AMC 10 · 2014 · #20

Grade 11 geometry-2d

Archetype Paper Fold-and-Cut Symmetry └ Sub-archetype Composition of Reflections of a Pictured Figure

reflection-symmetryreflection-unfoldingpath-length-comparisonlaw-of-cosines reflection-unfoldingspatial-visualizationextreme-principle ↑ Prerequisites: reflection-symmetrylaw-of-cosines

📏 Long solution 💡 3 insights

Problem

A bent path bounces between two sides of an angle with fixed endpoints. Find its smallest total length.

Pick an answer.

(A)

$6\sqrt 3+3$

(B)

$\dfrac{27}2$

(C)

$8\sqrt 3$

(D)

(E)

$3\sqrt 3+9$

How to solve

Strategy Visualize Spatial Relationships

The sum BE+DE+CD is one path from the fixed point B to the fixed point C that is forced to touch side AC (at E) and side AB (at D). A path that bounces off lines is shortest when you flatten the bounces out, so the move is to reflect the two fixed endpoints across the two sides (Tool #17): reflection keeps every length the same but replaces the bent path by a path between two new fixed points, where straight-is-shortest applies. That gives a lower bound. A lower bound alone is not the answer, though: it counts only if some legal D and E actually reach it (Tool #14 for the extreme case). So the plan has two halves — bound the sum below by one segment, then prove that segment really does cut across AB and AC inside them. The lengths come from the Law of Cosines and the Law of Sines applied to a single triangle (Tools #7 and #4).

1STEP 1

See it as one bouncing path

The three segments form one path.

BE + DE + CD = length of the path B → E → D → C

2STEP 2

Reflect the two fixed endpoints

Reflecting the endpoints keeps every length.

EB = EB' for all E ∈ AC, DC = DC' for all D ∈ AB → BE + DE + CD = B'E + ED + DC'

3STEP 3

Unfold the angle to 120 degrees

That unfolds the angle to 120 degrees.

∠ C'AB' = ∠ C'AB + ∠ BAC + ∠ CAB' = 40° + 40° + 40° = 120° ＜ 180° AC' = 6, AB' = 10

4STEP 4

Straight beats bent: a lower bound

A straight line gives the lower bound.

BE + DE + CD = B'E + ED + DC' ≥ B'D + DC' ≥ B'C'

5STEP 5

Measure the straight segment

The law of cosines measures it as 14.

B'C'² = 10² + 6² - 2 · 10 · 6 cos 120° = 100 + 36 + 60 = 196 B'C' = 14 → BE + DE + CD ≥ 14

6STEP 6

Check the bound is reachable in order

The crossings come in the right order.

Directions from A: AC' = 0°, AB = 40°, AC = 80°, AB' = 120° Segment C'B' meets ray AB at D₀, then ray AC at E₀

7STEP 7

Confirm the crossings land on the sides

They land inside the sides, so 14 is reached.

AE₀ = (10sin∠ AB'C')/(sin(40° + ∠ AB'C')) = 30√3/(13sin 40° + 3√3cos 40°) ＜ 60√3/(13 + 3√6) ≈ 5.11 ＜ 6 AD₀ = (6sin∠ AC'B')/(sin(40° + ∠ AC'B')) = 30√3/(11sin 40° + 5√3cos 40°) ＜ 60√3/(11 + 5√6) ≈ 4.47 ＜ 10 → min(BE+DE+CD) = 14 → (D)

Answer

Sanity-bound the answer from both directions. Taking D=E=A is legal and gives BE+DE+CD = 10+0+6 = 16, so the true minimum is at most 16; the winning value 14 sits just under that, which is believable for a path that is still forced to touch both sides. From below, any route from B to C is at least BC = √(100+36-120cos 40°) ≈ 6.64, and 14 is comfortably above that, as it must be since the bouncing costs extra. Then plug the optimal positions back in: AD₀ ≈ 3.791 and AE₀ ≈ 4.212 give BE ≈ 7.294, DE ≈ 2.766, CD ≈ 3.940, totalling 14.000 — the bound is met, not just approached. One warning the checks make loud: the choices are 13.39, 13.50, 13.86, 14, 14.20, all within a hair of each other, so a decimal estimate cannot pick between them. The exact 196 = 14² is what decides it, and that perfect square is itself a signal the 120° configuration is the intended one.

💡Key takeaway

Mirror the fixed ends across the two sides, and the bouncing path unfolds into one straight segment whose length is the answer — but only after you check that segment really does cross both sides.

See it as one bouncing path
Reflect the two fixed endpoints
Unfold the angle to 120 degrees
Straight beats bent: a lower bound
Measure the straight segment
Check the bound is reachable in order
Confirm the crossings land on the sides