2018 AMC 12B Problems/Problem 12

Revision as of 05:09, 21 September 2021 by MRENTHUSIASM (talk | contribs) (Changed the variable definition so it is more manageable. Also, used the ordered list command for the three inequalities.)

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. $AC+BC>AB$

    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*}

  2. $AB+BC>AC$

    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*}

  3. $AB+AC>BC$

    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

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

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