Difference between revisions of "2021 AMC 10A Problems/Problem 21"
MRENTHUSIASM (talk | contribs) m (→Solution) |
MRENTHUSIASM (talk | contribs) (→Diagram: Remade diagram using Asy.) |
||
| Line 5: | Line 5: | ||
==Diagram== | ==Diagram== | ||
| − | [[ | + | <asy> |
| − | + | size(250); | |
| − | ~MRENTHUSIASM | + | /* Made by MRENTHUSIASM */ |
| + | 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)); | ||
| + | </asy> | ||
| + | ~MRENTHUSIASM | ||
==Solution== | ==Solution== | ||
Revision as of 11:22, 29 September 2021
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
![[asy] size(250); /* Made by MRENTHUSIASM */ 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)); [/asy]](http://latex.artofproblemsolving.com/0/1/a/01a19051ff7bb0b7b2cb6df0ebc510c789f08c39.png)
~MRENTHUSIASM
Solution
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
![\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*}](http://latex.artofproblemsolving.com/5/f/5/5f5b55f5b4fa69ea2c5266b0c2cb7fc1f0ca9f17.png)
so we get 
and 
respectively.
By the equilateral triangles and the Segment Addition Postulate, we find the perimeter of hexagon 
Finally, the answer is 
~sugar_rush (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
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. 
