2018 AMC 12B Problem 12

Attempt Problem 12 of the 2018 AMC 12B below, then check your answer against the professionally curated solution 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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12.

Side AB\overline{AB} of ABC\triangle ABC has length 10.10. The bisector of angle AA meets BC\overline{BC} at D,D, and CD=3.CD=3. The set of all possible values of ACAC is an open interval (m,n).(m, n). What is m+n?m+n?

1616

1717

1818

1919

2020

Answer: C
Concepts:angle bisector theoremtriangle inequality
Difficulty rating: 1820
Solution:

Let q=ACq=AC and r=BD.r=BD. The angle bisector theorem gives q3=10r,\tfrac{q}{3}=\tfrac{10}{r}, so r=30q.r=\tfrac{30}{q}.

Applying the triangle inequalities to sides q,q, 10,10, and 3+r3+r and substituting r=30qr=\tfrac{30}{q} yields (q15)(q+2)<0(q-15)(q+2)\lt0 and (q3)(q+10)>0(q-3)(q+10)\gt0 (the third inequality holds automatically). Together these force 3<q<15.3\lt q\lt15.

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

Thus, the correct answer is C.

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