2014 AMC 12A Problem 10

Question

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?

\dfrac{\sqrt3}{4}

\dfrac{\sqrt3}{3}

\dfrac{2}{3}

\dfrac{\sqrt2}{2}

\dfrac{\sqrt3}{2}

Answer: B

Concepts:equilateral triangle triangle area Pythagorean Theorem

Difficulty rating: 1560

Small hint:

Each isosceles triangle has base

1

; let its height be

h

Big hint:

Three triangles of area

\tfrac12 h

sum to

\tfrac{\sqrt3}{4}

; solve for

h

, then use the Pythagorean theorem

Solution:

The equilateral triangle has area

\dfrac{\sqrt3}{4}.

Each isosceles triangle has base

1

and height

h,

so

3\cdot\dfrac12 h=\dfrac{\sqrt3}{4},

giving

h=\dfrac{\sqrt3}{6}.

A congruent side is the hypotenuse from the apex to a base endpoint:

\begin{gathered} \sqrt{\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{6}\right)^2}\\ =\sqrt{\dfrac14+\dfrac{1}{12}}\\ =\sqrt{\dfrac13}=\dfrac{\sqrt3}{3}. \end{gathered}

Thus, the correct answer is B.