Can You Solve A Cambridge Math Olympiad Problem?

#onlymathlovers

[dugong]

I'm not usually good with geometry, but I think I found an easier way to get to this result. The construction creates 3 new triangles. For each of them, the base is a times the base of the original, and the height is (a+1) times the original height (height is proportional to the parallel side), so the area of each is ar(ABC) * a * (a+1). The area of A'B'C' is the sum of the 3 new triangles plus the original, 3 * ar(ABC) * a * (a+1) + ar(ABC). So ar(A'B'C') = ar(ABC) * (3a^2 + 3a + 1).

[madhukushwah]

This is also an amazing magnificent video like your all videos.
I hope you will also going through like that.😘😘😘

[RichardChen]

Very cool! I got a very hard geometric prove problem using the same diagram as this video:
If triangle A'B'C' is an equilaterial triangle and satisfied A'B=B'C=C'A, prove triangle ABC is also an equilaterial triangle.
This is a very very hard challenge

[tushroy]

He has done it very hard... I had a easy explanation of 4 steps, just using the 9th Grade Theorems. Ratio of area of two triangles with same height is equal to the ratio of their bases. Using this theorem you can easily find the answer.

[rtfacts]

It could have been solved very easily by just using triangle with same height.

[ilk]

Maaan your channel is underrated that makes me sad :(