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

m (Problem)
(10 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
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>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>
  
 
<math>
 
<math>
Line 14: Line 14:
 
</math>
 
</math>
  
==Solution==
+
==Solution 1==
 +
 
 +
Since <math>\angle ABO=90^\circ</math>, and <math>\angle AOB=30^\circ</math>, we know that this triangle is one of the [[Special Right Triangles]].
 +
 
 +
We also know that <math>A</math> is <math>(5,0)</math>, so <math>A</math> lies on the x-axis. Therefore, <math>OA = 5</math>.
 +
 
 +
Then, since we know that this is a Special Right Triangle(30-60-90 triangle), we can use the proportion <cmath>\frac{5}{\sqrt 3}=\frac{x}{1}</cmath> to find <math>AB</math>.
 +
 
 +
We find that <cmath>AB=\frac{5\sqrt 3}{3}</cmath>
 +
 
 +
That means that the coordinates of <math>A</math> are <math>\left(5,\frac{5\sqrt 3}3\right)</math>.
 +
 
 +
Rotate this triangle <math>90^\circ</math> counterclockwise around <math>O</math>, and you will find that <math>A</math> will end up in the second quadrant with the coordinates
 +
<math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) \text{or B.}}</math>.
 +
 
 +
==Solution 2==
  
 
As <math>\angle ABO=90^\circ</math> and <math>A</math> in the first quadrant, we know that the <math>x</math> coordinate of <math>A</math> is <math>5</math>. We now need to pick a positive <math>y</math> coordinate for <math>A</math> so that we'll have <math>\angle AOB=30^\circ</math>.  
 
As <math>\angle ABO=90^\circ</math> and <math>A</math> in the first quadrant, we know that the <math>x</math> coordinate of <math>A</math> is <math>5</math>. We now need to pick a positive <math>y</math> coordinate for <math>A</math> so that we'll have <math>\angle AOB=30^\circ</math>.  
Line 28: Line 43:
 
After we rotate <math>A</math> <math>90^\circ</math> counterclockwise about <math>O</math>, it will get into the second quadrant and have the coordinates  
 
After we rotate <math>A</math> <math>90^\circ</math> counterclockwise about <math>O</math>, it will get into the second quadrant and have the coordinates  
 
<math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }</math>.
 
<math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }</math>.
 +
So the answer is <math>\boxed{\text{B}}</math>.
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2008|ab=B|num-b=13|num-a=15}}
 +
{{MAA Notice}}

Revision as of 11:19, 18 May 2020

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 $A$?

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

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 $A$ is $(5,0)$, so $A$ lies on the x-axis. Therefore, $OA = 5$.

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

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

That means that 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 $\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) \text{or B.}}$.

Solution 2

As $\angle ABO=90^\circ$ and $A$ 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 $A$ $90^\circ$ counterclockwise about $O$, it will get into the second quadrant and have the coordinates $\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }$. So the answer is $\boxed{\text{B}}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png