Track / Stage 3 / 115 of 260 #115 of 2000

Problem 115

AMC 10/12, early questions

Geometry Difficulty 3.4 Find the answer

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

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

Your answer

Official solution

We use the square bracket notation $[\cdot]$ to denote area.
WLOG, we can assume $[\triangle BTD] = 1$ . Then $[\triangle BTA] = 3$ , and $[\triangle ATE] = 3/4$ . We have $CD/BD = [\triangle ACD]/[\triangle ABD]$ , 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 $x = [\triangle TDC]$ , and $y = [\triangle TEC]$ . 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 $\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.