Difference between revisions of "2002 AIME II Problems/Problem 15"
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<cmath>k^2 r^2 - (18k + 12) r + 117 = 0.</cmath>We are told the product of the circles is 68, so by Vieta's formulas, <math>\frac{117}{k^2} = 68.</math> Hence, <math>k^2 = \frac{117}{68},</math> and <math>k = \sqrt{\frac{117}{68}}.</math> | <cmath>k^2 r^2 - (18k + 12) r + 117 = 0.</cmath>We are told the product of the circles is 68, so by Vieta's formulas, <math>\frac{117}{k^2} = 68.</math> Hence, <math>k^2 = \frac{117}{68},</math> and <math>k = \sqrt{\frac{117}{68}}.</math> | ||
− | + | <asy> | |
unitsize(0.25 cm); | unitsize(0.25 cm); | ||
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draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2); | draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2); | ||
− | label(" | + | label("$y = mx$", (8,12*sqrt(221)/49*8), N); |
− | dot(" | + | dot("$(9,6)$", (9,6), NE); |
− | dot(" | + | dot("$(kr,r)$", O[1], N); |
dot(P,red); | dot(P,red); | ||
− | + | </asy> | |
Since the circle is tangent to the line <math>y = mx,</math> the distance from the center <math>(kr,r)</math> to the line is <math>r.</math> We can write <math>y = mx</math> as <math>y - mx = 0,</math> so from the distance formula, | Since the circle is tangent to the line <math>y = mx,</math> the distance from the center <math>(kr,r)</math> to the line is <math>r.</math> We can write <math>y = mx</math> as <math>y - mx = 0,</math> so from the distance formula, |
Revision as of 14:17, 1 March 2021
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 0, to get Then so Since Hence,
See also
2002 AIME II (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.