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

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==Solution 1==
 
==Solution 1==
  
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Because <math>\overline{BD} + \overline{CD} = 5,</math> we can see that when we draw a line from point <math>B</math> to imaginary point <math>D</math> that line applies to both triangles. Let us say that <math>x</math> is that line. Perimeter of <math>\triangle{ABD}</math> would be <math>\overline{AD} + 4 + x</math>, while the perimeter of <math>\triangle{ACD}</math> would be <math>\overline{AD} + 3 + (5 - x)</math>. Notice that we can find <math>x</math> from these two equations by setting them equal and then canceling <math>\overline{AD}</math>. We find that <math>x = 2</math>, and because the height of the triangles is the same, the ratio of the areas is <math>2:3</math>, so that means that the area of <math>\triangle ABD = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>.
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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==
 
==Solution 2==
  
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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>.
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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>
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==Solution 3==
 
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Since point <math>D</math> is on line <math>BC</math>, it will split it into <math>CD</math> and <math>DB</math>. Let <math>CD = 5 - x</math> and <math>DB = x</math>. Triangle <math>CAD</math> has side lengths <math>3, 5 - x, AD</math> and triangle <math>DAB</math> has side lengths <math>x, 4, AD</math>. Since both perimeters are equal, we have the equation <math>3 + 5 - x + AD = 4 + x + AD</math>. Eliminating <math>AD</math> and solving the resulting linear equation gives <math>x = 2</math>. Draw a perpendicular from point <math>D</math> to <math>AB</math>. Call the point of intersection <math>F</math>. Because angle <math>ABC</math> is common to both triangles <math>DBF</math> and <math>ABC</math>, and both are right triangles, both are similar. The hypotenuse of triangle <math>DBF</math> is 2, so the altitude must be <math>6/5</math> Because <math>DBF</math> and <math>ABD</math> share the same altitude, the height of <math>ABD</math> therefore must be <math>6/5</math>. The base of <math>ABD</math> is 4, so <math>[ABD] = \frac{1}{2} \cdot 4 \cdot \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}</math>.
 
  
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==Solution 4==
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~megaboy6679
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Using any preferred method, realize <math>BD = 2</math>. Since we are given a 3-4-5 right triangle, we know the value of <math>\sin(\angle ABC) = \frac{3}{5}</math>. Since we are given <math>AB = 4</math>, apply the Sine Area Formula to get <math>\frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>.
 
  
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==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
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==Video Solution by the RMM Club==
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https://youtu.be/TRY34buacYU
 
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~Education, the Study of Everything
 
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== Video Solution by OmegaLearn ==
 
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https://youtu.be/51K3uCzntWs?t=1663
 
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~ pi_is_3.14
 
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==Video Solutions==
 
 
https://youtu.be/itz3JyoZQYg
 
https://youtu.be/itz3JyoZQYg
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https://youtu.be/IVHTUjOPePY
 
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~savannahsolver
 
  
 
==See Also==
 
==See Also==
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{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 16:37, 8 June 2025

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 Solution by the RMM Club

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

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