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Problem 869

AMC 12 late, AIME early

Geometry Difficulty 4.6 Prove it Harvard-MIT Math Tournament · United States

Parallelogram $A E C F$ is inscribed in square $A B C D$ . It is reflected across diagonal $A C$ to form another parallelogram $A E^{\prime} C F^{\prime}$ . The region common to both parallelograms has area $m$ and perimeter $n$ . Compute the value of $\frac{m}{n^{2}}$ if $A F: A D=1: 4$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

By symmetry, the region is a rhombus, $AXCY$ , centered at the center of the square, $O$ . Consider isoceles right triangle $ACD$ . By the technique of mass points, we find that $DO: YO=7: 1$ . Therefore, the rhombus is composed of four triangles, whose sides are in the ratio $1: 7: 5 \sqrt{2}$ . The perimeter of the rhombus is $20 \sqrt{2} N$ , and the area is $14 N^{2}$ . The required ratio is thus $\frac{7}{400}$ .

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