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Difference between revisions of "2003 AIME I Problems/Problem 15"

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(solution (mass points) by me@home, Asymptote [myself])
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== Problem ==
 
== Problem ==
−
In <math> \triangle ABC, AB = 360, BC = 507, </math> and <math> CA = 780. </math> Let <math> M </math> be the midpoint of <math> \overline{CA}, </math> and let <math> D </math> be the point on <math> \overline{CA} </math> such that <math> \overline{BD} </math> bisects angle <math> ABC. </math> Let <math> F </math> be the point on <math> \overline{BC} </math> such that <math> \overline{DF} \perp \overline{BD}. </math> Suppose that <math> \overline{DF} </math> meets <math> \overline{BM} </math> at <math> E. </math> The ratio <math> DE: EF </math> can be written in the form <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m + n. </math>
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In <math> \triangle ABC, AB = 360, BC = 507, </math> and <math> CA = 780. </math> Let <math> M </math> be the [[midpoint]] of <math> \overline{CA}, </math> and let <math> D </math> be the point on <math> \overline{CA} </math> such that <math> \overline{BD} </math> bisects angle <math> ABC. </math> Let <math> F </math> be the point on <math> \overline{BC} </math> such that <math> \overline{DF} \perp \overline{BD}. </math> Suppose that <math> \overline{DF} </math> meets <math> \overline{BM} </math> at <math> E. </math> The ratio <math> DE: EF </math> can be written in the form <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m + n. </math>
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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<center><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);
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</asy></center>
 +
 
 +
For computation, instead consider the triangle as above except <math>AB = 120,BC = 169,CA = 260</math>. 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 <math>C,D,A</math> of <math>120,\,289,\,169</math> respectively. Thus, a mass of <math>\frac {289}{2}</math> belongs at <math>F</math> (seen by reflecting <math>F</math> across <math>BD</math>, to an image which lies on <math>AB</math>).
 +
Having determined <math>CB/CF</math>, we reassign mass points to determine <math>FE/FD</math>. This setup involves <math>\tri CFD</math> and [[transversal]] <math>MEB</math>. For simplicity, put masses of <math>240,289</math> at <math>C,F</math>. To find the mass we should put at <math>D</math>, we compute <math>CM/MD</math>: applying the Angle Bisector Theorem again and using the fact <math>M</math> is a midpoint, we find
 +
<cmath>
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\frac {CM}{MD} = \frac {169\cdot\frac {260}{289} - 130}{130} = \frac {49}{289}
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</cmath>
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At this point we could find the mass at <math>D</math> but it's unnecessary.
 +
<cmath>
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\frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\frac {C}{D} = \frac {289}{240}\frac {49}{289} = \boxed{\frac {49}{240}}
 +
</cmath>
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and the answer is <math>49 + 240 = \boxed{289}</math>.
 +
 
 
== See also ==
 
== See also ==
−
* [[2003 AIME I Problems/Problem 14 | Previous problem]]
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{{AIME box|year=2003|n=I|num-b=14|after=Last question}}
−
* [[2003 AIME I Problems]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 16:18, 11 June 2008

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

[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. Unknown error_msg) 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 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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