Thank you Andy. I am a British maths teacher (well, I was until yesterday, when I retired) and watch lots of math solutions - I have given up watching many channels similar to this, as they take so long to get to the point. I like Numberphile and similar channels but this is the only geometry problem one I follow now as Andy zips through the explanation very quickly and clearly.
Same her without the being or having been a math teacher part (well, except for my girlfriend who needs math for her studying medicine and has a math phobia).
I solved it a different way: 1. Draw a radius to the point the circle touches the square bottom left of it. 2. Make this the hypotenuse of a 45-45-90 triangle within the circle. 3. If we call the radius x, both legs of the triangle must be x/√2 (because of the property of a 45-45-90 triangle where the hypotenuse = leg√2). 4. Draw another radius directly right, to the point the circle touches the outer box. 5. If you look at the horizontal leg of your triangle and the radius drawn in step four, you can see that the two distances add up to create the distance labelled underneath to be 2. 6. With this, you can create the equation: x/√2 + x = 2 7. solve for x and you get x = 4/(2 + √2), or x = 1.172
There's another way for solving this. If you draw a line between the top-left and bottom-right vertex of the square you get a right-angled triangle with the circle being its inscribed circle. Let P be the perimeter of the triangle and A its Area. The radius of the inscribed Circle (inside a right-angled triangle) is equal to 2A/P
I did something similar, but instead of the inscribed circle property, I used the top letf corner going right until it tangents the circle, and then a diagonal line from the same corner to the center of the big square, creating another tangent line that must equal each other in length, since they have the same starting point. From there, 4 = 2sqrt2 [which is the tangent segment length] + r So R = 4-2sqrt2 =~1.17
One way to prove this is: Inscribe a circle in the triangle with sides a, b, c. Connect the vertices to the center, and draw the perpendicular radio from the center to the tangent points. There are three triangles with height r and bases a, b, c, so the total area is r(a+b+c)/2. For a right triangle, the area is also ab/2, so: r(a+b+c)/2 = ab/2 r = ab/(a+b+c) r = ab(a+b-c)/((a+b)^2-c^2) r = ab(a+b-c)/(a^2+2ab+b^2-c^2) r = ab(a+b-c)/(2ab) r = (a+b-c)/2 In a general triangle, we have r(a+b+c)/2 = ab sin(C) / 2 r = ab sin(C) / (a+b+c) r = ab sin(C) (a+b-c) / ((a+b)^2-c^2) r = ab sin(C) (a+b-c) / (a^2+2ab+b^2-c^2) r = ab sin(C) (a+b-c) / (2ab(1+cos(C)) r = [(a+b-c)/2] * [sin(C)/(1+cos(C))] Which from Wolfram is: r = [(a+b-c)/2] * tan(C/2)
I caluculated it differently. We know that 2 lines originating from the same point, touching the same cirlce have the same length. If we draw a diagonal from the bottom right to touch the circle, we know that its 2√2. That means the verticle line that touches the circle is also 2√2. Then we just subtract that from the full length of the side giving us 4-2√2 which comes out to the correct number
By touching you mean perpendicular to the circle's radius, right? You could draw a line from the bottom right to intersect with the circle anywhere between the points the circle touches the little square and the big square, and that wouldn't be 2√2, right?
I only discovered your channel a week ago, you've ignited a love for maths I never knew I had. Thanks for the enjoyable videos. all the best, from Australia :)
r = 2√2/(1+√2) = 4 - 2√2 (rationalizing the denominator) ≈ 1.1716 I solved it somewhat convolutedly I drew the square in the top left with radius r at the tangent points, then extended the vertical and horizontal lines of that square to make a + shape. I also extended the lines of the bottom-left square to make a + shape I end up with two small triangles of sides (2-r):(2-r):r, with radius r as the common hypotenuse then I used a²+b²=c² on one of the triangles to solve for r. (2-r)² + (2-r)² = r² you can also compare it to a similar 1:1:√2 triangle: (2-r)/r = 1/√2
I solved this an (i think) easier way. From the center of the circle, you can draw a line straight to one of the edges of the outer square, then another line at 45° to the corner of the inner square. Since we know that the height/width of these lines as one shape is 2, you can use the equation r + rsin(45°) = 2 to solve for r and get the same answer
I did it similarly to this, but instead of using sin(45°), I just plugged the value of sqrt(2)/2 because the pithagorean theorem gave me that number. Then rationalized to: 4 - 2sqrt(2), which gave me the answer
Ok, now I think I understood your solution. You can also draw a line from the center of the circle to the common angle of the squares and another one parallel with a side of little square and toward a common line of two squares. You get again 2 Δ that are proportional. So you get (4-r)/2 = (sqr8+r)/sqr8 .
It's been decades since I finished high school, and I'm still in love with these kind of problems. Better than real world problems. Keep up the great work.
I like the notion of a general solution with such problems. Looking at the sketch, the half of the square's diagonal is s·√2/2, which is equal to r + r√2, so one gets s·√2/2 = r(1 + √2) → r = s·√2/(2·(1 + √2)). Then we can rationalize this to get r = (s/2)·√2/(1 + √2)·[(1 - √2)/(1 - √2)], hence r = (s/2)·(√2 - 2)/(1-2) → r = s·(2-√2)/2. Et voilà!
Here’s my thinking before watching yours. Set origin at upper right of 2x2 square. Center position can either be represented (in both horizontal and vertical) as 2-r or r/√2, so these are equal statements. Solving 2-r=r/√2 gives r=4-2√2
After calculating 2√2 as half of the diameter, we should remember that the two tangents drawn from a point to a circle are equal. So the tangent line drawn from the top left corner of the square to the circle is also 2√2. Finally, subtract 4 - 2√2, and you'll get the same answer.
I'm a big believer in rationalizing the denominator. Sometimes it makes it better. When you start factoring cubics and bigger, it's a good skill to have.
I totally trig'd it knowing it is just a 45 deg angle, thus basically immediately got r(1 + sqrt(1/2)) = 2 and thus simplified to r = 4 - 2 sqrt(2). Obviously same result as you got, just with the usual trick of expanding the fraction by the conjugate of the denominator.
It is easier to draw a line from the upper left vertex to the central point 2,2 which is tangent to the circle. Knowing that two tangents to a circle from a common point are equal, the upper horizontal tangent will be equal to the diagonal 2sqrt(2). Thus, 2sqrt(2)+r=4. Then r=4-2sqrt(2).
Pretty simple task. We have two squares: one with side 2, one with side r. Their diagonals are square root of 2 times 2 and square root of 2 times r. Add up these diagonals, add another r and you get the diagonal of the biggest square which is square root of 2 times 4. So we have: 4•2^1/2 = 2•2^1/2 + r•2^1/2 + r r(1 + 2^1/2) = 4•2^1/2 - 2•2^1/2 r = (2•2^1/2)/(1 + 2^1/2) r = 2,818 / 2,414 = 1,167
connect diagonal from top left to bottom right, it will be tangent to the circle and form a triangle then use formula rp=S where p is a semiperimeter, S is an area of a triangle and r is radius you can find the area of triangle by using heron's formula and a semiperimeter by summing up three sides and dividing them by 2.
Please always rationalize the denominator. You'll get rid of the fraction and you'll only have one root in your answer: 4 - 2sqrt(2) is so much nicer and simpler.
