AMC 10 · 2021 · #11

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

reflection-unfoldingcoordinate-geometrypythagorean-theoremline-symmetry easier-related-problemwork-backwards ↑ Prerequisites: coordinate-geometry

📏 Long solution 💡 2 insights

Problem

A laser sits at one point. Its beam goes straight, bounces once off the y-axis, bounces once off the x-axis, and then arrives at another point. The path is three straight segments joined at the two bounce points. Find the total length of those three segments.

Pick an answer.

(A)

$2\sqrt{10}$

(B)

$5\sqrt2$

(C)

$10\sqrt2$

(D)

$15\sqrt2$

(E)

$10\sqrt5$

How to solve

Strategy Solve an Easier Related Problem

A three-piece bent path is hard to measure directly. Tool #9 turns it into an easier problem: unfold the bounces. A mirror bounce off the y-axis has exactly the same length as the straight continuation into the mirror image, so instead of bending the beam we bend the target. Reflect the endpoint D across the x-axis, then across the y-axis, and the whole zig-zag straightens into ONE segment from A to the twice-reflected target. Tool #1 (Draw a Diagram) keeps the four points straight; Tool #4 (Introduce a Variable) names the bounce points B=(0,b) and C=(c,0); Tool #7 (Identify Subproblems) splits the job into "unfold" then "measure one distance"; Tool #11 (Work Backwards) recovers those actual bounce points from the straightened line to confirm the path is legal; Tool #3 (Eliminate Possibilities) matches the result to the five choices.

1STEP 1

Name the two bounce points

The path is a sum of three segments.

B = (0,b), C = (c,0), total = AB + BC + CD

2STEP 2

Reflect instead of bending

Reflect instead of bending the beam.

reflect over x-axis: (x,y) ↦ (x,-y) reflect over y-axis: (x,y) ↦ (-x,y)

3STEP 3

Unfold both bounces

Two reflections make it one straight line.

D = (7,5) → D' = (7,-5) → D'' = (-7,-5) AB + BC + CD = AD''

4STEP 4

Measure the straightened path

Measure between the two points.

AD'' = √((3-(-7))² + (5-(-5))²) = √(10² + 10²) = √(200) = 10√(2)

5STEP 5

Find the real bounce points

Work backwards to the bounces.

y = x + 2 → B = (0,2), C = (2,0)

6STEP 6

Add the three segments

Adding confirms ten root two.

AB = 3√(2), BC = 2√(2), CD = 5√(2) AB + BC + CD = (3 + 2 + 5)√(2) = 10√(2) → (C)

Answer

10√2

Three checks agree. First, size: the beam must dive from height 5 down to the x-axis and climb back up, so it travels far more than the 4 units separating A and D directly; 10√(2) ≈ 14.1 is the right order of magnitude, while 5√(2) ≈ 7.07 is too small to reach the x-axis and return. Second, the mirror law holds at both bounces: arriving at B = (0,2) the direction is (-3,-3) and leaving it is (3,-3) — the y-axis flipped the horizontal part and left the vertical part alone, which is exactly a reflection; arriving at C = (2,0) the direction is (2,-2) and leaving it is (2,2), the x-axis flipping the vertical part. Third, the two computations match: unfolding gave AD'' = 10√(2) and adding the real segments gave 3√(2) + 2√(2) + 5√(2) = 10√(2).

💡Key takeaway

This AMC 12 problem needs nothing past Grade 8: reflections on the coordinate plane and the distance formula. A bouncing beam is just a straight beam seen in mirrors — so mirror the TARGET instead of bending the beam. Reflect (7,5) over the x-axis, then over the y-axis, to get (-7,-5), and the whole zig-zag becomes one straight line from (3,5) of length √(10²+10²) = 10√(2), choice (C).

Name the two bounce points
Mirror the beam instead of bending it
Unfold both bounces into one line
Measure the straightened path
Work backwards to find the real bounces
Add the three segments and pick the choice