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Difference between revisions of "2008 AMC 10B Problems/Problem 14"

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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Triangle <math>OAB</math> has <math>O=(0,0)</math>, <math>B=(5,0)</math>, and <math>A</math> in the first quadrant.  In addition, <math>\angle ABO=90^\circ</math> and <math>\angle AOB=30^\circ</math>. Suppose that <math>OA</math> is rotated <math>90^\circ</math> counterclockwise about <math>O</math>. What are the coordinates of the image of <math>A</math>? <!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Triangle [katex]OAB[/katex] has [katex]O=(0,0)[/katex], [katex]B=(5,0)[/katex], and [katex]A[/katex] in the first quadrant.  In addition, [katex]\angle ABO=90^\circ[/katex] and [katex]\angle AOB=30^\circ[/katex]. Suppose that [katex]OA[/katex] is rotated [katex]90^\circ[/katex] counterclockwise about [katex]O[/katex]. What are the coordinates of the image of [katex]A[/katex]? <!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
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Revision as of 23:46, 1 November 2025

Problem

Triangle [katex]OAB[/katex] has [katex]O=(0,0)[/katex], [katex]B=(5,0)[/katex], and [katex]A[/katex] in the first quadrant. In addition, [katex]\angle ABO=90^\circ[/katex] and [katex]\angle AOB=30^\circ[/katex]. Suppose that [katex]OA[/katex] is rotated [katex]90^\circ[/katex] counterclockwise about [katex]O[/katex]. What are the coordinates of the image of [katex]A[/katex]?

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

Solution 1

Since $\angle ABO=90^\circ$, and $\angle AOB=30^\circ$, we know that this triangle is one of the Special Right Triangles.

We also know that $B$ is $(5,0)$, so $B$ lies on the x-axis. Therefore, $OB = 5$.

Then, since we know that this is a Special Right Triangle ($30$-$60$-$90$ triangle), we can use the proportion \[\frac{5}{\sqrt 3}=\frac{AB}{1}\] to find $AB$.

We find that \[AB=\frac{5\sqrt 3}{3}\]

That means the coordinates of $A$ are $\left(5,\frac{5\sqrt 3}3\right)$.

Rotate this triangle $90^\circ$ counterclockwise around $O$, and you will find that $A$ will end up in the second quadrant with the coordinates $\left( -\frac{5\sqrt 3}3, 5\right)$, or $\boxed{(B)}$

Note: To better visualize this, one can sketch a diagram.

Solution 2

As $\angle ABO=90^\circ$ and $A$ is in the first quadrant, we know that the $x$ coordinate of $A$ is $5$. We now need to pick a positive $y$ coordinate for $A$ so that we'll have $\angle AOB=30^\circ$.

By the Pythagorean theorem we have $AO^2 = AB^2 + BO^2 = AB^2 + 25$.

By the definition of sine, we have $\frac{AB}{AO} = \sin AOB = \sin 30^\circ = \frac 12$, hence $AO=2\cdot AB$.

Substituting into the previous equation, we get $AB^2 = \frac{25}3$, hence $AB=\frac{5\sqrt 3}3$.

This means that the coordinates of $A$ are $\left(5,\frac{5\sqrt 3}3\right)$.

After we rotate $OA$ $90^\circ$ counterclockwise about $O$, it will be in the second quadrant and have the coordinates $\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }$.

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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