Difference between revisions of "2021 AMC 10A Problems/Problem 21"
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| + | ==Problem 21== | ||
| + | Let <math>ABCDEF</math> be an equiangular hexagon. The lines <math>AB, CD,</math> and <math>EF</math> determine a triangle with area <math>192\sqrt{3}</math>, and the lines <math>BC, DE,</math> and <math>FA</math> determine a triangle with area <math>324\sqrt{3}</math>. The perimeter of hexagon <math>ABCDEF</math> can be expressed as <math>m +n\sqrt{p}</math>, where <math>m, n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the square of any prime. What is <math>m + n + p</math>? | ||
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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> | ||
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==Solution (Misplaced problem?)== | ==Solution (Misplaced problem?)== | ||
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> | 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> | ||
Revision as of 23:41, 11 February 2021
Problem 21
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 
?
Solution (Misplaced problem?)
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 , so the side length is 
. The area of the second triangle is , so the side length is 
. We can set the first value equal to 
and the second equal to 
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 
and 
Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles)
~ pi_is_3.14
