2018 AMC 12B Problem 12

Below is the professionally curated solution for Problem 12 of the 2018 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12B solutions, or check the answer key.

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Concepts:angle bisector theorem triangle inequality

Difficulty rating: 1820

12.

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 $m+n?$

$16$

$17$

$18$

$19$

$20$

Solution:

Let $q=AC$ and $r=BD.$ The angle bisector theorem gives $\tfrac{q}{3}=\tfrac{10}{r},$ so $r=\tfrac{30}{q}.$

Applying the triangle inequalities to sides $q,$ $10,$ and $3+r$ and substituting $r=\tfrac{30}{q}$ yields $(q-15)(q+2)\lt0$ and $(q-3)(q+10)\gt0$ (the third inequality holds automatically). Together these force $3\lt q\lt15.$

So $(m,n)=(3,15)$ and $m+n=18.$

Thus, the correct answer is C.

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