Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m (→‎Solution 1)
 
(5 intermediate revisions by 2 users not shown)
Line 20: Line 20:
 
We also know that <math>B</math> is <math>(5,0)</math>, so <math>B</math> lies on the x-axis. Therefore, <math>OB = 5</math>.
 
We also know that <math>B</math> is <math>(5,0)</math>, so <math>B</math> lies on the x-axis. Therefore, <math>OB = 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{AB}{1}</cmath> to find <math>AB</math>.
+
Then, since we know that this is a Special Right Triangle (<math>30</math>-<math>60</math>-<math>90</math> triangle), we can use the proportion <cmath>\frac{5}{\sqrt 3}=\frac{AB}{1}</cmath> to find <math>AB</math>.
  
 
We find that <cmath>AB=\frac{5\sqrt 3}{3}</cmath>
 
We find that <cmath>AB=\frac{5\sqrt 3}{3}</cmath>
Line 27: Line 27:
  
 
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
 
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)}</math>.
+
<math>\left( -\frac{5\sqrt 3}3, 5\right)</math>, or <math>\boxed{(B)}</math>
  
 
Note: To better visualize this, one can sketch a diagram.
 
Note: To better visualize this, one can sketch a diagram.

Latest revision as of 16:25, 28 October 2024

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 $B$ 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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_14&oldid=230758"