Before watching the video: Two tangent theorem. Top-left corner to tangent points is 2*sqrt(2). It is also 4-r. 2*sqrt(2) = 4-r Square both sides: 8 = 16-8r+r^2 r^2 - 8r + 16 = 8 r^2 - 8r + 8 = 0 Quadratic formula: (8+or-sqrt(64 - 32))/2 = r (8+or-sqrt(32))/2 = r (8+or-4*sqrt(2))/2 = r (4+or-2*sqrt(2) = r Discard the + result as that is larger than the square's side length. r = 4 - 2*sqrt(2) In decimal that approximates to 1.172... EDIT: I think Andy overcomplicated this one a bit, but some might say that about my way :)
You could apply the rule that says for an isosceles right triangle, the length of the hypotenuse is the length of another leg times the square root of 2. (48 second mark)
I like to take shortcuts when possible in math, by using known values. So, with a right triangle with base and height equal, the hypotenuse is basically those values multiplied by Root 2. IE, base and height of 4, hypotenuse is 4(sqrt 2). Taking that, and the intersecting point being 2 units in from the sides of the square, 2 = R + R/(sqrt 2). R/(sqrt 2) being essentially the base of a triangle with R as its hypotenuse, based on the above. Multiply the fraction by (sqrt 2)/(sqrt 2), to make it 2 = R + R(sqrt 2)/2. Multiply both sides by 2, to get 4 = 2R + R(sqrt 2). Square root of 2 is approximately 1.414, so 4 = 2R + 1.414R = 3.414R. Divide both sides by 3.414, to get 1.17 = R.
i solved it differently, i made a right angled triangle using from the center of the circle to the top right diagonal of the square. my hypotenuse was r and my side lengths were 2-r. then afterwards i quad formula the quadratic and got 4-2root2 which gives the right answer
Cool! I solved it in a different way: 1. Determined that the total length of one side of the large square is equal to 4. 2. Knowing this, I can say that this length 4 is equal to 2 + a radius + the horizontal component of a second radius. 3. Based on the geometry and positioning of the little square relative to the circle, I know that the vertical and horizontal components of the radius I'm discussing are the same. Therefore, r^2 = x^2+y^2 -> x=y -> r^2=2x^2-> r=sqrt(2)x -> x= r/sqrt(2) as the horizontal component that makes part of the full length 4. 4. Final equation: 2+r+r/sqrt(2)=4. Clean up steps: r+r/sqrt(2)=2 -> 2=(sqrt(2)r+r)/sqrt(2) -> 2sqrt(2) = sqrt(2)r+r -> 2sqrt(2) = r(sqrt(2)+1) -> r=(sqrt(2))/(sqrt(2)+1).
Solved it a different way: 1. Rotated the whole thing 180 degrees 2. Now it looks like the co-ordinate system 3. One side of square as X-axis and the other as Y-axis 4. We can clearly see that the circle is passing through the point (2,2) 5. Now it's just a simple question that is taught to us "A circle is passing through a point (2,2) and touches both the axes, find the radius" 6. If we take x^2 + y^2 + 2gx + 2fy + c as the general equation of the circle, the radius become sqrt(g^2+f^2-c) 7. The circle touches both the axes, so g^2 = f^2 = c, Hence the radius = √(g^2+g^2-g^2) = g = f = √(c) 8. Put that in the equation, x^2 + y^2 + 2√cx + 2√cy + c = 0 9. Substitute the point (2,2) in the equation and take squaring on both sides. You get a quadratic and put that in calculator and you get c. Now we need √c so just sqrt the number you get and that's the answer.
I did it a bit different, we can ignore the big square and just consider the right top quarter, and since the circle is limited on one side by the edges of that square and on the other side by the corner of that square, then we can draw 2 more circles: 1. draw a circle that goes thru all corners of the square, find the diameter, here ofc the diameter of the circle is the diagonal of the square, so same as in the video, using pytagora -> 2 root(2) 2. draw another circle, that fits entirely inside the square, a.k.a. inscribed in the square -> here we can see the diameter is the length of the circle = 2 -> and we can see the difference between the diameter of these 2 new circles is 'x' on both sides, so big circle diameter = small circle diameter + 2x => 2 root(2) = 2 + 2x => x = root(2) - 1 -> we can also see that the difference between the diameter of the new smaller circle and the original circle is just 'x', so original circle diameter = small circle diameter + x => d = 2 + x => d = 2 + root(2) - 1 => d = 1 + root(2) -> now we can find the radius by dividing the diameter with 2 => r = (1 + root(2)) / 2 = 1.2071
Just draw the diagonal of the big square from top left to bottom right. You get an isosceles triangle and the circle will be it's incircle. Then it's easy to calculate the radius.
More cleanly, 2*sqrt(2)/(1+sqrt(2)) = 4-2*sqrt(2). So... that means the diagonal between that bottom right corner and the center point of the circle is exactly 4. In case anyone was interested in a more intuitive way of looking at the problem.
1. Consider lower left corner to be (0,0). Draw line from (0,0) to (4,4). 2. Mark circle center as (x,y). Draw radius R from (2,2) to (x,y). 3. Draw horizontal line from (2,2) to (x,2). Label as A' 4. Draw vertical line from (x,2) to (x,y). Label as A 5. Draw horizontal line from (x,y) to the right to circle square tangency point. Label as R. 6. Angle at (2,2) = 45. Solve for A. A = R*cos(45) 7. A + R = 2. R*cos(45) + R = 2. 8. R(cos(45) +1) = 2. 9. R = 2/cos(45) + 1 = 1.1715
Faster way: The point of the circle that touches the small square is 2 units from the right, and we have a radius going horizontally plus a radius at a 45deg diagonal to make that 2-unit horizontal, meaning 1r + r/sqrt(2) = 2 => r = 2/(1+1/sqrt(2)) = 1.17157
I went about this differently and got the same result. Draw a diagonal line from the upper left corner(call it A) to the lower right corner of the large square. Find the point of the circle that touches this diagonal (call it B). The length from the upper corner to point A is the diagonal of the square, which is 2 x sqrt(2). Find the point where the circle meets the upper side of the square (call it C). AB = AC. r = 4 - AC = 4 - 2 x sqrt(2) = 1.17.157
i took a different approach as follows: First, i took the point of contact between the circle and the top right corner of the inner square, and make it the origin point of a X,Y coordinate system. Then the general equation for that circle in that X,Y coordinate system would be (X-a)^2 + (Y-b)^2=r^2 where "a" and "b" are the horizontal and vertical distances from the center of the circle to the origin point. And because the circle is bisected by the diagonal of the big square, we can asume "a" and "b" are equal, and we call this value c. Thus, the equation now comes to (X-c)^2+(Y-c)^2=r^2 Now, we know when x=0 then y=0 wich leads to r=c√2 (1) Then, we also know that for x=c then y=2 wich leads to c=2-r (2) Then replacing (2) in (1) we get r=(2√2)/(1+V2) ---> r=1,17
Hi sir, I highly admire the way you solve math problems it emphasizes my love toward solving complex problems. Thereby can you make a pdf (free for veiwers) based on some questions like these amd other of your math problems for guys like me to solve.
Each time you solve one of these, I almost always don't figure it out before you do... And that both irritates me, and humbles me at the same time 😜.. You rock... I'm showing this to my students as well.
