches a flat line better and better the further you "zoom in"). A hint to get started, I did this by looking at the coordinates of an outer corner of each following square. There is an obvious recurrence relation between subsequent y-coordinates, d_i determines the x-coordinates in a straightforward way, and the (x|y) couples must satisfy a specific condition so that they lie on the circle.
But your equation for getting the side length of the (i)th square needs the side length of the (i-1)th side length, therefore you need to calculate (i-1)th side length and to get that you need the (i-2) side length and so on. I’ve tried to get the side length of the (i)th square without using previous one, but i failed. So my question is “is there a possible way to derive such equation without the need for the side length of the previous one?”
@@baqirhussein1109 That's what a recursive formula is, yes. When using it you have to calculate the values in order, and you can do that for as many iterative squares as you like. Solving this recurrence relation to obtain a closed formula for the values may be very hard or even impossible here, because this recurrence relation is non-linear. It certainly exceeds my capabilities but it may very well also just be genuinely impossible. I only claimed to get a recursive formula, and I did.
Is it weird that I got it right just by looking at it and thinking, "Yeah, that small square looks like it could fit 2 more on each side, so the small square is 1/25th the size of the big one."? Like, I know the math to do it the right way, but it just seemed so _obvious._
I had a very similar approach, but I think mine was a little bit more elegant. I denoted half the large square side length as "a" and half the small square side length as "b", which also means that radius of the circle is equal to a*sqrt(2). In that same triangle using Pythagorean theorem I got this equation: b^2 + (a+2b)^2 = 2a^2 b^2+2^2+4a*b+4b^2=2a^2 a^2-4ab-5b^2=0 - I divided this equation by b^2 (a/b)^2-4(a/b)-5=0 a/b = 5 or a/b = -1 Ratio of area would be square of side length ratio, which gives me 25.
In my experience, lengths are usually related to irrational numbers like pi or prime roots, but ratios (like this one) are usually integers or a simple rational numbers.
yes, you are right. if it would have been for instance 5,000012 ,nobody would have noticed the difference ... but it is exactly 5. sound weird!...nice Day from Spain .
I have to say that x25 seemed obvious from a visual point of view- the lower side of the small square looked like a fifth of the side of the larger square. The good thing about geometry- circles, squares etc- is that proportion remains the same throughout, regardless of size, so proportional and initial calculations can be made. Square and circle examples are interesting- inscribed or circumscribed- fascinating stuff but one needs to start from basic principles and work forwards........ NOT what I did when I worked backwards in this example! Haha The ongoing thing with any circle problem is good old pi....... because it's irrational and r is the only quantity available. Thanks again Presh for an interesting example.
Thing is, you can look at it and say "That looks like 1/25", but until you do the math, you don't know if it actually is 1/25, or if it's some Lovecraftian horror of a trigonometric formula that's just really close to 1/25...
There's something magical about forms like this having perfect integer solutions. And out of all integers, I wouldn't expect 5 to show up in a question about squares and circles.
treesarecool12345678 its true and if in the interview was printed then folding the large square on the small one will give the answer in less than a minute
I guess 25:1 by eyeballing it too. And in the real world, that would have been good enough. Although, it is nice to know I was right in the fantasy world of perfect math also.
Ok. It’s like this. You cut out a template of the smaller square, then you use it to make lots of small squares. Then you lay the small squares like tiles down onto the larger square. Finally you simply count how many smaller squares it took to completely cover the larger square. My method, while being less mathematical, allows me to demonstrate my scissor skills. Top marks to me:-)
I just took a ruler, measured the small top square side, roughly 4 cm at my screen resolution at full window. Then measured the large square at 20cm - thereby 20/4=5, then 5*5 =25
Hm.. Quick solution, but in other math problems it can be wrong. In this case you can't proof, that correct answer is not some close value, like, for example 2*sqrt(2*pi) ~ 5.013, which easily could occur in such kind of math problems.
I think the easiest way to do it is: observe that the side that the small square shares with the big square can fit 2 times in each side, so this is a total if 5 small square in a side of the the big square, do the maffs, and you can get 25 small squares in the big square
That's an excellent drawing, so I started by estimating the answer, then I went and figured out that you find the smaller square by taking a 60 degree line off the center of a big square edge, then I found that when you solved for the intersection of that line and the circle, you get a nice simple linear ratio of r, and from there it was easy to calculate the real ratio, which was surprisingly exactly the same as the estimate. Wow, what a neat problem!
You can get the same conclusion, by considering the angle of the small pointy triangle, and then determining the sin and cosine of that angle and using the sin**2 + cos**2 = 1 identity,
The question ask for ration, not precise numbers and scale won't matter either as long as the circle is... a circle. You can easily solve this by vision alone
You can also use trigonometry. Consider the triangle formed by: Line from large circle center to midpoint of side common with small square, From there to small square corner on large circle Line returning to middle of large circle. The first and third sides are easily calculated. The third side, can be found using law of sines. The side of the small square can then be found using trig or Pythagorean theorem.
I had a history with this problem. A few years ago, I wanted to find a way to describe the sequence of side lengths of these squares as you added smaller and smaller inscribed squares. The sum of which had to sum to the radius of the circle (minus half the first square side length), so it seemed really interesting. I never figured that out. This problem however, is much easier I'm glad to say! So, set up some variables: r is the radius of the circle, and x is the side length of the smaller square. You can work out very easily that the side length of the larger square is r√2. Draw a line from the center of the circle to the top right corner of the smaller square that touches the circle. Draw another line from the center of the circle to the midpoint of the right edge of the smaller square. There you have a right triangle. The hypotenuse is r, the height is x/2, and the base is r√2/2+x. Now we set up the Pythagorean theorem, (r√2/2+x)²+(x/2)²=r². Now expand and get everything on one side, (5/4)x²+(r√2)x-r²/2=0. Now set up the quadratic formula and solve for x, x=((-r√2)±(3√2/2)r)/(5/2). Since x and r are positive, we replace the ± with + and simplify, x=(√2/5)r. Now find the area of the larger square, (√2r)²=2r², and the area of the smaller square, ((√2/5)r)²=2r²/25. Now find the ratio, 2r²/(2r²/25)=25. There's the final answer. The ratio of the areas of the larger and smaller squares is 25.
The only possible explanation for you not finding the solution, if I had to guess, is that you thought that there was some trick that you had to use, while all you have to use is high school math, so you didn't even try latter. That happens too often.
I looked at it and thought okay so this probably involves the radius...then you draw lines to the corners...then you...! And the second line confused me. Then I didn’t know what to do. I thought okay so if you draw a line to the bottom of the second square...and then I didn’t know how to deal with that little part underneath. Turns out Pythagoras was all you needed
I could tell the smaller square was 1/25th the size of the big one right away, just by inspection. I'm no math whiz, but I do understand the math and really appreciate the confirmation of my intuition given by the narrator.
I eyeballed it and thought... looks to be about 1/5 proportion in its side lengths. Then I did some actual math. It's a bit ironic though, funny and frustrating, that I could "eyeball it" and guess it correctly because this is what many of my students tend to do and I'm always telling them not to.
Here's another way (the description here sounds complicated, but if you draw a diagram it's pretty easy to see): Since we're looking for a ratio, we can just set the size of the larger square to something convenient, so make it a square from (-1,-1) to (1,1). The radius of the circle is then sqrt(2). The parametric equation of the line from the lower left corner of the smaller square to (0,-1) is x=t, y=2t-1. Using the equation of a circle (x^2 + y^2 = r^2), we get t^2 + (2t-1)^2 = 2, which gives 5t^2 - 4t - 1 = 0. The quadratic formula gives solutions t = { -1/5, 1 }. Since x=t, the side of the smaller square is 2/5 and its area is 4/25; the area of the larger square is 4. The ratio is 1:25.
Funny that this channel used the intersecting chords theorem to solve another puzzle, when the same theorem can be applied here and you can solve it very easily even without a paper. If you extend one of the vertical lines of the small square all the way to the top you can arrive at the following ratio: a/b = 2 * (a+b)/(a+b) => a/b = 2 2a+b is the side of the bigger square b is the side of the smaller one, therefore the bigger square has side 5x greater than the smaller one, hence 25x the area of the smaller one.
The Power of a Point theorem, combined with setting the side length of the square equal to 1(which is legitamte, because if isnt one, we can scale everything proportionally) trivializes the problem.
I just used the intersecting chord theorem to solve it, it makes it very trivial, didn't even need a piece of paper for that, did just a few calculations in my head.
Spolierspace I did the same proof, but starting with the half side lengths of the squares being S and s respectively. Then the square of the radius of the circle is 2S^2, and the square of the length of the vertical can be written as either (S+2s)^2 or as 2S^2 - s^2, giving S^2 +4sS + 4s^2 = 2S^2 - s^2, or S^2 - 4sS - 5s^2 = 0 or (S-5s)(S+s) = 0, giving S = 5s and hence the same ratio of 25 for the area. It saves introducing the radius of the circle, or using any square roots - always useful when writing up answers in comments.
This is a high school problem. A college level problem would be to keep adding smaller and smaller squares along the same lines and calculate the infinite sum of their areas. Much more fun -- I hope you will make a video about it.
I was able to solve it as a 10th grader going into calculus AB, but it seemed more difficult than other high school problems... I think it is a challenging high school question designed to find applicants that “think outside the box”
Solved it exactly your way. A bit easier I think if you assume the large square has unit length of 1, and you can go from there, working with real numbers, leading you to the area of the small number and then you just take its reciprocal to get the ratio you want.
I believe it is 25. Here is my work. I defined the radius of the circle as r. I placed the circle on coordinate axis where the center of the circle was at the origin. That means the equation of the circle is x^2 + y^2 = r^2 First, I found the coordinate of the top right corner of the square. I calculated the intercept of y=x and the circle. x^2 + x^2 = r^2 x = r / sqrt 2 That means the area of the big square is 2 times that number, squared. Big square area = (r sqrt 2)^2 = 2r^2 I place my small square at the top of the big square, rather than the bottom, which would not change the ratio due to symmetry. I defined the length of the small square's sides as a. That means that when x = a/2, y = r /sqrt 2 + a Plugging that into the equation for the circle: (a/2)^2 + (r/sqrt 2 + a)^2 = r^2 a^2/4 + r^2/2 + (r sqrt 2)a + a^2 = r^2 5a^2/4 + (r sqrt 2)a - r^2/2 = 0 5a^2 + (4r sqrt 2)a - 2r^2 = 0 Applying the quadratic formula: a = ( - 4r sqrt 2 + - sqrt(32r^2 + 40r^2))/10 I chose the positive solution. a = (sqrt(72r^2) - 4r sqrt 2) / 10 a = (6r sqrt 2 - 4r sqrt 2)/10 a = 2r sqrt 2/10 a = r sqrt 2/5 Since I know the length of the side of the small square, I just square it to get the area of the small square. Small square area = (r sqrt 2/5)^2 = 2r^2/25 The ratio of the 2 squares areas is: 2r^2/(2r^2/25) = 25 Edit: I am amazed I got the same answer, yet my method was very different from the one in the video.
