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Difference between revisions of "1986 AHSME Problems/Problem 28"

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\textbf{(E)}\ 5  </math>   
 
\textbf{(E)}\ 5  </math>   
 
    
 
    
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==Solution==
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==Solution 1==
  
 
To solve the problem, we compute the area of regular pentagon <math>ABCDE</math> in two different ways. First, we can divide regular pentagon <math>ABCDE</math> into five congruent triangles.
 
To solve the problem, we compute the area of regular pentagon <math>ABCDE</math> in two different ways. First, we can divide regular pentagon <math>ABCDE</math> into five congruent triangles.
Line 111: Line 111:
 
<cmath>\frac{s}{2} (AO + AQ + AR + 1) = \frac{5s}{2},</cmath>
 
<cmath>\frac{s}{2} (AO + AQ + AR + 1) = \frac{5s}{2},</cmath>
 
which means <math>AO + AQ + AR + 1 = 5</math>, or <math>AO + AQ + AR = \boxed{4}</math>. The answer is <math>\boxed{(C)}</math>.
 
which means <math>AO + AQ + AR + 1 = 5</math>, or <math>AO + AQ + AR = \boxed{4}</math>. The answer is <math>\boxed{(C)}</math>.
 +
 +
==Solution 2==
 +
Now, we know that angle <math>D</math> has measure <math>\frac{180 \cdot 3}{5} = 108</math>. Since
 +
<cmath>\sin 54 = \frac{OP}{DO} = \frac{1}{DO}, DO = \frac{1}{\sin 54}</cmath><cmath>\tan 54 = \frac{OP}{DP} = \frac{1}{DP}, DP = \frac{1}{\tan 54}</cmath>Therefore, <math>AB = 2DP = \frac{2}{\tan 54}</math>.
 +
<cmath>\sin 72 = \frac{AQ}{AB} = AQ \tan 54 \cdot \frac{1}{2}, AQ = \frac{2 \sin 72}{\tan 54}</cmath>Therefore, <math>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)</math>. Recalling that <math>\cos 36 = \frac{1 + \sqrt{5}}{4}</math> gives a final answer of <math>\boxed{4}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 16:13, 17 April 2020

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 $\boxed{4}$.

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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