2019 AIME II Problems/Problem 1

Problem

Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ , $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

[asy] size(10cm); pair A=(0,0), B=(9,0), C=(15,8), D=(-6,8), C_1=(15,0), P=(9/2,12/5); draw(B--C--D--A); draw(C--P--D); draw(B--C_1--C,dashed); draw((A+B)/2--(C+D)/2,dashed); label(" $A$ ",A,SW); label(" $B$ ",B,S); label(" $C'$ ",C_1,SE); label(" $C$ ",C,NE); label(" $D$ ",D,NW); label(" $P$ ",P+(0.5,0),E); draw(A--B--P--cycle,red+linewidth(1.1)); dot(A); dot(B); dot(C); dot(D); dot(P); dot(C_1); label(" $9$ ",(A+B)/2,dir(A--B)*dir(-90)); label(" $10$ ",(B+C)/2,dir(B-C)*dir(90)); label(" $6$ ",(B+C_1)/2,dir(B-C_1)*dir(90)); label(" $8$ ",(C+C_1)/2,dir(C_1-C)*dir(90)); label(" $21$ ",(C+D)/2,dir(C--D)*dir(-90)); label(" $\frac{12}{5}$ ",(2*P+A+B)/4+(-0.1,-0.3),E); [/asy]

Note that $17^2-15^2=8^2=10^2-6^2$ and $15-6=9$ so if $C'$ is the projection of $C$ onto $AB$ then $AC'=15,BC'=6,CC'=8$ . It follows that $CD=21$ . Let the diagonals $AC$ and $BD$ intersect at $P$ , then $\triangle{PAB}\sim\triangle{PCD}$ with similarity factor $\frac{9}{21}$ . Thus the height of $\triangle{PAB}$ is $\frac{9}{9+21}\cdot8=\frac{12}{5}$ so $\left[PAB\right]=\frac{1}{2}\cdot9\cdot\frac{12}{5}=\frac{54}{5}$ and hence the answer is $\boxed{059}$ as desired.

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2019 AIME II Problems/Problem 1

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