Difference between revisions of "1983 AIME Problems/Problem 15"
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Next, notice that <math>\angle AOB = \angle AOM - \angle BOM</math>. We can therefore apply the tangent subtraction formula to obtain , <math>\tan AOB = \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 AOB =\displaystyle \frac{7^2}{\sqrt{7^2+24^2}} = \frac{7}{25}</math>, resulting in an answer of <math>7 \cdot 25=175</math>. | Next, notice that <math>\angle AOB = \angle AOM - \angle BOM</math>. We can therefore apply the tangent subtraction formula to obtain , <math>\tan AOB = \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 AOB =\displaystyle \frac{7^2}{\sqrt{7^2+24^2}} = \frac{7}{25}</math>, resulting in an answer of <math>7 \cdot 25=175</math>. | ||
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== See also == | == See also == | ||
+ | {{AIME box|year=1983|num-b=14|after=Last Question}} | ||
* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] |
Revision as of 13:18, 6 May 2007
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
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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 such that . From right angle 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 |