Thanks for a wonderful 2018! Here are my most popular and best videos from the year! Can You Solve Amazon's Hanging Cable Interview Question? (4.8 million views) th-cam.com/video/l_ffdarcJiQ/w-d-xo.html The REAL Answer To The Viral Chinese Math Problem "How Old Is The Captain?" (3 million views) th-cam.com/video/uyS1cXrsgIg/w-d-xo.html What Is The Area? "You Should Be Able To Solve This" (1 million views) th-cam.com/video/BgrWHOocYZA/w-d-xo.html Homework From China: How Tall Is The Table? (1 million views) th-cam.com/video/BPRueCu3fXU/w-d-xo.html Can You Solve The 6s Challenge? (796,000 views) th-cam.com/video/h2vkrxvh76c/w-d-xo.html How Many Holes Does A Straw Have? The Correct Answer Explained (666,000 views) th-cam.com/video/l-PQ-pUVeho/w-d-xo.html "Prove" 3 = 0. Can You Spot The Mistake? (530,000 views) th-cam.com/video/SGUZ-8u1OxM/w-d-xo.html A Chinese 5th Grader Solved This In Just 1 Minute! HARD Geometry Problem (390,000 views) th-cam.com/video/OuJQaxZvlYs/w-d-xo.html What Is 60÷5(7-5) = ? The Viral Problem Everyone Is Arguing About (330,000 views) th-cam.com/video/wzchhbrqIBI/w-d-xo.html Viral Facebook Puzzle. What Is AB = ? (330,000 views) th-cam.com/video/ah7hwQgJ_zE/w-d-xo.html INCREDIBLE DISCOVERY! Multiply By Lines Group Theory th-cam.com/video/LmVt_KiyDC4/w-d-xo.html This was my first ever worldwide premiere of a mathematical discovery. I want to thank everyone who gave positive feedback and I'll remember you when this method becomes more famous! Line multiplication deserves an entry in Wikipedia: 11 years with no recognition is long enough!
MindYourDecisions Hi, I love your chanel and I have a question that my complex calculator couldn't specify it. The question is... Function i (n) = Log*i (i+n) (N starts being 1) Can you solve it for me? Thank you for your awesome videos.
If you want more tricky math problems, I suggest browsing for math olympiades on the internet. There are tons of these kinds of problems. Just as impossible to solve as this one...
geometry trignomerty question asked for std 10 students in class. a straight highway lead to the foot of the tower. a man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower at uniform speed. six seconds later the angle of depression of the car is found to be 60°. find the time by the car to reach the foot of the tower
@@DylanSargesson really depends on the choices. I did UKMT last year and got gold award. If the options are close to 2/3, like 3/5, it is not easy to tell.
@@jayi31 Not the maths you do for tests at GCSE level (UK). It's just method learning. But you are correct, if you're a problem solver then complicated maths is made much easier.
David Patel we don’t really do that kind of thing here, what we learn in-class alone is a C-grade. To get an A or a B, you need to be able to correctly apply the maths to answer questions, aka problem solving
@@jayi31 Ahh I see. I'm currently doing my further maths at GCSE and that's way more problem solving but the normal GCSE isn't. I guess some of the circle theorems require you to decipher which is needed but generally I wouldn't consider it problem solving.
It’s getting really close to the screen to count each red pixel, then counting each white pixel on the monitor & adding them up & dividing them. You should get & answer very close to 2/3. Like .66666666666667
I feel like, even if, during the most intense focus I had during my Geometry classes, when I actually knew stuff about circles, I wouldn’t have the imagination to piece my way through this, even though the steps he used seemed blatantly obvious when he did them. It’s like I have the understanding, but not the mindset needed to comprehend the problem
I loved this problem. Even though I didn't figure it out, I feel like I gained more of an understanding of the geometry behind it. We need more videos like this.
I used vector addition to solve this. Once you realize you have a square in the lower left corner, assuming s=1, then you can write the vector equation... + = The vector on the right side of this equation describes a right triangle with hypotenuse q. Now you can use Pythagoras's theorem to solve for q. q^2 = (1+1/sqrt(2))^2 + (1-1/sqrt(2))^2 q = sqrt(3). Now it's just a simple matter of plugging s and q into the appropriate area formulas (quarter-circle and semi-circle) and calculating their ratio.
I used the intersecting chord theorem for this one. Let the big radius be R and the little one be r. First extend the quarter circle to make a full circle. Now draw a line through the center of both circles and extend it to hit the large circle on both sides. Now we have two intersecting chords of length 2r (r and r) and 2R (R+sqrt(2)*r and R-sqrt(2)*r). By the intersecting chord theorem, we get r^2 = (R+sqrt(2)*r)(R-sqrt(2)*r) -> 3r^2 = R^2. Thus we get (1/2)r^2/(1/4)R^2 = (1/2)r^2/(3/4)r^2 = 2/3. Nice problem!
@@aniruddhvasishta8334 Neither had I but a quick google came up with the answer: In any circle where two chords intersect the product of the lengths of each chord are equal. Let chords AC and BD intersect at point Y then the theorem states: AY × YC = BY × YD If ABCD are the vertices of a quadrilateral and its diagonals intersect such that the product above holds then ABCD is a cyclic quad. In this problem if AC is the diameter of the semicircle (=2r) then we can work out the lengths of BY and YD since the semicircle is tangent to the radii (R) of the quarter circle - the distance from the centre of the large circle to the centre of the semicircle is found using Pythagoras to be r sqrt(2). Thus the segments of the large circle's diameter are R - r sqrt(2) = the length between the semicircle's diameter and the circumference of the quarter circle; and R + r sqrt(2) = the length between the circumference of the full large circle opposite to the semicircle and the centre of the large circle plus the distance between the two centres.
I tried doing it in my head by setting the quarter circle radius to 1 and figuring out the radius of the semicircle by Cartesian geometry, by finding expressions for the coordinates of the end of the diameter assuming that the diameter is at 45 degrees by symmetry. Squares of those coordinates sum to 1 because it's on a unit circle. Then I messed up the mental algebra, got that the radius was the square root of a sixth, squared that for the area factor, multiplied by 2 for being twice as much of the circle to get 1/3. Then I looked at the diagram, which is obviously mostly red, and said the answer had to be 2/3 instead. Doing the algebra correctly ((r(1+1/√2))^2+(r(1-1/√2))^2=1 means r is 3) gives the correct answer. It's really handy to pick the unit you will find convenient when doing these, and it's what Euclid would do.
