AMC 10 Step by Step

2023 AMC 10A Problem 19

P16-20~5 minPrint

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

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Key insight

A rotation keeps each point the same distance from the center. So the center is on the perpendicular bisector of A and A', and of B and B'.

Solution

1. What does a rotation keep? When you spin a shape around a point PP, every point stays the same distance from PP. So PP is as far from AA as from A′A', and as far from BB as from B′B'.
2. Where are such points? The points that are the same distance from two given points form a line: the perpendicular bisector (it goes through the midpoint, at a right angle). So PP is on the perpendicular bisector of BB′BB' and on the perpendicular bisector of AA′AA'.
3. Start with the easy pair, BB and B′B'. B=(3,3)B = (3, 3) and B′=(4,3)B' = (4, 3) are side by side on the line y=3y = 3. The midpoint is (72,3)\left(\frac72, 3\right), and the line crossing at a right angle is straight up and down:
x=72.x = \tfrac72.
4. Now A=(1,2)A = (1, 2) and A′=(3,1)A' = (3, 1). The midpoint is (2,32)\left(2, \frac32\right). Going from AA to A′A' is 22 right and 11 down, so the slope is −12-\frac12. A perpendicular line has slope 22 (flip it and change the sign). So the bisector is
y=32+2(x−2).y = \tfrac32 + 2(x - 2).
5. Where do they cross? Put x=72x = \frac72 in: y=32+2×32=32+3=92y = \frac32 + 2 \times \frac32 = \frac32 + 3 = \frac92. So P=(72,92)P = \left(\frac72, \frac92\right).
6. Check one pair of distances (squared): from PP to AA, (52)2+(52)2=252\left(\frac52\right)^2 + \left(\frac52\right)^2 = \frac{25}{2}; from PP to A′A', (12)2+(72)2=14+494=252\left(\frac12\right)^2 + \left(\frac72\right)^2 = \frac{1}{4} + \frac{49}{4} = \frac{25}{2}. ✓
7. Finally ∣r−s∣=∣72−92∣=1|r - s| = \left|\frac72 - \frac92\right| = 1.

The answer is (E) 1\boxed{\textbf{(E)}\ 1}.

Why this works

The center of a rotation is the one point that does not move, and every other point keeps its distance from it. So each point and its image give one line the center must lie on, and two such lines pin the center down. You never need to know the angle of the turn.

Alternative approach

Write the equal distances as equations with P=(r,s)P = (r, s). From PB2=PB′2PB^2 = PB'^2:
(r−3)2+(s−3)2=(r−4)2+(s−3)2.(r-3)^2 + (s-3)^2 = (r-4)^2 + (s-3)^2.
The ss parts are the same on both sides, and the r2r^2 terms cancel: −6r+9=−8r+16-6r + 9 = -8r + 16, so r=72r = \frac72. Doing the same with PA2=PA′2PA^2 = PA'^2 gives 4r−2s=54r - 2s = 5, so s=92s = \frac92, and ∣r−s∣=1|r - s| = 1.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

A rotation about PP keeps every point at the same distance from PP. So PA=PA′PA = PA', meaning PP lies on the perpendicular bisector of AA′AA', and likewise PP lies on the perpendicular bisector of BB′BB'.

Bisector of BB′BB': B(3,3)B(3,3) and B′(4,3)B'(4,3) are on a horizontal line with midpoint (72,3)\left(\frac72, 3\right), so the bisector is the vertical line x=72x = \frac72.

Bisector of AA′AA': A(1,2)A(1,2) and A′(3,1)A'(3,1) have midpoint (2,32)\left(2, \frac32\right) and slope −12-\frac12, so the bisector has slope 22:
y−32=2(x−2).y - \tfrac32 = 2(x - 2).
At x=72x = \frac72 this gives y=32+3=92y = \frac32 + 3 = \frac92.

So P=(72,92)P = \left(\frac72, \frac92\right) and ∣r−s∣=∣72−92∣=1|r - s| = \left|\frac72 - \frac92\right| = 1.

Check: PA2=(52)2+(52)2=252PA^2 = \left(\frac52\right)^2 + \left(\frac52\right)^2 = \frac{25}{2} and PA′2=(12)2+(72)2=252PA'^2 = \left(\frac12\right)^2 + \left(\frac72\right)^2 = \frac{25}{2}.

The answer is (E) 1\boxed{\textbf{(E)}\ 1}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.