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Difference between revisions of "2021 AMC 10A Problems/Problem 21"

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<math>\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63</math>
 
<math>\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63</math>
  
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==Solution (Misplaced problem?)==
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==Diagram==
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Note that the extensions of the given lines will determine an equilateral triangle because the hexagon is equiangular. The area of the first triangle is <math>192\sqrt{3}</math>, so the side length is <math>\sqrt{192\cdot 4}=16\sqrt{3}</math>. The area of the second triangle is <math>324\sqrt{3}</math>, so the side length is <math>\sqrt{4\cdot 324}=36</math>. We can set the first value equal to <math>AB+CD+EF</math> and the second equal to <math>BC+DE+FA</math> by substituting some lengths in with different sides of the same equilateral triangle. The perimeter of the hexagon is just the sum of these two, which is <math>16\sqrt{3}+36</math> and <math>16+3+36=\boxed{55~\textbf{(C)}}</math>
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[[File:2021 AMC 10A Problem 21 Diagram.png|center|500px]]
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~MRENTHUSIASM (by Geometry Expressions)
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 +
==Solution==
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Let <math>P,Q,R,X,Y,</math> and <math>Z</math> be the intersections <math>\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},</math> and <math>\overleftrightarrow{FA}\cap\overleftrightarrow{BC},</math> respectively.
 +
 
 +
The sum of the interior angles of any hexagon is <math>720^\circ.</math> Since hexagon <math>ABCDEF</math> is equiangular, each of its interior angles is <math>720^\circ/6=120^\circ.</math> By angle chasing, we conclude that the interior angles of <math>\triangle PBC,\triangle QDE,\triangle RFA,\triangle XCD,\triangle YEF,</math> and <math>\triangle ZAB</math> are all <math>60^\circ.</math> Therefore, these triangles are all equilateral triangles, from which <math>\triangle PQR</math> and <math>\triangle XYZ</math> are both equilateral triangles.
 +
 
 +
We are given that
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<cmath>\begin{alignat*}{8}
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[PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\
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[XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3,
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\end{alignat*}</cmath>
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so we get <math>PQ=16\sqrt3</math> and <math>YZ=36,</math> respectively.
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 +
By the equilateral triangles and the Segment Addition Postulate, we find the perimeter of hexagon <math>ABCDEF:</math>
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<cmath>\begin{align*}
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AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\
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&=(YF+FA+AZ)+(PC+CD+DQ) \\
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&=YZ+PQ \\
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&=36+16\sqrt{3}.
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\end{align*}</cmath>
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Finally, the answer is <math>36+16+3=\boxed{\textbf{(C)} ~55}.</math>
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 +
~Sugar rush (Fundamental Logic)
 +
 
 +
~MRENTHUSIASM (Reconstruction)
  
 
== Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles) ==
 
== Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles) ==
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~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution by MRENTHUSIASM (English & Chinese)==
 +
https://www.youtube.com/watch?v=0n8EAu2VAiM
 +
 +
~MRENTHUSIASM
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2021|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:46, 19 July 2021

Problem

Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$, and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$. The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$, where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p$?

$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$

Diagram

500px

~MRENTHUSIASM (by Geometry Expressions)

Solution

Let $P,Q,R,X,Y,$ and $Z$ be the intersections $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},$ and $\overleftrightarrow{FA}\cap\overleftrightarrow{BC},$ respectively.

The sum of the interior angles of any hexagon is $720^\circ.$ Since hexagon $ABCDEF$ is equiangular, each of its interior angles is $720^\circ/6=120^\circ.$ By angle chasing, we conclude that the interior angles of $\triangle PBC,\triangle QDE,\triangle RFA,\triangle XCD,\triangle YEF,$ and $\triangle ZAB$ are all $60^\circ.$ Therefore, these triangles are all equilateral triangles, from which $\triangle PQR$ and $\triangle XYZ$ are both equilateral triangles.

We are given that \begin{alignat*}{8} [PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\ [XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3, \end{alignat*} so we get $PQ=16\sqrt3$ and $YZ=36,$ respectively.

By the equilateral triangles and the Segment Addition Postulate, we find the perimeter of hexagon $ABCDEF:$ \begin{align*} AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ &=(YF+FA+AZ)+(PC+CD+DQ) \\ &=YZ+PQ \\ &=36+16\sqrt{3}. \end{align*} Finally, the answer is $36+16+3=\boxed{\textbf{(C)} ~55}.$

~Sugar rush (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles)

https://youtu.be/ptBwDcmDaLA

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/8qcbZ8c7fHg

~IceMatrix

Video Solution by MRENTHUSIASM (English & Chinese)

https://www.youtube.com/watch?v=0n8EAu2VAiM

~MRENTHUSIASM

See Also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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