2004 AMC 10A Problems/Problem 20
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 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 . 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.