NecroLord I got the same answer too. All I did was estimate how many squares could fit on either side of the square shown and saw that 5 squares fit on the big squares line. So 5*5 = 25. I did it in about 20 seconds.
That is not a scientifically or mathematically sound method though. What if the answer was 4.956412354123^2. It would still look like 25, but it would not be 25. Also, what if the picture was not drawn to scale, let's say it was drawn by hand for example, it is very difficult to draw an accurate image by hand without knowing some of the information we solved for in the first place.
I figured it out using an online graphing calculator. Graphing a circle with radius 1, translate it down so that the top large side of the square is in the x axis. Then draw a horizontal line y=2x, where y is the translated circle equation. The y value of where they intersect is the side of the smaller square. The ratio will end up 24.99 to 25.01.
jimmosio It's not. He just wasn't too clear with his notation. It's actually have the big square side length *plus* the small square side length. (The smaller leg is the small square side length, and the hypotenuse is the radius of the circle or half the diagonal of the large square.)
It's an error, actually. It should be half the big square's side length + the little square's side length, which is what he shows at 2:39. You should probably mention that in the error feed he pinned.
I was so close. I drew the exact same diagram and had everything except realizing that the longer leg of the blue right triangle was the side length of the smaller square plus half the side length of the larger square. Excellent video.
-_- I meant my previous comment in the literal sense. China teaches math at a higher level than Singapore when I was in secondary school. not sure if that has changed.
I got this in my GDR math book from 1981. Former Warsaw pact math books are a gold mine with lots of very interesting math problems (almost half of the problems are in there. There might be someone who actually posts stuff and says it´s new while these are already 40 years old.). This is from an 8th grade math exam. Awarded with 6 points.
I figured this out with a trigonometric system: s/2=L/2*sin(alpha)s/2=r*sin(beta)r/2*cos(beta)-L/2*cos(alpha)=swhere:• alpha is the angle between vertical axis (passing for the center of the circle) and one of the vertexes of the small square closet to the centre of the circle.• beta is the angle between vertical axis and one of the vertexes of the small square touching the circumference of the circle.• L is the side of the big square.from these equation it is possible to find out s and therefore the L^2/s^2 ratio. It is just a bit more complicated than the version in the video but is 100% possible.
You can make the algebra in this solution a bit easier by carefully naming your unknowns. No fractions, no square roots. Preparation: call R radius of the circle X half the side of the big square Y half the side of the small square Part 1: the big square a square inscribed in a circle has diagonal R in the rectangular triangle formed by the centre, a corner and the middle of a side, Pythagoras' formula holds: X² + X² = R², which gives 2X² = R² (1st result) Part 2: the small square We take the narrow rectangular triangle formed by dropping a line from the circle's centre, once to the corner and once to the middle of the small square's lower side. Pythagoras holds again: Y² + (X+2Y)² = R² this gives Y² + X² + 4XY + 4Y² = R², and this gives X² + 4XY + 5Y² = R² (2nd result) Part 3: combining both results by equating the two left parts of the results, because they both equal R² X² + 4XY + 5Y² = 2X² or 5Y² + 4XY - X² = 0. Using the standard solution for equations of second degree, or by educated guesswork about the factorisation, this has solutions X = 5Y or X = -Y. Only the first solution can be real in this context. Half the sides have ratio 5, so the same is true for the full sides. The surfaces have a square ratio: 25.
What do you need numbers for, when the question is about relative measures, as in this case a ratio? Why on earth would one need numbers for solving it?
Jaakko Fagerlund If you had the numbers of the sides of the squares, you could simply calculate their areas separately, then calculate the ratio. or you could do as I did and calculate the ratio of the sides of the squares, then square it.
I found it a fairly straightforward problem. I used essentially the same technique as shown, but I set up Cartesian coordinates and used the equation for the circle and the coordinates for the lower right corner of the small square and ended up getting the same quadratic equation as shown in the video and the same solution. I'm not sure why Presh found this difficult. He's gotten much harder problems than this. It is a nice little problem, though.
Is it? I didn't finish watching the video yet but when I first looked at it, it looked like the side length of the smaller square was exactly 5 times smaller than the larger square.
I had a similar, but a bit different solution: Since the side of the big square equals r*sqrt2 and the side of the small square equals s, we are looking for the proportion r^2*2/s^2. I then used the pythagorean theorem like you, but from 2:46 on, I went a different way. First I multiplied through with (-4), and then divided everything by s^2 to get the following: 2r^2/s^2-4sqrt(2)*r/s-5=0 I then substituted u=sqrt(2)*r/s , so u^2 is the ratio we were searching for: u^2-4u-5=0 Using the p-q-formula: u_1/2=+4/2+-sqrt((4/2)^2-(-5))= =2+-sqrt(4+5) u_1=-1 u_2=5 We‘ll choose u=5, so u^2=25.
I started the way you did, and I got stuck after getting the side of the large square. I thought about the problem for a very long time, and eventually I realized I could solve it algebraically. I put the circle on the coordinate plane centered at the origin (to simplify things, you can even make it a unit circle), so any point on the circle fits the formula x^2+y^2=r^2. The side of the larger square is 2x and x=y. That gave me the x=r*sqrt(1/2) and area of 2r^2 for the larger square. Then solved for the corner of the smaller square by setting 2x=y-r*sqrt(1/2) and plugging it into the circle equation, which got me to the same quadratic formula.
I used a kind of similar approach using trigonometry to solve this: I wanted the circle to be the unit circle, so I set its radius to 1. I mentally flipped the image upside down so that the angle to the small square's corner touching the circle would be in the first quadrant. The height from the center of the circle (and large square) to the top of the large square (and bottom of the small one) is sin(pi/4)=1/sqrt(2). The height from the bottom of the small square to the top of the small square, s = sin(pi/4+a)-1/sqrt(2), where _a_ is the angle to the top-left corner of the small square, from the same corner of the larger square. s can also be expressed by cosine (from the unit circle identities), as s = 2cos(pi/4+a). Let x=sin(pi/4+a). From the first formula for s we get s=x-1/sqrt(2). From the second formula we get s=2sqrt(1-x^2) since sin^2(x)+cos^2(x)=1. Set these equal to each other x-1/sqrt(2) = 2sqrt(1-x^2), square both sides and simplify to x^2-sqrt(2)/5*x-7/10=0. Use the quadratic formula to get x = sqrt(2)/10 +/- 3/5*sqrt(2). Only the positive answer is valid since we cannot have negative side lengths. Now solve for x-1/sqrt(2) to get s = sqrt(2)/5. The area of the small square is A1 = s^2 and the area of the larger is 2 which gives that the small square is 1/25th the area of the large square.
Here is a somewhat easier method, which is a little difficult to describe without a diagram, but I'll try: First draw a 10x10 square grid, then draw a 2x2 grid half way along the bottom edge. Now draw a line connecting the centre of the 10x10 grid with its bottom left corner. This length is sqrt(5^2 + 5^2) = sqrt(50) (in grid units). Finally draw a line connecting the centre of the 10x10 grid with the bottom left corner of the 2x2 grid. The length of this line is then sqrt(7^2 + 1^2), which is also sqrt(50). Since these two lines are exactly the same length, they are both the radius of the circle circumscribing the larger grid, and the bottom corners of the small grid also lie precisely on this circle... which is exactly the figure we are after. From here it's then trivial to see that the ratio of the areas of the two squares is 100:4, or 25:1.
That's a good place to start, but it's not a solution. The answer could conceivably be something like 8pi/5, which is about 5.02, and there is no way your eye is good enough to see the difference. The visual estimate is good for checking if your mathematical solution is correct, though. On my first try, I made a mistake and got an answer of about 1 to 20, which I knew was wrong because visually it was obviously 1 to 25 or very close to it.
You can keep it all in integers... Let the side length of the large square be x, small square be, y, radius of circle be, r, and the perpendicular from the centre of the circle to centre of base of the small square be, z. As in Presh's diagram, this forms a right triangle with hypotenuse, r, and legs, z and y / 2. First, using the Pythagorean theorem in the large square, with the diameter being the hypotenuse, we get, (2r)^2 = x^2 + x^2 => r^2 = x^2 / 2. In the right triangle we formed, z^2 = r^2 - (y/2)^2 = x^2 / 2 - y^2 / 4 => 4z^2 = 2x^2 - y^2. As z = x / 2 + y => z^2 = x^2 / 4 + xy + y^2 => 4z^2 = x^2 + 4xy + 4y^2. Equating both 4z^2's, we get, x^2 - 4xy - 5y^2 = 0 => (x - 5y)(x + y) = 0. As both x and y are positive, x + y != 0, so x = 5y => x / y = 5 => x^2 / y^2 = 25.
Another way of solving it I found before I saw Presh's method. Assuming the circle is a unit circle, modify the standard circle formula (x^2+y^2=1) so that it's displaced along a coordinate plane to the point where the top edge of the small square is lying on the x-axis (i.e., x^2 + (y-(√2÷2))^2 = 1). All we know about the small square so far is that there can be only one way to have each side length equal within the bottom semi-circle. Since the height of the semi-circle is greatest at the y-axis and also mirrors the y-axis, it implies that the height of the small square (side s) is twice the distance between the y-axis and either side of the small square, so we'd want to find a valid solution to |y| = 2|x|, as well as x^2 + (y-(√2÷2))^2 = 1 from before. Solving for y of a given vertex for the small square, after discarding the extraneous solutions at irrelevant sectors of the circle, we find that y = -(√2÷5), so for that vertex, x is either one of ±(√2÷10), confirming that the full doubled/mirrored width is also √2÷5. Now we just square (√2÷5), which gives us an area of 2/25 squared units for the small square. As for the large square, since from the Pythagorean theorem the side length must be √2, the large square's area is 2 square units. Dividing the large square area by the small one: 2 ÷ (2/25) = *25*, which is the ratio of the squared areas of the large square by the small square.
I got it, but with a different line of reasoning, though it naturally produced the same maths. I found half the side length of the big square as you did, but then considered the small square at the top of the big one, and treated the circle as being centred at (0, -big_square_side/2). The x-axis then runs along the top of the big square. The y-axis splits everything down the middle, ie all the geometry is symmetrical about the y-axis. A point (x,y) on the top of the arc then defines a rectangle sitting on top of the big square, inscribed in the circle. This rectangle is the square we want when 2x=y, since that little rectangle is split down the middle by the y-axis. When y=2x, y is the side-length of the small square, so plug x=y/2 into the equation for the circle centred as described above. Solve the quadratic for y in terms of r. It's the same quadratic you had. Proceed to the ratio of areas.
