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

(Created page with "== Problem == 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>...")
 
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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>?
  
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<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>
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\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>
  
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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.
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Recall that <math>x>0.</math> We apply the Triangle Inequality to <math>\triangle ABC:</math>
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<ol style="margin-left: 1.5em;">
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  <li><math>AC+BC>AB \iff x+\left(\frac{30}{x}+3\right)>10</math> <p>
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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>
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</li><p>
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  <li><math>AB+BC>AC \iff 10+\left(\frac{30}{x}+3\right)>x</math> <p>
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We simplify and factor to get <math>(x+2)(x-15)<0,</math> from which <math>0<x<15.</math>
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</li><p>
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  <li><math>AB+AC>BC \iff 10+x>\frac{30}{x}+3</math> <p>
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We simplify and factor to get <math>(x+10)(x-3)>0,</math> from which <math>x>3.</math>
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</li><p>
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</ol>
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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>
  
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* <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>.
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~quinnanyc ~MRENTHUSIASM
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* <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>.
 
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* The third inequality can be disregarded, because <math>30/x > 7-x</math> has no real roots.
 
  
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Then our interval is simply <math>(3,15)</math> to get  <math>18</math>  <math>\boxed{C}</math>. (quinnanyc)
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==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/OfnHE-KxZJI
 +
 
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
Line 23: Line 37:
 
{{AMC12 box|year=2018|ab=B|num-b=11|num-a=13}}
 
{{AMC12 box|year=2018|ab=B|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 12:25, 29 May 2023

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

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/OfnHE-KxZJI

~Education, the Study of Everything

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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