Here's the hack (soft) way of solving it: 1. The size of the squares doesn't matter - their total area will always be the same (try chopping off an "L" from the bigger square and sticking it on the other one) 2. Therefore we can deduce that both squares can be the same dimension, say x*x 3. Therefore the diagonal of each square is 8 units 4. Therefore by Pythagoras: 8^2 = x^2 + x^2, which also happens to be the area of both squares! 5. Therefore Total Area = 2x^2 = 64 units^2
i realised that when he found that the radii of 8 came out to 90 degrees even though it reached from the same points of the outer circle as the corner of the two boxes, and the center was not visually indicated at the corner point. it means that the dimensions don't matter, they come out to some ratio that equals 64.
How is that hack or soft..i thonk its smarter or more lrga ic because his metbod ises onscribed angle thing thst not a lotnof ppl.will remember or know
@@QUASARCREATIVE_YTthe title has a shared naming convention with a viral gross out video from like 2008. I hope you'll leave it at that, but if your intrusive thoughts are winning, Google will show you the way
Got to the end and suddenly realised: "wait - 64 is 8 squared, and 8 is half 16, the diameter - so the sum of the area of the squares is just radius squared! That worked out nicely!!"
where did you come with the equation? did you work only on the specific case where both squares are equals, thus having their border ending exactly on the center of the circle ?
@@tharnator6018 Of course, i don't work only on the specific case where both squares are equals (x=y) I work with general case when x can be different from y. I don't explain this simple equation : sqrt(8^2-x^2)+sqrt(8^2-y^2)=x+y Look at the video at 2:47 and you will understand.
While I think the solution is wonderful, I am a little confused as to why you took that roundabout way of reaching 64? Since the sizes of the squares aren't set there is a case where both squares are the same size, when their corners perfectly intersect the center of the circle. Then you know that the diagonal is the same as the radius, i.e. 8. From there using a bare minimum of trigonometry will very easily end up with the area of the squares. The fact that the squares are different sizes in the image feels a bit like a red herring.
For a test that only asks for the numerical answer, that would be a quick way to do it. But you are assuming that the placement of the squares does not matter for the area. Rather than assuming, it is better to prove it, which is what he did in the video.
@@oliverbutterfield9844 heck, I just thought of an even easier solution. If both squares are equal they make up half of an imaginary square fit into a full circle. And there's a rule that a square in a circle will have a side length of r√2, leading to a full square area of 128. Just half that and you're done.
@@SwitchAndLever It also did not specify that the placement of the squares does not matter. So you would just be assuming that with no good reason other than guessing about the intentions of the person who wrote the question.
If all the configurations work, what do you think about constraining it to make it convenient? Make the line between the squares the center, so that the square diagonals are also a radius. Then you get area 64 immediately. Proving it doesn't change with different circles is another problem.
since it will work the same in all scenarios, we can work with two equal squares as it will be far easier. mirror the image to complete the circle. we got a large square inscribed inside a circle. evidently, this large square is twice as large as the sum of the two original squares. since it's an inscribed square, it's extremely easy to find the area and halve it to get our answer. how exciting.
well i think the fact that the answer is the same regardless of the sidelengths is not so trivial you can just state it as fact. of course if you do set them the same you can get the area much quicker, but you havent proved the general case
@@WilliamWizer yes, that does not make it trivial, it's more like a spoiler for the solution he found in the video. it doesn't prove anything. that is why he solved the problem for any arbitrary x and y
@@oniondesu9633 it's completely trivial. as the video explains, "whatever work we do, should work for any one of those scenarios" but, since I had a bit more time to prepare something more funny: rotate the semicircle to complete a circle. (similar to what the video does but rotating instead of mirroring so we have a blue and a yellow square vertically too) join the corners to form c, like shown in the video, you get 4 instances of c that form an inscribed square. that square will have the same size (since it's inscribed) no matter what x or y are. the size of that square depends only on the radius of the circle. there. the values of x and y are completely trivial
Diameter = 16. Radius = 8. If when we draw a square in the one right quarter circle touching the quarter circles perimeter, the distance from the rightmost extreme of the quarter circle to the side of the square is z, which can be found by symmetry, then the side of the larger square is, a = 8 - z.
Nice! I got it by reasoning that the lack of information means we can make the squares equally sized WLOG. The diagonal of both squares would be 8, so using 45-45-90 the side lengths would have to be 8/sqrt2. Squaring this is 64/2=32, so if each square is 32 then both make 64.
@@DSTUSEV while you bring up a valid point, a point of observation is that you can reflect the squares to the opposite side (left is bigger than the right) and still get the same answer. However, reflection can also be interpreted as size transformation, and therefore there must be a continuous function from one size to another. There HAS to be a point on this continuous function such that both squares are indeed equal in size.
I like the Calc. There 2 easier ways to do it in my opinion. 1. squares are equally big and diagonal equals radius. 2. One square is in the middle and other small one you get with analysis (y=sqroot(64-x^2) )
The radius is 8. The relative sizes of the squares is immaterial, so make them equal. They have side lengths of a. Draw the other half of the circle. Intersecting chord theorem : (8+a) * (8-a) = a*a so 64 - a^2 = a^2. ==>2a^2=64 ==> The 2 squares area is a^2 + a^2 = 2a^2 = 64
@@oniondesu9633 I see what you are saying. The problem wouldn't be solvable if you get different answers depending on square size ratios. i.e there would be infinite solutions. The assumption that there is one solution for the ratio of all square sizes seems logical, otherwise the problem would have been stated as, "prove that varying square size ratios matter."
@thewolfdoctor761 sure, you can find the area when they are the same size, but then you have only solved the problem for that one special case. the thing that is interesting about his solution is that he solves it for every case at once, and if you don't do that, you don't find out that the problem gives the same solution for every possible pair of squares.
what i did was (since it didn't specify) made them both the same size, then reflect below to make one big square inside the circle. The square's diagonal was 16, so its side is 16/sqrt(2). square that to get 128. divide that by 2 (because it was reflected) to get 64.
