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Difference between revisions of "2024 AIME II Problems/Problem 5"

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==Solution 1==
 
==Solution 1==
 
(Sorry i have zero idea how to make drawings)
 
(Sorry i have zero idea how to make drawings)
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Draw a good diagram!
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Let <math>AB \cap DC</math>, <math>CD \cap FE</math>, and <math>BA \cap EF</math> be P, Q, and R, respectively. Let <math>QR=200, RP=300, PQ=240</math>. Notice that all smaller triangles formed are all similar to the larger <math>(200,240,300)</math> triangle. Let the side length of the hexagon be x. Triangle <math> \triangle BCP \sim \triangle RQP</math>, so <math>\frac{BC}{BP} =\frac{x}{BP} =\frac{200}{300} \implies BP=\frac{3x}{2}</math>. Triangle <math>\triangle AFR \sim \triangle PQR</math>, so <math>\frac{AF}{AR}=\frac{x}{AR} = \frac{240}{300} \implies AR=\frac{5x}{4}</math>. We know <math>RA+AB+BP=300</math>, so <math>\frac{5}{4}x + x + \frac{3}{2}x = 300</math>. Solving, we get <math>x=\boxed{080}</math>. -westwoodmonster
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==See also==
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{{AIME box|year=2024|num-b=4|num-a=6|n=II}}
  
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Let <math>AB \cap DC</math>, <math>CD \cap FE</math>, and <math>BA \cap EF</math> be P, Q, and R, respectively. Let <math>QR=200, RP=300, PQ=240</math>. Notice that all smaller triangles formed are all similar to the larger <math>(200,240,300)</math> triangle. Let the side length of the hexagon be x. Triangle <math> \triangle BCP \sim \triangle RQP</math>, so <math>\frac{BC}{BP} =\frac{x}{BP} =\frac{200}{300} \implies BP=\frac{3x}{2}</math>. Triangle <math>\triangle AFR \sim \triangle PQR</math>, so <math>\frac{AF}{AR}=\frac{x}{AR} = \frac{240}{300} \implies AR=\frac{5x}{4}</math>. We know <math>RA+AB+BP=300</math>, so <math>\frac{5}{4}x + x + \frac{3}{2}x = 300</math>. Solving, we get <math>x=\boxed{080}</math>. -westwoodmonster
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[[Category:]]
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{{MAA Notice}}

Revision as of 20:52, 8 February 2024

Let ABCDEF be a convex equilateral hexagon in which all pairs of opposite sides are parallel. The triangle whose sides are extensions of segments AB, CD, and EF has side lengths 200, 240, and 300. Find the side length of the hexagon.

Solution 1

(Sorry i have zero idea how to make drawings) Draw a good diagram! Let $AB \cap DC$, $CD \cap FE$, and $BA \cap EF$ be P, Q, and R, respectively. Let $QR=200, RP=300, PQ=240$. Notice that all smaller triangles formed are all similar to the larger $(200,240,300)$ triangle. Let the side length of the hexagon be x. Triangle $\triangle BCP \sim \triangle RQP$, so $\frac{BC}{BP} =\frac{x}{BP} =\frac{200}{300} \implies BP=\frac{3x}{2}$. Triangle $\triangle AFR \sim \triangle PQR$, so $\frac{AF}{AR}=\frac{x}{AR} = \frac{240}{300} \implies AR=\frac{5x}{4}$. We know $RA+AB+BP=300$, so $\frac{5}{4}x + x + \frac{3}{2}x = 300$. Solving, we get $x=\boxed{080}$. -westwoodmonster


See also

2024 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

[[Category:]] The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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