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Difference between revisions of "2021 AIME I Problems/Problem 2"

(→‎Solution 2 (Coordinate Geometry))
(→‎Solution 2 (Coordinate Geometry))
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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>
  
−
We now have a system of two equations with two variables. Expanding, rearranging, and simplifying gives
+
We now have a system of two equations with two variables. Expanding, rearranging, and simplifying respectively give
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
x^2+y^2-6y&=72, \\
+
x^2+y^2-6y&=72, \ &(1) \\
−
x^2+y^2-22x&=-72.
+
x^2+y^2-22x&=-72. \ &(2)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 +
Subtracting <math>(2)</math> from <math>(1)</math> then simplifying, we get <math>11x-3y=72.</math> Rearranging produces <cmath>x=\frac{3y+72}{11}. \ \ \ \ \ \ \ \ \ (*)</cmath>
  
 
I will be filling the rest later. Wait for me.
 
I will be filling the rest later. Wait for me.

Revision as of 18:55, 11 March 2021

Problem

In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. [asy] pair A, B, C, D, E, F; A = (0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]

Solution 1 (Similar Triangles)

Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\angle FGA= \angle DGC$. Also, given that $ABCD$ and $AFCE$ are rectangles, we know that $\angle AFG= \angle CDG=90 ^{\circ}$. Therefore, by AA similarity, we know that triangles $AFG$ and $CDG$ are similar.

Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\frac{7}{3}(11-x)$ and $CG=\frac{3}{7}x$. We have $\frac{7}{3}(11-x)+\frac{3}{7}x=FC=9$.

Solving for $x$, we have $x=\frac{35}{4}$. The area of the shaded region is just $3\cdot \frac{35}{4}=\frac{105}{4}$. Thus, the answer is $105+4=\framebox{109}$. ~yuanyuanC

Solution 2 (Coordinate Geometry)

Suppose $B=(0,0).$ It follows that \begin{align*} A&=(0,3), \\ C&=(11,0), \\ D&=(11,3). \end{align*}

Since $AECF$ is a rectangle, we have $AE=FC=9$ and $EC=AF=7.$ The equation of the circle with center $A$ and radius $\overline{AE}$ is $x^2+(y-3)^2=81,$ and the equation of the circle with center $C$ and radius $\overline{CE}$ is $(x-11)^2+y^2=49.$

We now have a system of two equations with two variables. Expanding, rearranging, and simplifying respectively give \begin{align*} x^2+y^2-6y&=72, \ &(1) \\ x^2+y^2-22x&=-72. \ &(2) \end{align*}

Subtracting $(2)$ from $(1)$ then simplifying, we get $11x-3y=72.$ Rearranging produces \[x=\frac{3y+72}{11}. \ \ \ \ \ \ \ \ \ (*)\]

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

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