There's a much MUCH easier way. If there's nothing defining where the squares are located (as mentioned in the video) then we can change the shape of the two squares to make them both the same size and it looks like one larger rectangle. This means that the side length of the two squares will be the same as each other (x=y) and the diagonal length of the two squares will be equal to the radius of the semicircle, 8. Therefore, using Pythagoras, x*(sqrt 2) = 8 and so x=8/(sqrt 2). Remembering that x is now equal to y we can simply square it and multiply it by 2. x^2 = 8^2 / 2 = 32 Which, when multiplied by 2, equals 64
if the squares can vary in size. the easiest way is to make them equal. Then you have 2 equal squares with a diagonal of 8. x and y are 8* cos(45) so entire area is 2 times sqr(8*cos(45)).
First the size of 2 inscribed squares doesn't matter so they could be the same size which gives you the diagonal of 8 immediately. Second you can skip the whole proving the angle of 90 at 2:50 since you already know that the arc of c is 90 degrees at 1:50
You can simplify you method even further based on your first observation. Since there is no information given as to the actual size of these squares, them the squares can be any size. As demonstrated by your animation, there has to be an instance where the area of both the left and right (as we increase and decrease their sizes), where they are exactly the same. At that point, we can actually perform our calculations with the assumption that both squares are equal in area. In doing so, we know that the diagonals of both squares are equal to the radius of 8. Since these are 45° Isosceles Triangles, we know the length of these diagonal is equal to x√2. 8 = x√2 -> x = 8/√2. A_left = A_right = 64/2 A = A_left + A_right = 2(A_left) = 2*(64/2) = 64 Therefore, the area is equal to 64 square units.
If the problem depends on the relative size of the two squares, then it is not solvable. If it does not, then the solution is the same for any relative size of the two squares, so pick the particular case where the two squares are the same size. Their bottom sides meet at the circle's center (symmetry), so the surface is 2 * the area of one square which diagonal is the radius of the circle.
I had no idea how to solve this at first. But once he pointed out that the answer must be true for ANY sized blue and yellow squares, as long as they touch each other, and the curve, and the diameter, then you can just assume the two squares are identical. If they are, then the point where both squares meet on the diameter also must be the center of the circle. And thus the squares both have a diagonal of 8 units. So the end result is the same as what andy figured out, but with a few less steps because we can skip right to the end.
My first approach was that since x and y will always be unknow this is most likely a question that gives the same answer for all sizes. So I would have started by setting x=y. This would lead to both square's diagonals to be the radius of the circle which is 8. Pythagora's theorem: a²+b²=c² Adjusting for a square: 2a²=c², c=8 Side of a square is a=√(c²/2) Area of a square is a=√(c²/2)² Area of two squares is A=2a=2√(8²/2)², Simplify we get 2√(8²/2)² = 2*8²/2 = 8² = 64 square units Then after that to verify I would scale one square to have area 0 and thus side lenght of 0, and the other square to have area 64 and side length of √(64)=8. Although one square is not visibile and one is extending outside of the cirle, their respective relevant corners still match upp with the edge of the circle. Small square's top left corner is straight left of the circle's center. Big square's top right corner is straight above the circle's center.
so, the area of the 2 squares inside a hemicircle… radius squared. Because, if this works equally for any two squares, in the set is always two equally described squares with a diagonal of the radius.
Interesting, I did it different. The middel of the circle is 8 away from the outline so since it is off center I made a new distance c which is the difference from a( the small square) to the middle. So a+c = b-c. Then I do pytagoras with both and since the slope is both 8^2 I can set them equal. After simplifying I get b-a=c. Plugging c in one of the formulas made with the pytagoras formula I can simplify again to a^2 + b^2 = 64. I was just flabbergasted to see what you did.
No brainer solution: If x and y are the sides of the two squares, and a is the offset of the lower middle corner from the center of the circle, the formulae for the radii that go to the corners of each square on the circle are : (x+a)^2+x^2 = 8^2 and (y-a)^2+y^2 = 8^2. Expanding, one gets: 2x^2 + 2ax + a^2 = 8^2 [1] and 2y^2 - 2ay + a^2 = 8^2 [2] Subtracting [2] from [1] and dividing by 2: x^2 - y^2 + a (x + y) = 0 As y^2 - y^2 = (x - y)(x + y), we get: (x + y)(x - y) + a(x + y) = 0, or (x + y)(x - y + a) = 0. Since x + y > 0, we must have (x-y+a) = 0, i.e. x - y = - a [3] Adding [1] and [2] and dividing by 2, we get: x^2 + y^2 + a(x - y) + a^2 = 8^2; plugging in [3] we get: x^2 + y^2 -a^2 + a^2 = 8^2; thus x^2 + y^2 = 8^2
Andy, I hated all this when I was at school - I did it (even A-level maths) but I hated it - you made me love this stuff and I've actually joined Brilliant - first time a youtube sponsor has ever worked on me - its only a free trial but I'm probably going to keep it up - the way you (and brilliant) explain maths makes it makes sense and therefore (to me) fun. maths was always learn this formula apply it - I actually understand quadratic equations now - I can look at an equation and know where it will fall on the axis - absolutely crazy! (Maybe everyone can do that nowadays but for me, a child of the 80s, it was like 'learn this do this'! (If I read this post I would think hmmm is this a shill account for Brilliant? I promise, I'm not.. (EXACTLY WHAT A SHILL ACCOUNT WOULD SAY!!!)
Alternative solution: Imagine both the squares are same, so they meet in the center of the circle. So, Diagonal is the radias of the circle. According to Pythagoras, square's diagonal = x√2. So, radius = 8 = x√2. So, x=8/√2. Thus, 2x^2 = 64
You can save several steps because the area is independent of the relationship between x & y. So just use the case where x = y. Diagonal of each square is 8, and the rest is in the video.
@@oniondesu9633 The first hint is that if they weren't independent, the question would have been differently -- you can't solve it with the info given if they are not independent.
If the size of the squares can vary but the area remain same, why not consider squares that are exactly exactly equal in size? In this case both x = y = 32^0.5 (since the diagonal will be 8 unites as they become radius of the circle). Then x^2+y^2 = 32 + 32 = 64!
