So, I have adhd, and really struggled with calculus and trigonometry in high school. Thanks to your videos I’m working through online courses now to fix my knowledge gaps. So, thanks for doing what you do. 😁
That was a good solve. I figured it was going to have to involve similar triangles, but I’ve noticed one place I have trouble with these questions is drawing in pieces that aren’t provided and using those to solve the problem.
I may have found the simpler solution you're looking for involving similar triangles. I'm just gonna copy paste my solution here, though I don't know if it would be understood since it's insanely difficult to explain without visuals and English is not my first language. Let's just say that this method starts by proving that the two small triangles at the right of the figure are CONGRUENT. (instead of the video starting at left side) Continuing from that logic you can also prove that the larger triangle at the bottom and a hidden triangle (made by the large square and the diagonal of the green square) are also CONGRUENT. Using CPCTC (Corresponding Parts of Congruent Triangles are Congruent) you can connect that the diagonal of the green square is Congruent to the hypotenuse of the large bottom triangle, which has a length of (√12+b)+(√12-b) = *√24*. Therefore the area of the green square is *24*. The √12 is the side of the blue square and "b" is the extra bit that large square cut off from the blue square, which we proved is congruent to the extra bits between the height of the green square and the blue square.
I hated geometry in high school. Sucked at it. Did everything I could to get out of that class. Now, when I see a new video on this channel, I can't wait to watch it. That's the power of a great teacher. Fantastic job, Andy!
Hi @AndyMath , big fan of your puzzles! *EDIT: below was all incorrect of me. Managed to check it GeoGebra, and there is only one solution (24). But it was interesting to check :) ...but are you sure that this problem is properly constrained(?) I tried modeling this problem in Desmos, but I've never used it before so didn't get it to work to test my intuition about the constraints. And I don't have access to a good CAD-sw to try and model this either (If anyone has feel free to try! :) ) My intuition tells me that the size of the green square is variable(?) Or is it uniquely determined by the way it's connected/constrained to the other squares? I'm very un-scientific here but assuming that the blue square is "anchored" in space; it feels like the lower left corner of the green square can be "dragged" in a way that varies the size of the green square while still following all the other constraints of the image(?) I might be completely wrong but is the assumptions about the a,b,c angles entirely correct? It would be really interesting if someone could help me check this and the constraints :)
I think I got the simpler solution. But it's insanely difficult to explain without visuals. Let's just say that this method starts by proving that the two small triangles at the right of the figure are CONGRUENT. (instead of the video starting at left side) Continuing from that logic you can also prove that the larger triangle at the bottom and a hidden triangle (made by the large square and the diagonal of the green square) are also CONGRUENT. Using CPCTC (Corresponding Parts of Congruent Triangles are Congruent) you can connect that the diagonal of the green square is Congruent to the hypotenuse of the large bottom triangle, which has a length of (√12+b)+(√12-b) = *√24*. Therefore the area of the green square is *24*. The √12 is the side of the blue square and "b" is the extra bit that large square cut off from the blue square, which we proved is congruent to the extra bits between the height of the green square and the blue square.
Big square side length = w Green square side length = g The small area between green and blue has length g-12 The slope of the line of the white square is -(g-12)/12=1-g/12 The angle theta is arctan(1-g/12) The diagonal has angle 45°+ theta and slope 12/(g+12) 12/(g+12)=tan(45+theta) = (tan45+tan(theta))/(1-tan45tan(theta)) = (1+1-g/12)/(1-(1-g/12)) So: 12/(g+12) = (2-g/12)/(g/12) Cross multiply and get: 12(g/12) = (2-g/12)(g+12) 12g = (24-g)(g+12) 12g = 24g-g²+288-12g g²=288+24g-12g-12g=288 I now realise that the blue square has area 12, and not side length. So just scale everything down by sqrt(12) (g/sqrt(12))²=288/12=24, which is the area of the green square
I would love to see you complete one of these questions on paper, to see how you would clearly draw out your explaination as if on an exam...how exciting
You can also obtain the area via rearrangement, so that the final calculation is a simple multiplication. If you want to work that out for yourself, try moving the parts of the squares outside of the largest one into the big square by finding congruent triangles. For example, you can move the excess of the square with area 12 into the square by extending the top edge to the other side of the large square and dropping a perpendicular from that intersection. Visual inspection will tell you that the resulting triangle will be congruent to that excess, but proving it just requires recalling the properties of parallel lines. If you do everything right, you should get that the area of the green square is 2*sqrt(12)*sqrt(12) = 24.
