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 Fall AMC 12A Problems/Problem 14"

(→‎Solution (Law of Cosines and Equilateral Triangle Area))
(→‎Solution (Law of Cosines and Equilateral Triangle Area))
Line 1: Line 1:
 
==Solution (Law of Cosines and Equilateral Triangle Area)==
 
==Solution (Law of Cosines and Equilateral Triangle Area)==
  
−
Isosceles triangles <math>ABE</math>, <math>CBD</math>, and <math>EDF</math> are identical by SAS similarity. <math>BF=BD=DF</math> by CPCTC, and triangle <math>BDF</math> is equilateral.  
+
Isosceles triangles <math>ABE</math>, <math>CBD</math>, and <math>EDF</math> are identical by [[SAS_Similarity|SAS similarity]]. <math>BF=BD=DF</math> by [[CPCTC]], and triangle <math>BDF</math> is equilateral.  
  
 
Let the side length of the hexagon be <math>s</math>.  
 
Let the side length of the hexagon be <math>s</math>.  

Revision as of 18:11, 23 November 2021

Solution (Law of Cosines and Equilateral Triangle Area)

Isosceles triangles $ABE$, $CBD$, and $EDF$ are identical by SAS similarity. $BF=BD=DF$ by CPCTC, and triangle $BDF$ is equilateral.

Let the side length of the hexagon be $s$.

The area of each isosceles triangle is $\frac{1}{2}s^2\sin{30}=\frac{1}{4}s^2$ by the fourth formula here.

By the Law of Cosines, the square of the side length of equilateral triangle BDF is $2s^2-2x^2\cos{30}=(2-\sqrt{3})s^2$. Hence, the area of the equilateral triangle is $\frac{\sqrt{3}}{4}(2-\sqrt{3})s^2=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2$.

So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle or $\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}$. The perimeter is $6s=\boxed{12\sqrt{3}}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_14&oldid=165585"