There is a much simpler solution. WLOG that the large squire's side length is 1. Then you can find the small square's side length is 1/5 with the Pythagorean theorem.
matix676 No values are given, and all you're asked for is the ratio of the sizes of the two squares. It's perfectly legitimate to assume a unit circle, or a unit square (either of 'em), or any other dimension you find helpful.
I've roughly estimated that big square is 25 times bigger than small one in about 10 seconds just by comparing side length of small square to big one. You may fit 5 small squares in the length of the big side, which means you can put one level of 5 small squares horizontally and then add another 4 levels on top of it so it will be 5 x 5 = 25 small squares inside the big one. Your math confirmed it ;)
For me a way easier solution: square length = 1 then the “circle radius” = 1/2*√2 then the “small square length” = “circle radius” - “half of square length” 1/2*√2-0.5 So the “ratio of the square areas” = (1)^2 / (1/2*√2-0,5)^2
'the “small square length” = “circle radius” - “half of square length”' Nope, there is a space between the small square border and the circle. Would not be a square if it spouse the border of the circle, does it?
Instead of splitting the quadratic in s into its factors, you could divide the whole equation by s^2 and set r*sqrt(2)/s = x which would give a quadratic in s which is much easier to factorise and the ratio of areas would be x^2
No offence but as a Stanford maths major it is rather shameful that you were not able to do such an elementary geometry problem. Literally just Pythag. Not sure what level of problems you had to in your degree if this 'stumped' you... I am a terrible problem solver but even I could do it very quickly.
The edge of the big square is sqrt(2)*r, where r is the radius of the circle. (This can be shown easy by the pythagorean theorem.) So the area of the big square is A1 = 2r². The are of the small square is x, so we search for x. So the way from the middle of the circle to the corner of the small square is r. Lets consider this length the hypotenuse of a another pythgorean triangle. The vertical cathetus has then the length 0,5*sqrt(2)*r + x. The horizontal cathetus has the length 0,5*x. The pythagorean theorem gives us: (0,5x)² + (0,5sqrt(2)r + x)² = r ² 1/4x² + 0,5r² +sqrt(2)*r*x + x² = r² 5/4x² + sqrt(2)r*x - 0,5r² = 0 x² + 4/5*sqrt(2)r*x - 0,4r² = 0 x = - 2/5*sqrt(2)r +- sqrt(8/25r² + 0,4r²) x = sqrt(0,08) * r So the area of the small square is A2 = 0,08 r² And the ratio: A2 / A1 = 0,08r² / 2 r² = 0,04. :)
Interesting problem. I solved it by assuming larger square side length was 2 and using this equation Y=SQRT(R^2 - x^) but equated it to Y = 2x+1. This yielded the following Quadratic equation 5x^2+4x-1 with one of the results being x = 0.2, representing half a side length of smaller square. To get area ratio of smaller vs larger squares. Ratio = (2x)^2/2^2 = 1/25th or Larger square has 25x more area.
there's a somewhat more elegant geometric solution: construct another square (with the same centre point as the circle) rotated 45deg from the original squares and larger, such that it's sides touch the bottom 2 corners of the small square. you then have a series of 45deg right angled triangles. since they are 45deg triangles, side lengths are all the same so it is easily seen that the side length of the large square is 5 times the side length of the small square (not so elegant when written down but I couldn't be bothered doing a video response!!)
Let x be the side of the smaller square. We can assume w/o loss of generality that the circle has radius sqrt(2). The side of the large square is 2 and its area 4. Draw an auxiliary line from the center of the circle to the lower right hand corner of the small square and another from the center to the midpoint of the base of the smaller square. By Pythagors, 2 = (1 + x)^2 + (x/2)^2 2 = 1 + 2x + x^2 + (x^2)/4 0 = 5x^2 + 8x -4 x = (-8 +- sqrt(64+80))/10 x =( -8 + 12)/10 x = 2/5 area of the small square is 4/25. The ratio of the areas is 25.
I solved with analytic geometry. You impose that the corners of the smaller square must hold the circunference equation. You can do it in terms of the sides of both squares and then solve directly for the ratio.
Alternative (and not that elegant) way to solve this. I arbitrarily chose 1 as a half of the square size. Then I created functions for x and y of every rectangle that would fit in that area of small square. For x, it was easy, just x = 2t. For y, it was trickier. Using pythagoras, we'll get (1+y)^2 + t^2 = 2, so y = sqrt(2-t^2) - 1. Since we are looking for a square, we need rectangles where x = y. So 2t = sqrt(2-t^2) - 1. That leads to a quadratic equation 5t^2 + 4t - 1 = 0. The positive solution is t = 1/5, which means that x = y = 2/5, which when compared to size of the square (2) means it's a fifth of that size, leading to 1/25 of the area.
Was so simple, I got it in 5 minutes. Just extend one of the vertical sides of the small square till the other end of the circle and then connect it to the bottom right vertex of the small square, you will see that the line you just drew is a diameter because diameter subtends 90 deg. Then Pythagoras. Although, the method this guy used also was very similar.
There is a simpler way for the younger ones of us, I'm 15 and this is how I got it. After figuring out the larger square's side lengh, which I did with Pythagorean a^2 + b^2 = c^2 knowing that it has to be a perfect square it ended up as 2r^2 = x^2 + x^2 with x being the side lenght, then you can assume that the center of the larger square is the center point of the circle, so you can make it so that r - x/2 = y, y being the height of the smaller square, which gives us all the tools: then it's just x^2/ y^2 and that's our answer! So just by knowing r's value, you get to know the rest through Finnish 9th grade math and little bit of creativity to realize or remember that you can bend the radius to all those different points around the circle, two points in this scenario. Btw, there's a minor factor with this way as circles tend to be round and square sides are straight lines, but hey, calculating the surface area of a cone has the same minor factor playing in the pi times r times s way, so math just ignores it already.
Wow, I finally solved one of Presh's hard problems without any help! If the circle is a unit circle with radius 1, then the distance from the center of both the square and the circle to the point on the bottom center of the square is Sqrt(1/2) (from a quick application of the Pythagorean Theorem). Then the distance from the bottom center point on the square to the very bottom of the circle would be 1-Sqrt(1/2), keeping the assumption that the circle has radius 1. Keeping the previous two paragraphs in mind for later, we can construct two lines connected from the center of the circle to the bottom center of the smaller, bottom square, and then to the bottom left of the smaller, bottom square. The sine of the angle between the two lines would equal the distance between the bottom center point of the smaller square, and the bottom left point of the smaller square. In other words, it would equal 1/2 the length of a horizontal side of the bottom square. The cosine of the angle between the two lines would equal the distance between the center point of the smaller square and the center of the circle. Now if we keep in mind what I said a few paragraphs ago, we should realize that the length of a vertical side of the bottom square should be cosine(theta)-sqrt(1/2). Since this is a square we're talking about here, we need to have: sine(theta) * 2 = cosine(theta)-sqrt(1/2). So we just need to solve for theta, then solve for the area of the bottom square, then divide that by (2 * sqrt(1/2))^2 (which equals 2), which is the area of the bigger square, to get the ratio. ...darn. I don't have a graphing calculator or mathematical software with me right now, and I spent the last 3 hours trying clever manipulations of trigonometric identities to make this solvable, but I can't do it.
I figured it out with a slightly different method. Basically take the function for a semi circle, y= sqrt( r^2-x^2) subtract half the side length of the large circle, y= sqrt( r^2-x^2) - sqrt(2)/2*r then solve for y=2x After that you get the exact same polynomial.
As it is an interview question, only intuition and visual/graphic estimation can give a quick anticipated answer. Mentally transform the small square upwards and compliment it by 4 other similar squares assuming the drawing is made to scale. Then 5 x 5 little squares would fit in the big square as a result. 1/25 would be then an instantaneous intuitive respond that would impress the interviewer who is seemingly looking for a holistic solution.
It can be easily solved by using coordinate geometry. First we shift the center of circle on y axis such that origin lies on bigger square. Then the edge of smaller square on circle will lie on the line y=2x (if the center of circle is shifted downward) . Solving the line and circle equation we will get x which is half of the side of the smaller square in terms of r .
Another way: Turn this diagram upside down and only consider the upper hemisphere. Look at the point where the small square intersects the circle in the first quadrant. The x component is x=S/2. Visually we can see that the y component is half of the big square side (half of R times square root of 2) plus the small square side S. We also know by the equation for a simple circle about the origin that the y component is equal to the square roots of (R^2 - X^2). Set our two expressions for Y equal to each other and solve for S in terms of R. Finally compare the big square ratio to the small square ration and we get 25.
At first glance, by [00:20] the tendency is to do Pythagorean geometry: the corner point is at a radius, 2nd a small side plus half a large side (radius/√2), and 3rd, half a small side...
1. If we consider the radius of circle as x, we can alsoconsider diagonal of square as 2x. 2.that leaves length of squares as x*root(2) 3.area of this square= square of (x*root(2))=2x*x 4.if we draw a hypothetical square(around the circle) equal to length of 2x,we can find out length of small circle as (2x-x*root(2)))/2 5. area of smaller square =square of(2x-x*root(2)))/2 6.now , just solve it
I started with a unit circle, in which case the bottom-right corner of the large square has coordinates (1/sqrt(2),-1/sqrt(2)). I then reasoned that if x were half the width of the small square, that its bottom-right corner would be at (x, -1/sqrt(2)-2x). Since this corner lies on the unit circle, it satisfies x^2+y^2=1. So we get x^2 + (-1/sqrt(2) - 2x)^2 = 1. If we expand we get the quadratic 5x^2 + 2sqrt(2) x - 1/2 = 0. Solving for the positive value of x with the quadratic formula and simplifying, we get x = sqrt(2)/10. So the area of the large square is (2/sqrt(2))^2=2 and the area of the small square is (2x)^2 = 2/25, one 25th of the area of the large square.
