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]$

$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$

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 (C).

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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1986 AHSME Problems/Problem 28

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