Difference between revisions of "2019 AIME I Problems/Problem 3"
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==Problem 3== | ==Problem 3== | ||
− | ==Solution== | + | 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>. |
+ | |||
+ | ==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 <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> | ||
+ | |||
+ | ==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>. | ||
+ | Shoelace theorem:Suppose the polygon <math>P</math> has vertices <math>(a_1, b_1)</math>, <math>(a_2, b_2)</math>, ... , <math>(a_n, b_n)</math>, listed in clockwise order. Then the area of <math>P</math> is | ||
+ | |||
+ | <cmath>\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|</cmath> | ||
+ | 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 <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 | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==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== | ==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}} |
Latest revision as of 17:35, 30 July 2020
Contents
Problem 3
In , , , and . Points and lie on , points and lie on , and points and lie on , with . Find the area of hexagon .
Diagram
Solution 1
We know the area of the hexagon to be . Since , we know that is a right triangle. Thus the area of is . Another way to compute the area is Then the area of . Preceding in a similar fashion for , the area of is . Since , the area of . Thus our desired answer is
Solution 2
Let be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that , and . Using the shoelace theorem, the area is . Shoelace theorem:Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area of is
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 has area and is a -- right triangle. Then, by similar triangles, the altitude from to has length and the altitude from to has length , so , meaning that . -Stormersyle
Solution 4
Knowing that has area and is a -- triangle, we can find the area of the smaller triangles , , and and subtract them from to obtain our answer. First off, we know has area since it is a right triangle. To the find the areas of and , we can use Law of Cosines () to find the lengths of and , respectively. Computing gives and . Now, using Heron's Formula, we find and . Adding these and subtracting from , we get -Starsher
Video Solution
https://www.youtube.com/watch?v=4jOfXNiQ6WM
Video Solution 2
https://youtu.be/TSKcjht8Rfk?t=941
~IceMatrix
Video Solution 3
~Shreyas S
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.