1st British Mathematical Olympiad 1965 Problems



1st British Mathematical Olympiad 1965 Problems

1.  Sketch f(x) = (x2 + 1)/(x + 1). Find all points where f '(x) = 0 and describe the behaviour when x or f(x) is large.
2.  X, at the centre a circular pond. Y, at the edge, cannot swim, but can run at speed 4v. X can run faster than 4v and can swim at speed v. Can X escape?


3.  Show that np - n is divisible by p for p = 3, 7, 13 and any integer n.
4.  What is the largest power of 10 dividing 100 x 99 x 98 x ... x 1?
5.  Show that n(n + 1)(n + 2)(n + 3) + 1 is a square for n = 1, 2, 3, ... .
6.  The fractional part of a real is the real less the largest integer not exceeding it. Show that we can find n such that the fractional part of (2 + √2)n > 0.999 .
7.  What is the remainder on dividing x + x3 + x9 + x27 + x81 + x243 by x - 1? By x2 - 1?
8.  For what real b can we find x satisfying: x2 + bx + 1 = x2 + x + b = 0?
9.  Show that for any real, positive x, y, z, not all equal, we have:   (x + y)(y + z)(z + x) > 8 xyz.
10.  A chord length √3 divides a circle C into two arcs. R is the region bounded by the chord and the shorter arc. What is the largest area of rectangle than can be drawn in R? 

Solutions
Problem 1
Sketch f(x) = (x2 + 1)/(x + 1). Find all points where f '(x) = 0 and describe the behaviour when x or f(x) is large.
Solution
x = -1 and x - 1 are asymptotes. The curve has two branches in the two smaller sectors. The stationary points are at x = √2 - 1 and x = -√2 - 1.
Problem 2
X, at the centre a circular pond. Y, at the edge, cannot swim, but can run at speed 4v. X can run faster than 4v and can swim at speed v. Can X escape?
Solution
Answer: yes.
Let 4R be the radius. Let O be the centre of the pond, and C the circle centre O, radius 0.95R. X swims to C and then swims around C until Y is at the opposite side (with X, O, Y in a straight line in that order). That is possible because the length of C is less than 1/4 of the perimeter of the pond. X then heads straight for the edge. X gets there first, because 3.05R < 4πR/4 = 3.14R. X then runs directly away from Y to escape.   
Problem 3
Show that np - n is divisible by p for p = 3, 7, 13 and any integer n.
Solution
This is a straightforward piece of bookwork. The result is true for any prime p. 
Problem 4
What is the largest power of 10 dividing 100 x 99 x 98 x ... x 1?
Solution
Answer: 24.
There are 20 multiples of 5: 1·5, 2·5, ... , 20·5. Of these 4 are multiples of 25: 1·25, 2·25, 3·25, 4·25. None are multiples of 125. Hence the highest power of 5 dividing 100! is 24. The highest power of 2 is obviously higher. 
Problem 5
Show that n(n + 1)(n + 2)(n + 3) + 1 is a square for n = 1, 2, 3, ... .
Solution
n(n + 1)(n + 2)(n + 3) + 1 = n4 + 6n3 + 11n2 + 6n + 1 = (n2 + 3n + 1)2
Problem 6
The fractional part of a real is the real less the largest integer not exceeding it. Show that we can find n such that the fractional part of (2 + √2)n > 0.999 .
Solution
(2 + √2)n + (2 - √2)n is an integer. So the fractional part of (2 + √2)n is 1 - (2 - √2)n. But (2 - √2) lies between 0 and 1, so (2 - √2)n becomes arbitarily small for large n. 
Problem 7
What is the remainder on dividing x + x3 + x9 + x27 + x81 + x243 by x - 1? By x2 - 1?
Solution
Let f(x) = x + x3 + x9 + x27 + x81 + x243. The remainder on dividing by x - 1 is k, satisfying: f(x) = (x - 1) g(x) + k, for some polynomial g(x). Setting x = 1, k = f(1) = 6. Similarly, the remainder on dividing by x2 - 1 is ax + b, where f(x) = (x2 - 1) h(x) + ax + b, for some polynomial h(x). Hence 6 = f(1) = a + b, -6 = f(-1) = -a + b. So a = 6, b = 0, and the remainder is 6x. 
Problem 8
For what real b can we find x satisfying: x2 + bx + 1 = x2 + x + b = 0?
Solution
Answer: b = -2.
If b and x satisfy x2 + bx + 1 = x2 + x + b, then (b - 1)(x - 1) = 0, so either b = 1 or x = 1. If b = 1, then x2 + x + 1 = 0, which has no real roots. If x = 1, then b = -2, which is a solution.   
Problem 9
Show that for any real, positive x, y, z, not all equal, we have:   (x + y)(y + z)(z + x) > 8 xyz.
Solution
(x + y)/2 ≥ √(xy) with equality only if x = y. 
Problem 10
A chord length √3 divides a circle C radius 1 into two arcs. R is the region bounded by the chord and the shorter arc. What is the largest area of rectangle than can be drawn in R?
Solution
Answer:
Let the ends of the chord be A and B and its midpoint M. Let O be the centre of the circle. OM = 1/2 (recognize a 1/2, √3/2, 1 triangle or use Pythagoras). So if we extend OM to meet the arc at N, them MN = 1/2. Angles AOB and ANB = 120o.
First, we have to show that 2 vertices of a maximal rectangle lie on the chord. Although that is "obvious", proving it is the hardest part.
If P, Q are vertices of a maximal rectangle on the chord, then they must be equidistant from M. Otherwise we could enlarge the rectangle by moving the point close to M outwards to until it was equidistant with the other point.


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