2003 AIME I Problems/Problem 15

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

In the following, let the name of a point represent the mass located there. Since we are looking for a ratio, we assume that $AB=120$, $BC=169$, and $CA=260$ in order to simplify our computations.

First, reflect point $F$ over angle bisector $BD$ to a point $F'$.

[asy] 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 = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F);pair Fprime=2*D-F; /* scale down by 100x */ D(MP("A",A,NW)--MP("B",B,N)--MP("C",C)--cycle); D(B--D(MP("D",D))--D(MP("F",F,NE))); D(B--D(MP("M",M)));D(A--MP("F'",Fprime,SW)--D); MP("E",E,NE); D(rightanglemark(F,D,B,4)); MP("390",(M+C)/2); MP("390",(M+C)/2); MP("360",(A+B)/2,NW); MP("507",(B+C)/2,NE); [/asy]

As $BD$ is an angle bisector of both triangles $BAC$ and $BF'F$, we know that $F'$ lies on $AB$. We can now balance triangle $BF'C$ at point $D$ using mass points.

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 both $F$ and $F'$ because BD is a median of triangle $BF'F$ . Therefore, $CB/FB=\frac{289}{240}$.

Now, we reassign mass points to determine $FE/FD$. This setup involves $\triangle CFD$ and transversal $MEB$. For simplicity, put masses of $240$ and $289$ at $C$ and $F$ respectively. 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 of $AC$, we find \[\frac {MD}{CM} = \frac {\frac{169}{289}\cdot 260 - 130}{130} = \frac {49}{289}\] At this point we could find the mass at $D$ but it's unnecessary. \[\frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\cdot\frac {C}{D} = \frac {289}{240}\cdot\frac {49}{289} = \boxed{\frac {49}{240}}\] and the answer is $49 + 240 = \boxed{289}$.

Solution 2

By the Angle Bisector Theorem, we know that $[CBD]=\frac{169}{289}[ABC]$. Therefore, by finding the area of triangle $CBD$, we see that \[\frac{507\cdot BD}{2}\sin\frac{B}{2}=\frac{169}{289}[ABC].\] Solving for $BD$ yields \[BD=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}}.\] Furthermore, $\cos\frac{B}{2}=\frac{BD}{BF}$, so \[BF=\frac{BD}{\cos\frac{B}{2}}=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}\cos\frac{B}{2}}.\] Now by the identity $2\sin\frac{B}{2}\cos\frac{B}{2}=\sin B$, we get \[BF=\frac{4[ABC]}{3\cdot289\sin B}.\] But then $[ABC]=\frac{360\cdot 507}{2}\sin B$, so $BF=\frac{240}{289}\cdot 507$. Thus $BF:FC=240:49$.

Now by the Angle Bisector Theorem, $CD=\frac{169}{289}\cdot 780$, and we know that $MC=\frac{1}{2}\cdot 780$ so $DM:MC=\frac{169}{289}-\frac{1}{2}:\frac{1}{2}=49:289$.

We can now use mass points on triangle CBD. Assign a mass of $240\cdot 49$ to point $C$. Then $D$ must have mass $240\cdot 289$ and $B$ must have mass $49\cdot 49$. This gives $F$ a mass of $240\cdot 49+49\cdot 49=289\cdot 49$. Therefore, $DE:EF=\frac{289\cdot 49}{240\cdot 289}=\frac{49}{240}$, giving us an answer of $\boxed{289}.$

Solution 3

Let $\angle{DBM}=\theta$ and $\angle{DBC}=\alpha$. Then because $BM$ is a median we have $360\sin{(\alpha+\theta)}=507\sin{(\alpha-\theta)}$. Now we know \[\sin{(\alpha+\theta)}=\sin{\alpha}\cos{\theta}+\sin{\theta}\cos{\alpha}=\dfrac{DF\cdot BD}{BF\cdot BE}+\dfrac{DE\cdot BD}{BE\cdot BF}=\dfrac{BD(DF+DE)}{BF\cdot BE}\] Expressing the area of $\triangle{BEF}$ in two ways we have \[\dfrac{1}{2}BE\cdot BF\sin{(\alpha-\theta)}=\dfrac{1}{2}EF\cdot BD\] so \[\sin{(\alpha-\theta)}=\dfrac{EF\cdot BD}{BF\cdot BE}\] Plugging this in we have \[\dfrac{360\cdot BD(DF+DE)}{BF\cdot BE}=\dfrac{507\cdot BD\cdot EF}{BF\cdot BE}\] so $\dfrac{DF+DE}{EF}=\dfrac{507}{360}$. But $DF=DE+EF$, so this simplifies to $1+\dfrac{2DE}{EF}=\dfrac{507}{360}=\dfrac{169}{120}$, and thus $\dfrac{DE}{EF}=\dfrac{49}{240}$, and $m+n=\boxed{289}$.

Solution (Overpowered Projective Geometry!!)

Firstly, angle bisector theorem yields $\frac{CD}{AD} = \frac{507}{360} = \frac{169}{120}$. We're given that $AM=MC$. Therefore, the cross ratio

\[(A,C;M,D) = \frac{AM(CD)}{AD(MC)} = \frac{169}{120}\]

We need a fourth point for this cross ratio to be useful, so let $K = DF \cup BA$. Obviously, $\Delta BFK$ is isosceles and $BD$ is an altitude so $DF = DK$. Therefore,

\[(A,C;M,D) = (F,K;D,E) \implies \frac{FD(EK)}{EF(DK)} = \frac{EK}{EF} = \frac{169}{120}\]

All that's left is to screw around with the ratios:

\[\frac{EK}{EF} = \frac{ED+DF}{EF} = \frac{EF+2DF}{EF} = 1+2(\frac{DF}{EF}) \implies \frac{DF}{EF} = \frac{49}{240} \implies \boxed{289}\]

See also

2003 AIME I (ProblemsAnswer KeyResources)
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