2003 AIME I Problems/Problem 15

Revision as of 22:39, 14 July 2013 by Mathgeek2006 (talk | contribs) (Solution)

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

Solution 1

[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); /* scale down by 100x */ D(MP("A",A)--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))); 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]

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 $\tri CFD$ (Error compiling LaTeX. ! Undefined control sequence.) 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 \[\frac {MD}{CM} = \frac {169\cdot\frac {260}{289} - 130}{130} = \frac {49}{289}\] 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 $49 + 240 = \boxed{289}$.

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 $\boxed{289}.$

See also

2003 AIME I (ProblemsAnswer KeyResources)
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Problem 14
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