PROBLEMS FOR JULY
Solutions should be submitted to
Dr. Valeria Pandelieva
708 - 195 Clearview Avenue
Ottawa, ON K1Z 6S1
Solution to these problems should be
postmarked no later than August 30, 2000.
Notes: An acute triangle has all of its
angles less than 90°. The orthocentre
of a triangle is the intersection point of its altitudes.
Points are collinear iff they lie on a straight line.
Is it possible to divide the natural numbers
1, 2, ¼, n into two groups, such that the squares
of the members in each group have the same sum, if
(a) n = 40000; (b) n = 40002? Explain your answer.
Given any six irrational numbers, prove that
there are always three of them, say a, b, c, for which
a + b, b + c and c + a are irrational.
The natural numbers x1, x2, ¼, x100
are such that
Prove that at least two of the numbers are equal.
Let R be a rectangle with dimensions 11 ×12.
Find the least natural number n for which it is possible to cover
R with n rectangles, each of size 1 ×6 or 1 ×7,
with no two of these having a common interior point.
Given 21 points on the circumference of a circle,
prove that at least 100 of the arcs determined by pairs of
these points subtend an angle not exceeding 120° at
ABC is an acute triangle with orthocentre H.
Denote by M and N the midpoints of the respective segments
AB and CH, and by P the intersection point of the
bisectors of angles CAH and CBH. Prove that the points
M, N and P are collinear.