Difference between revisions of "1983 AIME Problems/Problem 15"
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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>AO</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 <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>AO</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 | + | Let M be the midpoint of the chord <math>BC</math>. From [[right 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 =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3</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 =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3</math> |
Revision as of 11:34, 14 June 2012
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 |