I did basically the same thing, but I let the radii of the circles be variables and they all cancelled out nicely in the end and I arrived at the answer 2/3. (I used a pen and paper though, I didn't try working in my head.) I often find when I get stuck on a geometry problem, I try setting up Cartesian coordinates and that often guides me to a solution. In this case, finding coordinates of the point where the cusp of the semi-circle meets the quarter circle and then substituting into the equation of the quarter circle gives the ratio of the square radii almost immediately.
I did the same thing as you in my head (about 5 mins) Not being able to draw lines on paper was a challenge and its probably why I couldnt get the elegant solution presented in the video.
I am so glad you told the viewers to put serious effort into this instead of giving up and watching the solution because it took me 10 minutes but I was finally able to see how I could solve it
These sort of math problems come in handy in engineering, like designing tanks to hold fluids, making components etc. That is why I like your channel. Thanks and thumbs up!
Oh interesting, I did use Pythagorean’s theorem at all ... I set up q=1 to work within the unit circle. I took the points of intersection between the 2 circles (the two extremities of the chord, let s say I and J) and gave then coordinates: I (a,b) and J (b,a) ; they are mirror image of each other along the x=y line (the bisector) Note that the coordinates of O the center of the semi circles is (a+b)/2 in both direction I (and J) are on the unit circle so a^2+b^2=1 The semicircle is tangent to the axis so it s radius s = (a+b)/2 The chord’s length (between I and J) is equal to root(2)*(b-a) and is also a diameter of the semicircle (=2s) Hence (a+b)=root(2)*(b-a) Then I solved the system of two equations for a and b (a lot of algebra) Finally we have s^2=(a+b)^2/4 S^2=(a^2+b^2+2ab)/4 S^2=(1+2ab )/4. As my algebra gave me 2ab=1/3 (I let you redo the work if you want won t type it) then S^2=4/3/4=1/3 The ratio (r) between s^2/2 and q^2/4 (always with q=1) is r=(1/6)/(1/4)=4/6=2/3 !!!! Surely this is more algebra heavy, but you can solve the problem even if you can t spot the hidden triangle ;)
I did it the same way, but set s=1. Then the coordinates of the center of the semicircle are (1,1). And then it's easy to find the coordinate of one of the points of intersection, because it's x=1-sqrt(.5) and y=1+sqrt(.5). Once you have that point (or its mirror), finding q is easy: q^2=x^2+y^2.
It was a nice one from you. The UK challenge is a fun. Mainly in such cases,theorems related to tangency and chord intersection of radius comes in aid.
I got 2/3 by extending the inner half circle to a full circle, drawing the square that circle is inscribed in and solving for the diagonal of the square. Then set the radius of the inner half circle equal to half the side length of the square. Then lots of algebra. I'm curious if Presh has a more elegant way.
Wow, I can't believe I got it! But Not as elegantly, I found the lower coordinates where both meet. Used Pythagorus to get (s - s/√2, s + s/√2), then subbed these x & y values into the large circle's equation, y² + x² = q². Magically simplified to q² = 3s², as you found more easily!
From the thumbnail: Obviously the scale of the drawing doesn't affect the ratio of areas; so just suppose the radius of the small circle is 1, and the radius of the large circle is unknown r. Consider a coordinate system with horizontal x-axis, vertical y-axis, and its origin (x,y) = (0,0) at the center of the small circle's diameter. The vertices of the small circle half that touch the large circle arc should then have coordinates that satisfy the following three equations: eq. 1 - arc of small circle: x² + y² = 1² eq. 2 - arc of large circle: (x+1)² + (y+1)² = r² eq. 3 - diameter of small circle: x + y = 0 (We know that eq. 3 holds because the line x = y is clearly a symmetry axis of the diagram.) From eq. 3, we derive y = -x; plug that into eqs. 1 and 2: eq. 1a: 2x² = 1 eq. 2a: (1+x)² + (1-x)² = r² ==> (1 + 2x + x²) + (1 - 2x + x²) = r² ==> 2 + 2x² = r² Plug result of eq. 1a into eq. 2a: 3 = r² ==> r = √3 Area (red) of half small circle : pi*1²/2 = pi/2 Area of quarter large circle : pi*r²/4 = pi*3/4 ratio of areas: (pi/2)/(pi*3/4) = (1/2)/(3/4) = 4/6 = 2/3
I solved this one a different way (based on approaches I learned in your other videos!). You can use intersecting chords on the large circle to get the following: (q + s*sqrt(2))(q - s*sqrt(2)) = s^2 q^2 - 2s^2 = s^2 q^2 = 3s^2 This substitutes very nicely into the equation for the ratios between the areas to give the final answer. Thanks for making these videos! I feel I'm becoming a better mathematician with each one.
... given how two-dimensional objects behave in that regard. 2/3 would be an immediate answer related to a gut feeling. If s was 1/2 q then the semicircle could fit in twice. (Ex: 0.5x2^2=2. 0.25x4^2=4. --> 2 fits twice in 4) Thus we find our interval in sq/2 bringing the solution closer to 0.666 then proceeding like you did. However the part above was merely a thought to estimate what would later become the solution.
I figured it out... Thanks again for your videos. Sometimes I stop them and try before continuing watching to check the result. It's fun and training for the brain, and improves the abilities. And success gives good feeling as a reward for the effort. And in case I'm wrong or not able to figure it out, your videos give us a lesson to learn from 👍
2/3. I see a lot of angles. The one important is 30°. The solution is tangent(30°), which is also the unit-radius of the red shaded area. Interesting note: a)If you square "tan(30)" you get exactly 1/3. That means that the area of a (full) red circles will fit exactly 3 times into one (full) white circle. b)It also means that the distance from centre of the white quarter circle to the point where red circle touches the tangent is tan(30)*Radius(white).
You can also write the equations of the two circles and then set the condition that the two points of intersections of the two circles belong to the line of equation y = -x +2 S, where S is the radius of the red circle. It simplifies out and you get the ratio of the two radii and then you find the ratio of the areas.
Took me a bit of time. Did it before I watched the Vid and ended up with the same method and answer :) Quite a challenge, having left high school for almost 20 years! Thanks!
I just took the radius of the red semicircle to be 1/2 the radius of the white quarter circle. made it pretty easy to just write one radius in terms of the other and solve for area, then compare. I think that the "small radius is half the large radius" was a wrong assumption to make and basically let me cheat the problem, but I don't get why it was wrong.
