1. September 11-15, 2000
Problem
A1 December 1999
Find
polynomials f(x), g(x), and h(x), if they exist, such that, for all x,
Solution:
Try
f(x), g(x), and h(x) as linear polynomials since the given function is
piece-wise linear. A bit of simple computation yields .
2.
September 18-22, 2000
Problem
A1 December 1975
Supposing
that an integer n is the sum of two triangular numbers,
write
4n+1 as the sum of two squares, , and show how x and y can be expressed in terms of a and b.
Show that, conversely, if , then n is the sum of two triangular numbers.
[Of course a, b, x, and y are understood
to be integers.]
Solution.
Suppose that a and b are integers and . Since are sums of two even
or two odd numbers then each sum is even.
Thus n is an integer. Compute
4n+1. We want to write
this as the sum of two squares of integers.
Experimenting with terms like (a + b), (a – b), and (a + b + 1) yields
Thus
we have the first part if we let x = a + b + 1 and y = a-b.
For the second part we simply solve the
equations above for a and b in terms of x and y. Adding x and y gives x + y = 2a + 1. Hence. We need to show that a is an integer. Since is odd it follows
that one of x and y is even and one is odd.
thus x + y –1 is even and a is an integer. Solving for b by subtracting y from x gives x – y = 2b + 1 or which is also an
integer by the same argument.
3.
September 25-29
In the additive group of ordered pairs of
integers (m,n) [with addition defined componentwise: (m,n) + (m’,n’) = (m + m’,n
+ n’)] consider the subgroup H generated by the three elements
(3,8) (4,-1)
(5,4).
then H has another set of generators of the form (1,b) and (0,a) for some integers a and b with a positive. Find a.
[Elements u and v generate H if every element
h of H can be written as mu + nv for some integers m and n.]
The desired generators must be combinations of the three given elements. (5,4) - (4,-1) yields the element (1,5) which is an element of the form (1,b). The generator of the form (0,a) with a positive is a bit harder to find. (3,8) + (5,4) – 2(4,-1) = (14,0). (0,14) wont work as a generator with (1,5). Try to express (3,8) with just (1,5) and (10,14). If (0,a) is a generator then a must be a factor of 14. Try (0,7).
(3,8) = 3(1,5) – (0,7)
(4,-1) = 4(1,5)-3(0,7)
(5,4) = 5(1,5)-4(0,7).
Thus all the given generators can be expressed in terms of (1,5) and (0,7). All we need to do to finish the problem is to express (0,7) in terms of the given generators.
2(3,8) + (4,-1) – 2(5,4) = (0,7).
4. October 2 – 6
Let there be given nine lattice points
(points with integral coordinates) in three dimensional Euclidean space. Show that there is a lattice point on the
interior of one of the line segments joining two of these points.
Solution.
Consider a lattice point (x, y, z). Classify it according to the parity (oddness
or evenness) of its entries. For
example (2, 4, 11) would be (even, even, odd).
There are exactly eight different ordered triples of even and odd. Thus two of the nine given points must have
the same parity. Suppose that (x, y, z)
and (a, b, c) are two points such that each coordinate has the same
parity. Then are all integers and
are the coordinates of the midpoint of the line joining (x, y, z) and (a, b,
c). This midpoint is a lattice point on the interior of the line segment
joining (x, y, z) and (a, b, c).
5.
October 9 -15
Let f(n) be the sum of the first n terms
of the sequence 0, 1, 1, 2, 2, 3, 3, 4,
4, 5,…, where the nth term is given by
Show
that if x and y are positive integers and then
Solution. Computing f(x) for a few small values of x yields the conjecture
that f(2n) = n2. Prove this
by induction. f(2) = 1 = 12
by direct computation. Assume that f(2n) = n2. Now compute f(2n+2) = f(2n) + a2n+1
+ a2n+2 = n2 + n + (n+1) = (n+1)2. For odd integers x = 2n + 1 we have f(2n+1)=
n2 + n by direct computation.
Case
1. If x+y is even, then so is x-y.
Assume that x is greater than or equal to y.
then we have f(x+y) – f(x-y) = .
Case
2. If x+y is odd, then so is x-y. Then f(x+y) – f(x-y) =
6.
October 16 – 20
Prove that among any ten consecutive
integers at least one is relatively prime to the others.
Solution. If two integers are not relatively prime then they share a prime
factor. If a and b are integers with then any common prime
factor must be less than 10, since any
common factor of a and b also is a factor of a – b. Thus we need to consider the
primes 2, 3, 5, and 7 as possible common factors. Let a set A of ten consecutive integers be given. Exactly five of these numbers are even and
hence none of these is relatively prime to all the rest. Of the remaining five odd integers at most 2
are divisible by 3. Removing these leaves at least three integers in A that do
not share 2 or 3 as a common factor with any other integers in A. Of the three remaining at most one is
divisible by five. Remove it. From the
remaining two integers at most one is divisible by 7. This leaves at least one integer in A which is not divisible by
2, or by 3, or by 5, or by 7. Thus that
integer can share no common factors with any other integer in A.
7. October 23 - 27
A right
circular cone has base radius 1 and height 3.
A cube is inscribed in the cone so that one face of the cube is
contained in the base of the cone. What
is the side length of the cube?
Solution. A cube whose base is in the base of the cone must
touch the cone at the four upper vertices.
If the side of the cube is x then at height x the cone must have
diameter equal to the face diagonal of the cube. Draw a picture of the cone and
set up similar triangles. This will
yield the proportion . Solving the
proportion for x yields .
8. October 30-November 3
Problem
A1 December 1974
Call a set of positive integers “conspiratorial” if no three of them are pairwise relatively prime. (A set of integers is “pairwise relatively prime” if no pair of them has a common divisor greater than 1.) What is the largest number of elements in any “conspiratorial” subset of the integers 1 through 16?
Solution.
Let S be a conspiratorial subset. Then S can contain at most two elements of
the set {1,2,3,5,7,11,13} since this set is pairwise relatively prime. Thus S can have at most two of these and the
other 9 numbers in the set {1, 2, 3, …,16}.
The maximum for S is thus 11 elements. Try the set {2,3,4,6,8,9,10,12,14,15,16}
that has 11 elements. It is
conspiratorial.
9.
November 6 - 10
Let ABC be a triangle with angle A <
Angle C < 90° < angle B. Consider the bisectors of the external
angles at A and B, each measured from the vertex to the opposite side
(extended). Suppose both these line segments are equal to AB. Compute the angle A.
Solution.
Let the measures of angles A, b, and C be
a, b, and c respectively. By drawing a picture and using supplementary angles
one sees that the bisected exterior angles at A and B have measure and . Let D be the point
of intersection of the bisector at A and side BC extended. Then by the condition of the problem ABD is
an isosceles triangle with AD=AB. Angle
ABD is supplementary to angle B so its measure is 180-b. Thus angle ADB is also 180-b (by the
property of base angles of isosceles triangles). Since the sum of the measures of the angles of a triangle is 180
we have 2(180-b) + angle DAB =180. Thus
angle DAB has measure 2b-180. This
angle is one-half of a supplement to angle A hence 2b-180 = . This yields the
equation a + 4b = 540. Very similarly
using the other angle bisector we develop the equation . This gives us a
system of three equations in three unknowns:
Solving
these yields a= 12.
Why
are the conditions on the sizes of A, B, and C given?
10.
November 13-17, 2000
Problem
B1 – December 1999
for
x>0.