I solved this a different way which was a very simplistic way of solving this problem: Take the smaller square and mark the size with your fingers. Then run that mark against the larger square's side. You'll find the side of the larger square to be 5x the side of the smaller square. Thus, the areas are in a ratio of 25:1. I also solved this problem is with a more complex mathematical method and arrived at the same answer. 🤷♂️
Yeah, I solved it... though not as simply. 1. First I took the circle to be a unit circle, and aligned the squares with the x and y axes to simplify calculations. I also put the small square on the top rather than the bottom. 2. This let me declare that one corner of the big square lands at 45° mark, meaning half the length is √2, so the whole square has area 2. 3. To find the small square area, I shifted the whole circle down by sin(45°). This makes the center of the face where the two squares touch to be the new origin. The shifted circle function is now x² + (y + 1/√2)² = 1 4. Rearranging this gives y = √(1-x²) - 1/√2, but where the corner of the small square touches the circle, y = 2x. 5. Replacing y with 2x in (4) and re-arranging / squaring sides eventually gives the quadratic: 5x² + 2√2x - 1/2 = 0 6. When using the quadratic formula, one result for x gives a point outside the unit circle. The other result is x = (√18 - √8)/10 6b. I plugged in this value for x and y=2x into the equation for step (3) to make sure the equation worked out. The radicals all wind up squaring out to integer values-none vanished by subtraction, which was unexpected. 7. The area of the small square is y²=(2x)² = 2/25. 8. The ratio of the areas of the large square and small square is 2/(2/25) = 25.
Alternative solution: Let radius of circle = 1 Find half the side-length of the big square via pythagorean theorem, call it x: 1=x^2 + x^2 = 2*x^2 x^2 = 1/2 (Then, the area of the big square must be 4*1/2 = 2) x= sqrt(1/2) Draw angle v between the bottom middle of the small square and the square's corner Since radius = 1, cos(v)*1 must be equal to x+y cos(v) = x+y cos(v)-x = y sin(v)*1 must be equal to half the side-length of the small square Thus 2*sin(v) = y Finally, you have: cos(v)-x = 2*sin(v) cos(v)-sqrt(1/2) = 2*sin(v) Solve for v with an advanced calculator (If several solutions, pick smallest positive number) : You get the answer v = 8.13010235° Input v in: y = cos(v)-sqrt(1/2) and you get y = 0.282842712 y^2 = 0.08 Thus, 2/0.08 = 25
another solution(might be simpler): plot the graph x^2 + y^2 = 2. Focus only on the first quadrant. The circle has a radius of root(2), so the side of the square is 1. The total area of the square will be 4*(1 x 1) = 4 For the small square, let's say its side has length 2L. Lets put that square on top of our big square on our graph. We need to find the point it's corner will touch the circle in the first quadrant. Graphically, we're looking for a point on x^2 + y^2 = 2 where: x = L y = 1 + 2L (1+2L)^2 + (L)^2 = 2 ... algebra ... L = 1/5 (positive solution) that means the side length of the small square is 2/5, and its area is 4/25. The ration between 4/25 (area of small square) and 4 (area of large square) is 1:25. Hence, the small square is 25 times smaller
I had an entirely different and possibly simpler method of solving than Presh did. First, I assumed the circle was a unit circle centered at the origin. This eliminates the need to carry the superfluous "r" variable that Presh did, since it's a ratio problem, the r will cancel out anyway. I moved the smaller square to the top of the diagram for ease of use in the Cartesian coordinate system. It's simple enough to find the area and side lengths of the large square- √2 side lengths and area = 2. Beyond this point, my method diverged completely from Presh's. The small square is half the side length of the large square from the center of the circle (distance = √(2)/2. I moved the circle so that the center of the top side of the large square was centered on the origin by rewriting the equation for the circle: f(x)=√(1-x^2) -√(2)/2. The trick to finding the side length of the square is to notice that the point on the square which touches the circle must be at the intersection of the graph of the circle and the line y=2x. So set 2x=√(1-x^2) -√(2)/2 and solve for x. The only positive value of x is 1/(5√2). The side length of the small square is 2x= 2/(5√2) and the area is 4/50. Therefore the ratio of small to big square is 25:1 This method can be easily extended to find the area of the third square and beyond which sit on top of the previous square. Simply add up the side lengths from the previous squares to find the distance from the center of the circle where the new square sits, and make that your new origin for the next calculation.
YAY!!! I got it. I just took advantage of the freedom to label the lengths to simplify the algebra so that I could do it in my head. I labeled the small box length as 2a, and the radius of the circle as sqrt(2) and the length of the large box as 2. This gives the length of the vertical segment a value of 1+2s. The equation is thus: (1+2s)^2 + s^2 = 2 5s^2 + 4s - 1 = 0 (5s-1)(s+1) = 0 the positive value of s is given by 5s-1 yielding s = .2 the ratio of the sides is 2/2s or 2/.4 or 5 thus the ratio of the areas is 25
I used coordinate geometry, and it wasn't too difficult. I solved the problem with the whole figure being flipped vertically so the small square sits atop the large one. Set the radius of the circle as one and the centre at the origin in the standard x-y plane. Then the point of the top right (north-east) corner of the big square inscribed in the circle is the solution of the pair: x^2 + y^2 = 1 and y = x, which gives (1/sqrt(2),1/sqrt(2)), allowing the side of the big square to be calculated by a simple difference as 2/sqrt(2) = sqrt(2). Next, to simplify the further computation, I recentred the axes to the transformed plane X-Y (note the upper case, indicating a transformed coordinate system) so that the circle was shifted vertically downward. The top edge of the large square would be flush with the new X-axis. The new equation of the translated circle is X^2 + (Y+1/sqrt(2))^2 = 1. The small square would be neatly bounded between the arc and the X-axis. The top right (north-east) corner of the small square can be found by solving the equation of the circle with Y = 2X. The resulting quadratic gives only a single positive X value, and the coordinates work out to be (sqrt(2)/10, sqrt(2)/5). Hence the side of the small square is sqrt(2)/5, and the ratio of the sides of the large and small square is 5, giving the ratio of the areas to be 5^2 = 25. Not an elegant method, but it's not too bad.
I always dread the moment these videos do factorization of polynomials because I was never taught this in high school and the results are often absolute magic to me. Couldn't even begin to try and cobble one together, it's like trying a dozen different factors of sums in my head until I stumble across the right one. Especially this factorization is _absolutely baffling_ to me and I really wish the video would have spent more time on it.
I did the same to get the side of the larger square (a = r*sqrt(2)) but to get the smaller square I used the intersecting chords theorem (that I learnt from your channel) to get (5/4)b^2 + (r*sqrt(2))b - r^2/2 = 0 (I used b for the side of the smaller square) which is a quadratic and solved to get b = r*sqrt(2)/5 which is the same as a/5 and squared to get a^2 : a^2/25 which if you divide by a^2 and times by 25 you get 25 : 1
I got it using equation of circle, x^2+y^2=r^2. Taking the centre of the circle as the origin, use the equation to get y when x=b (where little square has side length 2b). Then use the fact that 2b=y-a (where big square has side length 2a).
Aw, I didn't get to do this one with you because I already did this problem a few months ago on a graphic calculator. I also did the reverse, where a circle is inscribed inside a square, and a circle is then made inscribed the corner where it's tangent to both sides of the square it touches as well as the original circle. (The math behind that one didn't end up as cleanly as 25 like this problem had, the ratio is 1/(3-2√2)², and that one took me about a day to figure out, while this one took me an hour or so.) Great video, though!
Not to sound overly pragmatic but I guessed that it looked like 5 small square sides could fit along the big square and assumed the length was 10 and 2 and 4/100 = 1/25 boom.
McDonalds is really focusing on hiring the best of the best
JohnnyB xDD
Well *SOMEONE* has to program the order-taking machines.
...but programmers are just asked in interviews if they can google/lift code, this crazy problem is for the cooks. You know... to make sure they get portion sizes correct =)
I think you mean Wendy's. They're the ones with a square patty in a round bun.
tsgsjeremy Actualy it's pretty easy to program an ordering machine.
I came up with an alternative solution: To find the side length of the small circle I used the following equations: x^2 + y^2 = 1 (formula for a unit circle) and y = 2x + root(2)/2 because the circle should bisect the square sitting on the line y = root(2)/2 (the height of the large square). Using this method I arrived at the same result.
Most of your puzzles are WAY over my head. A few of the Pythagorean type puzzles are a delight because they boost my ego. Sunday is 'Daisy Day' the one day I can devote to my wife, my daughter and Little Daisy, my grandaughter.
Your video comes around that 'my cup runneth over' moment.
Thank you. You are a true friend.
A fun follow-up question on this is to look at the subsequent, even smaller squares you get between the circle and the small (and following) square(s). It's actually possible to obtain a recursive formula that describes the side length (or area) of the i-th square constructed this way. d_0 = r*√2 is our starting value, the biggest inner square. I'll leave the way to solve this open as a challenge. If you want to check your final result, here's what I got:
d_i = 2/5 * ( √(4*r^2 + 1/4 * d_{i-1}^2) - √(4*r^2 - d_{i-1}^2) )
Or the whole thing squared for the area. Taking the ratio to the previous term if you want to compare ratios. As you'd expect intuitively, the incremental ratios get bigger and bigger (because the circle segment approaches a flat line better and better the further you "zoom in").
A hint to get started, I did this by looking at the coordinates of an outer corner of each following square. There is an obvious recurrence relation between subsequent y-coordinates, d_i determines the x-coordinates in a straightforward way, and the (x|y) couples must satisfy a specific condition so that they lie on the circle.
But your equation for getting the side length of the (i)th square needs the side length of the (i-1)th side length, therefore you need to calculate (i-1)th side length and to get that you need the (i-2) side length and so on.
I’ve tried to get the side length of the (i)th square without using previous one, but i failed. So my question is “is there a possible way to derive such equation without the need for the side length of the previous one?”
@@baqirhussein1109 That's what a recursive formula is, yes. When using it you have to calculate the values in order, and you can do that for as many iterative squares as you like.
Solving this recurrence relation to obtain a closed formula for the values may be very hard or even impossible here, because this recurrence relation is non-linear. It certainly exceeds my capabilities but it may very well also just be genuinely impossible. I only claimed to get a recursive formula, and I did.
That feeling when you can't solve it but then you see the solution and realize how simple it is
Is it weird that I got it right just by looking at it and thinking, "Yeah, that small square looks like it could fit 2 more on each side, so the small square is 1/25th the size of the big one."? Like, I know the math to do it the right way, but it just seemed so _obvious._
I had a very similar approach, but I think mine was a little bit more elegant.
I denoted half the large square side length as "a" and half the small square side length as "b", which also means that radius of the circle is equal to a*sqrt(2). In that same triangle using Pythagorean theorem I got this equation:
b^2 + (a+2b)^2 = 2a^2
b^2+2^2+4a*b+4b^2=2a^2
a^2-4ab-5b^2=0 - I divided this equation by b^2
(a/b)^2-4(a/b)-5=0
a/b = 5 or a/b = -1
Ratio of area would be square of side length ratio, which gives me 25.
I honestly thought it would be some irrational number related to pi. It's not often you get nice whole numbers when talking about circles.
In my experience, lengths are usually related to irrational numbers like pi or prime roots, but ratios (like this one) are usually integers or a simple rational numbers.
