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

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m (Made the solution more aligned to the diagram. I will give credit to Sugar Rush as the primary author.)
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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.
 
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 wish to find <math>AB+BC+CD+DE+EF+FA.</math> We are given that
 +
<cmath>\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*}</cmath>
 +
so <math>PQ=16\sqrt3</math> and <math>YZ=36.</math>
 +
 +
By the equilateral triangles and the Segment Addition Postulate, we find the perimeter of hexagon <math>ABCDEF:</math>
 +
<cmath>\begin{align*}
 +
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}.
 +
\end{align*}</cmath>
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Finally, the answer is <math>36+16+3=\boxed{\textbf{(C)} ~55}.</math>
  
 
~Sugar rush (Fundamental Logic)
 
~Sugar rush (Fundamental Logic)

Revision as of 06:21, 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 $P,Q,R,X,Y,Z$ be the intersections $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},\overleftrightarrow{FA}\cap\overleftrightarrow{BC},$ 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 wish to find $AB+BC+CD+DE+EF+FA.$ 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 $PQ=16\sqrt3$ and $YZ=36.$

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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