@@IntrusiveThot420 I think the most basic reason why it is wrong is because it is not correct. Plug in your values for each radius and take the ratio of the areas. Do you get 2/3? There is no reason to suspect that the radius of the semicirle is 1/2 let alone that it is even rational.
This is the best geometry problem you ever had on this channel imo. The one from China with the parallelogram was also pretty cool though. Really like the approach of solving for the squares rather than the actual radii.
Yes. When caring for knowing the correct answer outweighs caring for knowing which and how to apply the (most) correct method to find the correct answer.
Or, extend the diagonal to the full large circle. Then by the Intersecting Chords theorem, s*s = (q+s√2)(q-s√2) = q² - 2s², and thereafter the same. From this, we deduce that the Intersecting Chords Theorem proves the Pythagorean Theorem :) The converse doesn't quite hold, since the two chords in the ICT don't have to be diameters of the large circle, so you may not get any right triangle.
I have another solution. Imagine the inner halfcircle as scalable and the quarter circle to have a radius of 1, where h is the vertical distance from its top left corner to the arc of the quarter circle above. The function of the quarter circle is sqrt(1-x^2), it gives us the vertical distance between a point on QC's arc and x line. Using this, you can derive the h function using geometry, where the input is the radius of inner HC: h(r) = sqrt(1-(r-(r/sqrt(2)))^2)-(r+(r/sqrt(2))) When we solve the above equation for 0 (when h is 0, meaning the HC got big as much as it could and its corners touched the QC's arc), we figure r has to be 1/sqrt(3). From there on, you can figure the areas of each shape. Area of the HC: (pi*(1/sqrt(3))^2)/2 = pi/6 Area of the QC: (pi*1^2)/4 = pi/4 Their ratio : 2/3
You can still figure out that r is 1/sqrt(3) if you follow the way in the video. What's even cooler is, this leads to a general formula where r is the HC radius, R is the QC radius: r*sqrt3 = R
It's pretty straight forward with analytic geometry. I called the center of the small circle P and defined it as L*(1|1). The crossbar through P has the formula P+µ*(1|-1). The quarter circle is defined by (X)²=1 (by similarity I am allowed to chose a radius to make things easier). The intersection of the crossbar and the quarter circle I called D. the lenght of PD is the radius of the smaller circle. The half circle is defined as (X-P)²=PD². A1 and A2 are the intersections of the half circle and the line (0|y). to only get a touching point we set A1=A2. Now we just solve µ for L, then solve A1 and A2 for L, set the discriminant to 0 to reduce A1 and A2 to one solution and finally we can use L to get PD, which is the radius of the small circle (the larger had radius 1) and then we can compare the areas.
This one is the first that I've been able to go down the right path to the solution. Past experience with these questions led me to look at the same triangle in your solution. I just bungled the math a bit once I found that the radius of the big circle was the square root of the sum of s squared & 2s squared.
This problem was in the maths challenge that I sat and I’m 16, you look about 35. Considering I go to a grammar school and only students who take further maths sit the paper, and people still found it hard. I think you should consider that you’re not his only demographic.
I did this differently, using algebra. Let radius of quarter circle be 1. Let s = radius of half-circle. Equation for quarter circle is X^2 + Y^2 = 1. Equation for half circle is (X-s)^2 + (Y-s)^2 = s^2. Boil down algebra for half-circle, you get equation containing X^2 and Y^2 terms on left side (along with others). Remove these 2 terms and replace with 1 (from the equation for the quarter circle) to get s^2 - 2sX - 2sY + 1 = 0. Equation for the diagonal (diameter of the half-circle) is Y = 2s - X. Substitute for Y into other equation resulting in -3s^2 -1 = 0, s = 1/((sqrt(3)). Area quarter circle = pi/4, area of half circle = pi/6. Ratio = 2/3. This is interesting because 3 equations were used: those of the 2 circles, and the diagonal diameter line of the half-circle to find s = 1/(sqrt(3)).
A more abstract way of looking at is if you tilt the semicircle (and increase its size) until it's straight line is perfectly vertical you could place it next to the quarter that it is in and create 3/4 of a full circle but then the max limit would be 3 pieces so x/3. Then you just take the same semicircle you rotated and measure how much it takes up of the incomplete circle, which is 2/3.
Alexander Corke it wasn’t just the Scottish one it’s was the whole of the uk, they’re were some good questions on the papers especially the cone one with the 2:1 ratio of arc lenghts
As soon as you visualise the radii meeting the tangents, which is a very obvious thing to do, the problem solves itself. I think that this is an 'O' level type problem, but what do I know.
Very elegant : "Because the radius bisects the chord, they meet at right angle" is the key here. I missed that, so i did it the hard way : intersecting cords and a lucky quadratic to find that R =sqrt(3)*r , lol .
In the exam I had like 5 mins left to do this and one other question. I drew out the diagram enlarged on rough paper. Ripped down the edge of the two regions and estimated that it seemed twice as big. Giving a ratio of 2/3. It worked!
RH Versity I actually did the paper this question was in. It was the last maths challenge paper I’d be able to take part in. Luckily I made it into the first round of the olympiad.
@@tpaxatb. It doesn't say that the diagram is not draw to scale. I doesn't matter if it is draw to scale or not the answer will still be 2/3. Don't believe me, then do some working out with different size and I bet you $100 that it is 2/3 for any size.
1. Measure the visible circumference of the semi and quarter circles. 2. Translate them in to full circles 3. Calculate the area of each 4. Divide the area by 4 for the quarter circle and 2 for the semi circle 5. Take away the semi circles area from the quarter circles area 6. Translate that number into a fraction
I used a similar but different method: Each s radius of the semicircle's base is at 45°, so it's extent in either x or y is s/√2. So assuming the corner of the quarter circle is at the origin, the coordinates of the upper end of the q radius is (s-s/√2, s+s/√2). Using Pythagorus, that makes q^2=3s^s. Putting that in the fraction for the area gives the answer: 2/3. Took less than 5 minutes.
A good generalization of this problem would be to draw a smaller semi circle inside an arc of a bigger circle where the arc subtends angle 2*theta at the center of the bigger circle. Calculate the ratio of the area of the smaller circle to the area of the sector formed by the bigger circle. This would need some basic trig.
