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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>.
+
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>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 19:42, 12 January 2023

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 Solutions

https://youtu.be/itz3JyoZQYg

https://youtu.be/IVHTUjOPePY

~savannahsolver

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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