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Difference between revisions of "2019 AIME I Problems/Problem 3"
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==Solution 2==
 
==Solution 2==
 
Let <math>R</math> be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that <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>.
 
Let <math>R</math> be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that <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>.
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==Solution 3==
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Note that <math>\triangle{PQR}</math> has area 120 and is a 3-4-5 right triangle. Then, by similar triangles, the altitude from <math>B</math> to <math>QC</math> has length 3 and the altitude from <math>A</math> to <math>FP</math> has length 4, so <math>[QBC]+[DRE]+[AFP]=\frac{15}{2}+\frac{25}{2}+\frac{20}{2}=30</math>, meaning that <math>[ABCDEF]=\boxed{150}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 16:32, 19 March 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. Noticing that the triangle is a 3-4-5 right triangle, we can see that $A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)$, and $F=(0,10)$. Using the shoelace theorem, the area is $\boxed{120}$.

Solution 3

Note that $\triangle{PQR}$ has area 120 and is a 3-4-5 right triangle. Then, by similar triangles, the altitude from $B$ to $QC$ has length 3 and the altitude from $A$ to $FP$ has length 4, so $[QBC]+[DRE]+[AFP]=\frac{15}{2}+\frac{25}{2}+\frac{20}{2}=30$, meaning that $[ABCDEF]=\boxed{150}$.

Video Solution

https://www.youtube.com/watch?v=4jOfXNiQ6WM


See Also

2019 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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