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Difference between revisions of "2019 AIME II Problems/Problem 1"

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==Solution==
 
==Solution==
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[asy]
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size(10cm);
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pair A=(0,0), B=(9,0), C=(15,8), D=(-6,8), C_1=(15,0), P=(9/2,12/5);
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draw(B--C--D--A); draw(C--P--D); draw(B--C_1--C,dashed); draw((A+B)/2--(C+D)/2,dashed);
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label("<math>A</math>",A,SW); label("<math>B</math>",B,S); label("<math>C'</math>",C_1,SE); label("<math>C</math>",C,NE); label("<math>D</math>",D,NW); label("<math>P</math>",P+(0.5,0),E);
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draw(A--B--P--cycle,red+linewidth(1.1));
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dot(A); dot(B); dot(C); dot(D); dot(P); dot(C_1);
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label("<math>9</math>",(A+B)/2,dir(A--B)*dir(-90)); label("<math>10</math>",(B+C)/2,dir(B-C)*dir(90)); label("<math>6</math>",(B+C_1)/2,dir(B-C_1)*dir(90)); label("<math>8</math>",(C+C_1)/2,dir(C_1-C)*dir(90)); label("<math>21</math>",(C+D)/2,dir(C--D)*dir(-90)); label("<math>\frac{12}{5}</math>",(2*P+A+B)/4+(-0.1,-0.3),E);
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[/asy]
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Note that <math>17^2-15^2=8^2=10^2-6^2</math> and <math>15-6=9</math> so if <math>C'</math> is the projection of <math>C</math> onto <math>AB</math> then <math>AC'=15,BC'=6,CC'=8</math>. It follows that <math>CD=21</math>. Let the diagonals <math>AC</math> and <math>BD</math> intersect at <math>P</math>, then <math>\triangle{PAB}\sim\triangle{PCD}</math> with similarity factor <math>\frac{9}{21}</math>. Thus the height of <math>\triangle{PAB}</math> is <math>\frac{9}{9+21}\cdot8=\frac{12}{5}</math> so <math>\left[PAB\right]=\frac{1}{2}\cdot9\cdot\frac{12}{5}=\frac{54}{5}</math> and hence the answer is <math>\boxed{059}</math> as desired.

Revision as of 17:35, 22 March 2019

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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