Difference between revisions of "1986 AHSME Problems/Problem 28"
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\textbf{(E)}\ 5 </math> | \textbf{(E)}\ 5 </math> | ||
− | ==Solution== | + | ==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. | ||
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<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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.