AMC 10 · 2008 · #14

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
thirty-sixty-ninety-trianglerotation-isometrycoordinate-geometry identify-subproblems ↑ Prerequisites: coordinate-geometry
📏 Medium solution 💡 2 insights
Problem
Triangle OAB has O = (0,0), B = (5,0), and A in the first quadrant. In addition, angle ABO = 90 degrees and angle AOB = 30 degrees. Segment OA is rotated 90 degrees counterclockwise about O. What are the coordinates of the image of A?

Pick an answer.

(A)
$\ \left( - \frac {10}{3}\sqrt {3},5\right)$
(B)
$\ \left( - \frac {5}{3}\sqrt {3},5\right)$
(C)
$\ \left(\sqrt {3},5\right)$
(D)
$\ \left(\frac {5}{3}\sqrt {3},5\right)$
(E)
$\ \left(\frac {10}{3}\sqrt {3},5\right)$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The problem lives on the coordinate plane, so drawing it out fixes where each point sits. It naturally splits into two subproblems: first pin down A using the right angle and the 30 degree angle, then apply the rule for a 90 degree counterclockwise rotation. Picturing that turn is what carries A into the second quadrant.

1STEP 1

Stack A above B

The right angle is at B and OB lies on the x-axis, so A stands directly above B: A = (5, h) with h ＞ 0.

A = (5, h), h ＞ 0
2STEP 2

Read the third angle

The three angles sum to 180 degrees, so the angle at A is 60 degrees and OAB is a 30-60-90 right triangle.

∠ OAB = 180° - 90° - 30° = 60°
3STEP 3

Find the height h

In a 30-60-90 triangle the hypotenuse doubles the short leg, so AB = h, OA = 2h, OB = 5, and Pythagoras gives 3h² = 25.

OA = 2h, OA² = OB² + AB² → (2h)² = 5² + h² → 3h² = 25
4STEP 4

Solve and place A

Take the positive root and rationalize: h = 5/√3 = 5√3/3, so A = (5, 5√3/3).

h = √(25/3) = 5/√3 = 5√3/3, A = (5, 5√3/3)
5STEP 5

Rotate 90 degrees CCW

A quarter turn counterclockwise sends (x, y) to (-y, x), so A moves to (-5√3/3, 5) in the second quadrant.

(x,y)↦(-y,x): (5, 5√3/3)↦(-5√3/3, 5)
Answer
( - 5/3√(3),5)
A first-quadrant point (positive x, positive y) rotated 90 degrees counterclockwise should land in the second quadrant (negative x, positive y), and (-5√3/3, 5) does exactly that. The distance from O is preserved by rotation: OA = 2h = 10/√3 ≈ 5.77, and the image also satisfies √((5√3/3)² + 5²) = √(25/3 + 25) ≈ 5.77. Only choices (A) and (B) have a negative x with y = 5, and the magnitude check selects (B) over (A).
💡Key takeaway

To turn a point 90 degrees counterclockwise around the origin, swap its two numbers and flip the sign of the new x: (x, y) becomes (-y, x).

  • Stack A above B
  • Read the third angle
  • Find the height h
  • Solve and place A
  • Rotate 90 degrees CCW

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