Difference between revisions of "1986 AHSME Problems/Problem 28"
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\textbf{(E)}\ 5 </math> | \textbf{(E)}\ 5 </math> | ||
− | + | ==Solution== | |
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. | ||
− | + | <asy> | |
unitsize(2 cm); | unitsize(2 cm); | ||
Line 56: | Line 56: | ||
draw((O--E),dashed); | draw((O--E),dashed); | ||
− | label(" | + | label("$A$", A, N); |
− | label(" | + | label("$B$", B, dir(0)); |
− | label(" | + | label("$C$", C, SE); |
− | label(" | + | label("$D$", D, SW); |
− | label(" | + | label("$E$", E, W); |
− | dot(" | + | dot("$O$", O, NE); |
− | label(" | + | label("$P$", P, S); |
− | label(" | + | label("$Q$", Q, dir(0)); |
− | label(" | + | label("$R$", R, W); |
− | label(" | + | 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>. | 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>. | Next, we divide regular pentagon <math>ABCDE</math> into triangles <math>ABC</math>, <math>ACD</math>, and <math>ADE</math>. | ||
− | + | ||
+ | <asy> | ||
unitsize(2 cm); | unitsize(2 cm); | ||
Line 93: | Line 95: | ||
draw(A--D,dashed); | draw(A--D,dashed); | ||
− | label(" | + | label("$A$", A, N); |
− | label(" | + | label("$B$", B, dir(0)); |
− | label(" | + | label("$C$", C, SE); |
− | label(" | + | label("$D$", D, SW); |
− | label(" | + | label("$E$", E, W); |
− | dot(" | + | dot("$O$", O, dir(0)); |
− | label(" | + | label("$P$", P, S); |
− | label(" | + | label("$Q$", Q, dir(0)); |
− | label(" | + | label("$R$", R, W); |
− | label(" | + | 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 | 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> | <cmath>\frac{s}{2} (AO + AQ + AR + 1).</cmath> |
Revision as of 09:51, 13 May 2017
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
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 (C).
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.