2017 AMC 8 Problems/Problem 16

Problem

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

Because $\overline{BD} + \overline{CD} = 5,$ we can see that when we draw a line from point $B$ to imaginary point $D$ that line applies to both triangles. Let us say that $x$ is that line. 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 $x$ from these two equations by setting them equal and then canceling $\overline{AD}$ . We find that $x = 2$ , and because the height of the triangles is the same, the ratio of the areas is $2:3$ , so that means that the area of $\triangle ABD = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}$ .

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 $\frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}$ .

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/TRY34buacYU

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Video Solution by OmegaLearn

https://youtu.be/51K3uCzntWs?t=1663

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Video Solutions

https://youtu.be/itz3JyoZQYg

https://youtu.be/IVHTUjOPePY

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2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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