Firstly, I think you should have more clearly stated that the top line of the diagonal square does touch the green square's top-right corner, since we aren't supposed to assume that the drawing is drawn to scale. Secondly, showing us that there's only one possible size for the green square by showing that making it bigger or smaller would make it impossible for the diagonal square to be a square (because its top side has to touch the green square's top-right corner) would go a long way towards helping people intuitively understand that an algebraic solution is in fact possible.
you can't assume that its drawn to scale, but what you CAN assume that things that touch each other do touch, in the same way you can assume that straight looking lines are straight
@@Hoolahups there is a little bump where the top right corner of the green square meets the tilted square, the drawing should be cleaned up so assumptions arent necessary.
This is the first time I've seen your intesecting vertices/overlapping angles method of discovering similar triangles. Something to keep in my for the next time.
My intuitive answer was 24 because points that touch blank square can be placen in circle with the middle where corners of two squares touch. But I dont know how to prove it, it's just intuition. So in this case x = diagonal of blue square.
Given that the only part where the numeric value of the area was relevant was for the actual final calculation, we can actually determine that the green area is always exactly twice as big as the blue area
If nobody believes me, that's fine, but 24 was my guess based on the size of the squares and the large square looking like it cuts of the same percentage of each of the smaller squares. Nice to know that common sense can at least help a little with things like these. I may not know the math, but i can guesstimate (as my elementary teachers once said) a little. I assumed I was gonna be off be +-2 with some decimals in there, but being dead on made me say "no way" out loud when he solved it 😂
You could solve it the same way, but in one step by noticing two triangles: 1. The same one you use with sides x + √12 and √12 (and that you call √2 y) 2. The smaller white one in the top left, with sides √12 and √12 - x Those are flipped but have the same angle, so (x + √12) / √12 = √12 / (√12 - x)
I have a challenge for you. Find all triples of three-digit natural numbers a, b, c for which is true: b²= a · c, b = a + 34.🙂I couldn't figure it out by myself.
"what's the green area?" Well, it's an area of a green square that seems to have been cut off by another larger, transparent square. I hope that helps. Good look with your colouring in.
You can also solve it by halving the 45 degree angle. If a = c, then the angles of the right triangles adjacent to both angles a and c must also be equal to a and c. Hence, a + a = 45 degrees; 2a = 45 degrees; a = 22.5 degrees. We can now use the side length of the green square that produces the equation, 2sqrt3 + n = x. Hence, (2sqrt3 + x)^2 = area of green square Using this equation, knowing the side length of the square with area 12, and knowing the narrow angle must of the right triangle must be 22.5 degrees, we get: (2sqrt3*tan(22.5 degrees) + 2sqrt3)^2 = 24 This can be properly explained if I have visuals.
It's always the ones where you have to add to the picture that get me... I try to do it without. I tried doing stuff with all of the similar triangles in the picture, but that didn't produce anything productive.
Great stuff, I think I may have found a slightly cleaner solution by getting the biggest right triangle (diagonal of largest square) similar to the white right triangle above the area 12 square, angles similar to your method! I wonder there are other interesting configurations?
Without really knowing anything about how to solve the problem, I just guessed the answer and decided to see how close I’d be. Apparently, right on the money.
