1986 AHSME Problems/Problem 28
Problem
is a regular pentagon. 
and 
are the perpendiculars dropped from 
onto 
extended and 
extended,
respectively. Let 
be the center of the pentagon. If 
, then 
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]](http://latex.artofproblemsolving.com/d/b/4/db4a282f655baace8a8bd02799266f80bb34198c.png)
Solution
C
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. 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 =... 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. 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 =... 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).
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
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