As there is no restriction on the combination of sizes of the two squares, we can set the left square to have an area of zero, thus forcing the right square's bottom left corner to be positioned at the left extremity of the semi-circle and the square's top right corner to be positioned at the top-most point of the semi-circle. This gives just one apparent square with a side length of 16/2 = 8. The area is then just 8 * 8 = 64.
Wait a minute. If we realize that there's nothing defining the location of these squares then we can try the shortcut and solve it for when they are equal. And if they are equal then so are their diagonals. And the only instance when they could have equal diagonals and be located the way they are is when their common vertex is the center of the semicircle. That would mean, they would form a rectangle and their diagonals would be equal to radius, which is 8. That would mean the height of the rectangle is 4sqrt(2) and the width is 8sqrt(2). So we multiply those and get 64 square units. I mean, that feels like cheating, because the method implies the task is solvable with just the given numbers, but hey, that's also an option if we follow the logic of the first statement. How exciting indeed.
It's not wrong. It's just not the rigorous way to prove it. If this was on a test and you had to choose an answer and you aren't asked to motivate your answer, do it your way every time. But Andy wants to prove it to us so there's no doubt.
@@Qermaq that's part of the reason I left the disclaimer in the end. But now that I think about it this is a legitimate method, though tricky to decide when applicable and even more tricky to debate with whoever is grading it: "if the problem is solvable with just the given data, then the answer is this. And if it's the wrong answer, then the task is flawed and needs more data."
@@WhooshWh0sh That's the risk. What if it looks like you can generalize but you miss something? What if the question has an error, so by giving an obvious answer you are wrong?
The angle between the 2 radii had to have been 90° - you originally had a 90° angle when you drew the 2 diagonals of the squares. You can move that angle along the diameter while maintaining the 2 points on the circle, and the angle will be constant at 90°
I "did it" diffirently. Since this problem "should" have the same answer indepent of the squares I took the case that both squares are the same in size. In that case the hypotenuse is equal to the radius of the circle, I then used the pythagorean theorem to figure out the lenght of the sides and multiplied the lenght of the height (1xside) by the base (2xside).
You don't even need to use pythagorean theorem. Once you split the squares by the radius, you have four identical isosceles right triangles that can be arranged in a square with a side of 8.
I found a much simpler solution :) If the area of the two squares will remain the same no matter their size, you can assume they are equal. Once you assume that, their diagonal is equal 8. From here it’s super simple solution
My method was a lot cheaper. Since it doesn't matter how big each square is, I chose to make them the same size, which puts them right in the middle making the diagonal the same as the radius. That means 2x²=8²=64 (pythagoras), which is already the area of both squares.
Dang. I was so excited cuz I thought I figured it out before he gave us the answer. It turns out I was right but only cuz I made an assumption which turned out to be true, I didn't prove it, I thought it was just a rule. I got lucky. Still I did a lot of the work on my own. Gonna keep practicing and I'll get there.
If it works wherever the squares are and is true for any case, then why don’t you just put the two squares by the center of the circle and solve in an easier way
The comments seem highly confused about this one. His animation at the start did not prove that the area we were aiming for was independent of the side lengths of the squares, think of it more like a spoiler, or something he suspected. He still had to solve the problem for an arbitrary x and y, the fact that his answer did not depend on x or y was the proof of the suspicion he layed out at the start. If he had just set x=y, yes, he would have got the area as 64 much easier, but he would NOT have solved the problem generally. He would have just solved one special case.
I got the same answer but since I knew the answer would work for any value of x and y as long as the squares touched the semicircle at the corner, I assumed x=y for the case where the squares were the same size. I knew x=(radius)sin(45) and there were two squares so 2((8)sin(45))^2=64= the area of the squares. Fun problem
Is it true then to say the area of two adjoining squares, which one edge the same the diameter of the semi circle, that has two corner points on the semicircles arc, and that are fully encompassed within said semicircle as pictured, The area always equals 4•D? Substitute that diameter of 16 with 12, 21, 196, etc, and the area of those two squares will always come out to be 4x that diameter. Correct? I’m not a math person, but looking at the problem, and as you slid the dimensions of those squares around, it would appear that that will always be the case. This could be a short formula for such problems that you ever come across.
Or, since it aplies for al inscribed squares, you can take the case where they colide in the center of the semi-circle. this would make the diagonal of both squares 8 (16/2) using pythagoras we know that, if x is the side of the square, x^2 + x^2 = 8^2 x^2 = 32. x^2 is the same as the area of that square, and we need it twice so the answer is 64.
This is how I did it before watching the video, I knew that because no information about the squares was given, it would be the same as if they were equal
Huh, I did it an easier way. I did have to assume the squares could be the same size (or indeed any ratio of sizes), but since they aren't given any dimensions that seemed valid
i just mirrored the semicircle to create a full circle with 4 squares then i drew a line from the yellow squares' corner thats touching the circle to the opposite blue squares' corner this is the diameter of the circle x*sqrt2 + y*sqrt2 = 16 x + y = 8*sqrt2 squaring this expression makes a bigger square that imperfectly matches the shape of the 4 smaller squares but thats okay here since the bit that sticks out perfectly slots into the bit thats missing so (x + y)^2 = (8*sqrt2)^2 (x + y)^2 = 128 128 is the area of all 4 squares so the area of the 2 starting squares is 64 not very rigorous but i thought it was an interesting solution
This figure is not defined, positions of squares are not given (vertex positions) We can modify the squares dimensions KEEPING the original conditions I decided to match both squares, and the common vertex becomes the center of the circle. Therefore, the diagonal of each square is the radius of semicircle A₁= A₂ = ½d² = ½R² A = 2A₁ = R² = 64cm²
Other solution, instead of matching both square, can be maximize area of large square, and minimize area of small square. And the original conditions are still fulfilled. In this case : R²=S²+(½S)² S² = 4/5 R² = 4/5 8² = 51,2cm² s² = ¼ S² = 12,8 cm² A = S² + s² = 64 cm² ( Solved √ )
You made it unnecessarily complicated. As you showed at the beginning, the two squares can be adjusted to be equal inside the semicircle. So then we know that the diagonal (and radii) is 8. Bing bang boom.