That is one long way to do it. Or draw a line from one corner to the other where that forms a triangle with the circle inside it, and the circle, that line (hypotenuse H), and the top right corner of the little square all touch. Then the formula for the max radius of a circle that can fit in a triangle is r=(P + B - H)/2 P and B (perpendicular and base) r= ( 4 + 4 - 4sqrt(2) ) / 2 r = ( 8 - 4sqrt(2) ) / 2 r = 4 - 2sqrt(2)
Or more precisely, 4 - sqrt(8). Co-ordinate type geometry is an easy way to see this. Centre of the circle is (x,y). We have two radii from (x,y) to (4,y) and (x,4). The radius is therefore 4-x (or 4-y), so we see x = y, as we'd expect from the symmetry of the construction. Now we have another radius from (2,2) to (x,y). So by Pythagoras we have 2(x-2)^2 = radius^2 = (4-x)^2. Solve this to get x = sqrt(8) (clearly x>0 by construction) and therefore the radius is 4-sqrt(8).
For any triangle S=p*r, where p - half perimeter, r - radius of inscribed circle. In our case, S = 0.5*4*4 = 8 (half the square); p = 0.5*(4+4+4*sqrt2). Thus, r = 8/(4+2*sqrt2) = 4/(2+sqrt2) = 2*(2-sqrt2) =~ 1.1715
The radius of a circle inscribed in a triangle is two times the area of the triangle divided by the perimeter of the triangle. The area of the triangle is 8. The perimeter of the triangle is 4 + 4 + 4xsqrt(2). The radius of the circle is 16/(8+4xsqrt(2))=1.17157.
I found a different way of solving it: the circle is tangent to a triangle when you divide the square in half. You can find the radius by using: r.p= A In this case, the area is 8, the semi perimeter is (sqrt(32)+8)/2. If you solve this and isolate the radius you'll have the same result. If there is a circle inscribed and tagent in any kind of convex polygon, the area of the polygon will be r.p This is true even if the circle isn't tagent to all sides of said polygon, it only needs to be tanget to the highest number of sides that it can touch while being inscripted. In this case, it was tangent to all 3 sides of the triangle
The top right point in the square is the director circle to the given circle now using coordinate geometry results we know distance of such point from center of circle is always r√2 Director circle : locus of points from which tangents drawn to a curve are mutually perpendicular Where r is the radius of given
I did it a bit differently. I completely ignored the smaller square, and instead I just focused on the fact that there's a right triangle with both of it's legs being equal to 4, whose incircle is that circle. First we find the hypotenuse, which using the Pythagorean theorem we find is equal to 4√2. Now we find the semi-perimeter p, which is half the perimeter. It is equal to 4 + 2√2. After we find p, we can now use Heron's formula to find the area S = √(p ⋅ (p - a) ⋅ (p - b) ⋅ (p - c)). We calculate the area to be equal to exactly 8 square units. Now we know that S = pr, therefore r = S / p, and when we calculate it and rationalize it we get 4 - 2√2 (same answer as Andy, just rationalized).
The diameter of the circle is larger than 2. Assuming the circle is tangent to the outer box in both parts of the corner that diagonal line would never intersect the center point of the circle. so your answer is close but not correct
I guess I did this one the hard way. First calculate the diagonal of the small square, as shown. Then extend that line to the center of the circle to create a right triangle with hypotenuse 2√2 + r. Each leg is 4 - r. The Pythagorean Theorem means that (4 - r)^2 + (4 - r)^2 = (2√2 + r)^2. Multiply it out and you get a knotty quadratic: r^2 - (16 + 4√2)r + 24 = 0. This actually is amenable to solving with the quadratic formula. You get a root within a root, but that can be handled. Finally, r = [(16 + 4√2) +/- (8 + 8√2)]/2. The larger root is way too big for the given parameters. But the smaller root, 4 - 2√2, does equal 1.17157.
Draw a Tangent to the circle from bottom right corner of Larger Square(where tangent touches the circle at centre of the larger square) and this length is half the diagonal of larger square = 2√2. Now tangents drawn from a point outside to the circle are equal in length(one along the diagonal and one along side from the bottom corner of larger square) hence 2√2 + r = 4 r = 4 - 2√2 OR Draw the diagonal of the larger square (connecting bottom right corner and top left corner vertices). Now we have a Right Triangle where the sides are 4, 4, 4√2 and circle is the incircle of this triangle whose radius is (4+4-4√2)/2 = 4-2√2 PS :: Inradius of a Right Triangle, whose sides are a,b,c with c being hypotenuse is (a+b-c)/2
For the radius of a circle to be expressed in p/q form, the denominator must be a rational number. In your case it is irrational, add 2 extra steps for componendo-dividendo and the final answer would be 1.51....
My answer: 4-2root2 My working: The vertical distance from the corner of the little square to the roof of the big square is 2 The vertical distance from where the circle touches the corner of the little square to the roof of the big square is radius + (radius/root(2)) since the angle is 45 degrees Then just do the algebra for 2 = r + r/root2
You should have rationalized the denominator by multiplying top and bottom by (1 - sqrt(2)). The denominator simplifies to 1, which eliminaties the fraction completely, and the numerator becomes 4 - 2(sqrt(2)). Evaluating that gives the same answer: 1.17157...etc. Also, much simpler way to do this: Extend the horizontal top of the lower left square to the right-hand vertical side, dividing it into 2 segments, each length 2 (just like the left side). The top part of the top segment is r, as you've shown. The bottom part of that top segment is the side of a square whose diagonal is r. That's r/sqrt(2). So now we have r + r/sqrt(2) = 2. Solving this equation gives the same result.
what i did was i drew a radius line from the center of the circle to the vertex of the top right corner of the small square. this line is the radius but anglesd 45 deg downwards, therefore, the horizontal component is r*cos45. drew another line horizontally from the center of the circle to the right side of the big square. that radius is r. so rcos(45) + r = 2. solve for r and you get 4-2√2
Exellent solution! I am trying to make understanding of this, and 1:30 you said you can sqaure root both sides of the equation. Unfortunately I don't think this works as 3^2 * 4^2 = 5^2, however 3 * 4 does not equal five. I would be gratefull if you could shine some insight on this.
Last advice to use calculator would lower the score when I was in school. My teachers always emphasized to use exact computation, in this case it means that fraction should be multiplied (upper and lower part) by (sqrt(2)-1). Then upper part becomes 4-2*sqrt(2) and lower part becomes 1 (because (a+b)*(a-b) = a^2-b^2, it his case a=sqrt(2), b=1 so a^2-b^2=2-1=1). Final value should be 4-2*sqrt(2).
I’m with you…how exciting! When I help my teenager with math problems, I’ve started saying “how exciting” when we get to a solution. Not sure it’s working but I keep trying!
An alternative form of the solution is 4-√8, now solve the 3D equivalent of the excercise (A half size cube inside a cube and a spehere touching its corner and the faces of the bigger cube, what is its radius?)
Its not the fact he solves these things - I dont think that I’ve ever seen him do maths that I didnt know - but its the ingenuity with which he solves it. Its one thing knowing how to get through the math, and another thing entirely on seeing a bunch of shapes and knowing what math to use to get you there.
Logic flaw here, if you rotate side 'c' down onto side side it 'must be longer' than side 'a', if it were actually 2 units the angle of 'ca' would be zero or side 'a'. Simple proof on paper, create the box, draw the diagonal 'c', get a protactor (Lead pencil) spead the legs the distance of 'a' and rotate the protractor to the verticle, 'c' should be bisected.
