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 15"

 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
The adjoining figure shows two intersecting chords in a circle, with <math>B</math> on minor arc <math>AD</math>. Suppose that the radius of the circle is <math>5</math>, that <math>BC=6</math>, and that <math>AD</math> is bisected by <math>BC</math>. Suppose further that <math>AD</math> is the only chord starting at <math>A</math> which is bisected by <math>BC</math>. It follows that the sine of the minor arc <math>AB</math> is a rational number. If this fraction is expressed as a fraction <math>\frac{m}{n}</math> in lowest terms, what is the product <math>mn</math>?
 +
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=794[/img]
 +
 +
{{image}}
  
 
== Solution ==
 
== Solution ==
 +
{{image}}
 +
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=796[/img]
 +
 +
Let <math>A</math> be any [[fixed point]] on [[circle]] <math>O</math> and let <math>AD</math> be a [[chord]] of circle <math>O</math>. The [[locus]] of [[midpoint]]s <math>N</math> of the chord <math>AD</math> is a circle <math>P</math>, with diameter <math>AD</math>. Generally, the circle <math>P</math> can intersect the chord <math>BC</math> at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle <math>P</math> is tangent to BC at point N.
 +
 +
Let M be the midpoint of the chord <math>BC</math> such that <math>BC=3</math>. From right angle triangle <math>OMB</math>, <math>OM = \sqrt{OB^2 - BM^2} =4</math>. Thus, <math>\tan \angle BOM = \frac{BM}{OM} = \frac 3 4</math>.
 +
 +
Notice that the distance <math>OM</math> equals <math>PN + PO \cos AOM = r(1 + \cos AOM)</math> (Where <math>r</math> is the radius of circle P). Evaluating this, <math>\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5</math>. From <math>\cos \angle AOM</math>, we see that <math>\tan \angle AOM = \displaystyle \frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3</math>
 +
 +
Next, notice that <math>\angle AOB = \angle AOM - \angle BOM</math>. We can therefore apply the tangent subtraction formula to obtain , <math>\tan AOM = \displaystyle \frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} = \displaystyle \frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}</math>. It follows that <math>\sin AOM =\displaystyle \frac{7^2}{\sqrt{7^2+24^2}} = \frac{7}{25}</math>, resulting in an answer of <math>7 \cdot 25=175</math>.
 +
----
 +
 +
* [[1983 AIME Problems/Problem 14|Previous 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 23:24, 23 July 2006

Problem

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is $5$, that $BC=6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the minor arc $AB$ is a rational number. If this fraction is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$? [img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=794[/img]

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=796[/img]

Let $A$ be any fixed point on circle $O$ and let $AD$ be a chord of circle $O$. The locus of midpoints $N$ of the chord $AD$ is a circle $P$, with diameter $AD$. Generally, the circle $P$ can intersect the chord $BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $P$ is tangent to BC at point N.

Let M be the midpoint of the chord $BC$ such that $BC=3$. From right angle triangle $OMB$, $OM = \sqrt{OB^2 - BM^2} =4$. Thus, $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$.

Notice that the distance $OM$ equals $PN + PO \cos AOM = r(1 + \cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5$. From $\cos \angle AOM$, we see that $\tan \angle AOM = \displaystyle \frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3$

Next, notice that $\angle AOB = \angle AOM - \angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\tan AOM = \displaystyle \frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} = \displaystyle \frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}$. It follows that $\sin AOM =\displaystyle \frac{7^2}{\sqrt{7^2+24^2}} = \frac{7}{25}$, resulting in an answer of $7 \cdot 25=175$.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_15&oldid=8239"