I don’t do the math but I usually pause to see what I would have to do to get the answer and man I would’ve struggled because I was looking in all the wrong places
Can someone explain to me what happened at 3:00, like what did he use tk determine that the top and bottom side has the same ratio between the two triangles
Ain't no way I just solved it by looking at the thumbnail two minutes and just guestimating that if the one with the area of 12 was 4x4 it just looks like the green square next to it would be 6x6. I love when the question models are proportional...
There is something wrong about it. I can't put my finger on it, but something is wrong.... If we just make green square bigger it will still be able to make the same figure.
I always try to solve these myself first.... i forgot about angle teuths and i forgot about proper distribution with a square root. But, my guess became something between 4.1 and 4.8.... using just geometry and shitty algebra. I could have trial and errored my way there but damn. I missed so much obvious. Worst part is, at the beginning my brain said the answer will be in the form of a square root since its gonnna be irrational...
What little math talent I have (nothing beyond SAT/ACT/GRE- testprep level) is skewed toward geometry. It was always easier than algebra for me. No idea why: zero visual/art talent.
Going to flex my intuition a bit here. look at the bottom left corner of the blue square. It lies on the line going from the top corner of the transparent box to the bottom corner of the transparent box. Due to symmetry over that line, the length of the bottom edge of the green box must be equal to the length of the diagonal of the blue box. a^2+b^2=c^2, and the diagonal of the blue box is sqrt(24), so the area of the green box is 24. Feeling so powerful rn
'i don't know what to do with it, so let's put a box around it' brilliant
Engineering school in a nutshell lol
I tried that, but I ended up with so many boxes, I didn't know what to do with them. So I put a big box around all of them.
"Let's enjoy all the things we know about... " is such a great line.
The mad lad, solving a squares problem in a square pattern shirt
My brain isn't braining
What's even a brain?
Brain?
Y he always has a hat
brain? What brain?
I keep thinking it’s impossible because green and white can grow forever, why can solve
love how elegant the solution feels when it all dissolves down at the end even to the answer being x^2
He defined the side length of the square to be x. Of course the square of side length x would have an area of x squared.
The loint being the solution gave x^2 before finding x and then squaring it to find the answer.
Literally how exciting
It is more exciting than having a pretty girl in front of me
@@User-jr7vf bro being too real
More than learning same things at school
@@User-jr7vf Fr. Don't chase girls.
So, I have adhd, and really struggled with calculus and trigonometry in high school. Thanks to your videos I’m working through online courses now to fix my knowledge gaps. So, thanks for doing what you do. 😁
Nice!
You can do it!
I highly recommend 3blue1brown for some top-notch math education content, especially for developing a "sense" for the math and how to approach it.
His delivery is brilliant for ADHD.
My mind melted with this solution... Amazing!
It's 00.30 am in Indonesia, and now I can sleep. Thank you
Sweet dreams.
Bro it's 3am I am watching it after a party
@@frankstrawnation ..... Zero dark thirty
It's 7h23 in Brazil, and now I can get ready to work. How exciting
I hope no teacher ever puts this on a test or he'd be accused of murder.
😂😂😂
That was a good solve. I figured it was going to have to involve similar triangles, but I’ve noticed one place I have trouble with these questions is drawing in pieces that aren’t provided and using those to solve the problem.
I may have found the simpler solution you're looking for involving similar triangles. I'm just gonna copy paste my solution here, though I don't know if it would be understood since it's insanely difficult to explain without visuals and English is not my first language.
Let's just say that this method starts by proving that the two small triangles at the right of the figure are CONGRUENT. (instead of the video starting at left side)
Continuing from that logic you can also prove that the larger triangle at the bottom and a hidden triangle (made by the large square and the diagonal of the green square) are also CONGRUENT.
Using CPCTC (Corresponding Parts of Congruent Triangles are Congruent) you can connect that the diagonal of the green square is Congruent to the hypotenuse of the large bottom triangle, which has a length of (√12+b)+(√12-b) = *√24*. Therefore the area of the green square is *24*.