I just put a diagonal in the square from the top left to the bottom right made the sides equal to n+r and the diagonal to 2n since we have 90° we know that the diagonal is equal to √(2×4²) meaning n is half of that than just 4-n and you have r. If you continue with this you can get that r is about equal to 0,29289[((2-√(2))÷2), 1-√0.5]×a side of the big square for all values if you have different side lengths it would obviously change, even tho you would still be able to generalize that.
My approach to this problem is different we don't need the square, divide the squared into 2 parts diagonally and this should give you a circle inscribed in a triangle then therefore we can use the formula r=(a+b-c)/2 a=b and c=4root2 r=(4+4-4root2)/2 or 1.1715...
My solution: - The large square has a side of 4. Take 2 from the small square. We have now 2. - From the circle we have r+r*cos(45°) which is equal to 2 Then r+r*cos(45°)=2 r(1+sqr(2)/2)=2 -> r≈1.1716
My solution: From the centre of the circle to the point it touches the square is r so the horizontal distance is r√2. Add that to r and you get the horizontal distance to the edge of the outer square therefore r + r√2 = 2 and r = 1/(1+√2)
I use BPT theorem to find radius of the circle by making two equal triangel with common hypotenious which is equal to the diameter of the square , and let the base of these two triangle by making one horizontal and vertical line from the side which is touching with the circle which makes the two triangle who follow the BPT rule because the length "r" both horizontal and vertical and "2units" both horzontal and vertical are prallei t each other . Now i put the values directly and indirectly and apply the BPT to get radius thanks for giving such cute question ❤🔥❤🔥 edit= my approx anser is |1.18| because i use"root2" as 1.14
lol I too a completely different approach (didn’t have pen/paper so I needed an easier way): basically, assume the center of the circle is (x,y). The center of the circle must lie on the main ascending diagonal of the square for symmetry reasons, so x=y. The lower square touches the circle at (2,2). Using this, the distance from (2,2) to (x,x) is sqrt(2)*(x-2)… But going from the center of the circle to the top of the large square is also a radius (4-x)… you can do the same horizontally. Thus, we get sqrt(2)*(x-2)=4-x. Notice that this is a simple linear equation in one variable… solve for x we immediately get: [sqrt(2)+1]*x = 4+sqrt(2)*2 x = 2[sqrt(2)+2]/[sqrt(2)+1] = 2*sqrt(2) … therefore, radius = 4 - 2*sqrt(2) 🥳 ✨as a bonus this mean’s the TH-camrs expression 2*sqrt(2)/[1+sqrt(2)] simplifies beautifully to 4 - 2*sqrt(2)
Just looking at the problem, I have no idea where to even start. well to be honest I got to 1:10, but no further. The center of the circle. How do I find that. Awesome dude!
If the goal was to solve this problem using the most complicated method possible it was a success. It's a lot less calculation intensive to solve this problem (starting from the drawing with the 3 radii) by simply adding a diagonal line from the top left corner of the square to the center of the square which is a point of tangency with the circle. Because we can see that this diagonal line is the hypotenuse of an isosceles right triangle with side lengths of 2, we know that this hypotenuse length will equal 2√2 and therefore the length of the square's top side left section (from the left corner to the point of tangency at the top of the circle) is also equal to 2√2. The drawing has already established that the square's top side right section (from the point of tangency at the top of the circle to the right corner) is equal to the length of the radius. The length of the radius equivalent section of the top side of the square (r) can be calculated from the difference between the total length of the top side of the square (4) and the length of the left section of the top side of the square (2√2). r = 4 - 2√2 r ≈ 1.17
Why are you using Pythagoras for all those diagonals ? It would be faster to use the formula d = s * √(2) where d is the diagonal of the square et s its side.
You need not be concerned with the big square. just draw the 2 x 2 square in the upper right corner. From that (summing along the horizontal OR the vertical) r/√2 + r = 2 & solve for r to get the same answer (but you should rationalize the denominator to get r = 4 - 2√2).
I suck at math but unless I am mistaken the radius of that circle has to be half the distance from the connecting corner of the lower right square to the square to the opposite corner of the upper right square and that is simply the square root of 2 squared + 2 squared = 2.828 no?
This seemed like one of those questions on a math test I would skip because it took way to many steps and not enough time to deal with it when you have 3 other questions that are just as long but are easier to solve.
He wasn’t as excited for this 😢😢🤧
Maybe because it was too easy for him and was not challenging? 😅😂😂😅
@@bebektoxic2136sad since it took me like 10 minutes to solve it
That's not the point, doesn't matter if he is excited or not.
@@arpieravidas7731 it is the point, I’m heartbroken
I hope Mr Andy Math makes a full recovery for the next one 😔💔
Thank you Andy. I am a British maths teacher (well, I was until yesterday, when I retired) and watch lots of math solutions - I have given up watching many channels similar to this, as they take so long to get to the point. I like Numberphile and similar channels but this is the only geometry problem one I follow now as Andy zips through the explanation very quickly and clearly.
isn't it the ultimate proof that math teachers retire from school, but never from math? absolute legend.
Congratulations and happy retirement! I'm 4 years in to maths teaching. Unsure if I'll carry on!
Amen he is alive!
Same her without the being or having been a math teacher part (well, except for my girlfriend who needs math for her studying medicine and has a math phobia).
Fake teacher , teacher don't watch TH-cam
I solved it a different way:
1. Draw a radius to the point the circle touches the square bottom left of it.
2. Make this the hypotenuse of a 45-45-90 triangle within the circle.
3. If we call the radius x, both legs of the triangle must be x/√2 (because of the property of a 45-45-90 triangle where the hypotenuse = leg√2).
4. Draw another radius directly right, to the point the circle touches the outer box.
5. If you look at the horizontal leg of your triangle and the radius drawn in step four, you can see that the two distances add up to create the distance labelled underneath to be 2.
6. With this, you can create the equation: x/√2 + x = 2
7. solve for x and you get x = 4/(2 + √2), or x = 1.172
Step 6: If you draw the "radius square" at the top right, you can ultimately see that x + x/√2 is half of the outer square's diagonal 😀
That's the method I used when I saw the thumbnail. Glad I'm not alone.
1. draw it in scale and measure the radius
I wish I could get high off potenuse 🤓
Here's how to solve x:
x/√2 + x = 2
x*√2 + x*√2*√2 = 2*√2*√2 multiply both sides by √2*√2
x*√2 + x*2 = 4 simplify (√2*√2 = 2)
x*(√2 + 2) = 4 factorize
x = 4/(√2 + 2) solve x
There's another way for solving this.
If you draw a line between the top-left and bottom-right vertex of the square you get a right-angled triangle with the circle being its inscribed circle.
Let P be the perimeter of the triangle and A its Area.
The radius of the inscribed Circle (inside a right-angled triangle) is equal to 2A/P
it's interesting how more knowledge leads to more elegant solutions
I did something similar, but instead of the inscribed circle property, I used the top letf corner going right until it tangents the circle, and then a diagonal line from the same corner to the center of the big square, creating another tangent line that must equal each other in length, since they have the same starting point. From there, 4 = 2sqrt2 [which is the tangent segment length] + r
So R = 4-2sqrt2 =~1.17
This is how I did it too and to me feels more direct
Whats the guarantee that this diagonal is tangent to the circle?
