Track / Stage 4 / 145 of 340 #405 of 1964

Problem 405

AMC 12 late, AIME early

Geometry Difficulty 4.8 Find the answer

28. $A B C D E$ is a
regular pentagon, $A P$ , $A Q$ and $A R$ are perpendiculars from $A$ to the extensions of $C D$ , $C B$ , and $D E$ respectively. Let $O$ be the center of the pentagon.
If $O P=1$ , then $A O+A Q+A R$ equals

Pick one

Official solution

$\frac{28}{C}$
28. Let $s$ denote the side length of a pentagon, and the areas of $\mathbb{C} \triangle O A B, \triangle O B C, \triangle O C D, \triangle O D E$ and $\triangle O E A$ are equal. Each of their bases is $s$ ,
1. The area of the pentagon is $\frac{5 \varepsilon}{2} \cdot \frac{1}{2}-$ (Note 2), and it is also the sum of the areas of $\triangle A B C, \triangle A C D$ and $\triangle A D E$ , with their respective heights being $A Q, A P$ and $A R$ . Therefore: the total is
$\begin{array}{l} \frac{s}{2}(A P+A Q+A R) \text {, so } A P+A Q+A R \\ =5, \text { then } A O+A Q+A R=4 . \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.