Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

m
m (→‎Solution 2: fixed comma)
 
(16 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
== 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 ==
−
{{solution}}
+
 
 +
=== Solution 1 ===
 +
<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
 +
<cmath>\begin{align*}\frac{[DEF]}{[ABC]} &= 1 - \frac{[ADF]}{[ABC]} - \frac{[BDE]}{[ABC]} - \frac{[CEF]}{[ABC]}
 +
\\ &= 1 - p(1-r) - q(1-p) - r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1 \end{align*}</cmath>
 +
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>.
 +
 
 +
=== Solution 2 ===
 +
 
 +
By the barycentric area formula, our desired ratio is equal to
 +
<cmath>\begin{align*}
 +
\begin{vmatrix}
 +
1-p & p & 0 \\
 +
0 & 1-q & q  \\
 +
r & 0 & 1-r \notag
 +
\end{vmatrix} &=1-p-q-r+pq+qr+pr\\
 +
&=1-(p+q+r)+\frac{(p+q+r)^2-(p^2+q^2+r^2)}{2}\\
 +
&=1-\frac{2}{3}+\frac{\frac{4}{9}-\frac{2}{5}}{2}\\
 +
&=\frac{16}{45},
 +
\end{align*}</cmath> so the answer is <math>\boxed{061}</math>
 +
 
 +
=== Solution 3 (Informal) ===
 +
 
 +
Since the only conditions are that <math>p + q + r = \frac{2}{3}</math> and <math>p^2 + q^2 + r^2 = \frac{2}{5}</math>, we can simply let one of the variables be equal to 0. In this case, let <math>p = 0</math>. Then, <math>q + r = \frac{2}{3}</math> and <math>q^2 + r^2</math> = <math>\frac{2}{5}</math>. Note that the ratio between the area of <math>DEF</math> and <math>ABC</math> is equivalent to <math>(1-q)(1-r)</math>. Solving this system of equations, we get <math>q = \frac{1}{3} \pm \sqrt{\frac{4}{45}}</math>, and <math>r = \frac{1}{3} \mp \sqrt{\frac{4}{45}}</math>. Plugging back into <math>(1-q)(1-r)</math>, we get <math>\frac{16}{45}</math>, so the answer is <math>\boxed{061}</math>
 +
 
 +
=== Note ===
 +
 
 +
Because the givens in the problem statement are all regarding the ratios of the sides, the side lengths of triangle <math>ABC</math>, namely <math>13, 15, 17</math>, are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all.
  
 
== 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]]
 +
{{MAA Notice}}

Latest revision as of 19:52, 28 June 2024

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

Solution 1

[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{[BDE]}{[ABC]} - \frac{[CEF]}{[ABC]} \\ &= 1 - p(1-r) - q(1-p) - r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1 \end{align*} 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 $m+n = \boxed{061}$.

Solution 2

By the barycentric area formula, our desired ratio is equal to \begin{align*} \begin{vmatrix} 1-p & p & 0 \\ 0 & 1-q & q \\ r & 0 & 1-r \notag \end{vmatrix} &=1-p-q-r+pq+qr+pr\\ &=1-(p+q+r)+\frac{(p+q+r)^2-(p^2+q^2+r^2)}{2}\\ &=1-\frac{2}{3}+\frac{\frac{4}{9}-\frac{2}{5}}{2}\\ &=\frac{16}{45}, \end{align*} so the answer is $\boxed{061}$

Solution 3 (Informal)

Since the only conditions are that $p + q + r = \frac{2}{3}$ and $p^2 + q^2 + r^2 = \frac{2}{5}$, we can simply let one of the variables be equal to 0. In this case, let $p = 0$. Then, $q + r = \frac{2}{3}$ and $q^2 + r^2$ = $\frac{2}{5}$. Note that the ratio between the area of $DEF$ and $ABC$ is equivalent to $(1-q)(1-r)$. Solving this system of equations, we get $q = \frac{1}{3} \pm \sqrt{\frac{4}{45}}$, and $r = \frac{1}{3} \mp \sqrt{\frac{4}{45}}$. Plugging back into $(1-q)(1-r)$, we get $\frac{16}{45}$, so the answer is $\boxed{061}$

Note

Because the givens in the problem statement are all regarding the ratios of the sides, the side lengths of triangle $ABC$, namely $13, 15, 17$, are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_9&oldid=221240"