2015 AMC 12A Problems/Problem 20

Problem

Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ , $5$ , and $8$ , while those of $T'$ have lengths $a$ , $a$ , and $b$ . Which of the following numbers is closest to $b$ ?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$

Solution

The area of $T$ is $\dfrac{1}{2} \cdot 8 \cdot 3 = 12$ and the perimeter is 18.

The area of $T$ is $\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ and the perimeter is $2a + b$ .

Thus $2a + b = 18$ , so $2a = 18 - a$ .

Thus $12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ , so $48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}$ .

We square and divide 36 from both sides to obtain $64 = b^2 (9 - b)$ , so $b^3 - 9b^2 + 64 = 0$ . This factors as $(b - 8)(b^2 - b - 8) = 0$ . Because clearly $b \neq 8$ but $b > 0$ , we have $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\textbf{(A)}$ .

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2015 AMC 12A Problems/Problem 20

Problem

Solution