New here, I liked this problem. I knew the key was to find the radius of the semicircle and find it’s relationship to the radius of the quarter circle, but forgot a lot of the geometry tricks that could be done to figure it out- I tried thinking I’m terms of calculus, but the answer was simpler than I thought.
Did it in a slightly simpler way by assuming the small radius is 1. Proceeded the same way as in the video from there and discovered the answer 2/3 pretty quickly. Nice problem.
You can use power of the point on the center of the semicircle in order to find the length of the space between semicircle and the quarter circle, and therefore the radius of the quartercircle. Everything works nice from there
As far as problem solving goes, I think if confronted with circles, assume the radius of one of them to be one, and from there look for relations to make with that radius. Here, you can try the outer quarter circle first, but then see it does not yield many easy relations between the two as there are two unknown distances if that radius goes through the center of the semicircle and two unknowns when it touches a tangent. Switch to the setting the semicircle radius to one. That radius touches the outer circle at each tangent and also twice on the circumference. From there it is just a matter of observing the right triangle whose hypotenuse is the radius of the outer circle and calculating its length.
This solution was so elegant that when I figured it out, it felt like I should have done so at least twice as fast. I was stuck with sqrt(2)s for a good while before realizing how easy it was to determine q to be sqrt(3)s. Except that I just used unity in place of s.
If you are for some reason not seeing that last triangle, you could also use the rule for intersecting cords: (2*sqrt(2)*s + x) * x = s^2, which resolves to (x + sqrt(2)s)^2=q^2=3s^2.
2/3. Split the quarter circle in halves, you will get two right triangles（the centres of two circles plus the end of the diameter of red circle, the centres of two circles plus point of tangency）. Apply the Pythagorean theorem in both triangles, you will find the ratio of squared radius between the two circles is 1:3. The rest is obvious.
I did figure it out, however witha different (and a more complicated) solution. Using trigonometry I found both horizontal and vertical components of the inclined bigger radius (which were 1-sqrt(2) and 1+sqrt(2)), used pythagorean theorem and got the same answer. I love it when unrelated solutions give the same answer. Cheers.
Geez, I did it the hard way: I completed the semi-circle, drew the square it would be inscribed in, and then added the line y = -x + 2r, where r = the smaller radius. Then found the x value where this line intersected the circle at the upper left, then found the distance that point was away from the origin, and that was R, the larger radius, in terms of r, the smaller radius. I also change a car's tire by first dropping the transmission.
coordinates of small circle: (r,r) area S0 = pi*r^2/2. Coordinates of big radius y = r + r*cos(45deg)=r+r/sqrt(2) x = r - r/sqrt(2). so big radius R = sqrt(x^2 + y*2) = r/sqrt(3). area S1 = pi*R^2/4 so S0/S1 = 2/3
Thank you for telling me that it took your 30 minutes to solve. I usually look at these problems for 5 minutes and if I can't solve them, I give up. But telling me that it took you 30 minutes, encouraged me to take a little more time.
Thanks for a wonderful 2018! Here are my most popular and best videos from the year!
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INCREDIBLE DISCOVERY! Multiply By Lines Group Theory
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This was my first ever worldwide premiere of a mathematical discovery. I want to thank everyone who gave positive feedback and I'll remember you when this method becomes more famous! Line multiplication deserves an entry in Wikipedia: 11 years with no recognition is long enough!
Can't we replace 60 as 5 (7+5) in that video where u discussed the problem
60÷5 (7-5)
Presh are you INDIAN??
Please please please reply!!
MindYourDecisions Hi, I love your chanel and I have a question that my complex calculator couldn't specify it.
The question is...
Function i (n) = Log*i (i+n)
(N starts being 1)
Can you solve it for me?
Thank you for your awesome videos.
If you want more tricky math problems, I suggest browsing for math olympiades on the internet. There are tons of these kinds of problems. Just as impossible to solve as this one...
geometry trignomerty question asked for std 10 students in class.
a straight highway lead to the foot of the tower. a man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower at uniform speed. six seconds later the angle of depression of the car is found to be 60°. find the time by the car to reach the foot of the tower
"Did you figure it out?"
No, Presh, I didn't
Hercules?
"No, I doubt".
I did...
Presh said calmly.
Doesn't look like half, so it has to be 2/3. That's my explanation.
lmao same
as a second guess I had 3/4 but my first was 2/3
i did that once in a math test and i got it right lol
Given that the UKMT is multiple choice, this would've been the rational of most test-takers
@@DylanSargesson really depends on the choices. I did UKMT last year and got gold award. If the options are close to 2/3, like 3/5, it is not easy to tell.
Lol my engineering professor saw this and said "cut out the semi and quarter circles on a piece of paper and get ratios of the weights."
Well thats one way of soLVING IT.
genius
I thought the same thing... I am an analytical chemist.
Jeremiah Halpin I saw it and said yea about 2/3 I always do this with fractions being lazy kills
i just looked at it, seemed like two thirds and i was right
Interesting, I'm pretty ok at maths, I can figure this out.
3 minutes later: I am not ok at maths.
😂 tbf the maths is quite basic with circle theorem and Pythagoras and such but the problem solving is hardddd
David Patel Problem solving is a big part of maths
@@jayi31 Not the maths you do for tests at GCSE level (UK). It's just method learning. But you are correct, if you're a problem solver then complicated maths is made much easier.
David Patel we don’t really do that kind of thing here, what we learn in-class alone is a C-grade. To get an A or a B, you need to be able to correctly apply the maths to answer questions, aka problem solving
@@jayi31 Ahh I see. I'm currently doing my further maths at GCSE and that's way more problem solving but the normal GCSE isn't. I guess some of the circle theorems require you to decipher which is needed but generally I wouldn't consider it problem solving.
eventhough i like his videos the part where he says "did you figure it out?" always triggers me
same feeling..
Everyone told me to keep saying "did you figure it out?" in this video: th-cam.com/video/Cjf_Mj29uLo/w-d-xo.html&lc=UgzYP1QMe70IuSqcFJZ4AaABAg
It's did *you* figure it out? 😂
@@MindYourDecisions yes keep saying that . I always like when you say *did you figure it out* . It suits you I guess.
It triggers me because the answer is always no lol
Pretty sure counting the pixels would’ve been easier
no counting the atoms is even easier
What’s that?
It’s getting really close to the screen to count each red pixel, then counting each white pixel on the monitor & adding them up & dividing them. You should get & answer very close to 2/3. Like .66666666666667
I put a ruler on my iPad. I guess I cheated.
