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.
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?
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
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.
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
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.
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.
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
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.
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 😂
"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.
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
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...
