Difference between revisions of "2021 AIME I Problems/Problem 2"
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== Solution 4 (Similar triangles and area) == | == Solution 4 (Similar triangles and area) == | ||
− | Again, let the intersection of <math>AE</math> and <math>BC</math> be <math>G</math>. By AA similarity, <math>AFG \sim CDG</math> with a <math>\frac{7}{3}</math> ratio. Define <math>x</math> as <math>\frac{[CDG]}{9}</math>. Because of similar triangles, <math>[AFG] = 49x</math>. Using <math>ABCD</math>, the area of the parallelogram is <math>33-18x</math>. Using <math>AECF</math>, the area of the parallelogram is <math>63-98x</math>. These equations are equal, so we can solve for <math>x</math> and obtain <math>x = \frac{3}{8}</math>. Thus, <math>18x = \frac{27}{4}</math>, so the area of the parallelogram is | + | Again, let the intersection of <math>AE</math> and <math>BC</math> be <math>G</math>. By AA similarity, <math>AFG \sim CDG</math> with a <math>\frac{7}{3}</math> ratio. Define <math>x</math> as <math>\frac{[CDG]}{9}</math>. Because of similar triangles, <math>[AFG] = 49x</math>. Using <math>ABCD</math>, the area of the parallelogram is <math>33-18x</math>. Using <math>AECF</math>, the area of the parallelogram is <math>63-98x</math>. These equations are equal, so we can solve for <math>x</math> and obtain <math>x = \frac{3}{8}</math>. Thus, <math>18x = \frac{27}{4}</math>, so the area of the parallelogram is <math>33 - \frac{27}{4} = \frac{105}{4}</math>. |
~mathboy100 | ~mathboy100 |
Revision as of 11:41, 12 March 2021
Contents
Problem
In the diagram below, is a rectangle with side lengths
and
, and
is a rectangle with side lengths
and
as shown. The area of the shaded region common to the interiors of both rectangles is
, where
and
are relatively prime positive integers. Find
.
Solution 1 (Similar Triangles)
Let be the intersection of
and
.
From vertical angles, we know that
. Also, given that
and
are rectangles, we know that
.
Therefore, by AA similarity, we know that triangles
and
are similar.
Let . Then, we have
. By similar triangles, we know that
and
. We have
.
Solving for , we have
.
The area of the shaded region is just
.
Thus, the answer is
.
~yuanyuanC
Solution 2 (Coordinate Geometry Bash)
Suppose It follows that
Since is a rectangle, we have
and
The equation of the circle with center
and radius
is
and the equation of the circle with center
and radius
is
We now have a system of two equations with two variables. Expanding and rearranging respectively give
Subtracting
from
we get
Simplifying and rearranging produce
Substituting
into
gives
which is a quadratic of
We clear fractions by multiplying both sides by
then solve by factoring:
Since
is in Quadrant IV, we have
It follows that the equation of
is
Let be the intersection of
and
and
be the intersection of
and
Since
is the
-intercept of
we obtain
By symmetry, quadrilateral is a parallelogram. Its area is
and the requested sum is
~MRENTHUSIASM
Solution 3 (Pythagorean Theorem)
Let the intersection of and
be
, and let
, so
.
By the Pythagorean theorem, , so
, and thus
.
By the Pythagorean theorem again, :
Solving, we get , so the area of the parallelogram is
, and
.
~JulianaL25
Solution 4 (Similar triangles and area)
Again, let the intersection of and
be
. By AA similarity,
with a
ratio. Define
as
. Because of similar triangles,
. Using
, the area of the parallelogram is
. Using
, the area of the parallelogram is
. These equations are equal, so we can solve for
and obtain
. Thus,
, so the area of the parallelogram is
.
~mathboy100
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY&t=289s
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.