RE pi is the ratio of a circles circumference to it's diameter
Yes, it is. So?
yes, you are right. if it would have been for instance 5,000012 ,nobody would have noticed the difference ... but it is exactly 5. sound weird!...nice Day from Spain .
In this case it is not talked about circles, but about squares (in a circle) - so do not need pi :-)
I have to say that x25 seemed obvious from a visual point of view- the lower side of the small square looked like a fifth of the side of the larger square. The good thing about geometry- circles, squares etc- is that proportion remains the same throughout, regardless of size, so proportional and initial calculations can be made. Square and circle examples are interesting- inscribed or circumscribed- fascinating stuff but one needs to start from basic principles and work forwards........ NOT what I did when I worked backwards in this example! Haha The ongoing thing with any circle problem is good old pi....... because it's irrational and r is the only quantity available. Thanks again Presh for an interesting example.
Thing is, you can look at it and say "That looks like 1/25", but until you do the math, you don't know if it actually is 1/25, or if it's some Lovecraftian horror of a trigonometric formula that's just really close to 1/25...
There's something magical about forms like this having perfect integer solutions. And out of all integers, I wouldn't expect 5 to show up in a question about squares and circles.
Yesssss I guessed it based on the appearance of the diagram. Full marks for me
treesarecool12345678 its true and if in the interview was printed then folding the large square on the small one will give the answer in less than a minute
treesarecool12345678 same if the interview printed it out it would be too easy
Rofl That's not how you do maths xD Still good effort, 10 points for Gryffondor !
I guess 25:1 by eyeballing it too. And in the real world, that would have been good enough. Although, it is nice to know I was right in the fantasy world of perfect math also.
We are the correct guessing crew
Ok. It’s like this. You cut out a template of the smaller square, then you use it to make lots of small squares. Then you lay the small squares like tiles down onto the larger square. Finally you simply count how many smaller squares it took to completely cover the larger square. My method, while being less mathematical, allows me to demonstrate my scissor skills. Top marks to me:-)
That's the first problem I actually figured out. I'm so happy
congrats !
I eye-balled it and guessed 25, but I like your solution better.
I just took a ruler, measured the small top square side, roughly 4 cm at my screen resolution at full window. Then measured the large square at 20cm - thereby 20/4=5, then 5*5 =25
Silversurfer 'roughly' isn't real maths though =)
Better not ever use Pi for anything than. You might have to roughly.
Hm.. Quick solution, but in other math problems it can be wrong. In this case you can't proof, that correct answer is not some close value, like, for example 2*sqrt(2*pi) ~ 5.013, which easily could occur in such kind of math problems.
Easy
engineer in da hauss
I think the easiest way to do it is: observe that the side that the small square shares with the big square can fit 2 times in each side, so this is a total if 5 small square in a side of the the big square, do the maffs, and you can get 25 small squares in the big square
"That stumped me" wait!? You get the answer to most of the questions on your channel?
That's an excellent drawing, so I started by estimating the answer, then I went and figured out that you find the smaller square by taking a 60 degree line off the center of a big square edge, then I found that when you solved for the intersection of that line and the circle, you get a nice simple linear ratio of r, and from there it was easy to calculate the real ratio, which was surprisingly exactly the same as the estimate. Wow, what a neat problem!
You can get the same conclusion, by considering the angle of the small pointy triangle, and then determining the sin and cosine of that angle and using the sin**2 + cos**2 = 1 identity,
The question ask for ration, not precise numbers and scale won't matter either as long as the circle is... a circle. You can easily solve this by vision alone
Or connect the center of the two squares and the corner of the smaller square, then use the law of cosine.
louisng114 nobody remembers cosines.
But they are easy to remember. And for those who don't, there is always wolfram alpha.
they r easy to remember if you haave done them recently, i did them 4 years ago, and i forgot
Yeah, that is exactly how I did it, although I will admit the solution in the video would have been easier for me.
Lol I remember thinking during high school "I'm never forgetting cosine! they are so useful!"
You can also use trigonometry. Consider the triangle formed by:
Line from large circle center to midpoint of side common with small square,
From there to small square corner on large circle
Line returning to middle of large circle.
The first and third sides are easily calculated. The third side, can be found using law of sines. The side of the small square can then be found using trig or Pythagorean theorem.
Eurgh, set r=1.
Aargh, set r = sqrt(2)
mmmhhh... set r=sqrt(2)/2
check mate
Eurgh, set r=∞
Sergio Korochinsky to the power of pi
I had a history with this problem. A few years ago, I wanted to find a way to describe the sequence of side lengths of these squares as you added smaller and smaller inscribed squares. The sum of which had to sum to the radius of the circle (minus half the first square side length), so it seemed really interesting. I never figured that out. This problem however, is much easier I'm glad to say!
So, set up some variables: r is the radius of the circle, and x is the side length of the smaller square. You can work out very easily that the side length of the larger square is r√2. Draw a line from the center of the circle to the top right corner of the smaller square that touches the circle. Draw another line from the center of the circle to the midpoint of the right edge of the smaller square. There you have a right triangle.
The hypotenuse is r, the height is x/2, and the base is r√2/2+x. Now we set up the Pythagorean theorem, (r√2/2+x)²+(x/2)²=r². Now expand and get everything on one side, (5/4)x²+(r√2)x-r²/2=0. Now set up the quadratic formula and solve for x, x=((-r√2)±(3√2/2)r)/(5/2). Since x and r are positive, we replace the ± with + and simplify, x=(√2/5)r.
Now find the area of the larger square, (√2r)²=2r², and the area of the smaller square, ((√2/5)r)²=2r²/25. Now find the ratio, 2r²/(2r²/25)=25. There's the final answer. The ratio of the areas of the larger and smaller squares is 25.
The only possible explanation for you not finding the solution, if I had to guess, is that you thought that there was some trick that you had to use, while all you have to use is high school math, so you didn't even try latter. That happens too often.
I looked at it and thought okay so this probably involves the radius...then you draw lines to the corners...then you...! And the second line confused me. Then I didn’t know what to do. I thought okay so if you draw a line to the bottom of the second square...and then I didn’t know how to deal with that little part underneath. Turns out Pythagoras was all you needed
I could tell the smaller square was 1/25th the size of the big one right away, just by inspection. I'm no math whiz, but I do understand the math and really appreciate the confirmation of my intuition given by the narrator.
Just use a ruler or guess how many times the length of the larger square to the smaller square is. I still got the right answer.
I eyeballed it and thought... looks to be about 1/5 proportion in its side lengths. Then I did some actual math. It's a bit ironic though, funny and frustrating, that I could "eyeball it" and guess it correctly because this is what many of my students tend to do and I'm always telling them not to.
Here's another way (the description here sounds complicated, but if you draw a diagram it's pretty easy to see):
Since we're looking for a ratio, we can just set the size of the larger square to something convenient, so make it a square from (-1,-1) to (1,1). The radius of the circle is then sqrt(2). The parametric equation of the line from the lower left corner of the smaller square to (0,-1) is x=t, y=2t-1.
Using the equation of a circle (x^2 + y^2 = r^2), we get t^2 + (2t-1)^2 = 2, which gives 5t^2 - 4t - 1 = 0. The quadratic formula gives solutions t = { -1/5, 1 }. Since x=t, the side of the smaller square is 2/5 and its area is 4/25; the area of the larger square is 4. The ratio is 1:25.
Funny that this channel used the intersecting chords theorem to solve another puzzle, when the same theorem can be applied here and you can solve it very easily even without a paper.
If you extend one of the vertical lines of the small square all the way to the top you can arrive at the following ratio:
a/b = 2 * (a+b)/(a+b) => a/b = 2
2a+b is the side of the bigger square
b is the side of the smaller one, therefore the bigger square has side 5x greater than the smaller one, hence 25x the area of the smaller one.
The best solution!
The Power of a Point theorem, combined with setting the side length of the square equal to 1(which is legitamte, because if isnt one, we can scale everything proportionally) trivializes the problem.
That's the way I did it. No quadratic equations to deal with.
I think it is easiest to let the small square have side length 2, and the large square a side length of 2x.
I just used the intersecting chord theorem to solve it, it makes it very trivial, didn't even need a piece of paper for that, did just a few calculations in my head.
Spolierspace
I did the same proof, but starting with the half side lengths of the squares being S and s respectively. Then the square of the radius of the circle is 2S^2, and the square of the length of the vertical can be written as either (S+2s)^2 or as 2S^2 - s^2, giving S^2 +4sS + 4s^2 = 2S^2 - s^2, or S^2 - 4sS - 5s^2 = 0 or (S-5s)(S+s) = 0, giving S = 5s and hence the same ratio of 25 for the area. It saves introducing the radius of the circle, or using any square roots - always useful when writing up answers in comments.
This is a high school problem. A college level problem would be to keep adding smaller and smaller squares along the same lines and calculate the infinite sum of their areas. Much more fun -- I hope you will make a video about it.
Mico Mrkaic this is not a high school problem lol
Oooh! That's a fun problem!!! I'll have to try that out!
I was able to solve it as a 10th grader going into calculus AB, but it seemed more difficult than other high school problems... I think it is a challenging high school question designed to find applicants that “think outside the box”
@@daviddye3191 ... but not outside the circle 😅
Solved it exactly your way. A bit easier I think if you assume the large square has unit length of 1, and you can go from there, working with real numbers, leading you to the area of the small number and then you just take its reciprocal to get the ratio you want.
I believe it is 25.
Here is my work.
I defined the radius of the circle as r. I placed the circle on coordinate axis where the center of the circle was at the origin. That means the equation of the circle is x^2 + y^2 = r^2
First, I found the coordinate of the top right corner of the square. I calculated the intercept of y=x and the circle.
x^2 + x^2 = r^2
x = r / sqrt 2
That means the area of the big square is 2 times that number, squared.
Big square area = (r sqrt 2)^2 = 2r^2
I place my small square at the top of the big square, rather than the bottom, which would not change the ratio due to symmetry.
I defined the length of the small square's sides as a.
That means that when x = a/2, y = r /sqrt 2 + a
Plugging that into the equation for the circle:
(a/2)^2 + (r/sqrt 2 + a)^2 = r^2
a^2/4 + r^2/2 + (r sqrt 2)a + a^2 = r^2
5a^2/4 + (r sqrt 2)a - r^2/2 = 0
5a^2 + (4r sqrt 2)a - 2r^2 = 0
Applying the quadratic formula:
a = ( - 4r sqrt 2 + - sqrt(32r^2 + 40r^2))/10
I chose the positive solution.
a = (sqrt(72r^2) - 4r sqrt 2) / 10
a = (6r sqrt 2 - 4r sqrt 2)/10
a = 2r sqrt 2/10
a = r sqrt 2/5
Since I know the length of the side of the small square, I just square it to get the area of the small square.