The √12 is the side of the blue square and "b" is the extra bit that large square cut off from the blue square, which we proved is congruent to the extra bits between the height of the green square and the blue square.
Smooth graphics transitions, excellent explanation, elegant solving... Great video!
My brain just malfunctioned, i will now resort to peeling potatoes for the rest of the week. I'm impressed.
I hated geometry in high school. Sucked at it. Did everything I could to get out of that class.
Now, when I see a new video on this channel, I can't wait to watch it. That's the power of a great teacher.
Fantastic job, Andy!
Hi @AndyMath , big fan of your puzzles!
*EDIT: below was all incorrect of me. Managed to check it GeoGebra, and there is only one solution (24). But it was interesting to check :)
...but are you sure that this problem is properly constrained(?) I tried modeling this problem in Desmos, but I've never used it before so didn't get it to work to test my intuition about the constraints. And I don't have access to a good CAD-sw to try and model this either (If anyone has feel free to try! :) )
My intuition tells me that the size of the green square is variable(?) Or is it uniquely determined by the way it's connected/constrained to the other squares? I'm very un-scientific here but assuming that the blue square is "anchored" in space; it feels like the lower left corner of the green square can be "dragged" in a way that varies the size of the green square while still following all the other constraints of the image(?)
I might be completely wrong but is the assumptions about the a,b,c angles entirely correct? It would be really interesting if someone could help me check this and the constraints :)
Done quickly, but I understood it on the first watch! Great job. Thanks!🎉
I think I got the simpler solution. But it's insanely difficult to explain without visuals. Let's just say that this method starts by proving that the two small triangles at the right of the figure are CONGRUENT. (instead of the video starting at left side)
Continuing from that logic you can also prove that the larger triangle at the bottom and a hidden triangle (made by the large square and the diagonal of the green square) are also CONGRUENT.
Using CPCTC (Corresponding Parts of Congruent Triangles are Congruent) you can connect that the diagonal of the green square is Congruent to the hypotenuse of the large bottom triangle, which has a length of (√12+b)+(√12-b) = *√24*. Therefore the area of the green square is *24*.
The √12 is the side of the blue square and "b" is the extra bit that large square cut off from the blue square, which we proved is congruent to the extra bits between the height of the green square and the blue square.
I did it by identifying 2 pairs of congruent triangles, which throws up (by CPCTC) that x√2 = 4√3, and so x² is simple to calculate.
At some point in every video I pause and think: "How the hell did we get here"
Absolutely lovin’ your proofs/derivations/playtime. How exciting! Thanks.
This was beautiful. The answer is literally just 2 times the blue square
I enjoyed all the things we learned about the triangles.
Big square side length = w
Green square side length = g
The small area between green and blue has length g-12
The slope of the line of the white square is -(g-12)/12=1-g/12
The angle theta is arctan(1-g/12)
The diagonal has angle 45°+ theta and slope 12/(g+12)
12/(g+12)=tan(45+theta)
= (tan45+tan(theta))/(1-tan45tan(theta))
= (1+1-g/12)/(1-(1-g/12))
So:
12/(g+12) = (2-g/12)/(g/12)
Cross multiply and get:
12(g/12) = (2-g/12)(g+12)
12g = (24-g)(g+12)
12g = 24g-g²+288-12g
g²=288+24g-12g-12g=288
I now realise that the blue square has area 12, and not side length. So just scale everything down by sqrt(12)
(g/sqrt(12))²=288/12=24, which is the area of the green square
I would love to see you complete one of these questions on paper, to see how you would clearly draw out your explaination as if on an exam...how exciting
I love how you explained it but this is definitely a hard one.
