Difference between revisions of "1997 USAMO Problems"
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<math>a_i+a_j \le a_{i+j} \le a_i+a_j+1</math> | <math>a_i+a_j \le a_{i+j} \le a_i+a_j+1</math> | ||
− | for all <math>i, j \ge 1</math> with <math>i+j \le 1997</math>. Show that there exists a real number <math>x</math> such that <math>a_n=\lfloor{nx}\rfloor</math> (the greatest integer <math>\ | + | for all <math>i, j \ge 1</math> with <math>i+j \le 1997</math>. Show that there exists a real number <math>x</math> such that <math>a_n=\lfloor{nx}\rfloor</math> (the greatest integer <math>\le x</math>) for all <math>1 \le n \le 1997</math>. |
[[1997 USAMO Problems/Problem 6|Solution]] | [[1997 USAMO Problems/Problem 6|Solution]] |
Revision as of 22:31, 29 March 2015
Contents
[hide]Day 1
Problem 1
Let be the prime numbers listed in increasing order, and let
be a real number between
and
. For positive integer
, define
where denotes the fractional part of
. (The fractional part of
is given by
where
is the greatest integer less than or equal to
.) Find, with proof, all
satisfying
for which the sequence
eventually becomes
.
Problem 2
Let be a triangle, and draw isosceles triangles
externally to
, with
as their respective bases. Prove that the lines through
perpendicular to the lines
, respectively, are concurrent.
Problem 3
Prove that for any integer , there exists a unique polynomial
with coefficients in
such that
.
Day 2
Problem 4
To clip a convex -gon means to choose a pair of consecutive sides
and to replace them by three segments
and
where
is the midpoint of
and
is the midpoint of
. In other words, one cuts off the triangle
to obtain a convex
-gon. A regular hexagon
of area
is clipped to obtain a heptagon
. Then
is clipped (in one of the seven possible ways) to obtain an octagon
, and so on. Prove that no matter how the clippings are done, the area of
is greater than
, for all
.
Problem 5
Prove that, for all positive real numbers
.
Problem 6
Suppose the sequence of nonnegative integers satisfies
for all with
. Show that there exists a real number
such that
(the greatest integer
) for all
.
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by 1996 USAMO |
Followed by 1998 USAMO | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.