2009 AMC 12A Problem 14

Attempt Problem 14 of the 2009 AMC 12A 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 2009 AMC 12A solutions, or check the answer key.

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14.

A triangle has vertices (0,0),(0, 0), (1,1),(1, 1), and (6m,0),(6m, 0), and the line y=mxy = mx divides the triangle into two triangles of equal area. What is the sum of all possible values of m?m?

13-\dfrac{1}{3}

16-\dfrac{1}{6}

16\dfrac{1}{6}

13\dfrac{1}{3}

12\dfrac{1}{2}

Answer: B
Concepts:coordinate geometrymedian (geometry)Vieta’s Formulas
Difficulty rating: 1820
Solution:

The line y=mxy = mx passes through the vertex (0,0),(0, 0), so it bisects the triangle's area exactly when it passes through the midpoint of the opposite side, joining (1,1)(1, 1) and (6m,0).(6m, 0). That midpoint is (6m+12,12).\left(\dfrac{6m + 1}{2}, \dfrac{1}{2}\right).

Requiring it to satisfy y=mxy = mx gives 12=m6m+12,\frac{1}{2} = m\cdot\frac{6m + 1}{2}, so 6m2+m1=0,6m^2 + m - 1 = 0, that is (3m1)(2m+1)=0.(3m - 1)(2m + 1) = 0.

The possible values are m=13m = \dfrac{1}{3} and m=12,m = -\dfrac{1}{2}, whose sum is 16.-\dfrac{1}{6}.

Thus, the correct answer is B.

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