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Difference between revisions of "2001 AIME I Problems/Problem 9"

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(solution and a totally misleading asymptote ... somebody please fix p,q,r if possible)
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== Problem ==
 
== Problem ==
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In triangle <math>ABC</math>, <math>AB=13</math>, <math>BC=15</math> and <math>CA=17</math>. Point <math>D</math> is on <math>\overline{AB}</math>, <math>E</math> is on <math>\overline{BC}</math>, and <math>F</math> is on <math>\overline{CA}</math>. Let <math>AD=p\cdot AB</math>, <math>BE=q\cdot BC</math>, and <math>CF=r\cdot CA</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are positive and satisfy <math>p+q+r=2/3</math> and <math>p^2+q^2+r^2=2/5</math>. The ratio of the area of triangle <math>DEF</math> to the area of triangle <math>ABC</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 [[triangle]] <math>ABC</math>, <math>AB=13</math>, <math>BC=15</math> and <math>CA=17</math>. Point <math>D</math> is on <math>\overline{AB}</math>, <math>E</math> is on <math>\overline{BC}</math>, and <math>F</math> is on <math>\overline{CA}</math>. Let <math>AD=p\cdot AB</math>, <math>BE=q\cdot BC</math>, and <math>CF=r\cdot CA</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are positive and satisfy <math>p+q+r=2/3</math> and <math>p^2+q^2+r^2=2/5</math>. The ratio of the area of triangle <math>DEF</math> to the area of triangle <math>ABC</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 ==
−
{{solution}}
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<center><asy>
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/* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */
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 +
real p = 0.5, q = 0.1, r = 0.05;
 +
 +
/* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */
 +
 
 +
pointpen = black; pathpen = linewidth(0.7) + black;
 +
pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C);
 +
D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,N))--cycle);
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D(D(MP("D",D))--D(MP("E",E,NE))--D(MP("F",F,NW))--cycle);
 +
</asy></center>
 +
 
 +
We let <math>[\ldots]</math> denote area; then the desired value is
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<center><math>\frac mn = \frac{[DEF]}{[ABC]} = \frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}</math></center>
 +
Using the [[Triangle#Related Formulae and Theorems|formula]] for the area of a triangle <math>\frac{1}{2}ab\sin C</math>, we find that
 +
<center><math>
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\frac{[ADF]}{[ABC]} = \frac{\frac 12 \cdot p \cdot AB \cdot (1-r) \cdot AC \cdot \sin \angle CAB}{\frac 12 \cdot AB \cdot AC \cdot \sin \angle CAB} = p(1-r)
 +
</math></center>
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and similarly that <math>\frac{[BDE]}{[ABC]} = q(1-p)</math> and <math>\frac{[CEF]}{[ABC]} = r(1-q)</math>. Thus, we wish to find
 +
<center><math>\begin{align*}\frac{[DEF]}{[ABC]} &= 1 - \frac{[ADF]}{[ABC]} - \frac{[DEF]}{[BDE]} - \frac{[CEF]}{[ABC]}
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\\ &= 1 - p(1-r) + q(1-p) + r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1</math></center>
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We know that <math>p + q + r = \frac 23</math>, and also that <math>(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \Longleftrightarrow pq + qr + rp = \frac{\left(\frac 23\right)^2 - \frac 25}{2} = \frac{1}{45}</math>. Substituting, the answer is <math>\frac 1{45} - \frac 23 + 1 = \frac{16}{45}</math>, and <math>m+n = \boxed{061}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2001|n=I|num-b=8|num-a=10}}
 +
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 17:53, 12 June 2008

Problem

In triangle $ABC$, $AB=13$, $BC=15$ and $CA=17$. Point $D$ is on $\overline{AB}$, $E$ is on $\overline{BC}$, and $F$ is on $\overline{CA}$. Let $AD=p\cdot AB$, $BE=q\cdot BC$, and $CF=r\cdot CA$, where $p$, $q$, and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$. The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

[asy] /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ real p = 0.5, q = 0.1, r = 0.05; /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ pointpen = black; pathpen = linewidth(0.7) + black; pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C); D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,N))--cycle); D(D(MP("D",D))--D(MP("E",E,NE))--D(MP("F",F,NW))--cycle); [/asy]

We let $[\ldots]$ denote area; then the desired value is

$\frac mn = \frac{[DEF]}{[ABC]} = \frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}$

Using the formula for the area of a triangle $\frac{1}{2}ab\sin C$, we find that

$\frac{[ADF]}{[ABC]} = \frac{\frac 12 \cdot p \cdot AB \cdot (1-r) \cdot AC \cdot \sin \angle CAB}{\frac 12 \cdot AB \cdot AC \cdot \sin \angle CAB} = p(1-r)$

and similarly that $\frac{[BDE]}{[ABC]} = q(1-p)$ and $\frac{[CEF]}{[ABC]} = r(1-q)$. Thus, we wish to find

$\begin{align*}\frac{[DEF]}{[ABC]} &= 1 - \frac{[ADF]}{[ABC]} - \frac{[DEF]}{[BDE]} - \frac{[CEF]}{[ABC]} \\ &= 1 - p(1-r) + q(1-p) + r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1$ (Error compiling LaTeX. Unknown error_msg)

We know that $p + q + r = \frac 23$, and also that $(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \Longleftrightarrow pq + qr + rp = \frac{\left(\frac 23\right)^2 - \frac 25}{2} = \frac{1}{45}$. Substituting, the answer is $\frac 1{45} - \frac 23 + 1 = \frac{16}{45}$, and $m+n = \boxed{061}$.

See also

2001 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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