Solutions to the 68 th William Lowell Putnam
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چکیده
If instead the tangents to C1 at P1, P2 are not perpendicular to L, then we claim there cannot be any point where C1 and C2 are tangent. Indeed, if we count intersections of C1 and C2 (by using C1 to substitute for y in C2, then solving for y), we get at most four solutions counting multiplicity. Two of these are P1 and P2, and any point of tangency counts for two more. However, off of L, any point of tangency would have a mirror image which is also a point of tangency, and there cannot be six solutions. Hence we have now found all possible α. Second solution: For any nonzero value of α, the two conics will intersect in four points in the complex projective plane P(C). To determine the y-coordinates of these intersection points, subtract the two equations to obtain
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Solutions to the 64 th William Lowell Putnam Mathematical Competition Saturday , December 6 , 2003
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A3 Start with a finite sequence a1, a2, . . . , an of positive integers. If possible, choose two indices j < k such that aj does not divide ak, and replace aj and ak by gcd(aj , ak) and lcm(aj , ak), respectively. Prove that if this process is repeated, it must eventually stop and the final sequence does not depend on the choices made. (Note: gcd means greatest common divisor and lcm means leas...
متن کاملSolutions to the 69 th William Lowell Putnam Mathematical Competition
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