Small square area = (r sqrt 2/5)^2 = 2r^2/25
The ratio of the 2 squares areas is:
2r^2/(2r^2/25) = 25
Edit: I am amazed I got the same answer, yet my method was very different from the one in the video.
NecroLord
I got the same answer too. All I did was estimate how many squares could fit on either side of the square shown and saw that 5 squares fit on the big squares line. So 5*5 = 25. I did it in about 20 seconds.
That is not a scientifically or mathematically sound method though.
What if the answer was 4.956412354123^2. It would still look like 25, but it would not be 25.
Also, what if the picture was not drawn to scale, let's say it was drawn by hand for example, it is very difficult to draw an accurate image by hand without knowing some of the information we solved for in the first place.
NecroLord but the ratio it's still the same
I figured it out using an online graphing calculator.
Graphing a circle with radius 1, translate it down so that the top large side of the square is in the x axis. Then draw a horizontal line y=2x, where y is the translated circle equation. The y value of where they intersect is the side of the smaller square. The ratio will end up 24.99 to 25.01.
Can someone explain why is the other leg's length half the big square's length at 2:08? I'm trying to figure it out but I have no clue.
jimmosio it is only from the middle to the square. The whole length would be +s
jimmosio It's not. He just wasn't too clear with his notation. It's actually have the big square side length *plus* the small square side length. (The smaller leg is the small square side length, and the hypotenuse is the radius of the circle or half the diagonal of the large square.)
Oh, that makes sense, thanks.
s = the length of one side of the small square, not the big one
It's an error, actually. It should be half the big square's side length + the little square's side length, which is what he shows at 2:39. You should probably mention that in the error feed he pinned.
I was so close. I drew the exact same diagram and had everything except realizing that the longer leg of the blue right triangle was the side length of the smaller square plus half the side length of the larger square. Excellent video.
I bet this is from a Singapore primary school. 😀😁
Bill Y. does it seems like a problem of primary school.... 😠
in China, you see this in kindergarten.
joshua old, it is capitalist China now. Those "communists" run capitalism better than US and UK.
-_- I meant my previous comment in the literal sense. China teaches math at a higher level than Singapore when I was in secondary school. not sure if that has changed.
I got this in my GDR math book from 1981. Former Warsaw pact math books are a gold mine with lots of very interesting math problems (almost half of the problems are in there. There might be someone who actually posts stuff and says it´s new while these are already 40 years old.). This is from an 8th grade math exam. Awarded with 6 points.
I figured this out with a trigonometric system: s/2=L/2*sin(alpha)s/2=r*sin(beta)r/2*cos(beta)-L/2*cos(alpha)=swhere:• alpha is the angle between vertical axis (passing for the center of the circle) and one of the vertexes of the small square closet to the centre of the circle.• beta is the angle between vertical axis and one of the vertexes of the small square touching the circumference of the circle.• L is the side of the big square.from these equation it is possible to find out s and therefore the L^2/s^2 ratio. It is just a bit more complicated than the version in the video but is 100% possible.
Before watching : (3-2*sqrt(2) )/4
You can make the algebra in this solution a bit easier by carefully naming your unknowns. No fractions, no square roots.
Preparation: call
R radius of the circle
X half the side of the big square
Y half the side of the small square
Part 1: the big square
a square inscribed in a circle has diagonal R
in the rectangular triangle formed by the centre, a corner and the middle of a side, Pythagoras' formula holds:
X² + X² = R²,
which gives 2X² = R² (1st result)
Part 2: the small square
We take the narrow rectangular triangle formed by dropping a line from the circle's centre, once to the corner and once to the middle of the small square's lower side. Pythagoras holds again:
Y² + (X+2Y)² = R²
this gives
Y² + X² + 4XY + 4Y² = R²,
and this gives
X² + 4XY + 5Y² = R² (2nd result)
Part 3: combining both results
by equating the two left parts of the results, because they both equal R²
X² + 4XY + 5Y² = 2X²
or 5Y² + 4XY - X² = 0.
Using the standard solution for equations of second degree, or by educated guesswork about the factorisation,
this has solutions X = 5Y or X = -Y. Only the first solution can be real in this context.
Half the sides have ratio 5, so the same is true for the full sides.
The surfaces have a square ratio: 25.
Why couldn't you figure it out?
where are the numbers
What do you need numbers for, when the question is about relative measures, as in this case a ratio? Why on earth would one need numbers for solving it?
Jaakko Fagerlund If you had the numbers of the sides of the squares, you could simply calculate their areas separately, then calculate the ratio. or you could do as I did and calculate the ratio of the sides of the squares, then square it.
I found it a fairly straightforward problem. I used essentially the same technique as shown, but I set up Cartesian coordinates and used the equation for the circle and the coordinates for the lower right corner of the small square and ended up getting the same quadratic equation as shown in the video and the same solution. I'm not sure why Presh found this difficult. He's gotten much harder problems than this. It is a nice little problem, though.
I somehow guessed 25:1 as the answer. I thought that it would be 5^2, but total luck.
Is it? I didn't finish watching the video yet but when I first looked at it, it looked like the side length of the smaller square was exactly 5 times smaller than the larger square.
I had a similar, but a bit different solution:
Since the side of the big square equals r*sqrt2 and the side of the small square equals s, we are looking for the proportion r^2*2/s^2.
I then used the pythagorean theorem like you, but from 2:46 on, I went a different way. First I multiplied through with (-4), and then divided everything by s^2 to get the following:
2r^2/s^2-4sqrt(2)*r/s-5=0
I then substituted u=sqrt(2)*r/s , so u^2 is the ratio we were searching for:
u^2-4u-5=0
Using the p-q-formula:
u_1/2=+4/2+-sqrt((4/2)^2-(-5))=
=2+-sqrt(4+5)
u_1=-1
u_2=5
We‘ll choose u=5, so u^2=25.
2:19 it is not (r√2 / 2) + s?
I started the way you did, and I got stuck after getting the side of the large square. I thought about the problem for a very long time, and eventually I realized I could solve it algebraically. I put the circle on the coordinate plane centered at the origin (to simplify things, you can even make it a unit circle), so any point on the circle fits the formula x^2+y^2=r^2. The side of the larger square is 2x and x=y. That gave me the x=r*sqrt(1/2) and area of 2r^2 for the larger square. Then solved for the corner of the smaller square by setting 2x=y-r*sqrt(1/2) and plugging it into the circle equation, which got me to the same quadratic formula.
What job includes this as a part of the interview 😱??
Unpaid internship, of course.
McDonald's. This theorem is used when placing a square of cheese on a burger.
@@Arcsecant underrated comment
I literally figured it out just by looking at it.
I didn't know how to prove it but I guesst it right.
Ok isn't that hard to see.
AUT Reini same happened to me.
I guessed 24 to 1 ..
pretty close for a first glance.
I used a kind of similar approach using trigonometry to solve this:
I wanted the circle to be the unit circle, so I set its radius to 1.
I mentally flipped the image upside down so that the angle to the small square's corner touching the circle would be in the first quadrant.
The height from the center of the circle (and large square) to the top of the large square (and bottom of the small one) is sin(pi/4)=1/sqrt(2).
The height from the bottom of the small square to the top of the small square, s = sin(pi/4+a)-1/sqrt(2), where _a_ is the angle to the top-left corner of the small square, from the same corner of the larger square.
s can also be expressed by cosine (from the unit circle identities), as s = 2cos(pi/4+a).
Let x=sin(pi/4+a).
From the first formula for s we get s=x-1/sqrt(2).
From the second formula we get s=2sqrt(1-x^2) since sin^2(x)+cos^2(x)=1.
Set these equal to each other x-1/sqrt(2) = 2sqrt(1-x^2), square both sides and simplify to x^2-sqrt(2)/5*x-7/10=0.
Use the quadratic formula to get x = sqrt(2)/10 +/- 3/5*sqrt(2). Only the positive answer is valid since we cannot have negative side lengths.
Now solve for x-1/sqrt(2) to get s = sqrt(2)/5.
The area of the small square is A1 = s^2 and the area of the larger is 2 which gives that the small square is 1/25th the area of the large square.
My initial guess from the visual was a 1 to 16 ratio, do I was wrong
How the fuck did you fit 16 small cubes on a side xD
it would be 4x4 actually Furrane... He was off by one, it is 5x5 not 4x4.
Yes, do you was wrong.
Here is a somewhat easier method, which is a little difficult to describe without a diagram, but I'll try: First draw a 10x10 square grid, then draw a 2x2 grid half way along the bottom edge. Now draw a line connecting the centre of the 10x10 grid with its bottom left corner. This length is sqrt(5^2 + 5^2) = sqrt(50) (in grid units). Finally draw a line connecting the centre of the 10x10 grid with the bottom left corner of the 2x2 grid. The length of this line is then sqrt(7^2 + 1^2), which is also sqrt(50). Since these two lines are exactly the same length, they are both the radius of the circle circumscribing the larger grid, and the bottom corners of the small grid also lie precisely on this circle... which is exactly the figure we are after. From here it's then trivial to see that the ratio of the areas of the two squares is 100:4, or 25:1.
Is it cheating that eyeballing it seemed like the little square was the middle of 5 across the bottom, and that gives 1/25 the area??
That's a good place to start, but it's not a solution. The answer could conceivably be something like 8pi/5, which is about 5.02, and there is no way your eye is good enough to see the difference. The visual estimate is good for checking if your mathematical solution is correct, though. On my first try, I made a mistake and got an answer of about 1 to 20, which I knew was wrong because visually it was obviously 1 to 25 or very close to it.
Eyeballing is just for you to know the approximate answer but you need solution to prove your answer exactly
You can keep it all in integers... Let the side length of the large square be x, small square be, y, radius of circle be, r, and the perpendicular from the centre of the circle to centre of base of the small square be, z. As in Presh's diagram, this forms a right triangle with hypotenuse, r, and legs, z and y / 2.
First, using the Pythagorean theorem in the large square, with the diameter being the hypotenuse, we get,
(2r)^2 = x^2 + x^2 => r^2 = x^2 / 2.
In the right triangle we formed, z^2 = r^2 - (y/2)^2 = x^2 / 2 - y^2 / 4 => 4z^2 = 2x^2 - y^2.
As z = x / 2 + y => z^2 = x^2 / 4 + xy + y^2 => 4z^2 = x^2 + 4xy + 4y^2.
Equating both 4z^2's, we get, x^2 - 4xy - 5y^2 = 0 => (x - 5y)(x + y) = 0.
As both x and y are positive, x + y != 0, so x = 5y => x / y = 5 => x^2 / y^2 = 25.
