Difference between revisions of "2004 AMC 10A Problems/Problem 20"
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− | + | <math>\bigtriangleup BEF</math> is equilateral, so <math>\angle EBF = 60^{\circ}</math>, and <math>\angle EBA = \angle FBC</math> so they must each be <math>15^{\circ}</math>. Then let <math>BE=EF=FB=1</math>, which gives <math>EA=\sin{15^{\circ}}</math> and <math>AB=\cos{15^{\circ}}</math>. | |
+ | The area of <math>\bigtriangleup ABE</math> is then <math>\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}</math>. | ||
+ | <math>\bigtriangleup DEF</math> is an isosceles right triangle with hypotenuse 1, so <math>DE=DF=\frac{1}{\sqrt{2}}</math> and therefore its area is <math>\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}</math>. | ||
+ | The ratio of areas is then <math>\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}</math> |
Revision as of 01:42, 11 April 2020
Contents
[hide]Problem
Points and
are located on square
so that
is equilateral. What is the ratio of the area of
to that of
?

Solution 1
Since triangle is equilateral,
, and
and
are
congruent. Thus, triangle
is an isosceles right triangle. So we let
. Thus
. If we go angle chasing, we find out that
, thus
.
. Thus
, or
. Thus
, and
, and
. Thus the ratio of the areas is
Solution 2 (Non-trig)
WLOG, let the side length of be 1. Let
. It suffices that
. Then triangles
and
are congruent by HL, so
and
. We find that
, and so, by the Pythagorean Theorem, we have
This yields
, so
. Thus, the desired ratio of areas is
Solution 3
is equilateral, so
, and
so they must each be
. Then let
, which gives
and
.
The area of
is then
.
is an isosceles right triangle with hypotenuse 1, so
and therefore its area is
.
The ratio of areas is then