HMMT, 2014

Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

Given a regular pentagon of area $1$ , a pivot line is a line not passing through any of the pentagon's vertices such that there are $3$ vertices of the pentagon on one side of the line and $2$ on the other. A pivot point is a point inside the pentagon with only finitely many non-pivot lines passing through it. Find the area of the region of pivot points.

Solution

Solution:

Let the pentagon be labeled $A B C D E$ . First, no pivot point can be on the same side of $A C$ as vertex $B$ . Any such point $P$ has the infinite set of non-pivot lines within the hourglass shape formed by the acute angles between lines $P A$ and $P C$ . Similar logic can be applied to points on the same side of $B D$ as $C$ , and so on. The set of pivot points is thus a small pentagon with sides on $A C$ , $B D$ , $C E$ , $D A$ , $E B$ . The side ratio of this small pentagon to the large pentagon is

$\left(2 \cos \left(72^{\circ}\right)\right)^{2}=\frac{3-\sqrt{5}}{2}$

so the area of the small pentagon is
$\left(\frac{3-\sqrt{5}}{2}\right)^{2}=\frac{1}{2}(7-3 \sqrt{5})$

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