The ratio seems in the picture 😂
Another way of solving it I found before I saw Presh's method.
Assuming the circle is a unit circle, modify the standard circle formula (x^2+y^2=1) so that it's displaced along a coordinate plane to the point where the top edge of the small square is lying on the x-axis (i.e., x^2 + (y-(√2÷2))^2 = 1).
All we know about the small square so far is that there can be only one way to have each side length equal within the bottom semi-circle. Since the height of the semi-circle is greatest at the y-axis and also mirrors the y-axis, it implies that the height of the small square (side s) is twice the distance between the y-axis and either side of the small square, so we'd want to find a valid solution to |y| = 2|x|, as well as x^2 + (y-(√2÷2))^2 = 1 from before.
Solving for y of a given vertex for the small square, after discarding the extraneous solutions at irrelevant sectors of the circle, we find that y = -(√2÷5), so for that vertex, x is either one of ±(√2÷10), confirming that the full doubled/mirrored width is also √2÷5.
Now we just square (√2÷5), which gives us an area of 2/25 squared units for the small square.
As for the large square, since from the Pythagorean theorem the side length must be √2, the large square's area is 2 square units.
Dividing the large square area by the small one: 2 ÷ (2/25) = *25*, which is the ratio of the squared areas of the large square by the small square.
This was an interview question? For what?😕
JScal5 McDonalds cashier
For a person who lays tiles on the floor in a round building ;)
Kindergarten XD
mathematician, maybe?
TheGinginator14 Not cashier the cook who bakes circular buns for McDonald's 🤣🤣😂
I got it, but with a different line of reasoning, though it naturally produced the same maths.
I found half the side length of the big square as you did, but then considered the small square at the top of the big one, and treated the circle as being centred at (0, -big_square_side/2). The x-axis then runs along the top of the big square. The y-axis splits everything down the middle, ie all the geometry is symmetrical about the y-axis.
A point (x,y) on the top of the arc then defines a rectangle sitting on top of the big square, inscribed in the circle. This rectangle is the square we want when 2x=y, since that little rectangle is split down the middle by the y-axis.
When y=2x, y is the side-length of the small square, so plug x=y/2 into the equation for the circle centred as described above. Solve the quadratic for y in terms of r. It's the same quadratic you had. Proceed to the ratio of areas.
There is a much simpler solution. WLOG that the large squire's side length is 1. Then you can find the small square's side length is 1/5 with the Pythagorean theorem.
Can you elaborate on this one, it sounds promissing but I can't figure it out ? Use mathjax (latex) syntax if needed, I'll process it on my side.
Only r is given. You can't just assume something is 1.
matix676 yes if you are looking for ratios. It doesen't depends on the value of R. I found convenient to set R=1/squrt(2) so I can simplify equations.
matix676 No values are given, and all you're asked for is the ratio of the sizes of the two squares. It's perfectly legitimate to assume a unit circle, or a unit square (either of 'em), or any other dimension you find helpful.
+Brennan the Nuker Other than simplifying by assuming side length 1, how is that different from the solution in the video?
I've roughly estimated that big square is 25 times bigger than small one in about 10 seconds just by comparing side length of small square to big one. You may fit 5 small squares in the length of the big side, which means you can put one level of 5 small squares horizontally and then add another 4 levels on top of it so it will be 5 x 5 = 25 small squares inside the big one. Your math confirmed it ;)
For me a way easier solution: square length = 1 then the “circle radius” = 1/2*√2 then the “small square length” = “circle radius” - “half of square length” 1/2*√2-0.5 So the “ratio of the square areas” = (1)^2 / (1/2*√2-0,5)^2
+echtfred Your answer is wrong. (1)^2 / (1/2*√2-0,5)^2 =/= 25
'the “small square length” = “circle radius” - “half of square length”'
Nope, there is a space between the small square border and the circle. Would not be a square if it spouse the border of the circle, does it?
Instead of splitting the quadratic in s into its factors, you could divide the whole equation by s^2 and set r*sqrt(2)/s = x which would give a quadratic in s which is much easier to factorise and the ratio of areas would be x^2
No offence but as a Stanford maths major it is rather shameful that you were not able to do such an elementary geometry problem. Literally just Pythag. Not sure what level of problems you had to in your degree if this 'stumped' you...
I am a terrible problem solver but even I could do it very quickly.
lmao youre such a bitch
The edge of the big square is sqrt(2)*r, where r is the radius of the circle. (This can be shown easy by the pythagorean theorem.)
So the area of the big square is A1 = 2r².
The are of the small square is x, so we search for x.
So the way from the middle of the circle to the corner of the small square is r. Lets consider this length the hypotenuse of a another pythgorean triangle.
The vertical cathetus has then the length 0,5*sqrt(2)*r + x.
The horizontal cathetus has the length 0,5*x.
The pythagorean theorem gives us:
(0,5x)² + (0,5sqrt(2)r + x)² = r ²
1/4x² + 0,5r² +sqrt(2)*r*x + x² = r²
5/4x² + sqrt(2)r*x - 0,5r² = 0
x² + 4/5*sqrt(2)r*x - 0,4r² = 0
x = - 2/5*sqrt(2)r +- sqrt(8/25r² + 0,4r²)
x = sqrt(0,08) * r
So the area of the small square is A2 = 0,08 r²
And the ratio: A2 / A1 = 0,08r² / 2 r² = 0,04.
:)
Now this was a nice riddle, none of the other rubbish that's posted here.
MrJason005 its not rubbish thats posted here (unless you mean yt as a whole, but you said it like you just meant this channel)
I like integral puzzles.
I don't like the "I found this on facebook, only geniuses can solve it!11!!!!!11!!!"
Interesting problem. I solved it by assuming larger square side length was 2 and using this equation Y=SQRT(R^2 - x^) but equated it to Y = 2x+1. This yielded the following Quadratic equation 5x^2+4x-1 with one of the results being x = 0.2, representing half a side length of smaller square. To get area ratio of smaller vs larger squares. Ratio = (2x)^2/2^2 = 1/25th or Larger square has 25x more area.
No "first" comments?
nope but 180th
Only yours.
630th, and 63rd "Like".
we had that in the 90ties...
This comment's number is imaginary: imagine it was first :q
there's a somewhat more elegant geometric solution: construct another square (with the same centre point as the circle) rotated 45deg from the original squares and larger, such that it's sides touch the bottom 2 corners of the small square. you then have a series of 45deg right angled triangles. since they are 45deg triangles, side lengths are all the same so it is easily seen that the side length of the large square is 5 times the side length of the small square (not so elegant when written down but I couldn't be bothered doing a video response!!)
Why the fuck does he say" hey this is PreshTallWalker" when his channel name is MindYourDecisions? It fucking drives me mad
PseudoQuelconque it's his actual name, he has written multiple books so it is to increase creditability of the channel
Any relation to Anakin Tallwallker?
*Presh Talwalkar
Let x be the side of the smaller square. We can assume w/o loss of generality that the circle has radius sqrt(2). The side of the large square is 2 and its area 4.
Draw an auxiliary line from the center of the circle to the lower right hand corner of the small square and another from the center to the midpoint of the base of the smaller square.
By Pythagors, 2 = (1 + x)^2 + (x/2)^2
2 = 1 + 2x + x^2 + (x^2)/4
0 = 5x^2 + 8x -4
x = (-8 +- sqrt(64+80))/10
x =( -8 + 12)/10
x = 2/5
area of the small square is 4/25.
The ratio of the areas is 25.
I solved with analytic geometry. You impose that the corners of the smaller square must hold the circunference equation. You can do it in terms of the sides of both squares and then solve directly for the ratio.
Alternative (and not that elegant) way to solve this. I arbitrarily chose 1 as a half of the square size. Then I created functions for x and y of every rectangle that would fit in that area of small square. For x, it was easy, just x = 2t. For y, it was trickier. Using pythagoras, we'll get (1+y)^2 + t^2 = 2, so y = sqrt(2-t^2) - 1. Since we are looking for a square, we need rectangles where x = y. So 2t = sqrt(2-t^2) - 1. That leads to a quadratic equation 5t^2 + 4t - 1 = 0. The positive solution is t = 1/5, which means that x = y = 2/5, which when compared to size of the square (2) means it's a fifth of that size, leading to 1/25 of the area.
Was so simple, I got it in 5 minutes. Just extend one of the vertical sides of the small square till the other end of the circle and then connect it to the bottom right vertex of the small square, you will see that the line you just drew is a diameter because diameter subtends 90 deg. Then Pythagoras. Although, the method this guy used also was very similar.
There is a simpler way for the younger ones of us, I'm 15 and this is how I got it. After figuring out the larger square's side lengh, which I did with Pythagorean a^2 + b^2 = c^2 knowing that it has to be a perfect square it ended up as 2r^2 = x^2 + x^2 with x being the side lenght, then you can assume that the center of the larger square is the center point of the circle, so you can make it so that r - x/2 = y, y being the height of the smaller square, which gives us all the tools: then it's just x^2/ y^2 and that's our answer! So just by knowing r's value, you get to know the rest through Finnish 9th grade math and little bit of creativity to realize or remember that you can bend the radius to all those different points around the circle, two points in this scenario. Btw, there's a minor factor with this way as circles tend to be round and square sides are straight lines, but hey, calculating the surface area of a cone has the same minor factor playing in the pi times r times s way, so math just ignores it already.
Wow, I finally solved one of Presh's hard problems without any help!
If the circle is a unit circle with radius 1, then the distance from the center of both the square and the circle to the point on the bottom center of the square is Sqrt(1/2) (from a quick application of the Pythagorean Theorem).
Then the distance from the bottom center point on the square to the very bottom of the circle would be 1-Sqrt(1/2), keeping the assumption that the circle has radius 1.
Keeping the previous two paragraphs in mind for later, we can construct two lines connected from the center of the circle to the bottom center of the smaller, bottom square, and then to the bottom left of the smaller, bottom square.
The sine of the angle between the two lines would equal the distance between the bottom center point of the smaller square, and the bottom left point of the smaller square. In other words, it would equal 1/2 the length of a horizontal side of the bottom square.
The cosine of the angle between the two lines would equal the distance between the center point of the smaller square and the center of the circle. Now if we keep in mind what I said a few paragraphs ago, we should realize that the length of a vertical side of the bottom square should be cosine(theta)-sqrt(1/2).
Since this is a square we're talking about here, we need to have: sine(theta) * 2 = cosine(theta)-sqrt(1/2). So we just need to solve for theta, then solve for the area of the bottom square, then divide that by (2 * sqrt(1/2))^2 (which equals 2), which is the area of the bigger square, to get the ratio.
