Difference between revisions of "2003 AIME I Problems/Problem 15"
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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> | + | 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 == | ||
− | {{ | + | <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); | ||
+ | </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> | ||
+ | \frac {CM}{MD} = \frac {169\cdot\frac {260}{289} - 130}{130} = \frac {49}{289} | ||
+ | </cmath> | ||
+ | At this point we could find the mass at <math>D</math> but it's unnecessary. | ||
+ | <cmath> | ||
+ | \frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\frac {C}{D} = \frac {289}{240}\frac {49}{289} = \boxed{\frac {49}{240}} | ||
+ | </cmath> | ||
+ | and the answer is <math>49 + 240 = \boxed{289}</math>. | ||
+ | |||
== See also == | == See also == | ||
− | + | {{AIME box|year=2003|n=I|num-b=14|after=Last question}} | |
− | |||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 16:18, 11 June 2008
Problem
In and
Let
be the midpoint of
and let
be the point on
such that
bisects angle
Let
be the point on
such that
Suppose that
meets
at
The ratio
can be written in the form
where
and
are relatively prime positive integers. Find
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]](http://latex.artofproblemsolving.com/b/0/7/b07153fb983b9e2b2209286c26983fb7db68ddb8.png)
For computation, instead consider the triangle as above except . 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 of
respectively. Thus, a mass of
belongs at
(seen by reflecting
across
, to an image which lies on
).
Having determined
, we reassign mass points to determine
. This setup involves $\tri CFD$ (Error compiling LaTeX. Unknown error_msg) and transversal
. For simplicity, put masses of
at
. To find the mass we should put at
, we compute
: applying the Angle Bisector Theorem again and using the fact
is a midpoint, we find
At this point we could find the mass at
but it's unnecessary.
and the answer is
.
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 |