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

Difference between revisions of "2004 AMC 10B Problems/Problem 20"
(Another way to solve the problem)
Line 22: Line 22:
 
label("$T$",T,2*WNW);
 
label("$T$",T,2*WNW);
 
</asy>
 
</asy>
 +
 +
== Solution (Triangle Areas) ==
 +
 +
Without loss of generality, we can assume that triangle <math>BTD</math> has area 1. Then triangle <math>BTA</math> has area 3, and triangle <math>ATE</math> has area <math>3/4</math>. The ratio <math>CD/BD</math> is equal to the ratio of the area of triangle <math>ACD</math> to that of triangle <math>ABD</math>, so we need to find the area of quadrilateral <math>TDCE</math>.
 +
 +
Draw the line segment <math>TC</math> to form the two triangles <math>TDC</math> and <math>TEC</math>. Let <math>x</math> be the area of triangle <math>TDC</math>, and <math>y</math> be the area of triangle <math>TEC</math>. By considering triangles <math>BTC</math> and <math>ETC</math>, we obtain <math>(1+x)/y=4</math>, and by considering triangles <math>ATC</math> and <math>DTC</math>, we obtain <math>(3/4+y)/x=3</math>. Solving, we get <math>x=4/11</math>, <math>y=15/44</math>, so the area of quadrilateral <math>TDEC</math> is <math>x+y=31/44</math>.
 +
 +
Therefore <math>\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\frac{4}{11}}</math>
 +
  
 
== Solution (Mass points) ==
 
== Solution (Mass points) ==

Revision as of 09:38, 24 January 2011

Problem

In $\triangle ABC$ points $D$ and $E$ lie on $BC$ and $AC$, respectively. If $AD$ and $BE$ intersect at $T$ so that $AT/DT=3$ and $BT/ET=4$, what is $CD/BD$?


$\mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12}$


[asy] unitsize(1cm); defaultpen(0.8); pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,S); label("$T$",T,2*WNW); [/asy]

Solution (Triangle Areas)

Without loss of generality, we can assume that triangle $BTD$ has area 1. Then triangle $BTA$ has area 3, and triangle $ATE$ has area $3/4$. The ratio $CD/BD$ is equal to the ratio of the area of triangle $ACD$ to that of triangle $ABD$, so we need to find the area of quadrilateral $TDCE$.

Draw the line segment $TC$ to form the two triangles $TDC$ and $TEC$. Let $x$ be the area of triangle $TDC$, and $y$ be the area of triangle $TEC$. By considering triangles $BTC$ and $ETC$, we obtain $(1+x)/y=4$, and by considering triangles $ATC$ and $DTC$, we obtain $(3/4+y)/x=3$. Solving, we get $x=4/11$, $y=15/44$, so the area of quadrilateral $TDEC$ is $x+y=31/44$.

Therefore $\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\frac{4}{11}}$


Solution (Mass points)

The presence of only ratios in the problem essentially cries out for mass points.

As per the problem, we assign a mass of $1$ to point $A$, and a mass of $3$ to $D$. Then, to balance $A$ and $D$ on $T$, $T$ has a mass of $4$.

Now, were we to assign a mass of $1$ to $B$ and a mass of $4$ to $E$, we'd have $5T$. Scaling this down by $4/5$ (to get $4T$, which puts $B$ and $E$ in terms of the masses of $A$ and $D$), we assign a mass of $\frac{4}{5}$ to $B$ and a mass of $\frac{16}{5}$ to $E$.

Now, to balance $A$ and $C$ on $E$, we must give $C$ a mass of $\frac{16}{5}-1=\frac{11}{5}$.

Finally, the ratio of $CD$ to $BD$ is given by the ratio of the mass of $B$ to the mass of $C$, which is $\frac{4}{5}\cdot\frac{5}{11}=\boxed{\textbf{(D)}\ \frac{4}{11}}$.

Solution (Coordinates)

Affine transformations preserve ratios of distances, and for any pair of triangles there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any $\triangle ABC$, and we just need to compute it for any single triangle.

We can choose the points $A=(-3,0)$, $B=(0,4)$, and $D=(1,0)$. This way we will have $T=(0,0)$, and $E=(0,-1)$. The situation is shown in the picture below:

[asy] unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,S); label("$T$",T,NW); label("$3$",A--T,N); label("$4$",B--T,W); label("$1$",D--T,N); label("$1$",E--T,W); [/asy]

The point $C$ is the intersection of the lines $BD$ and $AE$. The points on the first line have the form $(t,4-4t)$, the points on the second line have the form $(t,-1-t/3)$. Solving for $t$ we get $t=15/11$, hence $C=(15/11,-16/11)$.

The ratio $CD/BD$ can now be computed simply by observing the $x$ coordinates of $B$, $C$, and $D$:

\[\frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\frac 4{11}}\]

See also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_20&oldid=36406"