I think that was the best I've ever seen ❤
That excited me.
how exciting
You can also obtain the area via rearrangement, so that the final calculation is a simple multiplication. If you want to work that out for yourself, try moving the parts of the squares outside of the largest one into the big square by finding congruent triangles. For example, you can move the excess of the square with area 12 into the square by extending the top edge to the other side of the large square and dropping a perpendicular from that intersection. Visual inspection will tell you that the resulting triangle will be congruent to that excess, but proving it just requires recalling the properties of parallel lines. If you do everything right, you should get that the area of the green square is 2*sqrt(12)*sqrt(12) = 24.
Firstly, I think you should have more clearly stated that the top line of the diagonal square does touch the green square's top-right corner, since we aren't supposed to assume that the drawing is drawn to scale.
Secondly, showing us that there's only one possible size for the green square by showing that making it bigger or smaller would make it impossible for the diagonal square to be a square (because its top side has to touch the green square's top-right corner) would go a long way towards helping people intuitively understand that an algebraic solution is in fact possible.
you can't assume that its drawn to scale, but what you CAN assume that things that touch each other do touch, in the same way you can assume that straight looking lines are straight
@@Hoolahups there is a little bump where the top right corner of the green square meets the tilted square, the drawing should be cleaned up so assumptions arent necessary.
This was amazing to witness 😂 I love these videos!
I never finish watching an AndyMath vid unhappy
Young man, you earned that one!
Wow!
How exciting indeed!
So the green is twice the blue? Never would have guessed. Fantastic work.
I learned how to squint my eyes by watching this video.
@3:15 I just smiled when you said top over bottom, twice.😘😛
What a clever lad you are!
This is the first time I've seen your intesecting vertices/overlapping angles method of discovering similar triangles. Something to keep in my for the next time.
My intuitive answer was 24 because points that touch blank square can be placen in circle with the middle where corners of two squares touch. But I dont know how to prove it, it's just intuition. So in this case x = diagonal of blue square.
This feels like when a mathematician tries their hand at being a magician 😂
Given that the only part where the numeric value of the area was relevant was for the actual final calculation, we can actually determine that the green area is always exactly twice as big as the blue area
alright so that's the last time im underestimating an andy math problem...
What software is Andy using to demonstrate these problems, or is this just a presentation where he manually pieces together each step?
probably manim for the equations (which is a python library for maths animation)
If nobody believes me, that's fine, but 24 was my guess based on the size of the squares and the large square looking like it cuts of the same percentage of each of the smaller squares.
Nice to know that common sense can at least help a little with things like these. I may not know the math, but i can guesstimate (as my elementary teachers once said) a little. I assumed I was gonna be off be +-2 with some decimals in there, but being dead on made me say "no way" out loud when he solved it 😂
this was how i sped thru some problems on the math section of the ACT back when i took it in order to give myself more time on the harder ones :P
The solving is very satisfying, that’s why I love math
You could solve it the same way, but in one step by noticing two triangles:
1. The same one you use with sides x + √12 and √12 (and that you call √2 y)
2. The smaller white one in the top left, with sides √12 and √12 - x
Those are flipped but have the same angle, so (x + √12) / √12 = √12 / (√12 - x)
Exactly what I did. Makes it so much easier.
i dont get the sqrt12 - x
That was like watching the most beautiful dance.
I have a challenge for you. Find all triples of three-digit natural numbers a, b, c for which is true: b²= a · c, b = a + 34.🙂I couldn't figure it out by myself.
This one was really hard for me but you really make it look easy.
"what's the green area?"
Well, it's an area of a green square that seems to have been cut off by another larger, transparent square.
I hope that helps. Good look with your colouring in.
Let A be the lower left green corner. B,C the upper right green and blue corners. Let O be the common corner. Note
You can also solve it by halving the 45 degree angle. If a = c, then the angles of the right triangles adjacent to both angles a and c must also be equal to a and c. Hence,
a + a = 45 degrees;
2a = 45 degrees;
a = 22.5 degrees.
