AMC 10 · 2018 · #12

Grade 10 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

angle-bisector-theorempolygon-inequalityinterval-arithmeticratio-proportion convert-to-algebracaseworkextreme-principle ↑ Prerequisites: angle-bisector-theorempolygon-inequality

📏 Long solution 💡 3 insights

Problem

In a triangle the side AB has length 10. The ray that cuts angle A in half hits BC at D, and the piece CD has length 3. The side AC is not fixed — some lengths work and some do not. The lengths that work fill an open interval. Find the sum of that interval's two endpoints.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

Tool #4 (Introduce a Variable): the problem hands us one free length, so name it x = AC and force every other length to be written in x. The Angle Bisector Theorem does exactly that — it turns CD=3 into a formula for BD, so the third side BC becomes a function of x and the whole triangle is controlled by a single number. Tool #1 (Draw a Diagram): a labelled sketch makes it obvious that BC splits as BD+3, which is the fact the bisector theorem attaches to. Tool #13 (Convert to Algebra): once all three sides are known in x, "does this triangle exist?" becomes three triangle inequalities, i.e. three algebraic inequalities in x. Tool #14 (Extreme Principle): the interval is open because its endpoints are the flattened, degenerate triangles — checking those two extreme cases confirms the endpoints are excluded, which is why the answer is an open interval.

1STEP 1

Sketch and label the split side

The base splits into two pieces.

BC = BD + CD = BD + 3

2STEP 2

Name AC = x and find BD

The bisector's ratio gives the other piece.

BD/3 = 10/x → BD = 30/x → BC = 30/x + 3

3STEP 3

Reduce to a pure side-length question

All three sides now use one letter.

AB = 10, AC = x, BC = 30/x+3

4STEP 4

Inequality 1: AB + AC ＞ BC

The first inequality fixes the lower end.

x² + 7x - 30 ＞ 0 ⇔ (x+10)(x-3) ＞ 0 ⇔ x ＞ 3

5STEP 5

Inequality 2: AB + BC ＞ AC

The second fixes the upper end.

x² - 13x - 30 ＜ 0 ⇔ (x-15)(x+2) ＜ 0 ⇔ x ＜ 15

6STEP 6

Inequality 3: AC + BC ＞ AB

The third always holds.

x² - 7x + 30 = (x-7/2)² + 71/4 ＞ 0 for all x

7STEP 7

Intersect, then read m+n

Adding the ends gives 18.

(m,n) = (3,15) → m+n = 3+15 = 18 → (C)

Answer

Test an interior value: x=6 gives sides AB=10, AC=6, BC=30/6+3=8 — a 6-8-10 right triangle, which certainly exists, and its bisector from A cuts off DC = 8·6/16=3 as required. Test just outside: x=2.9 gives BC=30/2.9+3 ≈ 13.34, while AB+AC=12.9 ＜ 13.34, so no triangle — correctly excluded. And x=15.1 gives BC ≈ 4.99, while AB+BC ≈ 14.99 ＜ 15.1, again no triangle. The endpoints are also memorable: m=3 is exactly the given CD, and at x=n=15 the leftover piece BD=30/15=2 makes BC=5, so AB+BC=AC exactly. Both endpoints land on whole numbers, and 3+15=18 sits in the middle of the choice range 16 to 20 — no choice is close enough to be confused with it.

💡Key takeaway

Name the free side x: the angle bisector rule turns CD=3 into BC=30/x+3, and then the triangle inequality alone decides which x survive — the interval (3,15), so m+n=18.

Sketch and label the split side
Name AC = x and find BD
Reduce to a pure side-length question
Inequality 1: AB + AC ＞ BC
Inequality 2: AB + BC ＞ AC
Inequality 3: AC + BC ＞ AB
Intersect, then read m+n