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

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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} * 4 * \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}</math>
 
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} * 4 * \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}</math>
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==Solution 4 (trig)==
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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} * 4 * 2 * \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 19:56, 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

Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that x is the length of the line from B to D. So, the 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 out $x$ from these two equations. We can find out that $x = 2$, 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 $[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6$, 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 $[ABD] = \frac{1}{2} * 4 * \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}$

Solution 4 (trig)

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 $\frac{1}{2} * 4 * 2 * \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}$.

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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