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Source: titu andreescu

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#1 h Oct 11, 2005, 9:57 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ a triangle and $P$ a point such that the triangles $PAB, PBC, PCA$ have the same area and the same perimeter. Prove that if:
a) $P$ is in the interior of the triangle $ABC$ then $ABC$ is equilateral.
b) $P$ is in the exterior of the triangle $ABC$ then $ABC$ is right angled triangle.

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#2 h Oct 29, 2005, 10:41 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

The locus of points such that the areas $S(\triangle PAB) = S(\triangle PCA)$ are equal is the median line AD of the triangle $\triangle ABC$ , where D is the midpoint of the side a = BC. Denote the remaining sides b = CA, c = AB and WLOG, assume $b \ge c$ . From equality of the perimeters $p(\triangle PAB) = p(\triangle PCA)$ , we get

$AB + BP + PA = p(\triangle PAB) = p(\triangle PCA) = AC + CP + PA$

$BP - CP = AC - AB = b - c \ge 0$

This means that the points A, P lie on the opposite branches of a hyperbola with the focal points B, C. If b > c, this hyperbola intersects the median line AD at 2 points A, P. The 2nd intersection P is symmetrical to A with respect to the midpoint D, hence, it lies outside of the triangle $\triangle ABC$ . Thus we see that for the specified point P to be inside of the triangle $\triangle ABC$ , this hyperbola has to degenerate into the perpendicular bisector of the segment BC, which means that b - c = 0 and the triangle $\triangle ABC$ is isosceles with $AB = c = b = CA$ . Similarly, it can be shown that it must be BC = a = c = AB, i.e., that the triangle $\triangle ABC$ must be equilateral. The point P then obviously is the intersection of the medians, i.e., the centroid of the equilateral triangle $\triangle ABC$ .

Let b > c and let P be the other intersection of our hyperbola with the median line AD. The quadrilateral ABPC is a parallelogram, because its diagonals BC, AP cut each other in half. The area of the triangle $\triangle PBC$ is equal to the same areas of the triangles $\triangle PAB, \triangle PCA$ , because both the paralleogram diagonals BC, AP cut its area in half, hence, $S(\triangle PBC) = S(\triangle PAB)$ . But their perimeters can be equal only if the diagonals of the parallelogram ABPC are equal, because

$BP + PC + CB = p(\triangle PBC) = p(\triangle PAB) = AB + BP + PA$

implies PA = CB. But this means that the parallelogram ABPC has to be a rectangle and the angle $\angle CAB = 90^\circ$ right, which is what we are supposed to show.

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