2004 AMC 10B Problem 20

P16-20~6 minPrint

Triangle Centers & Cevians Triangles: Area & Pythagorean Set up the equation/formula and compute; no special trick needed

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 $\frac{CD}{BD}$ ?

Show solution without answering

Key insight

Put weights on the corners so the triangle balances at T along both lines. A weighs 5, B weighs 4, C weighs 11, and then CD/BD = 4/11.

Solution

Only ratios are given, so use "mass points": put weights on the points so that each segment balances like a seesaw. On a seesaw, the heavier end sits closer to the balance point. If $AT$ is $3$ times $DT$ , then $D$ must be $3$ times as heavy as $A$ .

1. Line $AD$ . $AT : TD = 3 : 1$ , so weight at $D$ is $3$ times weight at $A$ .
2. Line $BE$ . $BT : TE = 4 : 1$ , so weight at $E$ is $4$ times weight at $B$ .
3. Make the two lines agree at $T$ . The weight at $T$ is the total of the two ends, and it has to be the same for both lines. Try $A = 5$ , $D = 15$ : total $20$ . Try $B = 4$ , $E = 16$ : total $20$ . They match.
4. Find $C$ . $E$ sits on side $AC$ , and its weight is the total of $A$ and $C$ . So $5 + C = 16$ , and $C = 11$ .
5. Check with $D$ . $D$ sits on side $BC$ , so its weight should be $B + C = 4 + 11 = 15$ . That matches step 3.
6. Read off the answer. On side $BC$ , the balance point $D$ is closer to the heavier end. $C$ weighs $11$ and $B$ weighs $4$ , so
$\frac{CD}{BD} = \frac{\text{weight at } B}{\text{weight at } C} = \frac{4}{11}.$
Makes sense: $D$ is nearer to $C$ , the heavier point, so $CD$ is the shorter piece.

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

Why this works

When a triangle problem gives only ratios along lines from the corners to the opposite sides, mass points turn those length ratios into weights, and then back into the ratio you want. The only care point is scaling the weights so the crossing point gets the same total from both lines. Always do a quick sense check: the balance point is closer to the heavier weight.

Alternative approach

Use areas. Triangles that share a height have areas in the same ratio as their bases.

Triangles $ABT$ and $AET$ share the height from $A$ , and $BT : TE = 4 : 1$ . Say their areas are $4$ and $1$ . Then triangle $ABE$ has area $5$ .
Triangles $ATE$ and $ADE$ share the height from $E$ , and $AD$ is $\frac{4}{3}$ of $AT$ (since $AT : TD = 3 : 1$ ). So triangle $ADE$ has area $\frac{4}{3}$ .
Triangles $ABE$ and $ADE$ share the base $AE$ , so their areas compare like the distances from $B$ and from $D$ to line $AC$ . Those distances compare like $BC$ to $DC$ . So $BC : DC = 5 : \frac{4}{3} = 15 : 4$ .
Then $BD = 15 - 4 = 11$ parts and $CD = 4$ parts, so $\frac{CD}{BD} = \frac{4}{11}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 7: Special Parts of a Triangle
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 6: Right Triangles (§6.5)
Introduction to Geometry, Chapter 10: Geometric Inequalities (§10.3)