MathDB

28

Part of 1986 AMC 12/AHSME

Problems(1)

Pentagon Geometry

Source: AHSME 1986 problem 28

10/2/2011
ABCDEABCDE is a regular pentagon. APAP, AQAQ and ARAR are the perpendiculars dropped from AA onto CDCD, CBCB extended and DEDE extended, respectively. Let OO be the center of the pentagon. If OP=1OP = 1, then AO+AQ+ARAO + 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("OO",O,dir(B)); label("11",(O+P)/2,W); label("AA",A,dir(A)); label("BB",B,dir(B)); label("CC",C,dir(C)); label("DD",D,dir(D)); label("EE",E,dir(E)); label("PP",P,dir(P)); label("QQ",Q,dir(Q-A)); label("RR",R,dir(R-A)); [/asy]
<spanclass=′latex−bold′>(A)</span> 3<spanclass=′latex−bold′>(B)</span> 1+5<spanclass=′latex−bold′>(C)</span> 4<spanclass=′latex−bold′>(D)</span> 2+5<spanclass=′latex−bold′>(E)</span> 5 <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
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