2004 AMC 10B Problem 20

Use mass points. Hang weights on the vertices so that $T$ is the balance point of both cevians.

On $AD$: $AT : TD = 3 : 1$, so the weight at $D$ must be three times the weight at $A$. On $BE$: $BT : TE = 4 : 1$, so the weight at $E$ is four times the weight at $B$. The total weight at $T$ must agree along both cevians, so choose
$$A = 5,\ D = 15 \quad(\text{sum } 20), \qquad B = 4,\ E = 16 \quad(\text{sum } 20).$$

$E$ lies on $AC$ and its weight is the sum of the weights at $A$ and $C$, so $C = 16 - 5 = 11$. Consistency check: $D$ lies on $BC$ and $B + C = 4 + 11 = 15$, matching the weight already assigned to $D$.

On segment $BC$, the foot $D$ divides it inversely to the end weights:
$$\frac{CD}{BD} = \frac{\text{weight at } B}{\text{weight at } C} = \frac{4}{11}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{4}{11}}$.

A problem that gives only ratios along two cevians and asks for a ratio on a side is the textbook setting for mass points: each cevian is a lever balanced at $T$, and the lever law converts length ratios into weights and back. The one subtlety is scaling the two levers so that $T$ carries the same total weight from each.

Area ratios. Take $[ABT] = 4$ and $[AET] = 1$ (they share the altitude from $A$ and $BT : TE = 4 : 1$). Then $[ABE] = 5$. Since $AT : TD = 3 : 1$, $[ADE] = \frac{4}{3}[ATE] = \frac{4}{3}$. Triangles $ABE$ and $ADE$ share base $AE$, so their areas are in the ratio of the distances from $B$ and $D$ to line $AC$, which is $BC : DC$. Hence $\frac{BC}{DC} = \frac{5}{4/3} = \frac{15}{4}$, so $BD : DC = 11 : 4$.

• Using masses from the two cevians on inconsistent scales (T must receive the same total from both), or reading off BD/CD = 4/11 and inverting the requested ratio.
• Assuming $D$ and $E$ are midpoints or that $T$ is the centroid, and answering a ratio like $\frac{1}{2}$ or $1$ that is not even offered.