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

Revision as of 20:19, 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)

I WILL BE COMPLETING THE REST RIGHT AFTER TEACHING A CLASS. PLEASE DO NOT EDIT IT. THANKS A LOT! :)

Solution 1 (Slopes)

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 $(1),$ 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 $y.$

Clearing fractions

Solution 2 (Bash)

I WILL BE COMPLETING THE REST RIGHT AFTER TEACHING A CLASS. PLEASE DO NOT EDIT IT. THANKS A LOT! :)

~MRENTHUSIASM

@@ Line 33: / Line 33: @@
 ~yuanyuanC
-==Solution 2 (Coordinate Geometry Bash)==
+==Solution 2 (Coordinate Geometry)==
+I WILL BE COMPLETING THE REST RIGHT AFTER TEACHING A CLASS. PLEASE DO NOT EDIT IT. THANKS A LOT! :)
+===Solution 1 (Slopes)===
 Suppose <math>B=(0,0).</math> It follows that
 <cmath>\begin{align*}
@@ Line 52: / Line 54: @@
 Clearing fractions
+===Solution 2 (Bash)===
-I will be filling the rest later. Wait for me.
+I WILL BE COMPLETING THE REST RIGHT AFTER TEACHING A CLASS. PLEASE DO NOT EDIT IT. THANKS A LOT! :)
 ~MRENTHUSIASM

2021 AIME I (Problems • Answer Key • Resources)
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