AMC12 2015 A

Answer (A): Let $g$ and $h$ be the lengths of the altitudes of $T$ and $T'$ from the sides with lengths $8$ and $b$, respectively. The Pythagorean Theorem implies that $g=\sqrt{5^2-4^2}=3$, and so the area of $T$ is $\frac12\cdot 8\cdot 3=12$, and the perimeter is $5+5+8=18$. The Pythagorean Theorem implies that $h=\frac12\sqrt{4a^2-b^2}$. Thus $18=2a+b$ and $$ 12=\frac12 b\cdot \frac12\sqrt{4a^2-b^2}=\frac14 b\sqrt{4a^2-b^2}. $$ Solving for $a$ and substituting in the square of the second equation yields $$ 12^2=\frac{b^2}{16}(4a^2-b^2)=\frac{b^2}{16}\big((18-b)^2-b^2\big) =\frac{b^2}{16}\cdot 18\cdot (18-2b)=\frac94 b^2(9-b). $$ Thus $64-b^2(9-b)=b^3-9b^2+64=(b-8)(b^2-b-8)=0$. Because $T$ and $T'$ are not congruent, it follows that $b\ne 8$. Hence $b^2-b-8=0$ and the positive solution of this equation is $\frac12(\sqrt{33}+1)$. Because $25<33<36$, the solution is between $\frac12(5+1)=3$ and $\frac12(6+1)=3.5$, so the closest integer is $3$.