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 == | ||
− | 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>. | + | 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 == | ||
− | {{ | + | <center><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></center> | ||
+ | |||
+ | We let <math>[\ldots]</math> denote area; then the desired value is | ||
+ | <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> | ||
+ | \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> | ||
+ | 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]} | ||
+ | \\ &= 1 - p(1-r) + q(1-p) + r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1</math></center> | ||
+ | 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 ,
,
and
. Point
is on
,
is on
, and
is on
. Let
,
, and
, where
,
, and
are positive and satisfy
and
. The ratio of the area of triangle
to the area of triangle
can be written in the form
, where
and
are relatively prime positive integers. Find
.
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]](http://latex.artofproblemsolving.com/a/a/5/aa532b18510960c694fe8213246465f219c7d942.png)
We let denote area; then the desired value is
![$\frac mn = \frac{[DEF]}{[ABC]} = \frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}$](http://latex.artofproblemsolving.com/f/0/3/f031606bd0480fc8cb6ee9da5a70ec5dccdfcecb.png)
Using the formula for the area of a triangle , we find that
![$\frac{[ADF]}{[ABC]} = \frac{\frac 12 \cdot p \cdot AB \cdot (1-r) \cdot AC \cdot \sin \angle CAB}{\frac 12 \cdot AB \cdot AC \cdot \sin \angle CAB} = p(1-r)$](http://latex.artofproblemsolving.com/c/c/2/cc254e39ee583e6f0e7845a84cb9f1e473868472.png)
and similarly that and
. Thus, we wish to find
We know that , and also that
. Substituting, the answer is
, and
.
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 |