Difference between revisions of "1986 AHSME Problems/Problem 28"
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\textbf{(E)}\ 5 </math> | \textbf{(E)}\ 5 </math> | ||
+ | ==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. | ||
+ | <asy> | ||
+ | unitsize(2 cm); | ||
− | ==Solution== | + | 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 <math>s</math> is the side length of the regular pentagon, then each of the triangles <math>AOB</math>, <math>BOC</math>, <math>COD</math>, <math>DOE</math>, and <math>EOA</math> has base <math>s</math> and height 1, so the area of regular pentagon <math>ABCDE</math> is <math>5s/2</math>. | ||
+ | |||
+ | Next, we divide regular pentagon <math>ABCDE</math> into triangles <math>ABC</math>, <math>ACD</math>, and <math>ADE</math>. | ||
+ | |||
+ | <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 <math>ACD</math> has base <math>s</math> and height <math>AP = AO + 1</math>. Triangle <math>ABC</math> has base <math>s</math> and height <math>AQ</math>. Triangle <math>ADE</math> has base <math>s</math> and height <math>AR</math>. Therefore, the area of regular pentagon <math>ABCDE</math> is also | ||
+ | <cmath>\frac{s}{2} (AO + AQ + AR + 1).</cmath> | ||
+ | Hence, | ||
+ | <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>. | ||
+ | |||
+ | ==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
Contents
[hide]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
Solution 1
To solve the problem, we compute the area of regular pentagon in two different ways. First, we can divide regular pentagon into five congruent triangles.
If is the side length of the regular pentagon, then each of the triangles , , , , and has base and height 1, so the area of regular pentagon is .
Next, we divide regular pentagon into triangles , , and .
Triangle has base and height . Triangle has base and height . Triangle has base and height . Therefore, the area of regular pentagon is also Hence, which means , or . The answer is .
Solution 2
Now, we know that angle has measure . Since Therefore, . Therefore, . Recalling that gives a final answer of .
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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