@@aindeevarr because the circle touches the centre point of the big square (because we know the little square is half as wide).
The radius of a circle inscribed in a right triangle is (a+b-c)/2 = (4+4-4√2)/2 = 4-2√2.
One way to prove this is:
Inscribe a circle in the triangle with sides a, b, c.
Connect the vertices to the center, and draw the perpendicular radio from the center to the tangent points.
There are three triangles with height r and bases a, b, c, so the total area is r(a+b+c)/2.
For a right triangle, the area is also ab/2, so:
r(a+b+c)/2 = ab/2
r = ab/(a+b+c)
r = ab(a+b-c)/((a+b)^2-c^2)
r = ab(a+b-c)/(a^2+2ab+b^2-c^2)
r = ab(a+b-c)/(2ab)
r = (a+b-c)/2
In a general triangle, we have
r(a+b+c)/2 = ab sin(C) / 2
r = ab sin(C) / (a+b+c)
r = ab sin(C) (a+b-c) / ((a+b)^2-c^2)
r = ab sin(C) (a+b-c) / (a^2+2ab+b^2-c^2)
r = ab sin(C) (a+b-c) / (2ab(1+cos(C))
r = [(a+b-c)/2] * [sin(C)/(1+cos(C))]
Which from Wolfram is:
r = [(a+b-c)/2] * tan(C/2)
I caluculated it differently. We know that 2 lines originating from the same point, touching the same cirlce have the same length. If we draw a diagonal from the bottom right to touch the circle, we know that its 2√2. That means the verticle line that touches the circle is also 2√2. Then we just subtract that from the full length of the side giving us 4-2√2 which comes out to the correct number
Very clever
By touching you mean perpendicular to the circle's radius, right? You could draw a line from the bottom right to intersect with the circle anywhere between the points the circle touches the little square and the big square, and that wouldn't be 2√2, right?
That's pretty much how I did it, but by using top left: 2*sqrt(2) = 4-r etc.
Fun, indeed. Thanks. 80 y/o here who used to have fun doing math. I'll be watching you channel regularly.
Everything is temporary but "how exciting" is permanent 😂😂
I only discovered your channel a week ago, you've ignited a love for maths I never knew I had. Thanks for the enjoyable videos.
all the best, from Australia :)
my fellow upside down mate.
Exactly the same for me, I’m actually enjoying maths, which is something i never thought I’d say. Greetings from Germany!
r = 2√2/(1+√2)
= 4 - 2√2 (rationalizing the denominator)
≈ 1.1716
I solved it somewhat convolutedly
I drew the square in the top left with radius r at the tangent points, then extended the vertical and horizontal lines of that square to make a + shape. I also extended the lines of the bottom-left square to make a + shape
I end up with two small triangles of sides (2-r):(2-r):r, with radius r as the common hypotenuse
then I used a²+b²=c² on one of the triangles to solve for r. (2-r)² + (2-r)² = r²
you can also compare it to a similar 1:1:√2 triangle: (2-r)/r = 1/√2
I solved this an (i think) easier way. From the center of the circle, you can draw a line straight to one of the edges of the outer square, then another line at 45° to the corner of the inner square. Since we know that the height/width of these lines as one shape is 2, you can use the equation r + rsin(45°) = 2 to solve for r and get the same answer
I like that one, it makes sense to me though it didn't occur to me. I really need to brush up on my trig, haven't done it since the 90's
I did it similarly to this, but instead of using sin(45°), I just plugged the value of sqrt(2)/2 because the pithagorean theorem gave me that number. Then rationalized to:
4 - 2sqrt(2), which gave me the answer
By your method we can solve in other two ways:
1. The center of that circle is the intersections of the bisectors of your triangle. A bisector from a
Nevermind, I didn't read carefully your thing. I thought you refered to a triangle with its hypotenuse as the other diameter of the big square.
Ok, now I think I understood your solution. You can also draw a line from the center of the circle to the common angle of the squares and another one parallel with a side of little square and toward a common line of two squares. You get again 2 Δ that are proportional. So you get (4-r)/2 = (sqr8+r)/sqr8 .
It's been decades since I finished high school, and I'm still in love with these kind of problems. Better than real world problems. Keep up the great work.
seeing a math problem from you on my feed always makes me happy to try it
EXACTLY
Big diagonal equals big diagonal. Mmm, yes. Quite.
I like the notion of a general solution with such problems. Looking at the sketch, the half of the square's diagonal is s·√2/2, which is equal to r + r√2, so one gets
s·√2/2 = r(1 + √2) → r = s·√2/(2·(1 + √2)). Then we can rationalize this to get
r = (s/2)·√2/(1 + √2)·[(1 - √2)/(1 - √2)], hence
r = (s/2)·(√2 - 2)/(1-2) → r = s·(2-√2)/2. Et voilà!
Here’s my thinking before watching yours.
Set origin at upper right of 2x2 square.
Center position can either be represented (in both horizontal and vertical) as 2-r or r/√2, so these are equal statements.
Solving 2-r=r/√2 gives r=4-2√2
After calculating 2√2 as half of the diameter, we should remember that the two tangents drawn from a point to a circle are equal. So the tangent line drawn from the top left corner of the square to the circle is also 2√2. Finally, subtract 4 - 2√2, and you'll get the same answer.
I'm a big believer in rationalizing the denominator. Sometimes it makes it better. When you start factoring cubics and bigger, it's a good skill to have.
I totally trig'd it knowing it is just a 45 deg angle, thus basically immediately got r(1 + sqrt(1/2)) = 2 and thus simplified to r = 4 - 2 sqrt(2). Obviously same result as you got, just with the usual trick of expanding the fraction by the conjugate of the denominator.
We making it out of math class with this one 🔥🔥🔥💯💯💯
Facts fr
u got the whole squad laughing
remove yourself
It is easier to draw a line from the upper left vertex to the central point 2,2 which is tangent to the circle. Knowing that two tangents to a circle from a common point are equal, the upper horizontal tangent will be equal to the diagonal 2sqrt(2). Thus, 2sqrt(2)+r=4. Then r=4-2sqrt(2).
This is for me the most elegant way to solve it.
Pretty simple task. We have two squares: one with side 2, one with side r. Their diagonals are square root of 2 times 2 and square root of 2 times r. Add up these diagonals, add another r and you get the diagonal of the biggest square which is square root of 2 times 4. So we have:
4•2^1/2 = 2•2^1/2 + r•2^1/2 + r
r(1 + 2^1/2) = 4•2^1/2 - 2•2^1/2
r = (2•2^1/2)/(1 + 2^1/2)
r = 2,818 / 2,414 = 1,167
@Andy Math, You did not explain why the diagonal of big square passes thru the center of the circle?
One can also solve through equation of circle by taking bottom left as origin, centre lies on y=x and y=4 and x=4 are tgts to circle
Diagonal of square will always be a(underroot2)
a=side of square
Radius of incircle of right triangle = (sum of sides - hypotenuse)/2
Standard result from properties of triangles
Haha, good catch.
Haven't been in a maths classroom for over 20 years, loving your videos.
Can follow along easy. One day I'll solve one myself :D
connect diagonal from top left to bottom right, it will be tangent to the circle and form a triangle
then use formula rp=S where p is a semiperimeter, S is an area of a triangle and r is radius
you can find the area of triangle by using heron's formula and a semiperimeter by summing up three sides and dividing them by 2.
