Difference between revisions of "2004 AMC 10A Problems/Problem 20"
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==Solution 5== | ==Solution 5== | ||
− | First, since <math>\bigtriangleup BEF</math> is equilateral and <math>ABCD</math> is a square, by the Hypothenuse Leg Theorem, <math>\bigtriangleup ABE</math> is congruent to <math>\bigtriangleup CBF</math>. Then, assume length <math>AB = BC = x</math> and length <math>DE = DF = y</math>, then <math>AE = FC = x - y</math>. <math>\bigtriangleup BEF</math> is equilateral, so <math>EF = EB</math> and <math>EB^2 = EF^2</math>, it is given that <math>ABCD</math> is a square and <math>\bigtriangleup DEF</math> and <math>\bigtriangleup ABE</math> are right triangles. Then we use the Pythagorean theorem to prove that <math>AB^2 + AE^2 = EB^2</math> and since we know that <math>EB^2 = EF^2</math> and <math>EF^2 = DE^2 + DF^2</math>, which means <math>AB^2 + AE^2 = DE^2 + DF^2</math>. Now we plug in the variables and the equation becomes <math>x^2 + (x | + | First, since <math>\bigtriangleup BEF</math> is equilateral and <math>ABCD</math> is a square, by the Hypothenuse Leg Theorem, <math>\bigtriangleup ABE</math> is congruent to <math>\bigtriangleup CBF</math>. Then, assume length <math>AB = BC = x</math> and length <math>DE = DF = y</math>, then <math>AE = FC = x - y</math>. <math>\bigtriangleup BEF</math> is equilateral, so <math>EF = EB</math> and <math>EB^2 = EF^2</math>, it is given that <math>ABCD</math> is a square and <math>\bigtriangleup DEF</math> and <math>\bigtriangleup ABE</math> are right triangles. Then we use the Pythagorean theorem to prove that <math>AB^2 + AE^2 = EB^2</math> and since we know that <math>EB^2 = EF^2</math> and <math>EF^2 = DE^2 + DF^2</math>, which means <math>AB^2 + AE^2 = DE^2 + DF^2</math>. Now we plug in the variables and the equation becomes <math>x^2 + (x-y)^2 = 2y^2</math>, expand and simplify and you get <math>2x^2 - 2xy = y^2</math>. We want the ratio of area of <math>\bigtriangleup DEF</math> to <math>\bigtriangleup ABE</math>. Expressed in our variables, the ratio of the area is <math>\frac{y^2}{x^2 - xy}</math> and we know <math>2x^2 - 2xy = y^2</math>, so the ratio must be <math>2</math>. So, the answer is <math>\boxed{\text{(D) }2}</math> |
==Video Solution== | ==Video Solution== |
Revision as of 02:47, 18 April 2024
Contents
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
z
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 4 (System of Equations)
Assume . Then,
is
and
is
. We see that using
,
is congruent to EAB. Using Pythagoras of triangles
and
we get
. Expanding, we get
. Simplifying gives
solving using completing the square (or other methods) gives 2 answers:
and
. Because
,
. Using the areas, the answer is
Solution 5
First, since is equilateral and
is a square, by the Hypothenuse Leg Theorem,
is congruent to
. Then, assume length
and length
, then
.
is equilateral, so
and
, it is given that
is a square and
and
are right triangles. Then we use the Pythagorean theorem to prove that
and since we know that
and
, which means
. Now we plug in the variables and the equation becomes
, expand and simplify and you get
. We want the ratio of area of
to
. Expressed in our variables, the ratio of the area is
and we know
, so the ratio must be
. So, the answer is
Video Solution
Education, the Study of Everything
Video Solution by TheBeautyofMath
~IceMatrix
See also
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 | ||
All AMC 10 Problems and Solutions |
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