Difference between revisions of "2004 AMC 10A Problems/Problem 20"
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+ | ==Video Solution== | ||
+ | https://youtu.be/hwHIHRukYMk | ||
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+ | Education, the Study of Everything | ||
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==Problem== | ==Problem== | ||
Points <math>E</math> and <math>F</math> are located on square <math>ABCD</math> so that <math>\triangle BEF</math> is equilateral. What is the ratio of the area of <math>\triangle DEF</math> to that of <math>\triangle ABE</math>? | Points <math>E</math> and <math>F</math> are located on square <math>ABCD</math> so that <math>\triangle BEF</math> is equilateral. What is the ratio of the area of <math>\triangle DEF</math> to that of <math>\triangle ABE</math>? |
Revision as of 18:11, 30 December 2020
Contents
Video Solution
Education, the Study of Everything
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
Solution 4(system of equations)
Assume AB=1 then FC is x ED is then we see that using HL FCB is congruent is EAB. Using Pythagoras of triangles FCB and FDE we get
. Expanding we get
. Simplifying gives
solving using completing the square(or other methods) gives 2 answers
and
because
then
then using the areas we get the answer to be D
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 2. Choice D
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.