# 1986 AHSME Problems/Problem 28

## Problem $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended, respectively. Let $O$ be the center of the pentagon. If $OP = 1$, then $AO + AQ + AR$ equals $[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label("O",O,dir(B)); label("1",(O+P)/2,W); label("A",A,dir(A)); label("B",B,dir(B)); label("C",C,dir(C)); label("D",D,dir(D)); label("E",E,dir(E)); label("P",P,dir(P)); label("Q",Q,dir(Q)); label("R",R,dir(R)); [/asy]$ $\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$

## Solution 1

To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles. $[asy] unitsize(2 cm); pair A, B, C, D, E, O, P, Q, R; A = dir(90); B = dir(90 - 360/5); C = dir(90 - 2*360/5); D = dir(90 - 3*360/5); E = dir(90 - 4*360/5); O = (0,0); P = (C + D)/2; Q = (A + reflect(B,C)*(A))/2; R = (A + reflect(D,E)*(A))/2; draw((2*R - E)--D--C--(2*Q - B)); draw(A--P); draw(A--Q); draw(A--R); draw(B--A--E); draw((O--B),dashed); draw((O--C),dashed); draw((O--D),dashed); draw((O--E),dashed); label("A", A, N); label("B", B, dir(0)); label("C", C, SE); label("D", D, SW); label("E", E, W); dot("O", O, NE); label("P", P, S); label("Q", Q, dir(0)); label("R", R, W); label("1", (O + P)/2, dir(0)); [/asy]$

If $s$ is the side length of the regular pentagon, then each of the triangles $AOB$, $BOC$, $COD$, $DOE$, and $EOA$ has base $s$ and height 1, so the area of regular pentagon $ABCDE$ is $5s/2$.

Next, we divide regular pentagon $ABCDE$ into triangles $ABC$, $ACD$, and $ADE$. $[asy] unitsize(2 cm); pair A, B, C, D, E, O, P, Q, R; A = dir(90); B = dir(90 - 360/5); C = dir(90 - 2*360/5); D = dir(90 - 3*360/5); E = dir(90 - 4*360/5); O = (0,0); P = (C + D)/2; Q = (A + reflect(B,C)*(A))/2; R = (A + reflect(D,E)*(A))/2; draw((2*R - E)--D--C--(2*Q - B)); draw(A--P); draw(A--Q); draw(A--R); draw(B--A--E); draw(A--C,dashed); draw(A--D,dashed); label("A", A, N); label("B", B, dir(0)); label("C", C, SE); label("D", D, SW); label("E", E, W); dot("O", O, dir(0)); label("P", P, S); label("Q", Q, dir(0)); label("R", R, W); label("1", (O + P)/2, dir(0)); [/asy]$ Triangle $ACD$ has base $s$ and height $AP = AO + 1$. Triangle $ABC$ has base $s$ and height $AQ$. Triangle $ADE$ has base $s$ and height $AR$. Therefore, the area of regular pentagon $ABCDE$ is also $$\frac{s}{2} (AO + AQ + AR + 1).$$ Hence, $$\frac{s}{2} (AO + AQ + AR + 1) = \frac{5s}{2},$$ which means $AO + AQ + AR + 1 = 5$, or $AO + AQ + AR = \boxed{4}$. The answer is $\boxed{(C)}$.

## Solution 2

Now, we know that angle $D$ has measure $\frac{180 \cdot 3}{5} = 108$. Since $$\sin 54 = \frac{OP}{DO} = \frac{1}{DO}, DO = \frac{1}{\sin 54}$$ $$\tan 54 = \frac{OP}{DP} = \frac{1}{DP}, DP = \frac{1}{\tan 54}$$Therefore, $AB = 2DP = \frac{2}{\tan 54}$. $$\sin 72 = \frac{AQ}{AB} = AQ \tan 54 \cdot \frac{1}{2}, AQ = \frac{2 \sin 72}{\tan 54}$$Therefore, $AO + AQ + AR = AO + 2AQ = \frac{1}{\sin 54}+\frac{4 \sin 72}{\tan 54} = \frac{1}{\sin 54} + 8 \sin 36 \cos 54 = \frac{1}{\cos 36} + 8-8\cos^2(36)$. Recalling that $\cos 36 = \frac{1 + \sqrt{5}}{4}$ gives a final answer of $\boxed{4}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 