Difference between revisions of "2021 AIME I Problems/Problem 2"
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== Solution 5 == | == Solution 5 == | ||
− | + | Let <math>P = AD \cap FC</math>, and <math>K = AE \cap BC</math>. Also let <math>AP = x</math>. | |
− | + | <math>CK</math> also has to be <math>x</math> by parallelogram properties. Then <math>PD</math> and <math>BK</math> must be <math>11-x</math> because the sum of the segments has to be <math>11</math>. | |
− | + | We can easily solve for <math>PC</math> by the Pythagorean Theorem: | |
− | + | <cmath>\begin{align*} | |
+ | DC^2 + PD^2 &= PC^2\\ | ||
+ | 9 + (11-x)^2 &= PC^2 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | It follows shortly that <math>PC = \sqrt{x^2-22x+30}</math>. | ||
− | + | Also, <math>FC = 9</math>, and <math>FP + PC = 9</math>. We can then say that <math>PC = \sqrt{x^2-22x+30}</math>, so <math>FP = 9 - \sqrt{x^2-22x+30}</math>. | |
− | + | Now we can apply the Pythagorean Theorem to <math>\triangle AFP</math>. | |
− | + | <cmath>\begin{align*} | |
− | + | AF^2 + FP^2 = AP^2\\ | |
+ | 49 + \left(9 - \sqrt{x^2-22x+30}\right)^2 = x^2 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | This simplifies (not-as-shortly) to <math>x = \dfrac{35}{4}</math>. Now we have to solve for the area of <math>APCK</math>. We know that the height is <math>3</math> because the height of the parallelogram is the same as the height of the smaller rectangle. | ||
+ | From the area of a parallelogram (we know that the base is <math>\dfrac{35}{4}</math> and the height is <math>3</math>), it is clear that the area is <math>\dfrac{105}{4}</math>, giving an answer of <math>\boxed{109}</math>. ~ishanvannadil2008 for the solution sketch, Tuatara for rephrasing and <math>\LaTeX</math>. | ||
==Solution 6 (Trigbash)== | ==Solution 6 (Trigbash)== |
Revision as of 19:13, 13 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
Solution 5
Let , and
. Also let
.
also has to be
by parallelogram properties. Then
and
must be
because the sum of the segments has to be
.
We can easily solve for by the Pythagorean Theorem:
It follows shortly that
.
Also, , and
. We can then say that
, so
.
Now we can apply the Pythagorean Theorem to .
This simplifies (not-as-shortly) to . Now we have to solve for the area of
. We know that the height is
because the height of the parallelogram is the same as the height of the smaller rectangle.
From the area of a parallelogram (we know that the base is and the height is
), it is clear that the area is
, giving an answer of
. ~ishanvannadil2008 for the solution sketch, Tuatara for rephrasing and
.
Solution 6 (Trigbash)
Let the intersection of and
be
. It is useful to find
, because
and
. From there, subtracting the areas of the two triangles from the larger rectangle, we get Area =
.
let = α. Let
= β. Note, α+β=
.
α =
β =
tan()= tan
=
=
=
Area= =
. The answer is
.
~ twotothetenthis1024
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY&t=289s
Video Solution
https://youtu.be/M3DsERqhiDk?t=275
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.