Difference between revisions of "2014 AMC 12A Problems/Problem 10"

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(Solution by johnstucky)
 
(Solution by johnstucky)
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==Video Solution==
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https://youtu.be/rJytKoJzNBY
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solution by sugar_rush
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2014|ab=A|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:19, 25 November 2020

Problem

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1$. The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?

$\textbf{(A) }\dfrac{\sqrt3}4\qquad \textbf{(B) }\dfrac{\sqrt3}3\qquad \textbf{(C) }\dfrac23\qquad \textbf{(D) }\dfrac{\sqrt2}2\qquad \textbf{(E) }\dfrac{\sqrt3}2$

Solution 1

Reflect each of the triangles over its respective side. Then since the areas of the triangles total to the area of the equilateral triangle, it can be seen that the triangles fill up the equilateral one and the vertices of these triangles concur at the circumcenter of the equilateral triangle. Hence the desired answer is just its circumradius, or $\boxed{\dfrac{\sqrt3}3\textbf{ (B)}}$.

(Solution by djmathman)

Solution 2

Since the total area of each congruent isosceles triangle is the same, the area of each is $\dfrac{1}3$ the total area of the equilateral triangle of side length 1, or $\dfrac{1}3$ x $\dfrac{\sqrt3}4$. Likewise, the area of each can be defined as $\dfrac{bh}2$ with base $b$ equaling 1, meaning that $\dfrac{h}2$ = $\dfrac{1}3$ x $\dfrac{\sqrt3}4$, or $h$ = $\dfrac{\sqrt3}6$. A side length of the isosceles triangle is the hypotenuse with legs $\dfrac{b}2$ and $h$. Using the Pythagorean Theorem, the side length is $\sqrt{{(\dfrac{1}2)}^2 + {(\dfrac{\sqrt3}6)}^2}$, or $\boxed{\dfrac{\sqrt3}3\textbf{ (B)}}$.

(Solution by johnstucky)

Video Solution

https://youtu.be/rJytKoJzNBY

solution by sugar_rush

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 12 Problems and Solutions

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