@@DisposableSupervillainHenchman dude, you are acting smug when you are completely wrong. yes, the area is the same no matter what the relative sizes of the squares are, but this cannot just be stated, it must be proved, which he did in the video.
I have to admit, when I looked at this, the way forward was certainly not obvious to me. If it was not so easy to just click the play button to satisfy my curiosity, I might have figured it out. Maybe.
this is a ugly hack, but there is no constraint stopping me from assuming that the 2 squares are of equal area. now the question becomes really easy because the diagonal of each square is 8, and the are is (8*8)/2 for each square, since there are 2 squares, the area is 64. The reason this hack is ugly is because i most probably would not have guessed that the total area of the two squares are not dependant on the individual widths of the squares. However, it isnt an ugly hack if we do this, from the yellow square, you can cut half of the extra portion and append it to the right of the yellow square. Now take the extra left after appending and add it on top right of the blue square. Now, take the same width from the left of the blue square and move it on top, and we have 2 equal squares. This proves that sum of areas of the squares are not dependant on the individual sides, which makes this hack beautiful!
it's not just ugly, it would be incorrect. nowhere was it stated that the total area is independent of the side lengths, it was something he proved in this video. you cant just assume it
@oniondesu9633 I've written a geometric proof in the second para of the comment. Go over it again. Try plotting what I'm telling in the second para, it's a fun proof
@@XxFALCONxX- you can turn any two squares into two equally sized squares with that procedure, but how do you know your new squares perfectly fit inside the same semicircle?
@@oniondesu9633 ohh you arnt wrong, mb, I was wrong. I made the assumption that if I cut the top half rectangle from the square (half the difference of the sides of the square), then it will fit on the right, I made that assumption without proof because it looked so (I can prove it tho by showing that the distance from the center of the circle to the bottom left corner but thats gonna take more triangles and math than the solution in the video). Thanks for the correction!
How exciting.
I like you showed how the squares can vary but their area will always be same. Another thing to remember! How exciting!
meaning you could simplify the question by implying if the squares were equal sized from the beginning the diagonal would be the radius of the circle.
3:12 this move right here really got me
Me too. I would never thought in that way.😅
Here's the hack (soft) way of solving it:
1. The size of the squares doesn't matter - their total area will always be the same (try chopping off an "L" from the bigger square and sticking it on the other one)
2. Therefore we can deduce that both squares can be the same dimension, say x*x
3. Therefore the diagonal of each square is 8 units
4. Therefore by Pythagoras: 8^2 = x^2 + x^2, which also happens to be the area of both squares!
5. Therefore Total Area = 2x^2 = 64 units^2
elegant solution!
glad someone else saw that I just commented this onto tiktok lol area = x * x^2 when both squares are the same size
i realised that when he found that the radii of 8 came out to 90 degrees even though it reached from the same points of the outer circle as the corner of the two boxes, and the center was not visually indicated at the corner point. it means that the dimensions don't matter, they come out to some ratio that equals 64.
I don't understand how we know 1 is true. Is this a theorem or something I don't know about?
How is that hack or soft..i thonk its smarter or more lrga ic because his metbod ises onscribed angle thing thst not a lotnof ppl.will remember or know
I am having so much fun with these problems. Exciting. even at 71 years old , brings me back Keep the grey matter on its toes Thx Andy
Man what is that title🤨
wym
@@QUASARCREATIVE_YTbelieve me you don’t want to know
@@yamikamui i literally do thats why i asked
@@QUASARCREATIVE_YTthe title has a shared naming convention with a viral gross out video from like 2008.
I hope you'll leave it at that, but if your intrusive thoughts are winning, Google will show you the way
👧👧🥤💩
... genius as always. God bless ...
Really good, thanks for that.
It's honestly a clever proof, that the area sum of two inscribed squares of a semicircle are equal to r^2.
Got to the end and suddenly realised: "wait - 64 is 8 squared, and 8 is half 16, the diameter - so the sum of the area of the squares is just radius squared! That worked out nicely!!"
Real question: could I possibly just answer this in an exam?
No way@@PaulWegert-oc2me
Diameter=16
Radius=8
Equation : sqrt(8^2-x^2)+sqrt(8^2-y^2)=x+y
Solution : x^2+y^2=8^2
Demonstration :
sqrt(8^2-x^2)+sqrt(8^2-y^2)=x+y
(sqrt(8^2-x^2)+sqrt(8^2-y^2))^2=(x+y)^2
8^2-x^2+8^2-y^2+2*sqrt((8^2-x^2)*(8^2-y^2))=x^2+y^2+2*x*y
2*sqrt((8^2-x^2)*(8^2-y^2))=2*x^2+2*y^2+2*x*y-2*8^2
sqrt((8^2-x^2)*(8^2-y^2))=x^2+y^2+x*y-8^2
(8^2-x^2)*(8^2-y^2)=(x^2+y^2+x*y-8^2)^2
x^2*y^2-8^2*(x^2+y^2)+8^4=(x^2+y^2+x*y)^2-2*8^2*(x^2+y^2+x*y)+8^4
x^2*y^2-8^2*(x^2+y^2)+2*8^2*(x^2+y^2+x*y)=(x^2+y^2+x*y)^2
x^2*y^2+8^2*(x^2+y^2+2*x*y)=(x^2+y^2)^2+2*x*y*(x^2+y^2)+x^2*y^2
8^2*(x^2+y^2+2*x*y)=(x^2+y^2)^2+2*x*y*(x^2+y^2)
8^2*(x^2+y^2+2*x*y)=(x^2+y^2)*(x^2+y^2+2*x*y)
8^2=x^2+y^2
…thanks
where did you come with the equation? did you work only on the specific case where both squares are equals, thus having their border ending exactly on the center of the circle ?
@@tharnator6018
Of course, i don't work only on the specific case where both squares are equals (x=y)
I work with general case when x can be different from y.
I don't explain this simple equation : sqrt(8^2-x^2)+sqrt(8^2-y^2)=x+y
Look at the video at 2:47 and you will understand.
@@matthieudutriaux Thanks, got it now. Nice solution as well.