We can now use the side length of the green square that produces the equation, 2sqrt3 + n = x. Hence,
(2sqrt3 + x)^2 = area of green square
Using this equation, knowing the side length of the square with area 12, and knowing the narrow angle must of the right triangle must be 22.5 degrees, we get:
(2sqrt3*tan(22.5 degrees) + 2sqrt3)^2 = 24
This can be properly explained if I have visuals.
I understood the logic but I could never think about it
Just a quick one,
Which software do you use to pick out the geometric figures
I come here to enjoy math
WHERE WAS THIS WHEN I NEEDED IT- I JUST HAD A COMPETITION RECENTLY😭😭😭
How exciting! 😀
I am loving these videos, they're short but fun! Take my Like and Subscribe!
That's not the way I would've gone about this but is definitely better
Thank God is still making math nerds to help us all.😮
Very exciting!
What was your thought process when approaching this problem? I'd be interested to see you approach a problem you'd never seen before.
It's always the ones where you have to add to the picture that get me... I try to do it without. I tried doing stuff with all of the similar triangles in the picture, but that didn't produce anything productive.
Heck of a ride
I just found a line through the big square and subtracted the line from the blue square on that line and that was the side length of the green square
Great stuff, I think I may have found a slightly cleaner solution by getting the biggest right triangle (diagonal of largest square) similar to the white right triangle above the area 12 square, angles similar to your method! I wonder there are other interesting configurations?
Without really knowing anything about how to solve the problem, I just guessed the answer and decided to see how close I’d be. Apparently, right on the money.
Let’s enjoy.
can you do one that doesn't look fun
Love your videos bro keep it up!
How Exciting!
How exciting
Are you using any special software when manipulating the drawing and symbols?
I shue dont know what all he did but how exciting
How equalibrating.
it tickles my brain nice way
Which website or application he uses to do these stuff
I don’t do the math but I usually pause to see what I would have to do to get the answer and man I would’ve struggled because I was looking in all the wrong places
What software does Andy use for his presentation?
Amazing...
this one was just plain fun
even though watching the explanation my brain worked hard
Can someone explain to me what happened at 3:00, like what did he use tk determine that the top and bottom side has the same ratio between the two triangles
So satisfying
Ain't no way I just solved it by looking at the thumbnail two minutes and just guestimating that if the one with the area of 12 was 4x4 it just looks like the green square next to it would be 6x6.
I love when the question models are proportional...
4x4 is 16, and 6x6 is 36....
Does this assume that the large square perfectly meets the green and blue square on their corners? I think that’s what is confusing to me
Incredible
I have to ask, where do you get these questions from?
Damn I tried to do this differently and got 16 + 8√3. Am I cooked?
We're all cooked after that one 😅
Motivational
My dumbass would have been trying for 30 minutes and when I finally solved for x², I would take the square root just to realize I needed the area
Wow
There is something wrong about it. I can't put my finger on it, but something is wrong....
If we just make green square bigger it will still be able to make the same figure.
where i see triangles i see a solution
Nice :)
I always try to solve these myself first.... i forgot about angle teuths and i forgot about proper distribution with a square root. But, my guess became something between 4.1 and 4.8.... using just geometry and shitty algebra. I could have trial and errored my way there but damn. I missed so much obvious. Worst part is, at the beginning my brain said the answer will be in the form of a square root since its gonnna be irrational...
Anyone knows what software he uses for these vids?
What little math talent I have (nothing beyond SAT/ACT/GRE- testprep level) is skewed toward geometry. It was always easier than algebra for me. No idea why: zero visual/art talent.
X=0 is also a solution.
Going to flex my intuition a bit here. look at the bottom left corner of the blue square. It lies on the line going from the top corner of the transparent box to the bottom corner of the transparent box. Due to symmetry over that line, the length of the bottom edge of the green box must be equal to the length of the diagonal of the blue box. a^2+b^2=c^2, and the diagonal of the blue box is sqrt(24), so the area of the green box is 24. Feeling so powerful rn