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2008 AMC 10B Problem 14

2008 AMC 10B Problem 14AMC 10 Step by Step

Name: Date: September 13, 2026

Coordinate Geometry · difficulty 2 of 5 · about 3 min · position P11-15

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 $A$ ?

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)
$\left(\frac {10}{3}\sqrt {3},5\right)$

Answer

Key insight

A = (5, 5/sqrt3) from the 30-60-90 triangle, and a 90-degree counterclockwise rotation sends (x, y) to (-y, x).

Solution

The right angle is at $B$ , so $A$ sits directly above $B$ : $A = (5, h)$ with $h = AB$ . In the $30$ - $60$ - $90$ triangle $OBA$ , the leg $OB = 5$ is opposite the $60^\circ$ angle at $A$ , and the leg $AB$ is opposite the $30^\circ$ angle, so
$h = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}.$
Thus $A = \left(5, \frac{5\sqrt3}{3}\right)$ .

A $90^\circ$ counterclockwise rotation about the origin sends $(x, y)$ to $(-y, x)$ . Applying it:
$A \longmapsto \left(-\frac{5\sqrt{3}}{3},\ 5\right).$

The answer is $\boxed{\textbf{(B)}\ \left(-\frac{5}{3}\sqrt{3}, 5\right)}$ .

Why this works

Locate the point first, then transform. The $30$ - $60$ - $90$ ratios give $A$ with no trigonometry beyond $\tan 30^\circ = 1/\sqrt3$ , and the rotation rule $(x, y) \to (-y, x)$ is worth memorizing: a counterclockwise quarter turn sends the positive $x$ -axis to the positive $y$ -axis, which fixes the signs. A quick sanity check: the image must land in the second quadrant.

The trap

Using the hypotenuse length 10/sqrt3 as the height of A, or rotating clockwise, which flips the signs.

Common mistakes

Using the hypotenuse length 10/sqrt3 as the height of A, or rotating clockwise, which flips the signs.
Computing $AB = 5\sqrt{3}$ by matching the leg to the wrong angle (that would make $\angle AOB = 60^\circ$ ).

Techniques

Set up the equation/formula and compute; no special trick needed