Holly T that’s still probably easier
"What fraction of the quarter circle is shaded in red?"
(points to the red half circle) That bit is.
@mike casanave
Lol yeah. I'm always looking for the easy out. Ya know? 😏😁
I feel like, even if, during the most intense focus I had during my Geometry classes, when I actually knew stuff about circles, I wouldn’t have the imagination to piece my way through this, even though the steps he used seemed blatantly obvious when he did them. It’s like I have the understanding, but not the mindset needed to comprehend the problem
I loved this problem. Even though I didn't figure it out, I feel like I gained more of an understanding of the geometry behind it. We need more videos like this.
I used vector addition to solve this. Once you realize you have a square in the lower left corner, assuming s=1, then you can write the vector equation...
+ =
The vector on the right side of this equation describes a right triangle with hypotenuse q. Now you can use Pythagoras's theorem to solve for q.
q^2 = (1+1/sqrt(2))^2 + (1-1/sqrt(2))^2
q = sqrt(3).
Now it's just a simple matter of plugging s and q into the appropriate area formulas (quarter-circle and semi-circle) and calculating their ratio.
Yes it is a good problem between hard ones and easy ones
I used the intersecting chord theorem for this one. Let the big radius be R and the little one be r. First extend the quarter circle to make a full circle. Now draw a line through the center of both circles and extend it to hit the large circle on both sides. Now we have two intersecting chords of length 2r (r and r) and 2R (R+sqrt(2)*r and R-sqrt(2)*r). By the intersecting chord theorem, we get r^2 = (R+sqrt(2)*r)(R-sqrt(2)*r) -> 3r^2 = R^2. Thus we get (1/2)r^2/(1/4)R^2 = (1/2)r^2/(3/4)r^2 = 2/3. Nice problem!
I've never heard of this weird theorem. Can you pls explain?
@@aniruddhvasishta8334
Neither had I but a quick google came up with the answer:
In any circle where two chords intersect the product of the lengths of each chord are equal.
Let chords AC and BD intersect at point Y then the theorem states:
AY × YC = BY × YD
If ABCD are the vertices of a quadrilateral and its diagonals intersect such that the product above holds then ABCD is a cyclic quad.
In this problem if AC is the diameter of the semicircle (=2r) then we can work out the lengths of BY and YD since the semicircle is tangent to the radii (R) of the quarter circle - the distance from the centre of the large circle to the centre of the semicircle is found using Pythagoras to be r sqrt(2). Thus the segments of the large circle's diameter are
R - r sqrt(2) = the length between the semicircle's diameter and the circumference of the quarter circle; and
R + r sqrt(2) = the length between the circumference of the full large circle opposite to the semicircle and the centre of the large circle plus the distance between the two centres.
Cigmorfil ok makes sense
@@aniruddhvasishta8334 Are you unable to do a Google search ?
Steve Augustin I am not unable to but seeing as this person already knew the theorem I didn’t find it necessary
MYD: how much of this quarter circle is red
Me: I agree
Hahah
😂😂
Yes!
The feeling when you look at the comment for clues then found out you are too early...
I got “The war of 1812” as the answer. 🤔
I got green.
I grew up in Baltimore so I never would have guessed that.
I tried doing it in my head by setting the quarter circle radius to 1 and figuring out the radius of the semicircle by Cartesian geometry, by finding expressions for the coordinates of the end of the diameter assuming that the diameter is at 45 degrees by symmetry. Squares of those coordinates sum to 1 because it's on a unit circle. Then I messed up the mental algebra, got that the radius was the square root of a sixth, squared that for the area factor, multiplied by 2 for being twice as much of the circle to get 1/3. Then I looked at the diagram, which is obviously mostly red, and said the answer had to be 2/3 instead. Doing the algebra correctly ((r(1+1/√2))^2+(r(1-1/√2))^2=1 means r is 3) gives the correct answer. It's really handy to pick the unit you will find convenient when doing these, and it's what Euclid would do.
Just use your head lol... These riddles never go deep
I did basically the same thing, but I let the radii of the circles be variables and they all cancelled out nicely in the end and I arrived at the answer 2/3. (I used a pen and paper though, I didn't try working in my head.) I often find when I get stuck on a geometry problem, I try setting up Cartesian coordinates and that often guides me to a solution. In this case, finding coordinates of the point where the cusp of the semi-circle meets the quarter circle and then substituting into the equation of the quarter circle gives the ratio of the square radii almost immediately.
Using coordinate geometry is my worst nightmare 😂
@@twwc960 yes using coordinates is the best way
I did the same thing as you in my head (about 5 mins)
Not being able to draw lines on paper was a challenge and its probably why I couldnt get the elegant solution presented in the video.
I am so glad you told the viewers to put serious effort into this instead of giving up and watching the solution because it took me 10 minutes but I was finally able to see how I could solve it
This video made me fall in love with mathematics again thank you😍😍😍😍😍
These sort of math problems come in handy in engineering, like designing tanks to hold fluids, making components etc. That is why I like your channel. Thanks and thumbs up!
I don't know why I watch your videos because I know nothing about maths, but I watch them and enjoy them a lot
Do you still know nothing of maths afterwards ?
At the beginning after seeing it I just said “looks like 2/3”. First time I’ve been right about one of these videos
Oh interesting, I did use Pythagorean’s theorem at all ...
I set up q=1 to work within the unit circle.
I took the points of intersection between the 2 circles (the two extremities of the chord, let s say I and J) and gave then coordinates: I (a,b) and J (b,a) ; they are mirror image of each other along the x=y line (the bisector)
Note that the coordinates of O the center of the semi circles is (a+b)/2 in both direction
I (and J) are on the unit circle so a^2+b^2=1
The semicircle is tangent to the axis so it s radius s = (a+b)/2
The chord’s length (between I and J) is equal to root(2)*(b-a) and is also a diameter of the semicircle (=2s)
Hence (a+b)=root(2)*(b-a)
Then I solved the system of two equations for a and b (a lot of algebra)
Finally we have s^2=(a+b)^2/4
S^2=(a^2+b^2+2ab)/4
S^2=(1+2ab )/4. As my algebra gave me 2ab=1/3 (I let you redo the work if you want won t type it) then
S^2=4/3/4=1/3
The ratio (r) between s^2/2 and q^2/4 (always with q=1) is r=(1/6)/(1/4)=4/6=2/3 !!!!
