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, $\angle ABO=90^\circ$ and $\angle AOB=30^\circ$ . Suppose that $OA$ is rotated $90^\circ$ 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.

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

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Toolkit + CCSS Solution

Understand

Restated: A triangle has O=(0,0) and B=(5,0), with A in the first quadrant. The angle at B is a right angle and the angle at O is 30 degrees. Find A, then rotate segment OA by 90 degrees counterclockwise about O and give the new coordinates of A.

Givens: O = (0,0) and B = (5,0), so OB lies along the x-axis with length 5; Angle ABO = 90 degrees (right angle at B); Angle AOB = 30 degrees (angle at O); A is in the first quadrant; OA is rotated 90 degrees counterclockwise about O

Unknowns: The coordinates of the image of A after the rotation

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships

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.

Execute — Answer: B

#1 Draw a Diagram 6.G.A.3 Step 1

Stack A above B

The right angle is at B, and OB runs along the x-axis.
A line through B that is perpendicular to the x-axis is vertical, so A sits directly above B and shares B's x-coordinate.
Write A = (5, h) for some height h > 0.

$$A = (5,\ h),\quad h > 0$$

💡 Perpendicular to a horizontal line means straight up, so A stacks right on top of B.

#7 Identify Subproblems 8.G.A.5 Step 2

Read the third angle

The three angles of a triangle add to 180 degrees.
With 90 degrees at B and 30 degrees at O, the angle at A is 60 degrees.
So triangle OAB is a 30-60-90 right triangle, whose side lengths follow a fixed pattern.

$$\angle OAB = 180^\circ - 90^\circ - 30^\circ = 60^\circ$$

💡 Knowing two angles hands you the third for free, and it labels the triangle's shape.

#7 Identify Subproblems 8.G.B.7 Step 3

Find the height h

In a 30-60-90 triangle the hypotenuse is twice the shortest leg.
The shortest leg AB is opposite the 30 degree angle, so it equals h; the hypotenuse OA equals 2h; and OB = 5 is the third side.
Apply the Pythagorean theorem to the right triangle at B.

$$OA = 2h,\quad OA^2 = OB^2 + AB^2 \;\Rightarrow\; (2h)^2 = 5^2 + h^2 \;\Rightarrow\; 3h^2 = 25$$

💡 The right angle lets the Pythagorean theorem tie the three sides into one equation.

#7 Identify Subproblems 8.EE.A.2 Step 4

Solve and place A

Solve 3h^2 = 25 for the positive height, then rationalize the denominator.
This gives the exact coordinates of A in the first quadrant.

$$h = \sqrt{\tfrac{25}{3}} = \frac{5}{\sqrt3} = \frac{5\sqrt3}{3},\qquad A = \left(5,\ \frac{5\sqrt3}{3}\right)$$

💡 Taking the square root undoes the squaring, and clearing the root from the bottom keeps the value tidy.

#17 Visualize Spatial Relationships 8.G.A.3 Step 5

Rotate 90 degrees CCW

A 90 degree counterclockwise rotation about the origin sends any point (x, y) to (-y, x).
Apply this to A.
The image lands in the second quadrant, matching answer choice (B).

$$(x,y)\mapsto(-y,x):\quad \left(5,\ \tfrac{5\sqrt3}{3}\right)\mapsto\left(-\tfrac{5\sqrt3}{3},\ 5\right)$$

💡 A quarter turn counterclockwise swaps the coordinates and flips the sign of the new x-value.

[1] #1 6.G.A.3 The right angle is at B, and OB runs along the x-axis. A line through B that is

[2] #7 8.G.A.5 The three angles of a triangle add to 180 degrees. With 90 degrees at B and 30 d

[3] #7 8.G.B.7 In a 30-60-90 triangle the hypotenuse is twice the shortest leg. The shortest le

[4] #7 8.EE.A.2 Solve 3h^2 = 25 for the positive height, then rationalize the denominator. This

[5] #17 8.G.A.3 A 90 degree counterclockwise rotation about the origin sends any point (x, y) to

Review

Reasonableness: A first-quadrant point (positive x, positive y) rotated 90 degrees counterclockwise should land in the second quadrant (negative x, positive y), and (-5\sqrt3/3, 5) does exactly that. The distance from O is preserved by rotation: OA = 2h = 10/\sqrt3 \approx 5.77, and the image also satisfies \sqrt{(5\sqrt3/3)^2 + 5^2} = \sqrt{25/3 + 25} \approx 5.77. Only choices (A) and (B) have a negative x with y = 5, and the magnitude check selects (B) over (A).

Alternative: Use polar thinking. A makes angle arctan((5\sqrt3/3)/5) = 30 degrees with the x-axis at radius r = 10/\sqrt3. Rotating adds 90 degrees, giving angle 120 degrees at the same radius: (r\cos120^\circ, r\sin120^\circ) = (-5\sqrt3/3, 5), the same result.

CCSS standards used (min grade 8)

6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices (Placing O, B, and A on the coordinate plane and reading A's x-coordinate from the vertical side)
8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (Using the 180-degree angle sum to find the 60-degree angle and classify the 30-60-90 triangle)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Relating OB, AB, and the hypotenuse OA to solve for the height of A)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Solving 3h^2 = 25 and simplifying the square root to 5\sqrt3/3)
8.G.A.3 Describe the effect of rotations and reflections on figures using coordinates (Applying the 90-degree counterclockwise rule (x, y) to (-y, x) to find the image of A)

⭐ 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).

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Stack A above B

Read the third angle

Find the height h

Solve and place A

Rotate 90 degrees CCW

Review

CCSS standards used (min grade 8)

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