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Difference between revisions of "2004 AMC 10B Problems/Problem 20"

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m (Show the letter for the answer)
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Draw the line segment <math>TC</math> to form the two triangles <math>\triangle TDC</math> and <math>\triangle TEC</math>. Let <math>x</math> be the area of <math>\triangle TDC</math>, and <math>y</math> be the area of <math>\triangle TEC</math>. By considering triangles <math>\triangle BTC</math> and <math>\triangle ETC</math>, we obtain <math>(1+x)/y=4</math>, and by considering triangles <math>\triangle ATC</math> and <math>\triangle 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>.
 
Draw the line segment <math>TC</math> to form the two triangles <math>\triangle TDC</math> and <math>\triangle TEC</math>. Let <math>x</math> be the area of <math>\triangle TDC</math>, and <math>y</math> be the area of <math>\triangle TEC</math>. By considering triangles <math>\triangle BTC</math> and <math>\triangle ETC</math>, we obtain <math>(1+x)/y=4</math>, and by considering triangles <math>\triangle ATC</math> and <math>\triangle 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>
+
Therefore <math>\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}</math>
  
 
== Solution (Mass points) ==
 
== Solution (Mass points) ==

Revision as of 14:28, 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 $\triangle BTD$ 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 $\triangle TDC$ and $\triangle TEC$. Let $x$ be the area of $\triangle TDC$, and $y$ be the area of $\triangle TEC$. By considering triangles $\triangle BTC$ and $\triangle ETC$, we obtain $(1+x)/y=4$, and by considering triangles $\triangle ATC$ and $\triangle DTC$, 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 $\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}$

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
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