AMC 10 · 2023 · #19

Grade 8 arithmetic

coordinate-geometryperpendicular-bisectorrotation-isometryslope-intercept identify-subproblemsconvert-to-algebra ↑ Prerequisites: coordinate-geometryslope-intercept

📏 Medium solution 💡 3 insights

Problem

The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$ ?

Pick an answer.

(A)

$frac{1}{4}$

(B)

$frac{1}{2}$

(C)

$frac{3}{4}$

(D)

$frac{2}{3}$

(E)

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

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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 $|r - s|$.

Givens: $A = (1, 2)$ rotates to $A' = (3, 1)$; $B = (3, 3)$ rotates to $B' = (4, 3)$; Center of rotation is $P = (r, s)$; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{3}{4}$, (D) $\tfrac{2}{3}$, (E) $1$

Unknowns: $(r, s)$, the coordinates of the center; $|r - s|$

Understand

Restated: 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 $|r - s|$.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

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 $\overline{BB'}$ 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 $\overline{AA'}$ takes a little more arithmetic — midpoint, slope of $\overline{AA'}$, negative-reciprocal slope, point-slope form — which is Tool #13 (Convert to Algebra). Intersecting the two bisectors gives $(r, s)$, and the answer is $|r - s|$.

Execute — Answer: E

#7 Identify Subproblems 8.G.A.1 Step 1

Rotation preserves distances, so the center $P$ is equidistant from $A$ and $A'$, and equidistant from $B$ and $B'$.
The locus of points equidistant from two given points is the perpendicular bisector of the segment joining them.
So $P$ lies on $\ell_A := $ perpendicular bisector of $\overline{AA'}$ and on $\ell_B := $ perpendicular bisector of $\overline{BB'}$.
The center is the intersection $\ell_A \cap \ell_B$.

$$|PA| = |PA'| \;\Rightarrow\; P \in \ell_A; \qquad |PB| = |PB'| \;\Rightarrow\; P \in \ell_B$$

💡 Distance-preserving = equidistant from before and after. Equidistant = on the perpendicular bisector.

#1 Draw a Diagram 6.NS.C.8 Step 2

Subproblem A — find $\ell_B$.
The points $B = (3, 3)$ and $B' = (4, 3)$ have the same $y$-coordinate, so $\overline{BB'}$ is horizontal.
Its midpoint is $(3.5, 3)$, and a perpendicular to a horizontal segment is vertical.
Therefore $\ell_B$ is the vertical line $x = 3.5$, which means $r = 3.5$.

$$\text{mid}(B, B') = (3.5, 3),\quad \ell_B: x = 3.5 \;\Rightarrow\; r = \tfrac{7}{2}$$

💡 When two points share a coordinate, their perpendicular bisector is the axis-aligned line through their midpoint — no calculation needed.

#13 Convert to Algebra 8.EE.B.6 Step 3

Subproblem B — find $\ell_A$.
Midpoint of $\overline{AA'}$: $M = \left(\tfrac{1+3}{2}, \tfrac{2+1}{2}\right) = (2, \tfrac{3}{2})$.
Slope of $\overline{AA'}$: $\tfrac{1 - 2}{3 - 1} = -\tfrac{1}{2}$.
Perpendicular slope: negative reciprocal = $2$.
Using point-slope through $M$: $y - \tfrac{3}{2} = 2(x - 2)$, i.e.
$y = 2x - \tfrac{5}{2}$.

$$\ell_A: y = 2x - \tfrac{5}{2}$$

💡 Perpendicular bisector recipe: midpoint + negative-reciprocal slope. Point-slope form bottles it into one line.

#13 Convert to Algebra 8.EE.C.8 Step 4

Intersect $\ell_A$ and $\ell_B$.
Substitute $x = \tfrac{7}{2}$ into $y = 2x - \tfrac{5}{2}$: $y = 2 \cdot \tfrac{7}{2} - \tfrac{5}{2} = 7 - \tfrac{5}{2} = \tfrac{9}{2}$.
So $P = \left(\tfrac{7}{2}, \tfrac{9}{2}\right)$ and $s = \tfrac{9}{2}$.

$$P = \left(\tfrac{7}{2}, \tfrac{9}{2}\right) \;\Rightarrow\; r = \tfrac{7}{2},\ s = \tfrac{9}{2}$$

💡 Two lines, one shared point — substitute the known coordinate and read off the other.

#7 Identify Subproblems 6.NS.C.7 Step 5

Compute $|r - s|$.

$$|r - s| = \left|\tfrac{7}{2} - \tfrac{9}{2}\right| = |-1| = 1 \;\Rightarrow\; \textbf{(E) }1$$

💡 Difference, absolute value — the final step is a one-line subtraction.

[1] #7 8.G.A.1 Rotation preserves distances, so the center $P$ is equidistant from $A$ and $A'$

[2] #1 6.NS.C.8 Subproblem A — find $\ell_B$. The points $B = (3, 3)$ and $B' = (4, 3)$ have the

[3] #13 8.EE.B.6 Subproblem B — find $\ell_A$. Midpoint of $\overline{AA'}$: $M = \left(\tfrac{1+

[4] #13 8.EE.C.8 Intersect $\ell_A$ and $\ell_B$. Substitute $x = \tfrac{7}{2}$ into $y = 2x - \t

[5] #7 6.NS.C.7 Compute $|r - s|$.

Review

Reasonableness: 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 $\theta$ from $PA$ to $PA'$ matches the angle from $PB$ to $PB'$, confirming a single consistent rotation. The answer $|r - s| = 1$ corresponds to (E).

Alternative: Tool #13 (Convert to Algebra) used directly via the distance equations: $|PA|^2 = |PA'|^2$ expands to $(r - 1)^2 + (s - 2)^2 = (r - 3)^2 + (s - 1)^2$, which simplifies to $4r - 2s = 5$. Similarly $|PB|^2 = |PB'|^2$ expands to $(r - 3)^2 + (s - 3)^2 = (r - 4)^2 + (s - 3)^2$, which simplifies to $2r = 7$, i.e. $r = 3.5$. Substituting back: $14 - 2s = 5$, so $s = 4.5$. Same $|r - s| = 1$.

CCSS standards used (min grade 8)

8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations (Using that a rotation is an isometry, so the center is equidistant from each point and its image — placing $P$ on the perpendicular bisectors of $\overline{AA'}$ and $\overline{BB'}$.)
6.NS.C.8 Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane (Reading off that $\overline{BB'}$ is horizontal (same $y$-coordinate), so its perpendicular bisector is the vertical line $x = 3.5$.)
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points on a non-vertical line (Computing the slope of $\overline{AA'}$ and the negative-reciprocal slope of its perpendicular bisector, then writing the bisector in point-slope form.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Intersecting $x = \tfrac{7}{2}$ with $y = 2x - \tfrac{5}{2}$ to pin down $(r, s) = (\tfrac{7}{2}, \tfrac{9}{2})$.)
6.NS.C.7 Understand ordering and absolute value of rational numbers (Taking $|r - s| = |3.5 - 4.5| = 1$ for the final answer.)

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