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2003 AIME I Problems/Problem 15

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Problem

In \triangle ABC, AB = 360, BC = 507, and CA = 780. Let M be the midpoint of \overline{CA}, and let D be the point on \overline{CA} such that \overline{BD} bisects angle ABC. Let F be the point on \overline{BC} such that \overline{DF} \perp \overline{BD}. Suppose that \overline{DF} meets \overline{BM} at E. The ratio DE: EF can be written in the form m/n, where m and n are relatively prime positive integers. Find m + n.

Solution

size(400); pointpen = black; pathpen = black+linewidth(0.7); pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da...

For computation, instead consider the triangle as above except AB = 120,BC = 169,CA = 260. In the following, let the name of a point represent the mass located there.

By the Angle Bisector Theorem, we can place mass points on C,D,A of 120,\,289,\,169 respectively. Thus, a mass of \frac {289}{2} belongs at F (seen by reflecting F across BD, to an image which lies on AB). Having determined CB/CF, we reassign mass points to determine FE/FD. This setup involves \tri CFD and transversal MEB. For simplicity, put masses of 240,289 at C,F. To find the mass we should put at D, we compute CM/MD: applying the Angle Bisector Theorem again and using the fact M is a midpoint, we find \frac {CM}{MD} = \frac {169\cdot\frac {260}{289} - 130}{130} = \frac {49}{289} At this point we could find the mass at D but it's unnecessary. \frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\frac {C}{D} = \frac {289}{240}\frac {49}{289} = \boxed{\frac {49}{240}} and the answer is 49 + 240 = \boxed{289}.

See also

2003 AIME I (ProblemsResources)
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Problem 14
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