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Difference between revisions of "2019 AIME I Problems/Problem 3"

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In <math>\triangle PQR</math>, <math>PR=15</math>, <math>QR=20</math>, and <math>PQ=25</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and <math>D</math> lie on <math>\overline{QR}</math>, and points <math>E</math> and <math>F</math> lie on <math>\overline{PR}</math>, with <math>PA=QB=QC=RD=RE=PF=5</math>. Find the area of hexagon <math>ABCDEF</math>.
 
In <math>\triangle PQR</math>, <math>PR=15</math>, <math>QR=20</math>, and <math>PQ=25</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and <math>D</math> lie on <math>\overline{QR}</math>, and points <math>E</math> and <math>F</math> lie on <math>\overline{PR}</math>, with <math>PA=QB=QC=RD=RE=PF=5</math>. Find the area of hexagon <math>ABCDEF</math>.
  
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==Solution==
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==Solution 1==
 
We know the area of the hexagon <math>ABCDEF</math> to be <math>\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED </math>. Since <math>PR^2+RQ^2=PQ^2</math>, we know that <math>\triangle PRQ</math> is a right triangle. Thus the area of <math>\triangle PQR</math> is <math>150</math>. Another way to compute the area is <cmath>\frac12 \cdot PQ\cdot RQ \sin \angle PQR = \frac12 \cdot 500 \cdot \sin \angle PQR=150 \implies \sin \angle PQR = \frac35.</cmath> Then the area of <math>\triangle BCQ = \frac12 \cdot BQ \cdot CQ \cdot \sin \angle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}</math>. Preceding in a similar fashion for <math>\triangle PAF</math>, the area of <math>\triangle PAF</math> is <math>10</math>. Since <math>\angle ERD = 90^{\circ}</math>, the area of <math>\triangle RED=\frac{25}{2}</math>. Thus our desired answer is <math>150-\frac{15}{2}-10-\frac{25}{2}=\boxed{120}</math>
 
We know the area of the hexagon <math>ABCDEF</math> to be <math>\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED </math>. Since <math>PR^2+RQ^2=PQ^2</math>, we know that <math>\triangle PRQ</math> is a right triangle. Thus the area of <math>\triangle PQR</math> is <math>150</math>. Another way to compute the area is <cmath>\frac12 \cdot PQ\cdot RQ \sin \angle PQR = \frac12 \cdot 500 \cdot \sin \angle PQR=150 \implies \sin \angle PQR = \frac35.</cmath> Then the area of <math>\triangle BCQ = \frac12 \cdot BQ \cdot CQ \cdot \sin \angle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}</math>. Preceding in a similar fashion for <math>\triangle PAF</math>, the area of <math>\triangle PAF</math> is <math>10</math>. Since <math>\angle ERD = 90^{\circ}</math>, the area of <math>\triangle RED=\frac{25}{2}</math>. Thus our desired answer is <math>150-\frac{15}{2}-10-\frac{25}{2}=\boxed{120}</math>
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==Solution 2==
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Let <math>R</math> be the origin. Then <math>A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)</math>, and <math>F=(0,10)</math>. Using the shoelace theorem, the area is <math>\boxed{120}</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=2|num-a=4}}
 
{{AIME box|year=2019|n=I|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:08, 16 March 2019

The 2019 AIME I takes place on March 13, 2019.

Problem 3

In $\triangle PQR$, $PR=15$, $QR=20$, and $PQ=25$. Points $A$ and $B$ lie on $\overline{PQ}$, points $C$ and $D$ lie on $\overline{QR}$, and points $E$ and $F$ lie on $\overline{PR}$, with $PA=QB=QC=RD=RE=PF=5$. Find the area of hexagon $ABCDEF$.

Solution 1

We know the area of the hexagon $ABCDEF$ to be $\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED$. Since $PR^2+RQ^2=PQ^2$, we know that $\triangle PRQ$ is a right triangle. Thus the area of $\triangle PQR$ is $150$. Another way to compute the area is \[\frac12 \cdot PQ\cdot RQ \sin \angle PQR = \frac12 \cdot 500 \cdot \sin \angle PQR=150 \implies \sin \angle PQR = \frac35.\] Then the area of $\triangle BCQ = \frac12 \cdot BQ \cdot CQ \cdot \sin \angle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}$. Preceding in a similar fashion for $\triangle PAF$, the area of $\triangle PAF$ is $10$. Since $\angle ERD = 90^{\circ}$, the area of $\triangle RED=\frac{25}{2}$. Thus our desired answer is $150-\frac{15}{2}-10-\frac{25}{2}=\boxed{120}$

Solution 2

Let $R$ be the origin. Then $A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)$, and $F=(0,10)$. Using the shoelace theorem, the area is $\boxed{120}$.

See Also

2019 AIME I (Problems • Answer Key • Resources)
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Problem 4
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All AIME Problems and Solutions

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