Solved by Dave Newquist

Question:

Suppose that the cotangents of the angles of a triangle are in arithmetic sequence. Prove that the squares of the sides of the triangle are also in arithmetic sequence.

Answer:

We procede to show that if the cotangents of a triangle are in arithmetic sequence then the squares of the sides are in arithmetic sequence.

First, label the angles of the triangles A, B, C such that cot(A) < cot(B) Recall from high school geometry that cot(X)= 1/tan(X) = cos(X)/sin(X). So

cot(A) = cos(A)/sin(A) = (b^2 +c^2 - a^2)/(2bc sin(A))
cot(B) = cot(A) + d = (a^2 +c^2 - b^2)/(2ac sin(B))
cot(C) = cot(A) + 2d = (a^2 +b^2 - c^2)/(2ab sin(C))

solving for b^2 for the equations we have for cot(B), we have

b^2 + c^2 -a^2 + 2bcd sin(A) = (a^2 + c^2 -b^2 (2bc sin(A) ) )/(2ac sin(B)) => b^2 + c^2 -a^2 + 2bcd sin(A) = (a^2 + c^2 -b^2) * (b sin(A)/a sin(B) )

=> by the law of sines

b^2 + c^2 -a^2 + 2bcd sin(A) = a^2 + c^2 - b^2 => b^2 = a^2 -d*( bc sin(A) )
using a similar strategy for the equations we have for cot(C), we have c^2 = a^2 - 2d*( bc sin(A) )

letting d' = -d*(bc sin(A)) we have b^2 = a^2 + d' and c^2 = a^2 + d', which shows a^2, b^2 and c^2 are in arithmetic sequence.

QED

-- Dave Newquist