Difference between revisions of "1983 AIME Problems/Problem 15"
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− | Let <math>A</math> be any | + | 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</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>. | 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>. |
Revision as of 18:48, 30 June 2015
Contents
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
.
Solution 2
The above solution works, but is quite messy and somewhat difficult to follow. This solution provides a diagram to scale, and the motivation behind the solution.
First of all, where did the statement " is the only chord starting at
and bisected by
" come from? What is its significance in this problem? What is the criterion for this statement to be true?
We consider the locus of midpoints of the chords from . It is well known that this is the circle with diameter
, where
is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with ratio
with center
. Thus, the locus is the result of the dilation with ratio
of circle
with center
. Let the center of this circle be
.
Aha! Now we see. is bisected by
if they cross at some point
on the circle. Moreover, since
is the only chord,
must be tangent to the circle
.
The rest of this problem is straight forward.
Our goal is to find where
is the midpoint of
. Then we have
and
.
Let
be the projection of
onto
, and similarly
be the projection of
onto
. Then it remains to find
so we can use the sine addition formula.
As is a radius of circle
,
, and similarly,
. Since
,
. Thus,
.
From here, we see that is a dilation of
about center
with ratio
, so
.
Lastly, we apply the formula:
Thus, our answer is .
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 |