Competition problem — Geometry — Maths Olympiad Prep

Library / /2 of 15

Geometry Difficulty 4.7 AIME Prove it Estonia

Consider a parallelogram $ABCD$ .

a) Prove that if the incenter of the triangle $ABC$ is located on the diagonal $BD$ , then the parallelogram $ABCD$ is a rhombus.

b) Is the parallelogram $ABCD$ a rhombus whenever the circumcenter of the triangle $ABC$ is located on the diagonal $BD$ ?

Solution

a) As the incenter of the triangle $ABC$ is located on diagonal $BD$ (Fig. 1), we can conclude that $BD$ is the bisector of $\angle ABC$ . Therefore $\angle ABD = \angle CBD$ . However, since $ABCD$ is a parallelogram, $\angle ABD = \angle CDB$ . Hence the triangle $BCD$ is isosceles, i.e. $|BC| = |CD|$ . Thus, $ABCD$ is a rhombus.

Fig. 1

b) Let $ABCD$ be a rectangle with different side lengths. The circumcenter of triangle $ABC$ is located on the intersection of the diagonals of the rectangle. We see that all the required conditions are satisfied, however $ABCD$ is not a rhombus.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.