Surely this is more algebra heavy, but you can solve the problem even if you can t spot the hidden triangle ;)
I did it the same way, but set s=1. Then the coordinates of the center of the semicircle are (1,1). And then it's easy to find the coordinate of one of the points of intersection, because it's x=1-sqrt(.5) and y=1+sqrt(.5). Once you have that point (or its mirror), finding q is easy: q^2=x^2+y^2.
Jakob Inführ seems easier that way indeed!
It was a nice one from you. The UK challenge is a fun. Mainly in such cases,theorems related to tangency and chord intersection of radius comes in aid.
I got 2/3 by extending the inner half circle to a full circle, drawing the square that circle is inscribed in and solving for the diagonal of the square. Then set the radius of the inner half circle equal to half the side length of the square. Then lots of algebra.
I'm curious if Presh has a more elegant way.
This is how I believe I did it as well
What an elegant and simple solution for a seemingly hard question. Very nice video and explanation.
2/3 is such a common number......
I actually expected it to be π/2 or something more irrational.....
Had to be more than 1/2 and he never uses hard riddles...
Hungarians can relate
Yeah it appears a lot in physics, strange to see it in a maths question
So is a 1/2.
Very good geometry problem.
Wow, I can't believe I got it! But Not as elegantly, I found the lower coordinates where both meet. Used Pythagorus to get (s - s/√2, s + s/√2), then subbed these x & y values into the large circle's equation, y² + x² = q². Magically simplified to q² = 3s², as you found more easily!
I’m 3 months old and I figured this out
And yet you still don't have a job
wow you clever little boy, you be doctor
Rafe Tizer Thanks, Obama.
I really want one of this channel's vids to have more likes than the YT rewind.
From the thumbnail:
Obviously the scale of the drawing doesn't affect the ratio of areas; so just suppose the radius of the small circle is 1, and the radius of the large circle is unknown r. Consider a coordinate system with horizontal x-axis, vertical y-axis, and its origin (x,y) = (0,0) at the center of the small circle's diameter. The vertices of the small circle half that touch the large circle arc should then have coordinates that satisfy the following three equations:
eq. 1 - arc of small circle: x² + y² = 1²
eq. 2 - arc of large circle: (x+1)² + (y+1)² = r²
eq. 3 - diameter of small circle: x + y = 0
(We know that eq. 3 holds because the line x = y is clearly a symmetry axis of the diagram.)
From eq. 3, we derive y = -x; plug that into eqs. 1 and 2:
eq. 1a: 2x² = 1
eq. 2a: (1+x)² + (1-x)² = r² ==> (1 + 2x + x²) + (1 - 2x + x²) = r² ==> 2 + 2x² = r²
Plug result of eq. 1a into eq. 2a:
3 = r² ==> r = √3
Area (red) of half small circle : pi*1²/2 = pi/2
Area of quarter large circle : pi*r²/4 = pi*3/4
ratio of areas: (pi/2)/(pi*3/4) = (1/2)/(3/4) = 4/6 = 2/3
I solved this one a different way (based on approaches I learned in your other videos!). You can use intersecting chords on the large circle to get the following:
(q + s*sqrt(2))(q - s*sqrt(2)) = s^2
q^2 - 2s^2 = s^2
q^2 = 3s^2
This substitutes very nicely into the equation for the ratios between the areas to give the final answer.
Thanks for making these videos! I feel I'm becoming a better mathematician with each one.
The UKMT are full of great problems!
... given how two-dimensional objects behave in that regard. 2/3 would be an immediate answer related to a gut feeling. If s was 1/2 q then the semicircle could fit in twice. (Ex: 0.5x2^2=2. 0.25x4^2=4. --> 2 fits twice in 4)
Thus we find our interval in sq/2 bringing the solution closer to 0.666 then proceeding like you did. However the part above was merely a thought to estimate what would later become the solution.
I like when Presh says: Hi this is Presh Talwalkar.
TH-cam's auto captions were messing up so that's why I think he started to do it this way.
@@Morgiliath Didn't help: "pressed Elle Walker" 😆
I figured it out...
Thanks again for your videos.
Sometimes I stop them and try before continuing watching to check the result.
It's fun and training for the brain, and improves the abilities. And success gives good feeling as a reward for the effort.
And in case I'm wrong or not able to figure it out, your videos give us a lesson to learn from 👍
When you scroll pass and just guess
2/3
Me too!
2/3. I see a lot of angles. The one important is 30°. The solution is tangent(30°), which is also the unit-radius of the red shaded area. Interesting note: a)If you square "tan(30)" you get exactly 1/3. That means that the area of a (full) red circles will fit exactly 3 times into one (full) white circle. b)It also means that the distance from centre of the white quarter circle to the point where red circle touches the tangent is tan(30)*Radius(white).
me, doing my math hw: *struggles*
youtube recommendation: improvise, adapt, overcome
You can also write the equations of the two circles and then set the condition that the two points of intersections of the two circles belong to the line of equation y = -x +2 S, where S is the radius of the red circle. It simplifies out and you get the ratio of the two radii and then you find the ratio of the areas.
Every Time you upload it's 2 or 3 am in Eu and i'm too lazy to get up for some piece of paper and solve a problem
You can solve on another day because video is on the TH-cam man till the world ends😅
😭
@Tokisaki Kurumi shut up weeb
This one can be done mentally
Took me a bit of time. Did it before I watched the Vid and ended up with the same method and answer :)
Quite a challenge, having left high school for almost 20 years! Thanks!
I'm going to be honest, this is one of the easier ones I've seen. When in doubt, triangles.
I just took the radius of the red semicircle to be 1/2 the radius of the white quarter circle. made it pretty easy to just write one radius in terms of the other and solve for area, then compare.
I think that the "small radius is half the large radius" was a wrong assumption to make and basically let me cheat the problem, but I don't get why it was wrong.
Especially 3-4-5 triangles!
@warcrimes, the ratio is actually 1/sqrt(3)
@@IntrusiveThot420 I think the most basic reason why it is wrong is because it is not correct. Plug in your values for each radius and take the ratio of the areas. Do you get 2/3?
There is no reason to suspect that the radius of the semicirle is 1/2 let alone that it is even rational.
@NuncNuncNuncNunc @bificommander thanks!
This is the best geometry problem you ever had on this channel imo. The one from China with the parallelogram was also pretty cool though.
Really like the approach of solving for the squares rather than the actual radii.
