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 | + | <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> | |
− | |||
− | x+\left(\frac{30}{x}+3\right) | ||
− | |||
− | |||
− | \left(x-\frac72\right)^2+\frac{71}{4} | ||
− | x | ||
− | |||
</li><p> | </li><p> | ||
− | <li><math>AB+BC>AC | + | <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> | |
− | |||
− | 10+\left(\frac{30}{x}+3\right) | ||
− | |||
− | |||
− | (x+2)(x-15) | ||
− | 0<x | ||
− | |||
</li><p> | </li><p> | ||
− | <li><math>AB+AC>BC | + | <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> | |
− | |||
− | 10+x | ||
− | |||
− | |||
− | (x+10)(x-3) | ||
− | x | ||
− | |||
</li><p> | </li><p> | ||
</ol> | </ol> |
Revision as of 11:34, 12 October 2021
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.