Side AB of △ ABC has length 10. The bisector of angle A meets BC at D, and CD=3. The set of all possible values of AC is an open interval (m, n). What is m+n?

2018 AMC 12B Problem 12

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12.

Side

\overline{AB}

\triangle ABC

has length

10.

The bisector of angle

A

meets

\overline{BC}

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

By the angle bisector theorem,

\dfrac{AC}{CD}=\dfrac{AB}{BD}

Let

AC=q,

BD=\dfrac{30}{q};

then apply the three triangle inequalities

Let

q=AC

and

r=BD.

The angle bisector theorem gives

\tfrac{q}{3}=\tfrac{10}{r},

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.

(m,n)=(3,15)

and

m+n=18.

Thus, the correct answer is C.