Please always rationalize the denominator. You'll get rid of the fraction and you'll only have one root in your answer: 4 - 2sqrt(2) is so much nicer and simpler.
Multiply the top and bottom by (1-√2)(-1).
The denominator becomes 1, and the numerator becomes 4-2√2.
Change from 1+√2
To (√2)+1 because it's the same
Hi, awesome stuff 👍
What textbook do you recommend for this stuff?
Before watching the video:
Two tangent theorem.
Top-left corner to tangent points is 2*sqrt(2).
It is also 4-r.
2*sqrt(2) = 4-r
Square both sides: 8 = 16-8r+r^2
r^2 - 8r + 16 = 8
r^2 - 8r + 8 = 0
Quadratic formula:
(8+or-sqrt(64 - 32))/2 = r
(8+or-sqrt(32))/2 = r
(8+or-4*sqrt(2))/2 = r
(4+or-2*sqrt(2) = r
Discard the + result as that is larger than the square's side length.
r = 4 - 2*sqrt(2)
In decimal that approximates to 1.172...
EDIT: I think Andy overcomplicated this one a bit, but some might say that about my way :)
You could apply the rule that says for an isosceles right triangle, the length of the hypotenuse is the length of another leg times the square root of 2. (48 second mark)
As a draftsman and tinkerer, it was pretty immediately obvious that r+r/sqrt2=2.
I like to take shortcuts when possible in math, by using known values.
So, with a right triangle with base and height equal, the hypotenuse is basically those values multiplied by Root 2.
IE, base and height of 4, hypotenuse is 4(sqrt 2).
Taking that, and the intersecting point being 2 units in from the sides of the square, 2 = R + R/(sqrt 2).
R/(sqrt 2) being essentially the base of a triangle with R as its hypotenuse, based on the above.
Multiply the fraction by (sqrt 2)/(sqrt 2), to make it 2 = R + R(sqrt 2)/2.
Multiply both sides by 2, to get 4 = 2R + R(sqrt 2). Square root of 2 is approximately 1.414, so 4 = 2R + 1.414R = 3.414R. Divide both sides by 3.414, to get 1.17 = R.
i solved it differently, i made a right angled triangle using from the center of the circle to the top right diagonal of the square. my hypotenuse was r and my side lengths were 2-r. then afterwards i quad formula the quadratic and got 4-2root2 which gives the right answer
Same
@@Eagle_SFM Not same, but similar
Cool! I solved it in a different way:
1. Determined that the total length of one side of the large square is equal to 4.
2. Knowing this, I can say that this length 4 is equal to 2 + a radius + the horizontal component of a second radius.
3. Based on the geometry and positioning of the little square relative to the circle, I know that the vertical and horizontal components of the radius I'm discussing are the same. Therefore, r^2 = x^2+y^2 -> x=y -> r^2=2x^2-> r=sqrt(2)x -> x= r/sqrt(2) as the horizontal component that makes part of the full length 4.
4. Final equation: 2+r+r/sqrt(2)=4. Clean up steps: r+r/sqrt(2)=2 -> 2=(sqrt(2)r+r)/sqrt(2) -> 2sqrt(2) = sqrt(2)r+r -> 2sqrt(2) = r(sqrt(2)+1) -> r=(sqrt(2))/(sqrt(2)+1).
you could simplify the fraction through multiplying oth sides by 1-sqrt2, making -2sqrt2(1-sqrt2) or just 4-2sqrt2
Solved it a different way:
1. Rotated the whole thing 180 degrees
2. Now it looks like the co-ordinate system
3. One side of square as X-axis and the other as Y-axis
4. We can clearly see that the circle is passing through the point (2,2)
5. Now it's just a simple question that is taught to us "A circle is passing through a point (2,2) and touches both the axes, find the radius"
6. If we take x^2 + y^2 + 2gx + 2fy + c as the general equation of the circle, the radius become sqrt(g^2+f^2-c)
7. The circle touches both the axes, so g^2 = f^2 = c, Hence the radius = √(g^2+g^2-g^2) = g = f = √(c)
8. Put that in the equation, x^2 + y^2 + 2√cx + 2√cy + c = 0
9. Substitute the point (2,2) in the equation and take squaring on both sides. You get a quadratic and put that in calculator and you get c. Now we need √c so just sqrt the number you get and that's the answer.
I did it a bit different, we can ignore the big square and just consider the right top quarter, and since the circle is limited on one side by the edges of that square and on the other side by the corner of that square, then we can draw 2 more circles:
1. draw a circle that goes thru all corners of the square, find the diameter, here ofc the diameter of the circle is the diagonal of the square, so same as in the video, using pytagora -> 2 root(2)
2. draw another circle, that fits entirely inside the square, a.k.a. inscribed in the square
-> here we can see the diameter is the length of the circle = 2
-> and we can see the difference between the diameter of these 2 new circles is 'x' on both sides, so big circle diameter = small circle diameter + 2x => 2 root(2) = 2 + 2x => x = root(2) - 1
-> we can also see that the difference between the diameter of the new smaller circle and the original circle is just 'x', so original circle diameter = small circle diameter + x => d = 2 + x => d = 2 + root(2) - 1 => d = 1 + root(2)
-> now we can find the radius by dividing the diameter with 2 => r = (1 + root(2)) / 2 = 1.2071
Just draw the diagonal of the big square from top left to bottom right. You get an isosceles triangle and the circle will be it's incircle. Then it's easy to calculate the radius.
More cleanly, 2*sqrt(2)/(1+sqrt(2)) = 4-2*sqrt(2). So... that means the diagonal between that bottom right corner and the center point of the circle is exactly 4. In case anyone was interested in a more intuitive way of looking at the problem.
1. Consider lower left corner to be (0,0). Draw line from (0,0) to (4,4).
2. Mark circle center as (x,y). Draw radius R from (2,2) to (x,y).
3. Draw horizontal line from (2,2) to (x,2). Label as A'
4. Draw vertical line from (x,2) to (x,y). Label as A
5. Draw horizontal line from (x,y) to the right to circle square tangency point. Label as R.
6. Angle at (2,2) = 45. Solve for A. A = R*cos(45)
7. A + R = 2. R*cos(45) + R = 2.
8. R(cos(45) +1) = 2.
9. R = 2/cos(45) + 1 = 1.1715
Dude he does it yet again. And yet again, it is exciting.
Bro is recording this from an AQUARIUM 🐠🎣🐟🐠🐡🐬🐬🦈🧜🏽♂️.
Faster way: The point of the circle that touches the small square is 2 units from the right, and we have a radius going horizontally plus a radius at a 45deg diagonal to make that 2-unit horizontal, meaning 1r + r/sqrt(2) = 2 => r = 2/(1+1/sqrt(2)) = 1.17157
I went about this differently and got the same result. Draw a diagonal line from the upper left corner(call it A) to the lower right corner of the large square. Find the point of the circle that touches this diagonal (call it B). The length from the upper corner to point A is the diagonal of the square, which is 2 x sqrt(2). Find the point where the circle meets the upper side of the square (call it C). AB = AC. r = 4 - AC = 4 - 2 x sqrt(2) = 1.17.157
i took a different approach as follows:
First, i took the point of contact between the circle and the top right corner of the inner square, and make it the origin point of a X,Y coordinate system.
