Difference between revisions of "1983 AIME Problems/Problem 15"
(→Solution) |
|||
Line 14: | Line 14: | ||
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> | 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 | + | 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>. |
---- | ---- | ||
Revision as of 22:41, 13 March 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
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 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
.