# Difference between revisions of "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 1

Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that $x$ is the length of the line from B to D. So, the perimeter of $\triangle{ABD}$ would be $\overline{AD} + 4 + x$, while the perimeter of $\triangle{ACD}$ would be $\overline{AD} + 3 + (5 - x)$. Notice that we can find out $x$ from these two equations. We can find out that $x = 2$, so that means that the area of $\triangle{ABD} = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}$

THIS SOLUTION MAKES NO SENSE PLEASE MOVE ON


## Solution 2

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 $[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6$, we get $[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}$

## Solution 3

Since point $D$ is on line $BC$, it will split it into $CD$ and $DB$. Let $CD = 5 - x$ and $DB = x$. Triangle $CAD$ has side lengths $3, 5 - x, AD$ and triangle $DAB$ has side lengths $x, 4, AD$. Since both perimeters are equal, we have the equation $3 + 5 - x + AD = 4 + x + AD$. Eliminating $AD$ and solving the resulting linear equation gives $x = 2$. Draw a perpendicular from point $D$ to $AB$. Call the point of intersection $F$. Because angle $ABC$ is common to both triangles $DBF$ and $ABC$, and both are right triangles, both are similar. The hypotenuse of triangle $DBF$ is 2, so the altitude must be $6/5$ Because $DBF$ and $ABD$ share the same altitude, the height of $ABD$ therefore must be $6/5$. The base of $ABD$ is 4, so $[ABD] = \frac{1}{2} \cdot 4 \cdot \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}$

## Solution 4

Using any preferred method, realize $BD = 2$. Since we are given a 3-4-5 right triangle, we know the value of $\sin(\angle ABC) = \frac{3}{5}$. Since we are given $AB = 4$, apply the Sine Area Formula to get $\frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}$.