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

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 $\triangle ABC:$

$AC+BC>AB$
We get $\begin{align*} x+\left(\frac{30}{x}+3\right)&>10 \\ x-7+\frac{30}{x}&>0 \\ x^2-7x+30&>0 \\ \left(x-\frac72\right)^2+\frac{71}{4}&>0 \\ x&>0. \end{align*}$

$AB+BC>AC$
We get $\begin{align*} 10+\left(\frac{30}{x}+3\right)&>x \\ x-13-\frac{30}{x}&<0 \\ x^2-13x-30&<0 \\ (x+2)(x-15)&<0 \\ 0<x&<15. \end{align*}$

$AB+AC>BC$
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

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

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

@@ 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>?
-<cmath>\textbf{(A) }16 \qquad
+<math>\textbf{(A) }16 \qquad
 \textbf{(B) }17 \qquad
 \textbf{(C) }18 \qquad
 \textbf{(D) }19 \qquad
-\textbf{(E) }20 \qquad</cmath>
+\textbf{(E) }20 \qquad</math>
 == 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*}
++\left(\frac{30}{x}+3\right)&>x \\
+x-13-\frac{30}{x}&<0 \\
+x^2-13x-30&<0 \\
+(x+2)(x-15)&<0 \\
+<x&<15.
+\end{align*}</cmath>
+</li><p>
+  <li><math>AB+AC>BC</math> <p>
+We get
+<cmath>\begin{align*}
++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>.
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Difference between revisions of "2018 AMC 12B Problems/Problem 12"

Revision as of 05:09, 21 September 2021

Problem

Solution

See Also