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 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 :)
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.
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
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.
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?
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.
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
"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.
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.
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 😂
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.
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
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.
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)
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.
'i don't know what to do with it, so let's put a box around it' brilliant
Engineering school in a nutshell lol
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
"Let's enjoy all the things we know about... " is such a great line.
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 mad lad, solving a squares problem in a square pattern shirt
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.
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.
I hope no teacher ever puts this on a test or he'd be accused of murder.
😂😂😂
My mind melted with this solution... Amazing!
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.
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
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.
Absolutely lovin’ your proofs/derivations/playtime. How exciting! Thanks.
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!
Done quickly, but I understood it on the first watch! Great job. Thanks!🎉
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 :)
This was amazing to witness 😂 I love these videos!
At some point in every video I pause and think: "How the hell did we get here"
This was beautiful. The answer is literally just 2 times 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.
What a clever lad you are!
So the green is twice the blue? Never would have guessed. Fantastic work.
I love how you explained it but this is definitely a hard one.
That was like watching the most beautiful dance.
I think that was the best I've ever seen ❤
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.
Love your videos bro keep it up!
This one was really hard for me but you really make it look easy.
@3:15 I just smiled when you said top over bottom, twice.😘😛
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
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.
That excited me.
how exciting
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?
Very 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.
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)
I understood the logic but I could never think about it
Amazing...
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
How exciting! 😀
Just a quick one,
Which software do you use to pick out the geometric figures
The solving is very satisfying, that’s why I love math
Heck of a ride
I learned how to squint my eyes by watching this video.
alright so that's the last time im underestimating an andy math problem...
"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.
Incredible
I never finish watching an AndyMath vid unhappy
WHERE WAS THIS WHEN I NEEDED IT- I JUST HAD A COMPETITION RECENTLY😭😭😭
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.
What was your thought process when approaching this problem? I'd be interested to see you approach a problem you'd never seen before.
This feels like when a mathematician tries their hand at being a magician 😂
Thank God is still making math nerds to help us all.😮
So satisfying
That's not the way I would've gone about this but is definitely better
How Exciting!
How exciting
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
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.
I shue dont know what all he did but how exciting
Are you using any special software when manipulating the drawing and symbols?
I am loving these videos, they're short but fun! Take my Like and Subscribe!
What software does Andy use for his presentation?
How equalibrating.
Motivational
Let’s enjoy.
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
can you do one that doesn't look fun
this one was just plain fun
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.
Nice :)
I have to ask, where do you get these questions from?
even though watching the explanation my brain worked hard
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.
Let A be the lower left green corner. B,C the upper right green and blue corners. Let O be the common corner. Note
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
Wow
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....
where i see triangles i see a solution
SquuuuaaaaaRe
Pretty easy
Damn I tried to do this differently and got 16 + 8√3. Am I cooked?
We're all cooked after that one 😅
I think I'm in the wrong class...
Funny how it were just all silent when I tried to solve it
Some day you'll do one that isn't a "fun one". :)
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...
X=0 is also a solution.
Who up Andying they Math
The green area is a square.
How tiring!
play with angles, prove that the blue's diagonal equals the green's side and you have it.
I prefer not to know the answer to this problem
I'm good with it
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.
this one was bit hard
You don’t need to do all that to figure out it’s 24
huh?
o.O
Ans is 24