...darn. I don't have a graphing calculator or mathematical software with me right now, and I spent the last 3 hours trying clever manipulations of trigonometric identities to make this solvable, but I can't do it.
I figured it out with a slightly different method.
Basically take the function for a semi circle, y= sqrt( r^2-x^2)
subtract half the side length of the large circle, y= sqrt( r^2-x^2) - sqrt(2)/2*r
then solve for y=2x
After that you get the exact same polynomial.
Thats what I did. Pretty straight forward and nothing fancy involved. 2 minutes if you make no mistakes :)
As it is an interview question, only intuition and visual/graphic estimation can give a quick anticipated answer. Mentally transform the small square upwards and compliment it by 4 other similar squares assuming the drawing is made to scale. Then 5 x 5 little squares would fit in the big square as a result. 1/25 would be then an instantaneous intuitive respond that would impress the interviewer who is seemingly looking for a holistic solution.
Saw it instantly but enjoyed proving it
It can be easily solved by using coordinate geometry. First we shift the center of circle on y axis such that origin lies on bigger square. Then the edge of smaller square on circle will lie on the line y=2x (if the center of circle is shifted downward) . Solving the line and circle equation we will get x which is half of the side of the smaller square in terms of r .
Another way: Turn this diagram upside down and only consider the upper hemisphere. Look at the point where the small square intersects the circle in the first quadrant. The x component is x=S/2. Visually we can see that the y component is half of the big square side (half of R times square root of 2) plus the small square side S. We also know by the equation for a simple circle about the origin that the y component is equal to the square roots of (R^2 - X^2). Set our two expressions for Y equal to each other and solve for S in terms of R. Finally compare the big square ratio to the small square ration and we get 25.
At first glance, by [00:20] the tendency is to do Pythagorean geometry: the corner point is at a radius, 2nd a small side plus half a large side (radius/√2), and 3rd, half a small side...
1. If we consider the radius of circle as x, we can alsoconsider diagonal of square as 2x.
2.that leaves length of squares as x*root(2)
3.area of this square= square of (x*root(2))=2x*x
4.if we draw a hypothetical square(around the circle) equal to length of 2x,we can find out length of small circle as (2x-x*root(2)))/2
5. area of smaller square =square of(2x-x*root(2)))/2
6.now , just solve it
I started with a unit circle, in which case the bottom-right corner of the large square has coordinates (1/sqrt(2),-1/sqrt(2)). I then reasoned that if x were half the width of the small square, that its bottom-right corner would be at (x, -1/sqrt(2)-2x). Since this corner lies on the unit circle, it satisfies x^2+y^2=1. So we get x^2 + (-1/sqrt(2) - 2x)^2 = 1. If we expand we get the quadratic 5x^2 + 2sqrt(2) x - 1/2 = 0. Solving for the positive value of x with the quadratic formula and simplifying, we get x = sqrt(2)/10. So the area of the large square is (2/sqrt(2))^2=2 and the area of the small square is (2x)^2 = 2/25, one 25th of the area of the large square.
There saying they were able to eyeball it. I get measuring, but how can you see something as 25x bigger?
I solved this a different way which was a very simplistic way of solving this problem: Take the smaller square and mark the size with your fingers. Then run that mark against the larger square's side. You'll find the side of the larger square to be 5x the side of the smaller square. Thus, the areas are in a ratio of 25:1. I also solved this problem is with a more complex mathematical method and arrived at the same answer. 🤷♂️
A very elegant solution!
Yeah, I solved it... though not as simply.
1. First I took the circle to be a unit circle, and aligned the squares with the x and y axes to simplify calculations. I also put the small square on the top rather than the bottom.
2. This let me declare that one corner of the big square lands at 45° mark, meaning half the length is √2, so the whole square has area 2.
3. To find the small square area, I shifted the whole circle down by sin(45°). This makes the center of the face where the two squares touch to be the new origin. The shifted circle function is now x² + (y + 1/√2)² = 1
4. Rearranging this gives y = √(1-x²) - 1/√2, but where the corner of the small square touches the circle, y = 2x.
5. Replacing y with 2x in (4) and re-arranging / squaring sides eventually gives the quadratic: 5x² + 2√2x - 1/2 = 0
6. When using the quadratic formula, one result for x gives a point outside the unit circle. The other result is x = (√18 - √8)/10
6b. I plugged in this value for x and y=2x into the equation for step (3) to make sure the equation worked out. The radicals all wind up squaring out to integer values-none vanished by subtraction, which was unexpected.
7. The area of the small square is y²=(2x)² = 2/25.
8. The ratio of the areas of the large square and small square is 2/(2/25) = 25.
Alternative solution:
Let radius of circle = 1
Find half the side-length of the big square via pythagorean theorem, call it x:
1=x^2 + x^2 = 2*x^2
x^2 = 1/2
(Then, the area of the big square must be 4*1/2 = 2)
x= sqrt(1/2)
Draw angle v between the bottom middle of the small square and the square's corner
Since radius = 1, cos(v)*1 must be equal to x+y
cos(v) = x+y
cos(v)-x = y
sin(v)*1 must be equal to half the side-length of the small square
Thus 2*sin(v) = y
Finally, you have:
cos(v)-x = 2*sin(v)
cos(v)-sqrt(1/2) = 2*sin(v)
Solve for v with an advanced calculator (If several solutions, pick smallest positive number) : You get the answer v = 8.13010235°
Input v in: y = cos(v)-sqrt(1/2) and you get y = 0.282842712
y^2 = 0.08
Thus,
2/0.08 = 25
another solution(might be simpler):
plot the graph x^2 + y^2 = 2. Focus only on the first quadrant.
The circle has a radius of root(2), so the side of the square is 1. The total area of the square will be 4*(1 x 1) = 4
For the small square, let's say its side has length 2L. Lets put that square on top of our big square on our graph. We need to find the point it's corner will touch the circle in the first quadrant.
Graphically, we're looking for a point on x^2 + y^2 = 2 where:
x = L
y = 1 + 2L
(1+2L)^2 + (L)^2 = 2
... algebra ...
L = 1/5 (positive solution)
that means the side length of the small square is 2/5, and its area is 4/25.
The ration between 4/25 (area of small square) and 4 (area of large square) is 1:25.
Hence, the small square is 25 times smaller
I had an entirely different and possibly simpler method of solving than Presh did. First, I assumed the circle was a unit circle centered at the origin. This eliminates the need to carry the superfluous "r" variable that Presh did, since it's a ratio problem, the r will cancel out anyway. I moved the smaller square to the top of the diagram for ease of use in the Cartesian coordinate system. It's simple enough to find the area and side lengths of the large square- √2 side lengths and area = 2.
Beyond this point, my method diverged completely from Presh's. The small square is half the side length of the large square from the center of the circle (distance = √(2)/2. I moved the circle so that the center of the top side of the large square was centered on the origin by rewriting the equation for the circle: f(x)=√(1-x^2) -√(2)/2.
The trick to finding the side length of the square is to notice that the point on the square which touches the circle must be at the intersection of the graph of the circle and the line y=2x. So set 2x=√(1-x^2) -√(2)/2 and solve for x. The only positive value of x is 1/(5√2). The side length of the small square is 2x= 2/(5√2) and the area is 4/50. Therefore the ratio of small to big square is 25:1
This method can be easily extended to find the area of the third square and beyond which sit on top of the previous square. Simply add up the side lengths from the previous squares to find the distance from the center of the circle where the new square sits, and make that your new origin for the next calculation.
YAY!!! I got it. I just took advantage of the freedom to label the lengths to simplify the algebra so that I could do it in my head.
I labeled the small box length as 2a, and the radius of the circle as sqrt(2) and the length of the large box as 2. This gives the length of the vertical segment a value of 1+2s. The equation is thus:
(1+2s)^2 + s^2 = 2
5s^2 + 4s - 1 = 0
(5s-1)(s+1) = 0
the positive value of s is given by 5s-1 yielding s = .2
the ratio of the sides is 2/2s or 2/.4 or 5
thus the ratio of the areas is 25
I used coordinate geometry, and it wasn't too difficult. I solved the problem with the whole figure being flipped vertically so the small square sits atop the large one. Set the radius of the circle as one and the centre at the origin in the standard x-y plane. Then the point of the top right (north-east) corner of the big square inscribed in the circle is the solution of the pair: x^2 + y^2 = 1 and y = x, which gives (1/sqrt(2),1/sqrt(2)), allowing the side of the big square to be calculated by a simple difference as 2/sqrt(2) = sqrt(2).
Next, to simplify the further computation, I recentred the axes to the transformed plane X-Y (note the upper case, indicating a transformed coordinate system) so that the circle was shifted vertically downward. The top edge of the large square would be flush with the new X-axis. The new equation of the translated circle is X^2 + (Y+1/sqrt(2))^2 = 1. The small square would be neatly bounded between the arc and the X-axis. The top right (north-east) corner of the small square can be found by solving the equation of the circle with Y = 2X. The resulting quadratic gives only a single positive X value, and the coordinates work out to be (sqrt(2)/10, sqrt(2)/5). Hence the side of the small square is sqrt(2)/5, and the ratio of the sides of the large and small square is 5, giving the ratio of the areas to be 5^2 = 25.
Not an elegant method, but it's not too bad.
I always dread the moment these videos do factorization of polynomials because I was never taught this in high school and the results are often absolute magic to me. Couldn't even begin to try and cobble one together, it's like trying a dozen different factors of sums in my head until I stumble across the right one.
Especially this factorization is _absolutely baffling_ to me and I really wish the video would have spent more time on it.
I did the same to get the side of the larger square (a = r*sqrt(2)) but to get the smaller square I used the intersecting chords theorem (that I learnt from your channel) to get (5/4)b^2 + (r*sqrt(2))b - r^2/2 = 0 (I used b for the side of the smaller square) which is a quadratic and solved to get b = r*sqrt(2)/5 which is the same as a/5 and squared to get a^2 : a^2/25 which if you divide by a^2 and times by 25 you get 25 : 1
I got it using equation of circle, x^2+y^2=r^2. Taking the centre of the circle as the origin, use the equation to get y when x=b (where little square has side length 2b). Then use the fact that 2b=y-a (where big square has side length 2a).
Aw, I didn't get to do this one with you because I already did this problem a few months ago on a graphic calculator. I also did the reverse, where a circle is inscribed inside a square, and a circle is then made inscribed the corner where it's tangent to both sides of the square it touches as well as the original circle. (The math behind that one didn't end up as cleanly as 25 like this problem had, the ratio is 1/(3-2√2)², and that one took me about a day to figure out, while this one took me an hour or so.) Great video, though!
Not to sound overly pragmatic but I guessed that it looked like 5 small square sides could fit along the big square and assumed the length was 10 and 2 and 4/100 = 1/25 boom.