Then the general equation for that circle in that X,Y coordinate system would be (X-a)^2 + (Y-b)^2=r^2 where "a" and "b" are the horizontal and vertical distances from the center of the circle to the origin point. And because the circle is bisected by the diagonal of the big square, we can asume "a" and "b" are equal, and we call this value c.
Thus, the equation now comes to (X-c)^2+(Y-c)^2=r^2
Now, we know when x=0 then y=0 wich leads to r=c√2 (1)
Then, we also know that for x=c then y=2 wich leads to c=2-r (2)
Then replacing (2) in (1) we get r=(2√2)/(1+V2) ---> r=1,17
Hi sir, I highly admire the way you solve math problems it emphasizes my love toward solving complex problems. Thereby can you make a pdf (free for veiwers) based on some questions like these amd other of your math problems for guys like me to solve.
Each time you solve one of these, I almost always don't figure it out before you do... And that both irritates me, and humbles me at the same time 😜..
You rock... I'm showing this to my students as well.
That is one long way to do it. Or draw a line from one corner to the other where that forms a triangle with the circle inside it, and the circle, that line (hypotenuse H), and the top right corner of the little square all touch. Then the formula for the max radius of a circle that can fit in a triangle is r=(P + B - H)/2 P and B (perpendicular and base)
r= ( 4 + 4 - 4sqrt(2) ) / 2
r = ( 8 - 4sqrt(2) ) / 2
r = 4 - 2sqrt(2)
What software do you use for the visualization?
Your how exciting was rushed for this one! Thanks for another fun problem.
sqrt(2)r + 2x = 2r, while 2r-x=2. x is the side length of the small rectangle between the circle and the large rectangle.
Or more precisely, 4 - sqrt(8).
Co-ordinate type geometry is an easy way to see this. Centre of the circle is (x,y). We have two radii from (x,y) to (4,y) and (x,4). The radius is therefore 4-x (or 4-y), so we see x = y, as we'd expect from the symmetry of the construction.
Now we have another radius from (2,2) to (x,y). So by Pythagoras we have 2(x-2)^2 = radius^2 = (4-x)^2. Solve this to get x = sqrt(8) (clearly x>0 by construction) and therefore the radius is 4-sqrt(8).
For any triangle S=p*r, where p - half perimeter, r - radius of inscribed circle. In our case, S = 0.5*4*4 = 8 (half the square); p = 0.5*(4+4+4*sqrt2). Thus, r = 8/(4+2*sqrt2) = 4/(2+sqrt2) = 2*(2-sqrt2) =~ 1.1715
I also remember there was a formula to determine the radius of an inner circle of a triangle, which you could transform this problem into
The radius of a circle inscribed in a triangle is two times the area of the triangle divided by the perimeter of the triangle. The area of the triangle is 8. The perimeter of the triangle is 4 + 4 + 4xsqrt(2). The radius of the circle is 16/(8+4xsqrt(2))=1.17157.
I found a different way of solving it: the circle is tangent to a triangle when you divide the square in half.
You can find the radius by using: r.p= A
In this case, the area is 8, the semi perimeter is (sqrt(32)+8)/2. If you solve this and isolate the radius you'll have the same result.
If there is a circle inscribed and tagent in any kind of convex polygon, the area of the polygon will be r.p
This is true even if the circle isn't tagent to all sides of said polygon, it only needs to be tanget to the highest number of sides that it can touch while being inscripted. In this case, it was tangent to all 3 sides of the triangle
The top right point in the square is the director circle to the given circle now using coordinate geometry results we know distance of such point from center of circle is always r√2
Director circle : locus of points from which tangents drawn to a curve are mutually perpendicular
Where r is the radius of given
I did it a bit differently.
I completely ignored the smaller square, and instead I just focused on the fact that there's a right triangle with both of it's legs being equal to 4, whose incircle is that circle.
First we find the hypotenuse, which using the Pythagorean theorem we find is equal to 4√2.
Now we find the semi-perimeter p, which is half the perimeter. It is equal to 4 + 2√2.
After we find p, we can now use Heron's formula to find the area S = √(p ⋅ (p - a) ⋅ (p - b) ⋅ (p - c)). We calculate the area to be equal to exactly 8 square units.
Now we know that S = pr, therefore r = S / p, and when we calculate it and rationalize it we get 4 - 2√2 (same answer as Andy, just rationalized).
The diameter of the circle is larger than 2. Assuming the circle is tangent to the outer box in both parts of the corner that diagonal line would never intersect the center point of the circle. so your answer is close but not correct
I guess I did this one the hard way. First calculate the diagonal of the small square, as shown. Then extend that line to the center of the circle to create a right triangle with hypotenuse 2√2 + r. Each leg is 4 - r. The Pythagorean Theorem means that (4 - r)^2 + (4 - r)^2 = (2√2 + r)^2. Multiply it out and you get a knotty quadratic: r^2 - (16 + 4√2)r + 24 = 0. This actually is amenable to solving with the quadratic formula. You get a root within a root, but that can be handled. Finally, r = [(16 + 4√2) +/- (8 + 8√2)]/2. The larger root is way too big for the given parameters. But the smaller root, 4 - 2√2, does equal 1.17157.
Draw a Tangent to the circle from bottom right corner of Larger Square(where tangent touches the circle at centre of the larger square) and this length is half the diagonal of larger square = 2√2. Now tangents drawn from a point outside to the circle are equal in length(one along the diagonal and one along side from the bottom corner of larger square) hence 2√2 + r = 4
r = 4 - 2√2
OR
Draw the diagonal of the larger square (connecting bottom right corner and top left corner vertices). Now we have a Right Triangle where the sides are 4, 4, 4√2 and circle is the incircle of this triangle whose radius is (4+4-4√2)/2 = 4-2√2
PS :: Inradius of a Right Triangle, whose sides are a,b,c with c being hypotenuse is (a+b-c)/2
You can go a step further and get rid of irrational ratio by muliplying top and bottom by √2-1, thus r is 4-2√2, looks better.
For the radius of a circle to be expressed in p/q form, the denominator must be a rational number. In your case it is irrational, add 2 extra steps for componendo-dividendo and the final answer would be 1.51....
My answer: 4-2root2
My working:
The vertical distance from the corner of the little square to the roof of the big square is 2
The vertical distance from where the circle touches the corner of the little square to the roof of the big square is radius + (radius/root(2)) since the angle is 45 degrees
Then just do the algebra for 2 = r + r/root2
You should have rationalized the denominator by multiplying top and bottom by (1 - sqrt(2)). The denominator simplifies to 1, which eliminaties the fraction completely, and the numerator becomes 4 - 2(sqrt(2)). Evaluating that gives the same answer: 1.17157...etc.
Also, much simpler way to do this: Extend the horizontal top of the lower left square to the right-hand vertical side, dividing it into 2 segments, each length 2 (just like the left side). The top part of the top segment is r, as you've shown. The bottom part of that top segment is the side of a square whose diagonal is r. That's r/sqrt(2). So now we have r + r/sqrt(2) = 2. Solving this equation gives the same result.
what i did was i drew a radius line from the center of the circle to the vertex of the top right corner of the small square. this line is the radius but anglesd 45 deg downwards, therefore, the horizontal component is r*cos45. drew another line horizontally from the center of the circle to the right side of the big square. that radius is r. so rcos(45) + r = 2. solve for r and you get 4-2√2
Exellent solution! I am trying to make understanding of this, and 1:30 you said you can sqaure root both sides of the equation. Unfortunately I don't think this works as 3^2 * 4^2 = 5^2, however 3 * 4 does not equal five. I would be gratefull if you could shine some insight on this.
