AMC 10 · 2022 · #5

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
rotation-isometrycoordinate-geometrytransformations-compositionvector-addition easier-related-problemspatial-visualization ↑ Prerequisites: coordinate-geometryrotation-isometry
📏 Medium solution 💡 2 insights
Problem
A point is turned 270 degrees counterclockwise around another fixed point. Find the coordinates of where it lands.

Pick an answer.

(A)
(-3, -4)
(B)
(0,5)
(C)
(2,-1)
(D)
(4,3)
(E)
(6,-3)
How to solve
Strategy Draw a Diagram

Rotation rules on the coordinate plane are only simple when the center is the origin. So sketch the two points (Tool #1), then shift the whole plane so that C lands on the origin (Tool #9) — that turns this into the easier problem "rotate a point about the origin." Do the turn there, then shift back by the same amount. The only real trap is the direction, so before turning, replace 270^° counterclockwise with the equal-but-shorter 90^° clockwise and check that swap against a landmark point (Tool #17). Tool #3 then confirms the result, since most choices are the wrong distance from the center.

1STEP 1

Plot the point and the center

Plot the point and the centre.

P - C = (-1 - 3, -2 - 1) = (-4, -3)
2STEP 2

Shift the center to the origin

Shift the centre to the origin.

C ↦ (0,0), P ↦ P' = (-4,-3)
3STEP 3

Rewrite the turn as a quarter clockwise

270 counterclockwise is 90 clockwise.

270^° counterclockwise = 360^° - 90^° ≡ 90^° clockwise
4STEP 4

Apply the quarter-turn rule

Apply the quarter-turn rule.

(x,y) ↦ (y,-x): (-4,-3) ↦ (-3, 4)
5STEP 5

Shift back and read the answer

Shifting back gives (0,5).

(-3,4) + (3,1) = (-3+3, 4+1) = (0,5) → (B)
Answer
(0,5)
Two independent checks. First, distance: a rotation cannot change how far the point is from the center. P is 5 units from C, since √(4² + 3²) = 5. Measuring each choice against C = (3,1) gives √(61) for (A), 5 for (B), √(5) for (C), √(5) for (D), and 5 for (E) — so Tool #3 kills (A), (C), (D) immediately and leaves only (B) and (E), which differ exactly by turn direction. Second, direction: P starts down-left of the center; a quarter-turn clockwise should carry a down-left arrow to an up-left arrow, and (0,5) is indeed up and left of (3,1). Choice (E) (6,-3) is down-right of the center — that is what 90^° counterclockwise would give, the trap for anyone who reads 270^° counterclockwise as a quarter-turn the same way. Numerically, the rotation matrix for 270^° counterclockwise is cos 270^° & -sin 270^° ; sin 270^° & cos 270^° = 0 & 1 ; -1 & 0 , and applying it to (-4,-3) gives (0 · (-4) + 1 · (-3), -1 · (-4) + 0 · (-3)) = (-3,4), confirming the step-by-step work.
💡Key takeaway

Slide the plane so the center of rotation sits at the origin, turn there — remembering that 270^° counterclockwise is just 90^° clockwise — then slide back.

  • Plot the point and the center
  • Shift the center to the origin
  • Rewrite the turn as a quarter clockwise
  • Apply the quarter-turn rule
  • Shift back and read the answer

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