A point P lies inside the triangle ABC and the triangles PAB, PBC, PCA all have the same area and the same perimeter. Show that the triangle is equilateral. If P lies outside the triangle, show that the triangle is right-angled.
Solution
Suppose P is inside the triangle. Since area PAB = area PBC, BP must be the median (for if M is the midpoint of AC, then area PAB = 1/2 PB·AM sin AMB, area PBC = 1/2 PB·CM sin AMB). Similarly AP must be the median, so P must be the centroid. Now suppose ∠ABC < ∠ACB. Then AB > AC and PB > PC, so perimeter PAB > perimeter PAC. Hence ∠ABC = ∠ACB. Similarly, the other angles are equal, so the triangle must be equilateral. It is obvious that the centroid then satisfies the conditions.
Suppose P is outside the triangle. Relabeling if necessary, we may assume that A and B lie on opposite sides of PC. Hence A and C lie on the same side of PB (this is where we need P outside the triangle). Since area PBC = area PBA, A and C must be the same distance from PB, so AC is parallel to PB. Similarly PA is parallel to BC. So PACB is a parallelogram. But now the perimeter requirement implies that PC = AB and hence the parallelogram is a rectangle and angle ACB = 90o.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002