Competition · AMC preparation · step 4 of 4

AMC 10 · 2023A · #19

Grade 8 arithmetic
coordinate-geometryperpendicular-bisectorrotation-isometryslope-intercept identify-subproblemsconvert-to-algebra ↑ Prerequisites: coordinate-geometryslope-intercept
📏 Medium solution 💡 3 insights
Problem
The segment from A(1, 2) to B(3, 3) is rotated to the segment from A'(3, 1) to B'(4, 3) about a center P(r, s). Find |r - s|.

Pick an answer.

(A)
$\frac{1}{4}$
(B)
$\frac{1}{2}$
(C)
$\frac{3}{4}$
(D)
$\frac{2}{3}$
(E)
1

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Plot the four points on a grid (Tool #1 Draw a Diagram) — the picture shows that B and B' share the same y-coordinate, so the perpendicular bisector of BB' is the vertical line x = 3.5. That gives r instantly. Tool #7 (Identify Subproblems) splits the work into two perpendicular-bisector tasks. The bisector of AA' takes a little more arithmetic — midpoint, slope of AA', negative-reciprocal slope, point-slope form — which is Tool #13 (Convert to Algebra). Intersecting the two bisectors gives (r, s), and the answer is |r - s|.

1STEP 1

Use equal distances from the center

Rotation preserves distance, so P sits on the perpendicular bisector of AA' and of BB'; the center is where they cross.

|PA| = |PA'| → P ∈ ℓ_A; |PB| = |PB'| → P ∈ ℓ_B
2STEP 2

Find B's perpendicular bisector

B and B' share a y-coordinate, so BB' is horizontal; its perpendicular bisector is the vertical line x = 3.5, giving r = 3.5.

mid(B, B') = (3.5, 3), ℓ_B: x = 3.5 → r = 7/2
3STEP 3

Find A's perpendicular bisector

For AA', midpoint M = (2, 32\frac{3}{2}) and slope -12\frac{1}{2} give perpendicular slope 2; through M the bisector is y = 2x - 52\frac{5}{2}.

ℓ_A: y = 2x - 5/2
4STEP 4

Intersect the two lines

Substituting x = 72\frac{7}{2} into y = 2x - 52\frac{5}{2} gives y = 92\frac{9}{2}, so P = (72\frac{7}{2}, 92\frac{9}{2}) and s = 92\frac{9}{2}.

P = (7/2, 9/2) → r = 7/2, s = 9/2
5STEP 5

Subtract the coordinates

Finally, |r - s| = |72\frac{7}{2} - 92\frac{9}{2}| = 1, which is choice (E).

|r - s| = |7/2 - 9/2| = |-1| = 1 → (E) 1
Answer
1
Direct distance check at P = (3.5, 4.5): |PA|² = (1 - 3.5)² + (2 - 4.5)² = 6.25 + 6.25 = 12.5; |PA'|² = (3 - 3.5)² + (1 - 4.5)² = 0.25 + 12.25 = 12.5 — equal. |PB|² = 0.25 + 2.25 = 2.5; |PB'|² = 0.25 + 2.25 = 2.5 — equal. Both isometry conditions hold exactly. Plotting on graph paper, P sits above the rectangle bounded by the four points, and the rotation angle θ from PA to PA' matches the angle from PB to PB', confirming a single consistent rotation. The answer |r - s| = 1 corresponds to (E).
💡Key takeaway

Rotation preserves distance, so the center of rotation sits on the perpendicular bisector of AA' and on the perpendicular bisector of BB'. The BB' bisector is the vertical line x = 3.5 (free!), and the AA' bisector is y = 2x - 52\frac{5}{2}; their intersection (3.5, 4.5) gives |r - s| = (E) 1.

  • Use equal distances from the center
  • Find B's perpendicular bisector
  • Find A's perpendicular bisector
  • Intersect the two lines
  • Subtract the coordinates

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