Whats cool about these type of problems is that they arent too hard or too easy, and can be solved using different methods
Last advice to use calculator would lower the score when I was in school. My teachers always emphasized to use exact computation, in this case it means that fraction should be multiplied (upper and lower part) by (sqrt(2)-1). Then upper part becomes 4-2*sqrt(2) and lower part becomes 1 (because (a+b)*(a-b) = a^2-b^2, it his case a=sqrt(2), b=1 so a^2-b^2=2-1=1). Final value should be 4-2*sqrt(2).
I’m with you…how exciting! When I help my teenager with math problems, I’ve started saying “how exciting” when we get to a solution. Not sure it’s working but I keep trying!
That was a real pleasure to watch. Thank you!
An alternative form of the solution is 4-√8, now solve the 3D equivalent of the excercise (A half size cube inside a cube and a spehere touching its corner and the faces of the bigger cube, what is its radius?)
Its not the fact he solves these things - I dont think that I’ve ever seen him do maths that I didnt know - but its the ingenuity with which he solves it.
Its one thing knowing how to get through the math, and another thing entirely on seeing a bunch of shapes and knowing what math to use to get you there.
Logic flaw here, if you rotate side 'c' down onto side side it 'must be longer' than side 'a', if it were actually 2 units the angle of 'ca' would be zero or side 'a'. Simple proof on paper, create the box, draw the diagonal 'c', get a protactor (Lead pencil) spead the legs the distance of 'a' and rotate the protractor to the verticle, 'c' should be bisected.
Caou say hat in German?
I just put a diagonal in the square from the top left to the bottom right made the sides equal to n+r and the diagonal to 2n since we have 90° we know that the diagonal is equal to √(2×4²) meaning n is half of that than just 4-n and you have r. If you continue with this you can get that r is about equal to 0,29289[((2-√(2))÷2), 1-√0.5]×a side of the big square for all values if you have different side lengths it would obviously change, even tho you would still be able to generalize that.
My approach to this problem is different we don't need the square, divide the squared into 2 parts diagonally and this should give you a circle inscribed in a triangle then therefore we can use the formula r=(a+b-c)/2 a=b and c=4root2 r=(4+4-4root2)/2 or 1.1715...
My solution:
- The large square has a side of 4. Take 2 from the small square. We have now 2.
- From the circle we have r+r*cos(45°) which is equal to 2
Then r+r*cos(45°)=2
r(1+sqr(2)/2)=2 -> r≈1.1716
Math becomes much more fun once you do it voluntarily just for the sake of it and not under pressure for grades.
My solution:
From the centre of the circle to the point it touches the square is r so the horizontal distance is r√2. Add that to r and you get the horizontal distance to the edge of the outer square therefore r + r√2 = 2 and r =
1/(1+√2)
I use BPT theorem to find radius of the circle by making two equal triangel with common hypotenious which is equal to the diameter of the square , and let the base of these two triangle by making one horizontal and vertical line from the side which is touching with the circle which makes the two triangle who follow the BPT rule because the length "r" both horizontal and vertical and "2units" both horzontal and vertical are prallei t each other . Now i put the values directly and indirectly and apply the BPT to get radius
thanks for giving such cute question
❤🔥❤🔥
edit= my approx anser is |1.18| because i use"root2" as 1.14
I solved it using trigonometry. It's essentially a circle inside a right isosceles triangle.
So the radius comes to 2√2 x tan(π/8)
lol I too a completely different approach (didn’t have pen/paper so I needed an easier way):
basically, assume the center of the circle is (x,y). The center of the circle must lie on the main ascending diagonal of the square for symmetry reasons, so x=y. The lower square touches the circle at (2,2). Using this, the distance from (2,2) to (x,x) is sqrt(2)*(x-2)…
But going from the center of the circle to the top of the large square is also a radius (4-x)… you can do the same horizontally. Thus, we get sqrt(2)*(x-2)=4-x.
Notice that this is a simple linear equation in one variable… solve for x
we immediately get:
[sqrt(2)+1]*x = 4+sqrt(2)*2
x = 2[sqrt(2)+2]/[sqrt(2)+1] = 2*sqrt(2)
… therefore, radius = 4 - 2*sqrt(2) 🥳
✨as a bonus this mean’s the TH-camrs expression 2*sqrt(2)/[1+sqrt(2)] simplifies beautifully to 4 - 2*sqrt(2)
OOTB thinking at its max!!! Nice work!
Edit: This channel deserves a sub, and so I have!
Just looking at the problem, I have no idea where to even start.
well to be honest I got to 1:10, but no further. The center of the circle. How do I find that.
Awesome dude!
If the goal was to solve this problem using the most complicated method possible it was a success.
It's a lot less calculation intensive to solve this problem (starting from the drawing with the 3 radii) by simply adding a diagonal line from the top left corner of the square to the center of the square which is a point of tangency with the circle.
Because we can see that this diagonal line is the hypotenuse of an isosceles right triangle with side lengths of 2, we know that this hypotenuse length will equal 2√2 and therefore the length of the square's top side left section (from the left corner to the point of tangency at the top of the circle) is also equal to 2√2.
The drawing has already established that the square's top side right section (from the point of tangency at the top of the circle to the right corner) is equal to the length of the radius.
The length of the radius equivalent section of the top side of the square (r) can be calculated from the difference between the total length of the top side of the square (4) and the length of the left section of the top side of the square (2√2).
r = 4 - 2√2
r ≈ 1.17
At the end multiply the top and bottom of that fraction by (√2 - 1) to clean up the answer
Nice solution! If you want to find the exact answer, then multiply the denominator by the conjugate. (2√2 - 4)/3.
Double-check your denominator there, I think you added the squares instead of subtracted.
That was the most genuine "how exciting" I've heard you proclaim. Hyped me up to make a comment.
Can you explain how your circle comes to the centre point of the square, and has an arc that extends past the centre on two sides only?
Why are you using Pythagoras for all those diagonals ?
It would be faster to use the formula d = s * √(2) where d is the diagonal of the square et s its side.
And where do you take this formula from?
@@WK-5775 It's a basic formula that you learn in high school...
You need not be concerned with the big square. just draw the 2 x 2 square in the upper right corner. From that (summing along the horizontal OR the vertical) r/√2 + r = 2 & solve for r to get the same answer (but you should rationalize the denominator to get r = 4 - 2√2).
Well.. to find the diagonal. we can directly use the formula. Side×√2
Here side = 2, so diagonal is 2√2.
I suck at math but unless I am mistaken the radius of that circle has to be half the distance from the connecting corner of the lower right square to the square to the opposite corner of the upper right square and that is simply the square root of 2 squared + 2 squared = 2.828
no?
Another calculation for big diagonal (bd):
bd^2 = 4^2 + 4^2
bd = root(4^2 + 4^2)
bd = root(16+16)
bd = root(2*16)
bd = 4*root(2)
That was cool 😎. Thanks for sharing! How exciting indeed! 😂
This seemed like one of those questions on a math test I would skip because it took way to many steps and not enough time to deal with it when you have 3 other questions that are just as long but are easier to solve.
Helps if you flip the whole thing 180 degrees. r + root 2 / 2 * r = 2. r = something or other…. You get the gist.