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> | ||
− | == | + | ==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 <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\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. | ||
+ | |||
+ | 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 | ||
== Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles) == | == Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles) == | ||
Line 16: | Line 80: | ||
~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}} |
Latest revision as of 15:53, 14 October 2023
Contents
Problem
Let be an equiangular hexagon. The lines
and
determine a triangle with area
, and the lines
and
determine a triangle with area
. The perimeter of hexagon
can be expressed as
, where
and
are positive integers and
is not divisible by the square of any prime. What is
?
Diagram
~MRENTHUSIASM
Solution 1
Let and
be the intersections
and
respectively.
The sum of the interior angles of any hexagon is Since hexagon
is equiangular, each of its interior angles is
By angle chasing, we conclude that the interior angles of
and
are all
Therefore, these triangles are all equilateral triangles, from which
and
are both equilateral triangles.
We are given that
so we get
and
respectively.
By equilateral triangles and segment addition, we find the perimeter of hexagon
Finally, the answer is
~sugar_rush ~MRENTHUSIASM
Solution 2
Let the length Then, we have
We get
We want
and it follows that
Finally, the answer is
~mathboy282
Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~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.