That is the best explanation I have found - thank you for making it so simple to understand. Great teacher!
Just realized that the sum of the two squares will always be the radius squared. If the diameter is 20 the area of the two squares will be 100.
Andy: "And we're gonna reflect these two squares down he-"
Me: "okay, now I feel like he's just messing with me at this point"
I was like NO HE Di' int! 😆 His teaching and content is the same as ART 🎨 but with rules. 💘
Perpetually being humbled by this channel
This was actually really cool. I’m not even into math but I like these quizzes.
Outstanding.
Exciting proof!
While I think the solution is wonderful, I am a little confused as to why you took that roundabout way of reaching 64? Since the sizes of the squares aren't set there is a case where both squares are the same size, when their corners perfectly intersect the center of the circle. Then you know that the diagonal is the same as the radius, i.e. 8. From there using a bare minimum of trigonometry will very easily end up with the area of the squares. The fact that the squares are different sizes in the image feels a bit like a red herring.
That’s how I did it, but this is a bit more interesting for the general case.
For a test that only asks for the numerical answer, that would be a quick way to do it.
But you are assuming that the placement of the squares does not matter for the area. Rather than assuming, it is better to prove it, which is what he did in the video.
@@oliverbutterfield9844 heck, I just thought of an even easier solution. If both squares are equal they make up half of an imaginary square fit into a full circle. And there's a rule that a square in a circle will have a side length of r√2, leading to a full square area of 128. Just half that and you're done.
@@XJWill1 of course, I'm not at all putting that down. The original question didn't ask for proof though, only a solution. 🙂
@@SwitchAndLever It also did not specify that the placement of the squares does not matter. So you would just be assuming that with no good reason other than guessing about the intentions of the person who wrote the question.
If all the configurations work, what do you think about constraining it to make it convenient? Make the line between the squares the center, so that the square diagonals are also a radius. Then you get area 64 immediately. Proving it doesn't change with different circles is another problem.
since it will work the same in all scenarios, we can work with two equal squares as it will be far easier.
mirror the image to complete the circle.
we got a large square inscribed inside a circle.
evidently, this large square is twice as large as the sum of the two original squares.
since it's an inscribed square, it's extremely easy to find the area and halve it to get our answer.
how exciting.
well i think the fact that the answer is the same regardless of the sidelengths is not so trivial you can just state it as fact. of course if you do set them the same you can get the area much quicker, but you havent proved the general case
@@oniondesu9633 watch the start of the video. it's stated just there, even with a cute animation.
@@WilliamWizer yes, that does not make it trivial, it's more like a spoiler for the solution he found in the video. it doesn't prove anything. that is why he solved the problem for any arbitrary x and y
@@oniondesu9633 it's completely trivial. as the video explains, "whatever work we do, should work for any one of those scenarios"
but, since I had a bit more time to prepare something more funny:
rotate the semicircle to complete a circle. (similar to what the video does but rotating instead of mirroring so we have a blue and a yellow square vertically too) join the corners to form c, like shown in the video, you get 4 instances of c that form an inscribed square.
that square will have the same size (since it's inscribed) no matter what x or y are. the size of that square depends only on the radius of the circle.
there. the values of x and y are completely trivial
@@WilliamWizer that is a fun way of proving it! but it is distictly not trivial, are you mixing up the word trivial for arbitrary?
Diameter = 16. Radius = 8. If when we draw a square in the one right quarter circle touching the quarter circles perimeter, the distance from the rightmost extreme of the quarter circle to the side of the square is z, which can be found by symmetry, then the side of the larger square is,
a = 8 - z.
Ioqm 2025 aspirant??
your graphics are simple, clear, well-thought out, and effective
Nice! I got it by reasoning that the lack of information means we can make the squares equally sized WLOG. The diagonal of both squares would be 8, so using 45-45-90 the side lengths would have to be 8/sqrt2. Squaring this is 64/2=32, so if each square is 32 then both make 64.
But then you are assuming, that the size of the squares does not matter without proofing it.
This is what I did. Took like a minute.
Same i was surprised by his solution this is much simpler.
@@DSTUSEV while you bring up a valid point, a point of observation is that you can reflect the squares to the opposite side (left is bigger than the right) and still get the same answer. However, reflection can also be interpreted as size transformation, and therefore there must be a continuous function from one size to another. There HAS to be a point on this continuous function such that both squares are indeed equal in size.
that "reasoning" is more of a guess, his answer shows it is true that the relative size dont katter, you cant just assume it is true from the get go
That was a really tough problem. SO clearly explained. Wonderful. Thanks!😊
I like the Calc. There 2 easier ways to do it in my opinion. 1. squares are equally big and diagonal equals radius. 2. One square is in the middle and other small one you get with analysis (y=sqroot(64-x^2) )
The radius is 8. The relative sizes of the squares is immaterial, so make them equal. They have side lengths of a. Draw the other half of the circle. Intersecting chord theorem : (8+a) * (8-a) = a*a
so 64 - a^2 = a^2. ==>2a^2=64 ==> The 2 squares area is a^2 + a^2 = 2a^2 = 64
you have to prove that the relative sizes are immaterial, not just state it. it is not a trivial proof
@@oniondesu9633 I see what you are saying. The problem wouldn't be solvable if you get different answers depending on square size ratios. i.e there would be infinite solutions. The assumption that there is one solution for the ratio of all square sizes seems logical, otherwise the problem would have been stated as, "prove that varying square size ratios matter."
@thewolfdoctor761 sure, you can find the area when they are the same size, but then you have only solved the problem for that one special case. the thing that is interesting about his solution is that he solves it for every case at once, and if you don't do that, you don't find out that the problem gives the same solution for every possible pair of squares.
This is insane. Usually I would figure out most of you video.
But there's no way for this video!
“2 squares 1 semicircle”
*2 girls 1-* 💀
Boy❤
@@CosmicHase ❌
@@redsus8725 you don't like overflow
Finaly someone good at math on the internet
what i did was (since it didn't specify) made them both the same size, then reflect below to make one big square inside the circle. The square's diagonal was 16, so its side is 16/sqrt(2). square that to get 128. divide that by 2 (because it was reflected) to get 64.
you got the answer, but you didnt prove that the answer is the same no matter what the ratio of the side lengfhs are like his solution
There's a much MUCH easier way.
