Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See Also)
(Changed the variable definition so it is more manageable. Also, used the ordered list command for the three inequalities.)
Line 3: Line 3:
 
Side <math>\overline{AB}</math> of <math>\triangle ABC</math> has length <math>10</math>. The bisector of angle <math>A</math> meets <math>\overline{BC}</math> at <math>D</math>, and <math>CD = 3</math>. The set of all possible values of <math>AC</math> is an open interval <math>(m,n)</math>. What is <math>m+n</math>?
 
Side <math>\overline{AB}</math> of <math>\triangle ABC</math> has length <math>10</math>. The bisector of angle <math>A</math> meets <math>\overline{BC}</math> at <math>D</math>, and <math>CD = 3</math>. The set of all possible values of <math>AC</math> is an open interval <math>(m,n)</math>. What is <math>m+n</math>?
  
−
<cmath>\textbf{(A) }16 \qquad
+
<math>\textbf{(A) }16 \qquad
 
\textbf{(B) }17 \qquad
 
\textbf{(B) }17 \qquad
 
\textbf{(C) }18 \qquad
 
\textbf{(C) }18 \qquad
 
\textbf{(D) }19 \qquad
 
\textbf{(D) }19 \qquad
−
\textbf{(E) }20 \qquad</cmath>
+
\textbf{(E) }20 \qquad</math>
  
 
== Solution ==
 
== Solution ==
 +
Let <math>AC=x.</math> By Angle Bisector Theorem, we have <math>\frac{AB}{AC}=\frac{BD}{CD},</math> from which <math>BD=CD\cdot\frac{AB}{AC}=\frac{30}{x}.</math>
  
−
Let <math>BD = x</math>. Then by Angle Bisector Theorem, we have <math>AC = 30/x</math>. Now, by the triangle inequality, we have three inequalities.
+
Recall that <math>x>0.</math> We apply the Triangle Inequality to <math>\triangle ABC:</math>
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>AC+BC>AB</math> <p>
 +
We get
 +
<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><math>AB+BC>AC</math> <p>
 +
We get
 +
<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><math>AB+AC>BC</math> <p>
 +
We get
 +
<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>
 +
</ol>
 +
Taking the intersection of the solutions gives <cmath>(m,n)=(0,\infty)\cap(0,15)\cap(3,\infty)=(3,15),</cmath> so the answer is <math>m+n=\boxed{\textbf{(C) }18}.</math>
  
−
* <math>10+x+3 > AC</math>, so <math>13+x > 30/x</math>. Solve this to find that <math>x > 2</math>, so <math>AC < 15</math>.
+
~MRENTHUSIASM
−
* <math>AC+10 > x+3</math>, so <math>30/x > x-7</math>. Solve this to find that <math>x < 10</math>, so <math>AC > 3</math>.
 
−
* The third inequality can be disregarded, because <math>30/x > 7-x</math> has no real roots.
 
−
 
 
−
Then our interval is simply <math>(3,15)</math> to get  <math>18</math>  <math>\boxed{C}</math>.
 
  
 
==See Also==
 
==See Also==

Revision as of 05:09, 21 September 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. $AC+BC>AB$

    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 (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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_12&oldid=162554"