Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem 16)
m (→‎Solution 2)
 
(59 intermediate revisions by 29 users not shown)
Line 1: Line 1:
−
==Problem 16==
+
==Problem==
  
 
In the figure below, choose point <math>D</math> on <math>\overline{BC}</math> so that <math>\triangle ACD</math> and <math>\triangle ABD</math> have equal perimeters. What is the area of <math>\triangle ABD</math>?
 
In the figure below, choose point <math>D</math> on <math>\overline{BC}</math> so that <math>\triangle ACD</math> and <math>\triangle ABD</math> have equal perimeters. What is the area of <math>\triangle ABD</math>?
 +
 
<asy>draw((0,0)--(4,0)--(0,3)--(0,0));
 
<asy>draw((0,0)--(4,0)--(0,3)--(0,0));
 
label("$A$", (0,0), SW);
 
label("$A$", (0,0), SW);
Line 12: Line 13:
 
<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>
 
<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==
+
==Solution 1==
 +
 
 +
We know that the perimeters of the two small triangles are <math>3+CD+AD</math> and <math>4+BD+AD</math>. Setting both equal and using <math>BD+CD = 5</math>, we have <math>BD = 2</math> and <math>CD = 3</math>. Now, we simply have to find the area of <math>\triangle ABD</math>. Since <math>\frac{BD}{CD} = \frac{2}{3}</math>, we must have <math>\frac{[ABD]}{[ACD]} = 2/3</math>. Combining this with the fact that <math>[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}</math>.
 +
 
 +
==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>\frac{(2)(4)(\sin B)}{2}=4(\frac{3}{5})=\boxed{\textbf{(D) } \frac{12}{5}}</math>
 +
 
 +
~megaboy6679
  
−
We know that the perimeters of the two small triangles are <math>3+CD+AD</math> and <math>4+BD+AD.</math> Setting both equal and using <math>BD+CD = 5,</math> we have <math>BD = 2</math> and <math>CD = 3.</math> Now, we simply have to find the area of <math>\triangle ABD.</math> We can use <math>AB</math> as the base and the altitude from <math>D</math>. Let's call the foot of the altitude <math>E.</math> We have <math>\triangle BDE</math> similar to <math>BAC.</math>
+
==Video Solutions==
 +
https://youtu.be/itz3JyoZQYg
  
 
==See Also==
 
==See Also==

Latest revision as of 13:57, 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

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_16&oldid=231243"