Library / Nordic Mathematical Contest, NMC 8

Nordic Mathematical Contest, NMC 8

Geometry Difficulty 4.4 AIME Prove it Nordic Mathematical Olympiad

Problem:
A piece of paper is the square $ABCD$ . We fold it by placing the vertex $D$ on the point $D'$ of the side $BC$ . We assume that $AD$ moves on the segment $A'D'$ , and that $A'D'$ intersects $AB$ at $E$ . Prove that the perimeter of the triangle $EBD'$ is one half of the perimeter of the square.

Solution

Solution:
(See Figure 6.) The fold gives rise to an isosceles trapezium $ADHG$ . Because of symmetry, the distance of the vertex $D$ from the side $GH$ equals the distance of the vertex $H$ from side $AD$ ; the latter distance is the side length $a$ of the square. The line $GH$ thus is tangent to the circle with center $D$ and radius $a$ . The lines $AB$ and $BC$ are tangent to the same circle. If the point common to $GH$ and the circle is $F$ , then $AE = EF$ and $FH = HC$ . This implies $AB + BC = AE + EB + BH + HC = EF + EB + BH + HF = EH + EB + BH$ , which is equivalent to what we were asked to prove.

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.