I submitted this problem to Brilliant but it got rejected so I decided to share it here. Anyone willing to solve the problem is welcome. Enjoy!

I am sorry but there was a mistake in the problem. Thanks to Gabriel W. for pointing it out. I had a different approach but after getting the answers I did not verify them by triangle inequality.

A scalene triangle has an in-radius of 1 cm. Given that the altitudes are positive integers when measured in centimeters, what is the only possible value of the sum of the altitudes?

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TopNewestThe area of the triangle is (a+b+c)/2.

altitudes are integer:

(a+b+c) = ka = lb = mc, where k,l,m are integers

assume a >= b >= c

then k is either 3 (if they are all equal) or 2.

The equilateral case is easy to check: an equilateral triangle with inradius 1 has three altitudes of length 3.

When k = 2, a+b+c = 2a; b+c = a, which is impossible by triangle inequality. – Gabriel Wong · 4 years, 3 months ago

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– Sreejato Bhattacharya · 4 years, 3 months ago

Note that the lengths of the altitudes are integers, but the lengths of the sides of the triangle need not be integers. So, assuming that your reasoning is correct, I think \( k \) can take any real value between \( 2 \) and \( 3 \).Log in to reply

The area is indeed (a+b+c)/2

(or just divide the triangle into three chunks ABI BCI CAI; its clear, given inradius is 1) – Gabriel Wong · 4 years, 3 months ago

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– Sreejato Bhattacharya · 4 years, 3 months ago

Yes now I get it. Sorry for my mistake in the previous comment, I have edited it out. But note that the lengths of the sides of the triangle need not be integers themselves, so \( k \) can be any real number between \( 2 \) and \( 3 \).Log in to reply

(a+b+c) = a*altitude from A (which is integer)

a+b+c = ka for integer a – Gabriel Wong · 4 years, 3 months ago

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