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

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==Problem==
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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>.
  
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==Problem 3==
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==Diagram==
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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>.
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<asy>
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dot((0,0));
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dot((15,0));
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dot((15,20));
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draw((0,0)--(15,0)--(15,20)--cycle);
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dot((5,0));
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dot((10,0));
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dot((15,5));
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dot((15,15));
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dot((3,4));
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dot((12,16));
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draw((5,0)--(3,4));
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draw((10,0)--(15,5));
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draw((12,16)--(15,15));
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</asy>
  
 
==Solution 1==
 
==Solution 1==
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You can also go counterclockwise order, as long as you find the absolute value of the answer.
 
You can also go counterclockwise order, as long as you find the absolute value of the answer.
  
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The Shoelace Theorem gets its name because if one lists the coordinates in a column, <math>\begin{align*} (a_1 &, b_1) \\ (a_2 &, b_2) \\ & \vdots \\ (a_n &, b_n) \\ (a_1 &, b_1) \\ \end{align*}</math> and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes.
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.
  
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==Solution 3==
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==Solution 3 (Easiest, uses only basic geometry too)==
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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>.  
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Note that <math>\triangle{PQR}</math> has area <math>150</math> and is a <math>3</math>-<math>4</math>-<math>5</math> right triangle. Then, by similar triangles, the altitude from <math>B</math> to <math>QC</math> has length <math>3</math> and the altitude from <math>A</math> to <math>FP</math> has length <math>4</math>, 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
  
 
==Solution 4==
 
==Solution 4==
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Knowing that <math>\triangle{PQR}</math> has area 150 and is a 3-4-5 triangle, we can find the area of the smaller triangles <math>\triangle{DRE}</math>, <math>\triangle{APF}</math>, and <math>\triangle{CQB}</math> and subtract them from <math>\triangle{PQR}</math> to obtain our answer. First off, we know <math>\triangle{DRE}</math> has area <math>12.5</math> since it is a right triangle. To the find the areas of <math>\triangle{APF}</math> and <math>\triangle{CQB}</math> , we can use Law of Cosines (<math>c^2 = a^2 + b^2 - 2ab\cos C</math>) to find the lengths of <math>AF</math> and <math>CB</math>, respectively. Computing gives <math>AF = \sqrt{20}</math> and <math>CB = \sqrt{10}</math>. Now, using Heron's Formula, we find <math>\triangle{APF} = 10</math> and <math>\triangle{CQB} = 7.5</math>. Adding these and subtracting from <math>\triangle{PQR}</math>, we get <math>150 - (10 + 7.5 + 12.5) = \boxed{120}</math> -Starsher
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Knowing that <math>\triangle{PQR}</math> has area <math>150</math> and is a <math>3</math>-<math>4</math>-<math>5</math> triangle, we can find the area of the smaller triangles <math>\triangle{DRE}</math>, <math>\triangle{APF}</math>, and <math>\triangle{CQB}</math> and subtract them from <math>\triangle{PQR}</math> to obtain our answer. First off, we know <math>\triangle{DRE}</math> has area <math>12.5</math> since it is a right triangle. To the find the areas of <math>\triangle{APF}</math> and <math>\triangle{CQB}</math> , we can use Law of Cosines (<math>c^2 = a^2 + b^2 - 2ab\cos C</math>) to find the lengths of <math>AF</math> and <math>CB</math>, respectively. Computing gives <math>AF = \sqrt{20}</math> and <math>CB = \sqrt{10}</math>. Now, using Heron's Formula, we find <math>\triangle{APF} = 10</math> and <math>\triangle{CQB} = 7.5</math>. Adding these and subtracting from <math>\triangle{PQR}</math>, we get <math>150 - (10 + 7.5 + 12.5) = \boxed{120}</math> -Starsher
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==Solution 5 (Official MAA)==
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Triangle <math>PQR</math> is a right triangle with are <math>\tfrac12\cdot15\cdot20=150</math>. Each of <math>\triangle PAF,\triangle QCB,</math> and <math>\triangle RED</math> shares an angle with <math>\triangle PQR</math>. Because the area of a triangle with sides <math>a,\,b,</math> and included angle <math>\gamma</math> is <math>\tfrac12a\cdot b\cdot \sin\gamma,</math> it follows that the areas of <math>\triangle PAF,\triangle QCB,</math> and <math>\triangle RED</math> are each <math>5\cdot5\cdot\tfrac{150}{ab},</math> where <math>a</math> and <math>b</math> are the lengths of the sides of <math>\triangle PQR</math> adjacent to the shared angle. Thus the sum of the areas of <math>\triangle PAF,\triangle QCB,</math> and <math>\triangle RED</math> is <cmath>5\cdot5\cdot\frac{150}{15\cdot25}+5\cdot5\cdot\frac{150}{25\cdot20}+5\cdot5\cdot\frac{150}{20\cdot15}=25\left(\frac25+\frac3{10}+\frac12\right)=30.</cmath> Therefore <math>ABCDEF</math> has area <math>150-30=120</math>.
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==Video Solution #1(Complementary Area Counting?)==
 +
https://youtu.be/JQdad7APQG8?t=417
  
 
==Video Solution==
 
==Video Solution==
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https://www.youtube.com/watch?v=4jOfXNiQ6WM
 
https://www.youtube.com/watch?v=4jOfXNiQ6WM
  
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==Video Solution 2==
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https://youtu.be/TSKcjht8Rfk?t=941
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 +
~IceMatrix
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 +
==Video Solution 3==
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https://youtu.be/9X18wCiYw9M
 +
 +
~Shreyas S
  
 
==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 15:48, 3 March 2021

Problem

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 $ABCDEF$.

Diagram

[asy] dot((0,0)); dot((15,0)); dot((15,20)); draw((0,0)--(15,0)--(15,20)--cycle); dot((5,0)); dot((10,0)); dot((15,5)); dot((15,15)); dot((3,4)); dot((12,16)); draw((5,0)--(3,4)); draw((10,0)--(15,5)); draw((12,16)--(15,15)); [/asy]

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.

.

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

Solution 5 (Official MAA)

Triangle $PQR$ is a right triangle with are $\tfrac12\cdot15\cdot20=150$. Each of $\triangle PAF,\triangle QCB,$ and $\triangle RED$ shares an angle with $\triangle PQR$. Because the area of a triangle with sides $a,\,b,$ and included angle $\gamma$ is $\tfrac12a\cdot b\cdot \sin\gamma,$ it follows that the areas of $\triangle PAF,\triangle QCB,$ and $\triangle RED$ are each $5\cdot5\cdot\tfrac{150}{ab},$ where $a$ and $b$ are the lengths of the sides of $\triangle PQR$ adjacent to the shared angle. Thus the sum of the areas of $\triangle PAF,\triangle QCB,$ and $\triangle RED$ is \[5\cdot5\cdot\frac{150}{15\cdot25}+5\cdot5\cdot\frac{150}{25\cdot20}+5\cdot5\cdot\frac{150}{20\cdot15}=25\left(\frac25+\frac3{10}+\frac12\right)=30.\] Therefore $ABCDEF$ has area $150-30=120$.

Video Solution #1(Complementary Area Counting?)

https://youtu.be/JQdad7APQG8?t=417

Video Solution

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

Video Solution 2

https://youtu.be/TSKcjht8Rfk?t=941

~IceMatrix

Video Solution 3

https://youtu.be/9X18wCiYw9M

~Shreyas S

See Also

2019 AIME I (Problems • Answer Key • Resources)
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Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions

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