Library / Singapore Mathematical Olympiad (SMO) /7 of 52

Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 4.9 AIME Prove it Singapore

Two rhombi ABCD and AXYZ are external to each other with a common vertex A. The vertices of the rhombi are labelled clockwise. If $\angle DAX = \angle BAZ$ , prove that the centres of the rhombi and the midpoint of the segment BZ form an isosceles triangle.

Solution

Let the centres be $P$ , $Q$ and the midpoint of the segment $BZ$ be $R$ as shown in the figure. Since $\angle DAX = \angle BAZ$ , we have $DZ = BX$ . Since $P$ and $R$ are the midpoints of $XZ$ and $BZ$ , respectively, $PR = BX/2$ . Similarly $QR = DZ/2$ . Therefore $QR = PR$ and it follows that $\triangle PQR$ is isosceles.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.