If there's nothing defining where the squares are located (as mentioned in the video) then we can change the shape of the two squares to make them both the same size and it looks like one larger rectangle. This means that the side length of the two squares will be the same as each other (x=y) and the diagonal length of the two squares will be equal to the radius of the semicircle, 8.
Therefore, using Pythagoras, x*(sqrt 2) = 8 and so x=8/(sqrt 2).
Remembering that x is now equal to y we can simply square it and multiply it by 2.
x^2 = 8^2 / 2 = 32
Which, when multiplied by 2, equals 64
if the squares can vary in size. the easiest way is to make them equal. Then you have 2 equal squares with a diagonal of 8. x and y are 8* cos(45) so entire area is 2 times sqr(8*cos(45)).
Fun problem!
First the size of 2 inscribed squares doesn't matter so they could be the same size which gives you the diagonal of 8 immediately.
Second you can skip the whole proving the angle of 90 at 2:50 since you already know that the arc of c is 90 degrees at 1:50
You can simplify you method even further based on your first observation. Since there is no information given as to the actual size of these squares, them the squares can be any size.
As demonstrated by your animation, there has to be an instance where the area of both the left and right (as we increase and decrease their sizes), where they are exactly the same.
At that point, we can actually perform our calculations with the assumption that both squares are equal in area.
In doing so, we know that the diagonals of both squares are equal to the radius of 8.
Since these are 45° Isosceles Triangles, we know the length of these diagonal is equal to x√2.
8 = x√2 -> x = 8/√2.
A_left = A_right = 64/2
A = A_left + A_right = 2(A_left) = 2*(64/2) = 64
Therefore, the area is equal to 64 square units.
3:12 I think you need to prove that this chord always contains the diagonal of the squares first in order for the angle to be 45.
If the problem depends on the relative size of the two squares, then it is not solvable.
If it does not, then the solution is the same for any relative size of the two squares, so pick the particular case where the two squares are the same size.
Their bottom sides meet at the circle's center (symmetry), so the surface is 2 * the area of one square which diagonal is the radius of the circle.
did you not watch the video? he did not take a special case, he assume the squares where an arbitrary sixe and still solved it
I had no idea how to solve this at first. But once he pointed out that the answer must be true for ANY sized blue and yellow squares, as long as they touch each other, and the curve, and the diameter, then you can just assume the two squares are identical. If they are, then the point where both squares meet on the diameter also must be the center of the circle. And thus the squares both have a diagonal of 8 units.
So the end result is the same as what andy figured out, but with a few less steps because we can skip right to the end.
My first approach was that since x and y will always be unknow this is most likely a question that gives the same answer for all sizes.
So I would have started by setting x=y.
This would lead to both square's diagonals to be the radius of the circle which is 8.
Pythagora's theorem: a²+b²=c²
Adjusting for a square: 2a²=c², c=8
Side of a square is a=√(c²/2)
Area of a square is a=√(c²/2)²
Area of two squares is A=2a=2√(8²/2)²,
Simplify we get 2√(8²/2)² = 2*8²/2 = 8² = 64 square units
Then after that to verify I would scale one square to have area 0 and thus side lenght of 0, and the other square to have area 64 and side length of √(64)=8.
Although one square is not visibile and one is extending outside of the cirle, their respective relevant corners still match upp with the edge of the circle.
Small square's top left corner is straight left of the circle's center.
Big square's top right corner is straight above the circle's center.
so, the area of the 2 squares inside a hemicircle… radius squared. Because, if this works equally for any two squares, in the set is always two equally described squares with a diagonal of the radius.
Interesting, I did it different. The middel of the circle is 8 away from the outline so since it is off center I made a new distance c which is the difference from a( the small square) to the middle. So a+c = b-c.
Then I do pytagoras with both and since the slope is both 8^2 I can set them equal. After simplifying I get b-a=c.
Plugging c in one of the formulas made with the pytagoras formula I can simplify again to a^2 + b^2 = 64.
I was just flabbergasted to see what you did.
No brainer solution:
If x and y are the sides of the two squares, and a is the offset of the lower middle corner from the center of the circle, the formulae for the radii that go to the corners of each square on the circle are :
(x+a)^2+x^2 = 8^2 and (y-a)^2+y^2 = 8^2.
Expanding, one gets:
2x^2 + 2ax + a^2 = 8^2 [1] and 2y^2 - 2ay + a^2 = 8^2 [2]
Subtracting [2] from [1] and dividing by 2:
x^2 - y^2 + a (x + y) = 0
As y^2 - y^2 = (x - y)(x + y), we get:
(x + y)(x - y) + a(x + y) = 0, or (x + y)(x - y + a) = 0.
Since x + y > 0, we must have (x-y+a) = 0, i.e. x - y = - a [3]
Adding [1] and [2] and dividing by 2, we get:
x^2 + y^2 + a(x - y) + a^2 = 8^2; plugging in [3] we get:
x^2 + y^2 -a^2 + a^2 = 8^2; thus
x^2 + y^2 = 8^2
Andy, I hated all this when I was at school - I did it (even A-level maths) but I hated it - you made me love this stuff and I've actually joined Brilliant - first time a youtube sponsor has ever worked on me - its only a free trial but I'm probably going to keep it up - the way you (and brilliant) explain maths makes it makes sense and therefore (to me) fun. maths was always learn this formula apply it - I actually understand quadratic equations now - I can look at an equation and know where it will fall on the axis - absolutely crazy! (Maybe everyone can do that nowadays but for me, a child of the 80s, it was like 'learn this do this'!
(If I read this post I would think hmmm is this a shill account for Brilliant? I promise, I'm not.. (EXACTLY WHAT A SHILL ACCOUNT WOULD SAY!!!)
