2017 AMC 8 Problems/Problem 16

Problem 16

In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$ ? $[asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,0), ESE); label("$C$", (0, 3), N); label("$3$", (0, 1.5), W); label("$4$", (2, 0), S); label("$5$", (2, 1.5), NE);[/asy]$

$\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}$

Solution

We know that the perimeters of the two small triangles are $3+CD+AD$ and $4+BD+AD$ . Setting both equal and using $BD+CD = 5$ , we have $BD = 2$ and $CD = 3$ . Now, we simply have to find the area of $\triangle ABD$ . Since $\frac{BD}{CD} = \frac{2}{3}$ , we must have $\frac{[ABD]}{ACD]} = 2/3. Combining this with the fact that$ [ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6 $, we get$ [ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) }$ \frac{12}{5}}$

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2017 AMC 8 Problems/Problem 16

Problem 16

Solution

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