
The
Problems:

The
Solvers:

Question 19991
Attributed to Polya, "How To Solve It"
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1) Andrew Dudzik, Lynbrook HS
2) Peter Ruse, Stuyvesant
3) Vladimir Novakovski, TJHSST
4) Richard Eager, TJHSST
5) Gary Sivek, TJHSST
6) Seth Kleinerman, Hunter
7) Abhijit Mehta, Moeller

Question 19992
Contributed by Melanie Wood
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1) Andrew Dudzik, Lynbrook HS
2) Peter Ruse, Stuyvesant
3) Jason Chiu, Laramie HS
4) Vladimir Novakovski, TJHSST
5) Steven Sivek, TJHSST
6) Andrei Simion, Brooklyn Tech
7) Carl Mautner, Miramonte HS

Question 19993
Contributed by Peter Ruse
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1) Andrew Dudzik, Lynbrook HS
2) Seth Kleinerman, Hunter
3) Vladimir Novakovski, TJHSST
4) Andrei Simion, Brooklyn Tech
5) Richard Eager, TJHSST
6) Kamaldeep Gandhi, Stuyvesant
7) John Huss, Paideia School

Question 19994
Contributed by Andrew Dudzik
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1) Vladmir Novakovski, TJHSST
2) Steven Sivek, TJHSST
3) Richard Eager, TJHSST
4) Steve Byrnes, Roxbury Latin
5) Michiru Kaiou
6) Seth Kleinerman, Hunter
7) Gregory Price, TJHSST

Question 19995
Contributed by Santo Diano
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1) Kamaldeep Gandhi, Stuyvesant
2) Abhijit Mehta, Arch. Moeller
3) Jeremy Miller, S. Douglas
4) Vladmir Novakovski, TJHSST
5) Wing Mui, Brooklyn Tech
6) Steven Sivek, TJHSST
7) Richard Eager, TJHSST

Question 19996
Contributed by Anthony Phillips
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1) Peter Ruse, Stuyvesant
2) Andrew Dudzik, Lynbrook HS
3) Vladimir Novakovski, TJHSST
4) Steve Byrnes, Roxbury Latin
5) Gregory Price, TJHSST
6) John Basias, Brooklyn Tech
7) Richard Eager, TJHSST

Question 19997
Contributed by Steven Sivek
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1) Seth Kleinerman, Hunter
2) Peter Ruse, Stuyvesant
3) Andrei Simion, Brooklyn Tech
4) Vladimir Novakovski, TJHSST
5) Andrew Dudzik, Lynbrook HS
6) Richard Eager, TJHSST
7) Steve Byrnes, Roxbury Latin

Question 19998
Contributed by Melanie Wood,
also attributed to Crux Mathematicorum
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1) Andrew Dudzik, Lynbrook
2) Vladimir Novakovski, TJHSST
3) Seth Kleinerman, Hunter
4) Andrei Simion, Brooklyn Tech
5) Peter Ruse, Stuyvesant
6) Steve Byrnes, Roxbury Latin
7) Jennifer Li, Morgantown

ANSWERS TO 1999
PROBLEMS
Question 19991
 there are two possibilities: either M=6 or M=6/19.
Thanks to Steven Sivek of Thomas Jefferson who spotted this
one as an exercise in Polya's classic book, How to Solve
It.
Question 19992  the sum
is 25 C(100,50), i.e. A=25, B=100, and C=50.
Question 19993  the
circumradius is R=15. (Hint: the triangle formed by the
centers of the three original circles is a right triangle.)
Now show that the radius of the small circle nestled between
the others and tangent to all three is r=15/29.
Question 19994  the
sum is 2, which can be proved by induction or by using the
power series expansion for 1/(1x)^51.
Question 19995  a)
3sqrt(2)/2 b) 9/2 c) 1/8 Along with his solution the
proposer included the interesting observation that the roots
of the equation are 1+sin(10), 1+sin(50), and 1sin(70),
where the angles are measured in degrees. Can you verify
this fact?
Question 19996  in
general f(k) equals the number of divisors of k, so the
answers are a) f(20)=6 and b) f(k^2)=5
Question 19997  the
total sum is 3/5. One neat approach is to let S be the sum,
then compute 9S3SS using the Fibonacci recursion to
simplify the result.
Question 19998  one
quickly conjectures that the product is the square root of
three after multiplying them out on a calculator. The proof
is left to the reader.
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