2021 AIME I Problems/Problem 2

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 Bash)

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 and rearranging respectively give $\begin{align*} x^2+y^2-6y&=72, \ &(1) \\ x^2+y^2-22x&=-72. \ &(2) \end{align*}$ Subtracting $(2)$ from $(1),$ we get $22x-6y=144.$ Simplifying and rearranging produce $x=\frac{3y+72}{11}. \ \ \ \ \ \ \ \ \ (*)$ Substituting $(*)$ into $(1)$ gives $\left(\frac{3y+72}{11}\right)^2+y^2-6y=72,$ which is a quadratic of $y.$ We clear fractions by multiplying both sides by $11^2=121,$ then solve by factoring: $\begin{align*} \left(3y+72\right)^2+121y^2-726y&=8712 \\ \left(9y^2+432y+5184\right)+121y^2-726y&=8712 \\ 130y^2-294y-3528&=0 \\ 2(5x+21)(13x-84)&=0 \\ y&=-\frac{21}{5}, \ \frac{84}{13}. \end{align*}$ Since $E$ is in Quadrant IV, we have $E=\left(\frac{3\left(-\frac{21}{5}\right)+72}{11},-\frac{21}{5}\right)=\left(\frac{27}{5},-\frac{21}{5}\right).$ It follows that the equation of $\overleftrightarrow{AE}$ is $y=-\frac{4}{3}x+3.$

Let $G$ be the intersection of $\overline{AD}$ and $\overline{FC},$ and $H$ be the intersection of $\overline{AE}$ and $\overline{BC}.$ Since $H$ is the $x$ -intercept of $\overleftrightarrow{AE},$ we obtain $H=\left(\frac94,0\right).$

By symmetry, quadrilateral $AGCH$ is a parallelogram. Its area is $HC\cdot AB=\left(11-\frac94\right)\cdot3=\frac{105}{4},$ and the requested sum is $105+4=\boxed{109}.$

~MRENTHUSIASM

Solution 3 (Pythagorean Theorem)

Let the intersection of $AE$ and $BC$ be $G$ , and let $BG=x$ , so $CG=11-x$ .

By the Pythagorean theorem, ${AG}^2={AB}^2+{BG}^2$ , so $AG=\sqrt{x^2+9}$ , and thus $EG=9-\sqrt{x^2+9}$ .

By the Pythagorean theorem again, ${CG}^2={EG}^2+{CE}^2$ : $11-x=\sqrt{7^2+(9-\sqrt{x^2+9})^2}.$

Solving, we get $x=\frac{9}{4}$ , so the area of the parallelogram is $3\cdot(11-\frac{9}{4})=\frac{105}{4}$ , and $105+4=\framebox{109}$ .

~JulianaL25

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=H17E9n2nIyY&t=289s

2021 AIME I (Problems • Answer Key • Resources)
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