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

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==Problem 3==
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==Problem==
 
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 4==
 
==Solution 4==
 
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
 
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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==Solution 6 (Using simple trigonometry)==
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Let's say that angle APF is 'u'. Then, we know that sin(u) is 20/25 = 4/5. Therefore, the area of APF is 0.5*5*5*sin(u)=10. Now, let's do the same thing with triangle BQC. If we name angle BQC as 'w', we know that sin(w) is 15/25 = 3/5. Therefore, the area of BQC is 0.5*5*5*sin(w)=7.5. Since triangle ERD is a right-angled isosceles triangle, the area of ERD is 12.5. In conclusion, (area of the hexagon) = tri(PRQ)-{tri(APF)+tri(BQC)+tri(ERD)}=150-(10+7.5+12.5)=120
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==Video Solution #1(Complementary Area Counting?)==
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https://youtu.be/JQdad7APQG8?t=417
  
 
==Video Solution==
 
==Video Solution==
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~IceMatrix
 
~IceMatrix
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==Video Solution 3==
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https://youtu.be/9X18wCiYw9M
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~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}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 08:38, 10 May 2024

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 $150-30=120$.

Solution 6 (Using simple trigonometry)

Let's say that angle APF is 'u'. Then, we know that sin(u) is 20/25 = 4/5. Therefore, the area of APF is 0.5*5*5*sin(u)=10. Now, let's do the same thing with triangle BQC. If we name angle BQC as 'w', we know that sin(w) is 15/25 = 3/5. Therefore, the area of BQC is 0.5*5*5*sin(w)=7.5. Since triangle ERD is a right-angled isosceles triangle, the area of ERD is 12.5. In conclusion, (area of the hexagon) = tri(PRQ)-{tri(APF)+tri(BQC)+tri(ERD)}=150-(10+7.5+12.5)=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 4
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All AIME Problems and Solutions

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