Alternative solution:
Imagine both the squares are same, so they meet in the center of the circle. So, Diagonal is the radias of the circle. According to Pythagoras, square's diagonal = x√2. So, radius = 8 = x√2. So, x=8/√2. Thus, 2x^2 = 64
You can save several steps because the area is independent of the relationship between x & y. So just use the case where x = y. Diagonal of each square is 8, and the rest is in the video.
but how do you know the area is independent of x and y without doing what he did?
@@oniondesu9633 The first hint is that if they weren't independent, the question would have been differently -- you can't solve it with the info given if they are not independent.
@JamesRedekop watch the video, he literally did solve it without assuming their relative size.
@@oniondesu9633 Yes, I know. But he didn't need to go through all those steps if you skip straight to x = y.
@@JamesRedekop why not skip to the area then, why do any maths at all, just let someone else do the problem for you
3:05 - 'side of square' is fine but how did we conclude that other line is 'diagonal of square' ??
I actually learn math here. Wish school would have been like this.
If the size of the squares can vary but the area remain same, why not consider squares that are exactly exactly equal in size? In this case both x = y = 32^0.5 (since the diagonal will be 8 unites as they become radius of the circle). Then x^2+y^2 = 32 + 32 = 64!
What happens if another square is introduced?
As there is no restriction on the combination of sizes of the two squares, we can set the left square to have an area of zero, thus forcing the right square's bottom left corner to be positioned at the left extremity of the semi-circle and the square's top right corner to be positioned at the top-most point of the semi-circle. This gives just one apparent square with a side length of 16/2 = 8.
The area is then just 8 * 8 = 64.
Wait a minute.
If we realize that there's nothing defining the location of these squares then we can try the shortcut and solve it for when they are equal. And if they are equal then so are their diagonals. And the only instance when they could have equal diagonals and be located the way they are is when their common vertex is the center of the semicircle. That would mean, they would form a rectangle and their diagonals would be equal to radius, which is 8. That would mean the height of the rectangle is 4sqrt(2) and the width is 8sqrt(2). So we multiply those and get 64 square units.
I mean, that feels like cheating, because the method implies the task is solvable with just the given numbers, but hey, that's also an option if we follow the logic of the first statement. How exciting indeed.
That's what I did, before watching.
It's not wrong. It's just not the rigorous way to prove it. If this was on a test and you had to choose an answer and you aren't asked to motivate your answer, do it your way every time. But Andy wants to prove it to us so there's no doubt.
@@Qermaq that's part of the reason I left the disclaimer in the end.
But now that I think about it this is a legitimate method, though tricky to decide when applicable and even more tricky to debate with whoever is grading it: "if the problem is solvable with just the given data, then the answer is this. And if it's the wrong answer, then the task is flawed and needs more data."
@@WhooshWh0sh That's the risk. What if it looks like you can generalize but you miss something? What if the question has an error, so by giving an obvious answer you are wrong?
Me with a ruler beating the system
The angle between the 2 radii had to have been 90° - you originally had a 90° angle when you drew the 2 diagonals of the squares. You can move that angle along the diameter while maintaining the 2 points on the circle, and the angle will be constant at 90°
There are only two points on the diameter where the angle is 90°. The point where the two squares meet and the center of the circle.
nice title
can you pls tell the some book names for this type of riddles and problems?
drive.google.com/file/d/1hVP8tLURVDphmHsphz5BQLVzHCeTts29/view
In my other comment is a link to a bunch of other Catriona Agg Puzzles. I hope you love them!
@@AndyMath thanqu alot love you bro it helps me in study alot
Luvit 👍👍👍👌💯
I "did it" diffirently. Since this problem "should" have the same answer indepent of the squares I took the case that both squares are the same in size. In that case the hypotenuse is equal to the radius of the circle, I then used the pythagorean theorem to figure out the lenght of the sides and multiplied the lenght of the height (1xside) by the base (2xside).
You don't even need to use pythagorean theorem. Once you split the squares by the radius, you have four identical isosceles right triangles that can be arranged in a square with a side of 8.
@@Snorkl7879 Oh yeah I never thought of that :D
So will two inscribed squares in a semicircle always have an area equal to the square of the radius?
Its always fun
One thing that,can we find answer of all questions similar to it?
I found a much simpler solution :)
If the area of the two squares will remain the same no matter their size, you can assume they are equal. Once you assume that, their diagonal is equal 8. From here it’s super simple solution
My method was a lot cheaper. Since it doesn't matter how big each square is, I chose to make them the same size, which puts them right in the middle making the diagonal the same as the radius. That means 2x²=8²=64 (pythagoras), which is already the area of both squares.
Dang. I was so excited cuz I thought I figured it out before he gave us the answer. It turns out I was right but only cuz I made an assumption which turned out to be true, I didn't prove it, I thought it was just a rule. I got lucky. Still I did a lot of the work on my own. Gonna keep practicing and I'll get there.
If it works wherever the squares are and is true for any case, then why don’t you just put the two squares by the center of the circle and solve in an easier way
That is the way I did it, but this way is more fun.
nice point!
How can you prove if that case represent the other cases? (where the two squares are not by the center of the circle)
You would have to prove that first
Was just gonna comment that. I believe coz Im Indian and we give tests like IIT.
The comments seem highly confused about this one. His animation at the start did not prove that the area we were aiming for was independent of the side lengths of the squares, think of it more like a spoiler, or something he suspected.
He still had to solve the problem for an arbitrary x and y, the fact that his answer did not depend on x or y was the proof of the suspicion he layed out at the start.
If he had just set x=y, yes, he would have got the area as 64 much easier, but he would NOT have solved the problem generally. He would have just solved one special case.
Heh, that’s fantastic 😂
Beauty of mathematics 😮😮😮
Damn bro you are a genious
Since the squares sizes didn't matter I would make them equally big and solve for x2+x2=2x2.
Engineering approach: assume x=y. Area=r squared. 8x8=64.
3:02 if the second chord is just a line from the two points on the circle, how do we know it bisects the bottom yellow square? is that provable?
Did you take these off pre math?
That was super cute math.
How to find the area of individual squares?
