Question

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 $b?$

$3$

$4$

$5$

$6$

$8$

Solution:

The altitude of $T$ to its base of length $8$ is $\sqrt{5^2 - 4^2} = 3,$ so $T$ has area $\dfrac{1}{2}\cdot 8\cdot 3 = 12$ and perimeter $18.$

For $T'$ we need $2a + b = 18$ and area $\dfrac{1}{4}b\sqrt{4a^2 - b^2} = 12.$ Substituting $a = \dfrac{18 - b}{2}$ and squaring leads to $(b - 8)(b^2 - b - 8) = 0.$

Since $T$ and $T'$ are not congruent, $b \ne 8,$ so $b^2 - b - 8 = 0$ and $b = \dfrac{1 + \sqrt{33}}{2}.$ Because $25 \lt 33 \lt 36,$ this is between $3$ and $3.5,$ so the closest integer is $3.$

Thus, the correct answer is A.

2015 AMC 12A Problem 20

Solution:

Problem 20 in Other Years