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

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==Solution 3==
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==Solution 3 (Easiest, uses only basic geometry too)==
 
Note that <math>\triangle{PQR}</math> has area <math>150</math> 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{120}</math>.  
 
Note that <math>\triangle{PQR}</math> has area <math>150</math> 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{120}</math>.  
 
-Stormersyle
 
-Stormersyle

Revision as of 15:21, 6 October 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 $\boxed{120}$. Shoelace theorem:Suppose the polygon $P$ has vertices $(a_1, b_1)$, $(a_2, b_2)$, ... , $(a_n, b_n)$, listed in clockwise order. Then the area of $P$ is

\[\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|\] You can also go counterclockwise order, as long as you find the absolute value of the answer.

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Solution 3 (Easiest, uses only basic geometry too)

Note that $\triangle{PQR}$ has area $150$ 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 $[ABCDEF]=\boxed{120}$. -Stormersyle

Solution 4

Knowing that $\triangle{PQR}$ has area 150 and is a 3-4-5 triangle, we can find the area of the smaller triangles $\triangle{DRE}$, $\triangle{APF}$, and $\triangle{CQB}$ and subtract them from $\triangle{PQR}$ to obtain our answer. First off, we know $\triangle{DRE}$ has area $12.5$ since it is a right triangle. To the find the areas of $\triangle{APF}$ and $\triangle{CQB}$ , we can use Law of Cosines ($c^2 = a^2 + b^2 - 2ab\cos C$) to find the lengths of $AF$ and $CB$, respectively. Computing gives $AF = \sqrt{20}$ and $CB = \sqrt{10}$. Now, using Heron's Formula, we find $\triangle{APF} = 10$ and $\triangle{CQB} = 7.5$. Adding these and subtracting from $\triangle{PQR}$, we get $150 - (10 + 7.5 + 12.5) = \boxed{120}$ -Starsher

Video Solution

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


See Also

2019 AIME I (ProblemsAnswer KeyResources)
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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