AMC 10 · 2016 · #12

Grade 10 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
angle-bisector-theoremratio-proportionarea-trianglesinradius identify-subproblemsconvert-to-algebra ↑ Prerequisites: angle-bisector-theoremratio-proportion
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
Two angle bisectors of a known triangle cross inside it. Find how one of them is divided.

Pick an answer.

(A)
3:2
(B)
5:3
(C)
2:1
(D)
7:3
(E)
5:2
How to solve
Strategy Identify Subproblems

One theorem does all the work here, and it gets used twice. The angle bisector theorem says a bisector drawn from a vertex cuts the opposite side into pieces whose ratio equals the ratio of the two sides that meet at that vertex. So the problem splits into two copies of the same subproblem: first use the bisector from A inside triangle ABC to learn where D sits on BC, then re-read the picture so that AD is a side of the smaller triangle ABD and BF is the bisector from B in that triangle. The second copy of the theorem hands over AF : FD directly. Nothing about E or about the length of AD is needed.

1STEP 1

Split BC with the bisector from A

The first bisector splits a side by a ratio.

BD/DC = AB/AC = 6/8 = 3/4
2STEP 2

Turn the ratio into actual lengths

The perimeter turns that into real lengths.

BD = 3k, DC = 4k, 3k + 4k = 7 → k = 1 → BD = 3, DC = 4
3STEP 3

Re-read AD as a side, not a cevian

The cevian rereads as a side of a smaller triangle.

ray BD = ray BC → ∠ ABD = ∠ ABC → BF bisects ∠ ABD
4STEP 4

Use the same theorem a second time

The same theorem applies a second time.

AF/FD = AB/BD = 6/3 = 2/1
5STEP 5

Confirm with the incircle, then choose

An area check confirms 2 to 1, choice (D).

AF/FD = [ABF]/[DBF] = (1/2 · AB · r)/(1/2 · BD · r) = AB/BD = 6/3 = 2 → AF : FD = 2 : 1
Answer
2:1
The ratio should be larger than 1 because F is pulled toward the shorter of the two sides at B, and BD = 3 is much shorter than AB = 6; the answer 2 : 1 has the right direction and the right size. Exact coordinates confirm it: placing A = (0,0) and B = (6,0) forces C = (17/4, 7√(15)/4) from AC = 8 and BC = 7. Then D, three-sevenths of the way from B to C, is (21/4, 3√(15)/4), giving AD = 1/4√(441 + 135) = 6. The incenter is (7A + 8B + 6C)/21 = (7/2, √(15)/2), so AF = 1/2√(49 + 15) = 4 and FD = 6 - 4 = 2, i.e. 4 : 2 = 2 : 1. Every other choice is ruled out: AB/BD is exactly 2, so 3 : 2, 5 : 3, 7 : 3, and 5 : 2 all fail. A common trap is to use AB/BC = 6/7 or the untouched side CA = 8 in the second application; the correct second application uses BD = 3, not BC = 7.
💡Key takeaway

An angle bisector cuts the far side in the ratio of the two sides beside it, so find BD first and then use the very same rule inside the smaller triangle ABD.

  • Split BC with the bisector from A
  • Turn the ratio into actual lengths
  • Re-read AD as a side, not a cevian
  • Use the same theorem a second time
  • Confirm with the incircle, then choose

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