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

m (Diagram)
m (Solution: Deleted the reference to the Segment Addition Postulate)
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so we get <math>PQ=16\sqrt3</math> and <math>YZ=36,</math> respectively.
 
so we get <math>PQ=16\sqrt3</math> and <math>YZ=36,</math> respectively.
  
By the equilateral triangles and the Segment Addition Postulate, we find the perimeter of hexagon <math>ABCDEF:</math>
+
By the equilateral triangles, we find the perimeter of hexagon <math>ABCDEF:</math>
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\
 
AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\

Revision as of 10:45, 5 October 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

[asy] /* Made by MRENTHUSIASM */ size(250); path P1, P2; P1 = scale(16sqrt(3))*polygon(3); P2 = shift(3,3)*scale(36)*rotate(180)*polygon(3); draw(P1, dashed+black); draw(P2, dashed+black); pair A, B, C, D, E, F; E = intersectionpoints(P1,P2)[0]; F = intersectionpoints(P1,P2)[1]; A = intersectionpoints(P1,P2)[2]; B = intersectionpoints(P1,P2)[3]; C = intersectionpoints(P1,P2)[4]; D = intersectionpoints(P1,P2)[5]; filldraw(A--B--C--D--E--F--cycle,yellow); dot("$E$",E,1.5*dir(0),linewidth(4)); dot("$F$",F,1.5*dir(60),linewidth(4)); dot("$A$",A,1.5*dir(120),linewidth(4)); dot("$B$",B,1.5*dir(180),linewidth(4)); dot("$C$",C,1.5*dir(-120),linewidth(4)); dot("$D$",D,1.5*dir(-60),linewidth(4)); dot(16sqrt(3)*dir(90)^^16sqrt(3)*dir(210)^^16sqrt(3)*dir(330),linewidth(4)); dot((3,3)+36*dir(30)^^(3,3)+36*dir(150)^^(3,3)+36*dir(270),linewidth(4)); [/asy] ~MRENTHUSIASM

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 $720^\circ\div6=120^\circ.$ 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, 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 (ProblemsAnswer KeyResources)
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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