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 $AC=x.$ By Angle Bisector Theorem, we have $\frac{AB}{AC}=\frac{BD}{CD},$ from which $BD=CD\cdot\frac{AB}{AC}=\frac{30}{x}.$

Recall that $x>0.$ We apply the Triangle Inequality to $\triangle ABC:$

$AC+BC>AB$
We get $\begin{align*} x+\left(\frac{30}{x}+3\right)&>10 \\ x-7+\frac{30}{x}&>0 \\ x^2-7x+30&>0 \\ \left(x-\frac72\right)^2+\frac{71}{4}&>0 \\ x&>0. \end{align*}$

$AB+BC>AC$
We get $\begin{align*} 10+\left(\frac{30}{x}+3\right)&>x \\ x-13-\frac{30}{x}&<0 \\ x^2-13x-30&<0 \\ (x+2)(x-15)&<0 \\ 0<x&<15. \end{align*}$

$AB+AC>BC$
We get $\begin{align*} 10+x&>\frac{30}{x}+3 \\ x+7-\frac{30}{x}&>0 \\ x^2+7x-30&>0 \\ (x+10)(x-3)&>0 \\ x&>3. \end{align*}$

Taking the intersection of the solutions gives $(m,n)=(0,\infty)\cap(0,15)\cap(3,\infty)=(3,15),$ so the answer is $m+n=\boxed{\textbf{(C) }18}.$

~MRENTHUSIASM

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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