What is the overlapping area when you rotate a right triangle?

Can you solve this delightful problem?

Sources
Nèstor Abad proof
Ignacio Larrosa Cañestro animation

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[NestorAbad]

As usual, thanks for sharing these problems and thanks for the quote!

[jackhandma]

Once you realize that BDC is isosceles, it's over.

[bluerizlagirl]

This is literally the first step of what you have to do to find the centre of mass of triangle ABC: the areas of the blue triangle on either side of AD must be equal, so the blue triangle is half the area of ABC. Therefore, the answer must be 1/3 (since the white bit left over from the rotated triangle must have the same area as the other white triangle, which is the same as the blue triangle).

Still, it's a very nice visual proof of _why_ the areas must be equal (and therefore, why this method actually finds the centre of mass). Both the blue triangle and the white triangle have the same base angles, which makes them isosceles triangles; therefore, the common side and the blue side are equal, and the blue side and the white side are equal, so the white and common sides must be equal. Dropping a perpendicular and using similar triangles shows half the blue area is equal to a quarter of the area of ABC, so the whole blue triangle is half the area of ABC and the centre of mass lies on AD.

[draughtoflethe]

I think I did something similar, if a bit simpler. Angle C is congruent with C-prime, because they're the same angle in rotations of the same triangle. Therefore the shaded region is an isosceles triangle, and lines DB and DC are equal. Looking at triangle ADB, we can tell that it is also an isosceles triangle because angle A + angle C + 90 = 180 (a triangle), thus angle A + angle C = 90, and 90 - angle C = angle A. Since C-prime = C, 90 - DBC = angle A, making ADB isosceles, and sides DA and DB equal. So we know DA = DB = DC. Since DA = DC, line DB is the median of triangle ABC, meaning that it divides ABC into two triangles of equal area. That means the shaded portion is half of triangle ABC. Because the two large triangles are rotations of one another, the shaded portion must also be half of triangle (ABC)-prime. The two unshaded halves are equal to one another and to the shaded portion, making the shaded portion 1/3 the area of the entire figure.

[MaxMathGames]

A good question on similarity of triangles. The problem with these questions is that, most often, there is a particular way of solving these questions.

And as always, great presentation skills 👍👍👍❤️❤️❤️

[ravirajamadan]

I solved it with trig and much shorter solution. But I am mighty disappointed as Gougo's theorem did not make an appearance again :-(

[LegendaryRc]

since the question implies that every right triangle will get you the same answer
you can just start by taking a 90 45 45 triangle to make it simpler

[drozfarnyline]

First problem that I solve in my mind quickly.Really a nice problem.Thanks Presh!

[fran]

There were already 2019 likes. Haven't been able to like the video

[omrynagar]

Please upload more often if you can! Cant wait for those videos

[theunknownman]

Being a 14 year old, I am really happy that I solved it myself..

[yoav]

after shwing that BD=DC you could just use the fact that the height of BDC falls at the mid point of BC so the area of this triangle is half of the area of ABC

[everydayjokes]

Joke of the day:
What do you call an alligator in a vest?
*An in-vest-igator*

[yashrawat]

Me sees this question
Happy median from the 90 degree vertex noises

Could be generalized tho by similarity

[kaustavbauri]

You don't even need similar triangles, DB is a median of triangle ABC which divides it into two triangles of equal area, therefore the area of Triangles BDC and BDA will obviously be half the area of Triangles ABC.

[mathwithjanine]

thank you for this fun puzzle! I really enjoyed it! :)

[dormeur]

Saw the thumbnail, solved it in 15 sec in my head, then clicked on the video. It might be the easiest problem on your channel ^^

[edmundprager]

Another way of doing it: Flip the second triangle so that B' = C and C' = B still, areas will still be the same, you're left with a rectangle with two diagonals with one resulting triangle removed, each remaining triangle therefore 1/4 of area, 1/4 (shaded) out of 3/4 (whole shape) is 1/3.

[asvinsingh]

Maind your decisions, aaiam Presh Talwalkar~*goosebumps*

[JWentu]

Nice, very nice and elegant, but I solved it in my head in about one minute.