Difference between revisions of "1983 AIME Problems/Problem 15"
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The adjoining figure shows two intersecting [[chord]]s 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 [[bisect]]ed 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>? | The adjoining figure shows two intersecting [[chord]]s 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 [[bisect]]ed 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>? | ||
− | [[Image:1983_AIME-15.png]] | + | [[Image:1983_AIME-15.png|200px]] |
== Solution == | == Solution == |
Revision as of 00:41, 17 May 2015
Problem
The adjoining figure shows two intersecting chords in a circle, with on minor arc . Suppose that the radius of the circle is , that , and that is bisected by . Suppose further that is the only chord starting at which is bisected by . It follows that the sine of the minor arc is a rational number. If this fraction is expressed as a fraction in lowest terms, what is the product ?
Solution
Let be any fixed point on circle and let be a chord of circle . The locus of midpoints of the chord is a circle , with diameter . Generally, the circle can intersect the chord at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle is tangent to BC at point N.
Let M be the midpoint of the chord . From right triangle , . Thus, .
Notice that the distance equals (Where is the radius of circle P). Evaluating this, . From , we see that
Next, notice that . We can therefore apply the tangent subtraction formula to obtain , . It follows that , resulting in an answer of .
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |