2018 AMC 12B Problems/Problem 12

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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 $18$ $\boxed{C}$. (quinnanyc)