I got the same answer but since I knew the answer would work for any value of x and y as long as the squares touched the semicircle at the corner, I assumed x=y for the case where the squares were the same size. I knew x=(radius)sin(45) and there were two squares so 2((8)sin(45))^2=64= the area of the squares. Fun problem
How to find the side length of x and y? Please show me
Is it true then to say the area of two adjoining squares, which one edge the same the diameter of the semi circle, that has two corner points on the semicircles arc, and that are fully encompassed within said semicircle as pictured,
The area always equals 4•D?
Substitute that diameter of 16 with 12, 21, 196, etc, and the area of those two squares will always come out to be 4x that diameter.
Correct?
I’m not a math person, but looking at the problem, and as you slid the dimensions of those squares around, it would appear that that will always be the case. This could be a short formula for such problems that you ever come across.
Wait, how do we know it's a semicircle?
Or, since it aplies for al inscribed squares, you can take the case where they colide in the center of the semi-circle. this would make the diagonal of both squares 8 (16/2) using pythagoras we know that, if x is the side of the square, x^2 + x^2 = 8^2 x^2 = 32. x^2 is the same as the area of that square, and we need it twice so the answer is 64.
This is how I did it before watching the video, I knew that because no information about the squares was given, it would be the same as if they were equal
@@bubblyphysics yep
Really it's so hard. Could you tell me that teachers can explain it
Huh, I did it an easier way. I did have to assume the squares could be the same size (or indeed any ratio of sizes), but since they aren't given any dimensions that seemed valid
since this works for any x/y, you could set x to zero and solve this trivially, no?
VEERRRRRY COOOOLLL
i just mirrored the semicircle to create a full circle with 4 squares
then i drew a line from the yellow squares' corner thats touching the circle to the opposite blue squares' corner
this is the diameter of the circle
x*sqrt2 + y*sqrt2 = 16
x + y = 8*sqrt2
squaring this expression makes a bigger square that imperfectly matches the shape of the 4 smaller squares
but thats okay here since the bit that sticks out perfectly slots into the bit thats missing
so
(x + y)^2 = (8*sqrt2)^2
(x + y)^2 = 128
128 is the area of all 4 squares
so the area of the 2 starting squares is 64
not very rigorous but i thought it was an interesting solution
Just take one university-level course in Euclidean geometry and you can solve any problem I have seen this fellow solve.
so it's just r^2
wow
If all these versions of this is true when diameter is 16 meter or not please comment
Extremely easy:
R = ½ 16 = 8 cm
A = R² = 64 cm² ( Solved √)
This figure is not defined, positions of squares are not given (vertex positions)
We can modify the squares dimensions KEEPING the original conditions
I decided to match both squares, and the common vertex becomes the center of the circle.
Therefore, the diagonal of each square is the radius of semicircle
A₁= A₂ = ½d² = ½R²
A = 2A₁ = R² = 64cm²
Other solution, instead of matching both square, can be maximize area of large square, and minimize area of small square.
And the original conditions are still fulfilled.
In this case :
R²=S²+(½S)²
S² = 4/5 R² = 4/5 8² = 51,2cm²
s² = ¼ S² = 12,8 cm²
A = S² + s² = 64 cm² ( Solved √ )
You made it unnecessarily complicated. As you showed at the beginning, the two squares can be adjusted to be equal inside the semicircle. So then we know that the diagonal (and radii) is 8. Bing bang boom.
That’s what I thought
X^2 + X^2 = 8^2 = 64 sq units assuming one square shrank infinitely
no he did not "show" it at the start, he showed it at the end when his answer didnt depend on x or y
@@oniondesu9633 You must be slow to understand basic concepts. Bless you.
@@DisposableSupervillainHenchman dude, you are acting smug when you are completely wrong. yes, the area is the same no matter what the relative sizes of the squares are, but this cannot just be stated, it must be proved, which he did in the video.
I have to admit, when I looked at this, the way forward was certainly not obvious to me. If it was not so easy to just click the play button to satisfy my curiosity, I might have figured it out. Maybe.
@3:08 please look again, the 45 degree angle is not justified
u r assuming that side (chord) of length c can be the hypotenuse in 2 separate right angled triangles! u r missing something...
or just see the case of 2 even squares. turn them into 4 triangles with base 8 and height 4 leading (8*4/2)*4 => 64
If they were equal squares, the height and base of each would be 8/sqrt(2).
@@markdaniel8740 yes correct. I turned the squares into 4 isosceles triangles
this is a ugly hack, but there is no constraint stopping me from assuming that the 2 squares are of equal area. now the question becomes really easy because the diagonal of each square is 8, and the are is (8*8)/2 for each square, since there are 2 squares, the area is 64. The reason this hack is ugly is because i most probably would not have guessed that the total area of the two squares are not dependant on the individual widths of the squares.
However, it isnt an ugly hack if we do this, from the yellow square, you can cut half of the extra portion and append it to the right of the yellow square. Now take the extra left after appending and add it on top right of the blue square. Now, take the same width from the left of the blue square and move it on top, and we have 2 equal squares. This proves that sum of areas of the squares are not dependant on the individual sides, which makes this hack beautiful!
it's not just ugly, it would be incorrect. nowhere was it stated that the total area is independent of the side lengths, it was something he proved in this video. you cant just assume it
@oniondesu9633 I've written a geometric proof in the second para of the comment. Go over it again. Try plotting what I'm telling in the second para, it's a fun proof
@@XxFALCONxX- you can turn any two squares into two equally sized squares with that procedure, but how do you know your new squares perfectly fit inside the same semicircle?
@@oniondesu9633 ohh you arnt wrong, mb, I was wrong. I made the assumption that if I cut the top half rectangle from the square (half the difference of the sides of the square), then it will fit on the right, I made that assumption without proof because it looked so (I can prove it tho by showing that the distance from the center of the circle to the bottom left corner but thats gonna take more triangles and math than the solution in the video). Thanks for the correction!
Actually the answer will be 4 times of diameter
The shown computation is unnescessary complicated.
there are a lot of assumptions here so much that I am not sure this can be generalized
woah!!!!!!!!!!!!!!!!!!!!!
WTF IS THE TITLE NAH💀💀💀💀💀💀💀💀💀💀💀
🟥🟧 1/2🔵 but the square are girls and the semicircle is a cup