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 "1983 AIME Problems/Problem 4"

 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is 50 cm, the length of <math>AB</math> is 6 cm, and that of <math>BC</math> is 2 cm. The angle <math>ABC</math> is a right angle. Find the square of the distance (in centimeters) from <math>B</math> to the center of the circle.
 +
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=790&sid=cfd5dae222dd7b8944719b56de7b8bf7[/img]
 +
{{img}}
  
 
== Solution ==
 
== Solution ==
 +
Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. Extend a perpendicular from <math>O</math> to <math>AB</math> and label it <math>D</math>. Additionally, extend a perpendicular from <math>O</math> to the line <math>BC</math>, and label it <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>.
 +
 +
Applying the Pythagorean Theorem, <math>OA^2 = OD^2 + AD^2</math>, and <math>OC^2 = EC^2 + EO^2</math>.
 +
 +
Thus, <math>(\sqrt{50})^2 = y^2 + (6-x)^2</math>, and <math>(\sqrt{50})^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, resulting in an answer of <math>1^2 + 5^2 = 26</math>.
 +
 +
----
 +
 +
* [[1983 AIME Problems/Problem |Previous Problem]]
 +
* [[1983 AIME Problems/Problem |Next Problem]]
 +
* [[1983 AIME Problems|Back to Exam]]
  
 
== See also ==
 
== See also ==
−
* [[1983 AIME Problems]]
+
* [[AIME Problems and Solutions]]
 +
* [[American Invitational Mathematics Examination]]
 +
* [[Mathematics competition resources]]
 +
 
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 23:57, 23 July 2006

Problem

A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is 50 cm, the length of $AB$ is 6 cm, and that of $BC$ is 2 cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=790&sid=cfd5dae222dd7b8944719b56de7b8bf7[/img] Template:Img

Solution

Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Extend a perpendicular from $O$ to $AB$ and label it $D$. Additionally, extend a perpendicular from $O$ to the line $BC$, and label it $E$. Let $OE=x$ and $OD=y$. We're trying to find $x^2+y^2$.

Applying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$, and $OC^2 = EC^2 + EO^2$.

Thus, $(\sqrt{50})^2 = y^2 + (6-x)^2$, and $(\sqrt{50})^2 = x^2 + (y+2)^2$. We solve this system to get $x = 1$ and $y = 5$, resulting in an answer of $1^2 + 5^2 = 26$.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_4&oldid=8218"