Every comment here: I didn't even look at the video and I guessed 2/3
Yes. When caring for knowing the correct answer outweighs caring for knowing which and how to apply the (most) correct method to find the correct answer.
@@bosoerjadi2838 Oh, we didn't actually KNOW the answer, we just guessed and happened to be correct. There's a big difference.
I figured it out slightly differently but along the same lines. Very satisfying to get a simple rational number solution!
This should be the last question in the SAT. Answer this and you deserve a perfect score.
Or, extend the diagonal to the full large circle. Then by the Intersecting Chords theorem, s*s = (q+s√2)(q-s√2) = q² - 2s², and thereafter the same.
From this, we deduce that the Intersecting Chords Theorem proves the Pythagorean Theorem :) The converse doesn't quite hold, since the two chords in the ICT don't have to be diameters of the large circle, so you may not get any right triangle.
This is the way I did it.
I have another solution.
Imagine the inner halfcircle as scalable and the quarter circle to have a radius of 1, where h is the vertical distance from its top left corner to the arc of the quarter circle above.
The function of the quarter circle is sqrt(1-x^2), it gives us the vertical distance between a point on QC's arc and x line. Using this, you can derive the h function using geometry, where the input is the radius of inner HC: h(r) = sqrt(1-(r-(r/sqrt(2)))^2)-(r+(r/sqrt(2)))
When we solve the above equation for 0 (when h is 0, meaning the HC got big as much as it could and its corners touched the QC's arc), we figure r has to be 1/sqrt(3). From there on, you can figure the areas of each shape.
Area of the HC: (pi*(1/sqrt(3))^2)/2 = pi/6
Area of the QC: (pi*1^2)/4 = pi/4
Their ratio : 2/3
You can still figure out that r is 1/sqrt(3) if you follow the way in the video. What's even cooler is, this leads to a general formula where r is the HC radius, R is the QC radius: r*sqrt3 = R
It's pretty straight forward with analytic geometry. I called the center of the small circle P and defined it as L*(1|1). The crossbar through P has the formula P+µ*(1|-1). The quarter circle is defined by (X)²=1 (by similarity I am allowed to chose a radius to make things easier). The intersection of the crossbar and the quarter circle I called D. the lenght of PD is the radius of the smaller circle. The half circle is defined as (X-P)²=PD². A1 and A2 are the intersections of the half circle and the line (0|y). to only get a touching point we set A1=A2. Now we just solve µ for L, then solve A1 and A2 for L, set the discriminant to 0 to reduce A1 and A2 to one solution and finally we can use L to get PD, which is the radius of the small circle (the larger had radius 1) and then we can compare the areas.
Very nice and elegant, which is more than I can say about my ex. Thanks for posting!
This one is the first that I've been able to go down the right path to the solution. Past experience with these questions led me to look at the same triangle in your solution. I just bungled the math a bit once I found that the radius of the big circle was the square root of the sum of s squared & 2s squared.
nice geometry problem... enjoyed it!
Looked at the diagram. Quick relative calculation in my head and arrived at 2/3. Congratulations you are correct.
I used cosines rule, it took longer but also got 2/3
this was easy, though I am glad these videos keep sharp, keep them coming, though a little more difficult would be good too.
This problem was in the maths challenge that I sat and I’m 16, you look about 35. Considering I go to a grammar school and only students who take further maths sit the paper, and people still found it hard. I think you should consider that you’re not his only demographic.
Jack Brennan,
Giancarlo is just compensating.
I did this differently, using algebra. Let radius of quarter circle be 1. Let s = radius of half-circle. Equation for quarter circle is X^2 + Y^2 = 1. Equation for half circle is (X-s)^2 + (Y-s)^2 = s^2. Boil down algebra for half-circle, you get equation containing X^2 and Y^2 terms on left side (along with others). Remove these 2 terms and replace with 1 (from the equation for the quarter circle) to get s^2 - 2sX - 2sY + 1 = 0. Equation for the diagonal (diameter of the half-circle) is Y = 2s - X. Substitute for Y into other equation resulting in -3s^2 -1 = 0, s = 1/((sqrt(3)). Area quarter circle = pi/4, area of half circle = pi/6. Ratio = 2/3. This is interesting because 3 equations were used: those of the 2 circles, and the diagonal diameter line of the half-circle to find s = 1/(sqrt(3)).
02:25 how did you proof that the tangent line must pass through the center of the semi-circle?
By definition a tangent line to a circle intersects it at a 90° angle in relation to the radius of the point of tangency.
It's a property
A more abstract way of looking at is if you tilt the semicircle (and increase its size) until it's straight line is perfectly vertical you could place it next to the quarter that it is in and create 3/4 of a full circle but then the max limit would be 3 pieces so x/3. Then you just take the same semicircle you rotated and measure how much it takes up of the incomplete circle, which is 2/3.
I did this question in the senior Scottish Mathematical challenge. Got a silver award but had no clue on this one!
Alexander Corke it wasn’t just the Scottish one it’s was the whole of the uk, they’re were some good questions on the papers especially the cone one with the 2:1 ratio of arc lenghts
@@howtomaths8560 ah yes getting mixed up with one of the other competitions
Tbh my year is at the lower end so we were just thrown in to see how well we could do
Wow, this one was so simple yet so interesting.
Can’t talk, busy crashing my Lambo.
If I had to think for next 10 years I'd still not be able to solve this problem. Thank you for your efforts Presh.
I almost had a heart attack trying to solve this.
I wasn’t even close btw
I liked this problem a lot!!!! Interesting!!! Regards from the Republic of Panama!!!
Good problem. Took a couple minute though not 30. ;-p
Me too!
i took less than 5 minutes, simply because I watch the videos
I took a minute
As soon as you visualise the radii meeting the tangents, which is a very obvious thing to do, the problem solves itself. I think that this is an 'O' level type problem, but what do I know.
I took - 25 minutes, I was so fast I went back in time :-o
Very elegant : "Because the radius bisects the chord, they meet at right angle" is the key here.
I missed that, so i did it the hard way : intersecting cords and a lucky quadratic to find that R =sqrt(3)*r , lol .
r of quarter circle kinda looks like d of semi circle
Wonderful geometry problem. I couldn't solve it, but the solution is so nice. Thanks for sharing.
"Did you figure it out?"
.
.
.
.
no
This is what we come to MindYourDecisions for. More of this.
Every time he says: Did you figure it out?
Me: ... ...
