Library / Mongolian Mathematical Olympiad /5 of 97

Mongolian Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Mongolia

Let $A$ , $C$ be interior points of square $XOBD$ , such that $\angle AXC = \angle ABC = 45^\circ$ . Prove that,
$S_{\triangle AXO} + S_{\triangle ABC} + S_{\triangle CXD} = S_{\triangle ACX} + S_{\triangle AOB} + S_{\triangle CBD}.$

Solution

Let $M$ and $N$ be exterior points of $XOBD$ , such that $\triangle XOA = \triangle XDM$ and $\triangle AOB = \triangle NDB$ . Then we have $S_{\triangle ACX} = S_{\triangle MCX}$ , $S_{\triangle ABC} = S_{\triangle BCN}$ . Since $MD = DN$ and $M$ , $D$ , $N$ are collinear, we have $S_{\triangle MDC} = S_{\triangle NDC}$ . This implies the equality.

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.