# BMO1 A 2020 2

New BMO problems!

So, if the triangle has side lengths a, a, and b, it’s area is \frac{b\sqrt{a^2-\frac{b^2}{4}}}{2}, and thus we have the equality:
\frac{b^2(a^2-\frac{b^2}{4})}{4}=2a+b

Multiplying both sides by 16 we get:
4b^2a^2-b^4 = 32a+16b

Factoring, we have:
b^2(4a^2-b^2)=b^2(2a+b)(2a-b)=16(2a+b)

Dividing both sides by (2a+b) we get:
b^2(2a-b)=16

From here, since we know that b has to be an integer, we only have to test b=1, 2, 4 and we can get the following pairs of (a, b):
(\frac{17}{2}, 1), (3, 2), (\frac{5}{2}, 4)

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