Noooooooo...
In the exam I had like 5 mins left to do this and one other question. I drew out the diagram enlarged on rough paper. Ripped down the edge of the two regions and estimated that it seemed twice as big. Giving a ratio of 2/3. It worked!
Who else gave up and skipped to the end?
didn't skip....I watched, I just didn't want to wait 30 minutes to find out if it took my that long to get it.
Thank you for all the beautiful videos you have shared with us along 2018. I wish you a happy new year!
Lol I literally just saw this problem like 1 hour ago. I was doing ukmt past papers to practice for the upcoming ukmt challenging in 2019
RH Versity I actually did the paper this question was in. It was the last maths challenge paper I’d be able to take part in. Luckily I made it into the first round of the olympiad.
When is it in 2019?
Toby Insley the senior challenge is on November 7th. I don’t know about the junior or intermediate challenges, but they could be on the same date.
@@abeljudah6341
Oh okay, I was just wondering why people are doing past papers for the senior in a year's time?
Very neat and clean solution, sir.
Before the solution came up, just by looking at it, I guess that the answer was 2/3. I can't believe that I got it correct by looking at it only.
👍
"Diagram not to scale" 😂
@@tpaxatb. It doesn't say that the diagram is not draw to scale. I doesn't matter if it is draw to scale or not the answer will still be 2/3. Don't believe me, then do some working out with different size and I bet you $100 that it is 2/3 for any size.
@@cameronwebb7964 that's an old joke in geometry problems not to assume anything. "It looks like 2/3" is nowhere near a good answer
tpaxatb yeah it’s not too hard to guesstimate it’s 2/3. Being able to prove it is another skill entirely
1. Measure the visible circumference of the semi and quarter circles.
2. Translate them in to full circles
3. Calculate the area of each
4. Divide the area by 4 for the quarter circle and 2 for the semi circle
5. Take away the semi circles area from the quarter circles area
6. Translate that number into a fraction
"did you figure it out?"
"did i figure out the sum? or figure out when this equation will actually be needed in my life?"
the list of things you *actually* need in your life is very small. the fun stuff is optional
the second part is the better question........which is the one that rarely anyone asks.
I used a similar but different method: Each s radius of the semicircle's base is at 45°, so it's extent in either x or y is s/√2. So assuming the corner of the quarter circle is at the origin, the coordinates of the upper end of the q radius is (s-s/√2, s+s/√2). Using Pythagorus, that makes q^2=3s^s. Putting that in the fraction for the area gives the answer: 2/3. Took less than 5 minutes.
Don't have a clue. Don't even know where to start.
A good generalization of this problem would be to draw a smaller semi circle inside an arc of a bigger circle where the arc subtends angle 2*theta at the center of the bigger circle. Calculate the ratio of the area of the smaller circle to the area of the sector formed by the bigger circle. This would need some basic trig.
No offense...but the way you got to the answer is unnecessarily long
How did you get to the answer?
New here, I liked this problem. I knew the key was to find the radius of the semicircle and find it’s relationship to the radius of the quarter circle, but forgot a lot of the geometry tricks that could be done to figure it out- I tried thinking I’m terms of calculus, but the answer was simpler than I thought.
Videos are great but don't make them easier day by day
Oh come on, that was not easy cake
@@vellyxenya3970 ikr
Did it in a slightly simpler way by assuming the small radius is 1. Proceeded the same way as in the video from there and discovered the answer 2/3 pretty quickly. Nice problem.
Yayyy! First time to solve a problem from this channel😆😆
You can use power of the point on the center of the semicircle in order to find the length of the space between semicircle and the quarter circle, and therefore the radius of the quartercircle. Everything works nice from there
That is so beautiful. the 1st step is the most beautiful.
For the first time I think I am great at maths....I was able to think the exact method u said to us in the first 30 sec of thinking
Solved it in my head in one minute without calculating anything, but just looking at it. Surprisingly, my answer was indeed correct!
cool
As far as problem solving goes, I think if confronted with circles, assume the radius of one of them to be one, and from there look for relations to make with that radius. Here, you can try the outer quarter circle first, but then see it does not yield many easy relations between the two as there are two unknown distances if that radius goes through the center of the semicircle and two unknowns when it touches a tangent. Switch to the setting the semicircle radius to one. That radius touches the outer circle at each tangent and also twice on the circumference. From there it is just a matter of observing the right triangle whose hypotenuse is the radius of the outer circle and calculating its length.
Every time I think I am pretty clever, this channel puts me back in my place.
This solution was so elegant that when I figured it out, it felt like I should have done so at least twice as fast. I was stuck with sqrt(2)s for a good while before realizing how easy it was to determine q to be sqrt(3)s. Except that I just used unity in place of s.
I like how you get a nice rational answer like 2/3 even tho there are no numbers in this problem.
If you are for some reason not seeing that last triangle, you could also use the rule for intersecting cords: (2*sqrt(2)*s + x) * x = s^2, which resolves to (x + sqrt(2)s)^2=q^2=3s^2.
2/3. Split the quarter circle in halves, you will get two right triangles（the centres of two circles plus the end of the diameter of red circle, the centres of two circles plus point of tangency）. Apply the Pythagorean theorem in both triangles, you will find the ratio of squared radius between the two circles is 1:3. The rest is obvious.
I did figure it out, however witha different (and a more complicated) solution. Using trigonometry I found both horizontal and vertical components of the inclined bigger radius (which were 1-sqrt(2) and 1+sqrt(2)), used pythagorean theorem and got the same answer. I love it when unrelated solutions give the same answer.
Cheers.
Geez, I did it the hard way: I completed the semi-circle, drew the square it would be inscribed in, and then added the line y = -x + 2r, where r = the smaller radius. Then found the x value where this line intersected the circle at the upper left, then found the distance that point was away from the origin, and that was R, the larger radius, in terms of r, the smaller radius.
I also change a car's tire by first dropping the transmission.
coordinates of small circle: (r,r) area S0 = pi*r^2/2. Coordinates of big radius y = r + r*cos(45deg)=r+r/sqrt(2) x = r - r/sqrt(2). so big radius R = sqrt(x^2 + y*2) = r/sqrt(3). area S1 = pi*R^2/4 so S0/S1 = 2/3
Thank you for telling me that it took your 30 minutes to solve. I usually look at these problems for 5 minutes and if I can't solve them, I give up. But telling me that it took you 30 minutes, encouraged me to take a little more time.