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Difference between revisions of "2018 AMC 12B Problems/Problem 12"

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Recall that <math>x>0.</math> We apply the Triangle Inequality to <math>\triangle ABC:</math>
 
Recall that <math>x>0.</math> We apply the Triangle Inequality to <math>\triangle ABC:</math>
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
−
   <li><math>AC+BC>AB</math> <p>
+
   <li><math>AC+BC>AB \iff x+\left(\frac{30}{x}+3\right)>10</math> <p>
−
We get
+
We simplify and complete the square to get <math>\left(x-\frac72\right)^2+\frac{71}{4}>0,</math> from which <math>x>0.</math>
−
<cmath>\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*}</cmath>
 
 
</li><p>
 
</li><p>
−
   <li><math>AB+BC>AC</math> <p>
+
   <li><math>AB+BC>AC \iff 10+\left(\frac{30}{x}+3\right)>x</math> <p>
−
We get
+
We simplify and factor to get <math>(x+2)(x-15)<0,</math> from which <math>0<x<15.</math>
−
<cmath>\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*}</cmath>
 
 
</li><p>
 
</li><p>
−
   <li><math>AB+AC>BC</math> <p>
+
   <li><math>AB+AC>BC \iff 10+x>\frac{30}{x}+3</math> <p>
−
We get
+
We simplify and factor to get <math>(x+10)(x-3)>0,</math> from which <math>x>3.</math>
−
<cmath>\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*}</cmath>
 
 
</li><p>
 
</li><p>
 
</ol>
 
</ol>

Revision as of 11:34, 12 October 2021

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 (Problems • Answer Key • Resources)
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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