Difference between revisions of "2017 AMC 8 Problems/Problem 16"

Revision as of 17:10, 22 November 2017

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

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 $[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6$ , we get $[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) } \frac{12}{5}}$

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 $[ABD] = \frac{1}{2} * 4 * \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}$

@@ Line 11: / Line 11: @@
 <math>\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}</math>
+==Solution 1==
 ==Solution 2==

2017 AMC 8 (Problems • Answer Key • Resources)
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