Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (Added in diagram.)
(→‎Solution 2)
 
(26 intermediate revisions by 5 users not shown)
Line 5: Line 5:
  
 
==Diagram==
 
==Diagram==
−
[[File:2021 AMC 10A Problem 21 Diagram.png|center|500px]]
+
<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
  
−
~MRENTHUSIASM (by Geometry Expressions)
+
==Solution 1==
 +
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.
  
−
==Solution (Misplaced problem?)==
+
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\div6=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.
−
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>
+
 
 +
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 we get <math>PQ=16\sqrt3</math> and <math>YZ=36,</math> respectively.
 +
 
 +
By equilateral triangles and segment addition, we find the perimeter of hexagon <math>ABCDEF:</math>
 +
<cmath>\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*}</cmath>
 +
Finally, the answer is <math>36+16+3=\boxed{\textbf{(C)} ~55}.</math>
 +
 
 +
~sugar_rush ~MRENTHUSIASM
 +
 
 +
==Solution 2==
 +
Let the length <math>AB=x, BC=y.</math> Then, we have
 +
<cmath>\begin{align*}
 +
(y+2x)^2\cdot\frac{\sqrt 3}{4}&=324\sqrt3, \\
 +
(x+2y)^2\cdot\frac{\sqrt 3}{4}&=192\sqrt3.
 +
\end{align*}</cmath>
 +
We get
 +
<cmath>\begin{align*}
 +
y+2x&=36, \\
 +
x+2y&=16\sqrt3.
 +
\end{align*}</cmath>
 +
We want <math>3x+3y,</math> and it follows that <cmath>3x+3y=(y+2x)+(x+2y)=36+16\sqrt3.</cmath>
 +
Finally, the answer is <math>36+16+3=\boxed{\textbf{(C)} ~55}.</math>
 +
 
 +
~mathboy282
 +
 
 +
==Solution 3==
 +
 
 +
Since it is an equiangular hexagon, each interior angle measures 120 degrees, and by angle chasing you can conclude that all the triangles are equilateral. Since the area formula for an equiangular triangle is s^2 sqrt(3)/4, the side lengths of the triangles are 36 and 16 sqrt(3). In the above diagram the sum of the lengths of AF, AB, and FE is 36 and the sum of the lengths of BC, CD, and DE is 16sqrt(3). Since these are also the side lengths of the hexagon, the answer is 36+16+3=55 (C).
 +
 
 +
~NamyaB.
  
 
== Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles) ==
 
== Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles) ==

Latest revision as of 21:47, 17 June 2024

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 1

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 equilateral triangles and segment addition, 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 ~MRENTHUSIASM

Solution 2

Let the length $AB=x, BC=y.$ Then, we have \begin{align*} (y+2x)^2\cdot\frac{\sqrt 3}{4}&=324\sqrt3, \\ (x+2y)^2\cdot\frac{\sqrt 3}{4}&=192\sqrt3. \end{align*} We get \begin{align*} y+2x&=36, \\ x+2y&=16\sqrt3. \end{align*} We want $3x+3y,$ and it follows that \[3x+3y=(y+2x)+(x+2y)=36+16\sqrt3.\] Finally, the answer is $36+16+3=\boxed{\textbf{(C)} ~55}.$

~mathboy282

Solution 3

Since it is an equiangular hexagon, each interior angle measures 120 degrees, and by angle chasing you can conclude that all the triangles are equilateral. Since the area formula for an equiangular triangle is s^2 sqrt(3)/4, the side lengths of the triangles are 36 and 16 sqrt(3). In the above diagram the sum of the lengths of AF, AB, and FE is 36 and the sum of the lengths of BC, CD, and DE is 16sqrt(3). Since these are also the side lengths of the hexagon, the answer is 36+16+3=55 (C).

~NamyaB.

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_21&oldid=220702"