2002 AIME II Problems/Problem 15
Problem
Circles and intersect at two points, one of which is , and the product of the radii is . The x-axis and the line , where , are tangent to both circles. It is given that can be written in the form , where , , and are positive integers, is not divisible by the square of any prime, and and are relatively prime. Find .
Solution 1
Let the smaller angle between the -axis and the line be . Note that the centers of the two circles lie on the angle bisector of the angle between the -axis and the line . Also note that if is on said angle bisector, we have that . Let , for convenience. Therefore if is on the angle bisector, then . Now let the centers of the two relevant circles be and for some positive reals and . These two circles are tangent to the -axis, so the radii of the circles are and respectively. We know that the point is a point on both circles, so we have that
Expanding these and manipulating terms gives
It follows that and are the roots of the quadratic
It follows from Vieta's Formulas that the product of the roots of this quadratic is , but we were also given that the product of the radii was 68. Therefore , or . Note that the half-angle formula for tangents is
Therefore
Solving for gives that . It then follows that .
It then follows that . Therefore , , and . The desired answer is then .
Solution 2 (Alcumus)
Let and be the radii of the circles. Then the centers of the circles are of the form and for the same constant since the two centers are collinear with the origin. Since lies on both circles, where represents either radius. Expanding, we get We are told the product of the circles is 68, so by Vieta's formulas, Hence, and
Since the circle is tangent to the line the distance from the center to the line is We can write as so from the distance formula, Squaring both sides, we get so Since we can divide both sides by r, to get Then so Since Hence, Our answer is thus
Solution 3
Let the centers of and be and , respectively, and let the point be .
Because both and are tangent to the x-axis, and both of them pass through , both and must be equidistant from and the x-axis. Therefore, they must both be on the parabola with as the focus and the x-axis as the directrix. Since the coordinates of is , we see that this parabola is the graph of the function
Let be . Because and are both tangent to the x-axis, the y-coordinates of and are and , respectively, so the x-coordinates of and are and . But since and are also on the graph of the function , the x-coordinates of and are also the roots of the equation , and by Vieta's Formulas, their product is . So we have .
We are also given that , so , which means that . Note that the line is the angle bisector of the angle between the line and the x-axis. Therefore, we apply the double-angle formula for tangents and get Thus, the answer is .
Sidenote
The two circles are centered at
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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