2018 AMC 12B Problems/Problem 12

Problem

Side $\overline{AB}$ of $\triangle ABC$ has length $10$ . The bisector of angle $A$ meets $\overline{BC}$ at $D$ , and $CD = 3$ . The set of all possible values of $AC$ is an open interval $(m,n)$ . What is $m+n$ ?

$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }20 \qquad$

Solution

Let $BD = x$ . Then by Angle Bisector Theorem, we have $AC = 30/x$ . Now, by the triangle inequality, we have three inequalities.

$10+x+3 > AC$ , so $13+x > 30/x$ . Solve this to find that $x > 2$ , so $AC < 15$ .
$AC+10 > x+3$ , so $30/x > x-7$ . Solve this to find that $x < 10$ , so $AC > 3$ .
The third inequality can be disregarded, because $30/x > 7-x$ has no real roots.

Then our interval is simply $(3,15)$ to get $18$ $\boxed{C}$ .

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

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2018 AMC 12B Problems/Problem 12

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