
The
Problems:

The
Solvers:

Question 20001
submitted by Vladimir Novakovski
........

......
1) Andrei Simion, Brooklyn Tech
2) Richard Eager, TJHSST
3) Andrew Dudzik, Lynbrook HS

Question 20002
submitted by Andrei Simion
........

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1) Vladmir Novakovski, TJHSST
2) Gregory Price, TJHSST
3) Peter Ruse, Stuyvesant
4) Eugene Fridman, Glenbrook N
5) Razvan Visan, Romania

Question 20003
submitted by Kamaldeep Gandhi
........

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1) Muhammad Atif, Flushing HS
2) Austin Shapiro, Davis HS
3) Nicolo Menez, Bronx HS of Sci

Question 20004
submitted by David Berger
........

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1) Jeremy Miller, Stoneman Douglas
2) Eugene Fridman, Glenbrook N
3) Mark Bicket, North HS
4) Steve Byrnes, Roxbury Latin
5) Fen Zhao, Hunter College HS
6) Andrei Simion, Brooklyn Tech
7) Charles Wang, IMSA

Question 20005
submitted by Eugene Fridman
........

......
1) Steve Byrnes, Roxbury Latin
2) Jun Lu, Forest Hills
3) Andrei Simion, Brooklyn Tech

Question 20006
submitted by David Shin
........

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1) Eugene Fridman, Glenbrook N
2) Steve Byrnes, Roxbury Latin
3) Fen Zhao, Hunter College HS

Question 20007
submitted by Andrew Dudzik
........

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1) Eugene Fridman, Glenbrook N
2) Austin Shapiro, Davis HS
3) John Mangual, Stuyvesant
4) Maria Shah, Hillcrest HS

Question 20008
submitted by Richard Eager
........

......
1) Austin Shapiro, Davis HS
2) Steven Byrnes, Roxbury Latin

ANSWERS TO 2000
PROBLEMS
Question 20001
 The given sum reduces to 4R/r2, where R and r are the
radii of the circumcircle and incircle, respectively.
Therefore the answer is 22.
Question 20002  One
conjectures that the cosine of each angle must equal 1/2,
leading to the unique solution A=20, B=140, and C=20
degrees.
Question 20003  If we
locate point P on side AB of the triangle with PA=3 and
PB=4, then arrange for PC to pass through the center of the
circumcircle, a limiting value of R=3.7 can be achieved.
Question 20004  First
show that the left hand side can be reduced to abc/K^2,
where K stands for the area of the triangle. It then follows
that R=5.
Question 20005  The
given statement implies that the triangle must be a right
triangle. (Why?) It is now easy to argue that the maximal
area is 25.
Question 20006  An
isosceles triangle with sides of length 25, 25, and 30
produces an area of 300, the minimum possible.
Question 20007  The
cute answer to this question is {101, 103, 109, 127,
181}.
Question 20008  This
question was a difficult exercise in the principle of
inclusionexclusion. There are 7413 even permutations of
eight objects with no fixed points.
......
