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 $\triangle ABC:$

  1. $AC+BC>AB \iff x+\left(\frac{30}{x}+3\right)>10$

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

  2. $AB+BC>AC \iff 10+\left(\frac{30}{x}+3\right)>x$

    We simplify and factor to get $(x+2)(x-15)<0,$ from which $0<x<15.$

  3. $AB+AC>BC \iff 10+x>\frac{30}{x}+3$

    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

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS