Difference between revisions of "2021 AIME I Problems/Problem 2"
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~yuanyuanC | ~yuanyuanC | ||
− | ==Solution 2 (Coordinate Geometry)== | + | ==Solution 2 (Coordinate Geometry Bash)== |
Suppose <math>B=(0,0).</math> It follows that | Suppose <math>B=(0,0).</math> It follows that | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
Line 40: | Line 40: | ||
D&=(11,3). | D&=(11,3). | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | |||
Since <math>AECF</math> is a rectangle, we have <math>AE=FC=9</math> and <math>EC=AF=7.</math> The equation of the circle with center <math>A</math> and radius <math>\overline{AE}</math> is <math>x^2+(y-3)^2=81,</math> and the equation of the circle with center <math>C</math> and radius <math>\overline{CE}</math> is <math>(x-11)^2+y^2=49.</math> | Since <math>AECF</math> is a rectangle, we have <math>AE=FC=9</math> and <math>EC=AF=7.</math> The equation of the circle with center <math>A</math> and radius <math>\overline{AE}</math> is <math>x^2+(y-3)^2=81,</math> and the equation of the circle with center <math>C</math> and radius <math>\overline{CE}</math> is <math>(x-11)^2+y^2=49.</math> | ||
Line 48: | Line 47: | ||
x^2+y^2-22x&=-72. \ &(2) | x^2+y^2-22x&=-72. \ &(2) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | Subtracting <math>(2)</math> from <math>(1),</math> we get <math>22x-6y=144.</math> Simplifying and rearranging produce <cmath>x=\frac{3y+72}{11}. \ \ \ \ \ \ \ \ \ (*)</cmath> | ||
+ | Substituting <math>(*)</math> into <math>(1)</math> gives <cmath>\left(\frac{3y+72}{11}\right)^2+y^2-6y=72,</cmath> which is a quadratic of <math>y.</math> | ||
+ | |||
+ | Clearing fractions | ||
− | |||
I will be filling the rest later. Wait for me. | I will be filling the rest later. Wait for me. |
Revision as of 20:03, 11 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, rearranging, and simplifying respectively give
Subtracting
from
we get
Simplifying and rearranging produce
Substituting
into
gives
which is a quadratic of
Clearing fractions
I will be filling the rest later. Wait for me.
~MRENTHUSIASM
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.