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

Revision as of 14:57, 2 November 2024

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

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 2

Since $\overline{AC}$ is $1$ less than $\overline{BC}$ , $\overline{CD}$ must be $1$ more than $\overline{BD}$ to equate the perimeter. Hence, $\overline{BD}+\overline{BD}+1=5$ , so $\overline{BD}=2$ . Therefore, the area of $\triangle ABD$ is $\frac{(2)(4)(\sin B)}{2}=4(\frac{3}{5})=\boxed{\textbf{(D) } \frac{12}{5}}$

~megaboy6679

Video Solutions

https://youtu.be/itz3JyoZQYg

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 ==Solution 2==
-Since <math>\overline{AC}</math> is <math>1</math> less than <math>\overline{BC}</math>, <math>\overline{CD}</math> must be <math>1</math> more than <math>\overline{BD}</math> to equate the perimeter. Hence, <math>\overline{BD}+\overline{BD}+1=5</math>, so <math>\overline{BD}=2</math>. Therefore, the area of <math>\triangle ABD</math> is <math>(2)(4)(\sin B)=8(\frac{3}{5})=\boxed{\textbf{(D) } \frac{12}{5}}</math>
+Since <math>\overline{AC}</math> is <math>1</math> less than <math>\overline{BC}</math>, <math>\overline{CD}</math> must be <math>1</math> more than <math>\overline{BD}</math> to equate the perimeter. Hence, <math>\overline{BD}+\overline{BD}+1=5</math>, so <math>\overline{BD}=2</math>. Therefore, the area of <math>\triangle ABD</math> is <math>\frac{(2)(4)(\sin B)}{2}=4(\frac{3}{5})=\boxed{\textbf{(D) } \frac{12}{5}}</math>
 ~megaboy6679

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