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
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.
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!!"
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.
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.
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
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.
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.
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.
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.
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.
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
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.
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.
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!
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.
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
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!
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?
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.
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.
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
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.
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°
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 √ )
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
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!!!)
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
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
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
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.
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
Too much work. Since they're undefined, make them two equal squares, meeting at the center of the circle. Split each square by a radius, now you have four identical isoscelese right triangles with a base of 8. Arrange them all in a square with the right angles touching. New square has a side of eight. Eight squared is 64.
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How exciting.
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 like you showed how the squares can vary but their area will always be same. Another thing to remember! How exciting!
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
👧👧🥤💩
3:12 this move right here really got me
Me too. I would never thought in that way.😅
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.
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?
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.
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.
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
... genius as always. God bless ...
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.
your graphics are simple, clear, well-thought out, and effective
That is the best explanation I have found - thank you for making it so simple to understand. Great teacher!
the equation of a circle is x^2 + y^2 =r^2 , which is also the same as the addition of both areas, so the sum of the areas is r^2=8^2=64
Perpetually being humbled by this channel
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.
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.
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.
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.
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.
This was actually really cool. I’m not even into math but I like these quizzes.
That was a really tough problem. SO clearly explained. Wonderful. Thanks!😊
Outstanding.
Really good, thanks for that.
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
“2 squares 1 semicircle”
*2 girls 1-* 💀
Boy❤
@@CosmicHase ❌
@@redsus8725 you don't like overflow
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"
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
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.
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!
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.
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
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!
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?
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, 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.
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
Fun problem!
So will two inscribed squares in a semicircle always have an area equal to the square of the radius?
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.
How to find the area of individual squares?
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.
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 √ )
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.
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
nice title
I actually learn math here. Wish school would have been like this.
Since the squares sizes didn't matter I would make them equally big and solve for x2+x2=2x2.
3:05 - 'side of square' is fine but how did we conclude that other line is 'diagonal of square' ??
How to find the side length of x and y? Please show me
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!!!)
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?
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
Me with a ruler beating the system
Its always fun
Beauty of mathematics 😮😮😮
Damn bro you are a genious
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
Did you take these off pre math?
Heh, that’s fantastic 😂
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
since this works for any x/y, you could set x to zero and solve this trivially, no?
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
u r assuming that side (chord) of length c can be the hypotenuse in 2 separate right angled triangles! u r missing something...
@3:08 please look again, the 45 degree angle is not justified
Luvit 👍👍👍👌💯
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.
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
🟥🟧 1/2🔵 but the square are girls and the semicircle is a cup
I almost did... Until the video finished
Hey Andy I'm very early here
so it's just r^2
wow
Wait, how do we know it's a semicircle?
WTF IS THE TITLE NAH💀💀💀💀💀💀💀💀💀💀💀
VEERRRRRY COOOOLLL
The shown computation is unnescessary complicated.
The titles of these videos. 😅
there are a lot of assumptions here so much that I am not sure this can be generalized
Too much work.
Since they're undefined, make them two equal squares, meeting at the center of the circle. Split each square by a radius, now you have four identical isoscelese right triangles with a base of 8. Arrange them all in a square with the right angles touching. New square has a side of eight. Eight squared is 64.
there is a much simpler solution i believe
I feel like it should be 64 based on if they were of equal sizes, but let's see
woah!!!!!!!!!!!!!!!!!!!!!
64
Not gonna lie, I didn't trust the process...
2 squares 1 semicircle 🤮😭😭
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So over complicated
64unit.sq
youre so smart and handsome, are you single btw? nah just kidding, very cool video🫰🏻