A triangle has consecutive side and height lengths for some integer n, as shown above.
What is its perimeter?
Name the points. Let the point on the bottom left corner be A, that on the bottom right corner be B, and the one on top be C.
Let H be the point on AB such that CH is perpendicular to AB.
We calculate angle A in terms of n using Law of Cosines.
\(\cos A = \dfrac{(n + 1)^2 + (n + 2)^2 - (n + 3)^2}{2(n+1)(n+2)}\\ \cos A = \dfrac{n-2}{2(n+1)}\)
By this, AH = \(\dfrac{n - 2}2\).
Using Pythagorean theorem on triangle AHC,
\(\left(\dfrac n2 - 1\right)^2 + n^2 = (n + 1)^2\\ \left(\dfrac n2 - 1\right)^2 = 2n+1\\ \dfrac{n^2}4 - 3n = 0\\ n(n - 12) = 0\\ n = 12\)
The perimeter is (12 + 1) + (12 + 2) + (12 + 3) = 42.