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 $\triangle ABC:$

$AC+BC>AB \iff x+\left(\frac{30}{x}+3\right)>10$
We simplify and complete the square to get $\left(x-\frac72\right)^2+\frac{71}{4}>0,$ from which $x>0.$

$AB+BC>AC \iff 10+\left(\frac{30}{x}+3\right)>x$
We simplify and factor to get $(x+2)(x-15)<0,$ from which $0<x<15.$

$AB+AC>BC \iff 10+x>\frac{30}{x}+3$
We simplify and factor to get $(x+10)(x-3)>0,$ from which $x>3.$

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

~quinnanyc ~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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