MathDB

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] size(200); 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,7)^^rightanglemark(A,R,T,7)); 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-A)); label("

R

",R,dir(R-A)); [/asy]

<span class='latex-bold'>(A)</span>\ 3\qquad<span class='latex-bold'>(B)</span>\ 1 + \sqrt{5}\qquad<span class='latex-bold'>(C)</span>\ 4\qquad<span class='latex-bold'>(D)</span>\ 2 + \sqrt{5}\qquad<span class='latex-bold'>(E)</span>\ 5

Problem Statement