Difference between revisions of "2018 AMC 12B Problems/Problem 12"
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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>? | ||
− | < | + | <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</ | + | \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> | ||
− | + | 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 \iff x+\left(\frac{30}{x}+3\right)>10</math> <p> | ||
+ | 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> | ||
+ | </li><p> | ||
+ | <li><math>AB+BC>AC \iff 10+\left(\frac{30}{x}+3\right)>x</math> <p> | ||
+ | We simplify and factor to get <math>(x+2)(x-15)<0,</math> from which <math>0<x<15.</math> | ||
+ | </li><p> | ||
+ | <li><math>AB+AC>BC \iff 10+x>\frac{30}{x}+3</math> <p> | ||
+ | We simplify and factor to get <math>(x+10)(x-3)>0,</math> from which <math>x>3.</math> | ||
+ | </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> | ||
− | + | ~quinnanyc ~MRENTHUSIASM | |
− | |||
− | |||
− | + | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | |
+ | https://youtu.be/OfnHE-KxZJI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==See Also== | ==See Also== |
Latest revision as of 13:25, 29 May 2023
Problem
Side of
has length
. The bisector of angle
meets
at
, and
. The set of all possible values of
is an open interval
. What is
?
Solution
Let By Angle Bisector Theorem, we have
from which
Recall that We apply the Triangle Inequality to
We simplify and complete the square to get
from which
We simplify and factor to get
from which
We simplify and factor to get
from which
Taking the intersection of the solutions gives so the answer is
~quinnanyc ~MRENTHUSIASM
Video Solution (HOW TO THINK CRITICALLY!!!)
~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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