A1. Given a positive integer n and real k > 0, what is the largest possible value for (x1x2 + x2x3 + x3x4 + ... + xn-1xn), where xi are non-negative real numbers with sum k?
A2. What is the smallest tower of 100s that exceeds a tower of 100 threes? In other words, let a1 = 3, a2 = 33, and an+1 = 3an. Similarly, b1 = 100, b2 = 100100 etc. What is the smallest n for which bn > a100?
A3. Three chords of the circumcircle bisect the three angles of a triangle. Show that the sum of their lengths exceeds the perimeter of the triangle.
B1. C is a circle of unit radius. For any line L define d(L) to be the distance between the two points of intersection if L meets C in two points, and zero otherwise. For any point P define f(P) to be the maximum value of d(L) + d(L') for two perpendicular lines through P. Which P have f(P) > 2?
B2. Define the sequence a1, a2, a3, ... by a1 = 1, a2 = b, an+2 =2an+1 - an + 2, where b is a positive integer. Show that for any n we can find an m such that anan+1 = am.
B3. A real polynomial of degree n > 2 has all roots real and positive. The coefficient of xn is 1, the coefficient of xn-1 is -1. The coefficient of x1 is non-zero and -n2 times the coefficient of x0. Show that all the roots are equal.
Solutions
Problem A1
Given a positive integer n and real k > 0, what is the largest possible value for (x1x2 + x2x3 + x3x4 + ... + xn-1xn), where xi are non-negative real numbers with sum k?
Answer
k2/4
Solution
Let xh = max(xi). Then x1x2 + x2x3 + x3x4 + ... + xh-1xh + xhxh+1 + xh+1xh+2 + ... + xn-1xn ≤ x1xh + x2xh + ... + xh-1xh + xhxh+1 + xhxh+2 + ... + xhxn = xh(k - xh) ≤ k2/4 (last step AM/GM). But k2/4 can be realized by x1 = x2 = k/2 and others 0.
Problem A2
What is the smallest tower of 100s that exceeds a tower of 100 threes? In other words, let a1 = 3, a2 = 33, and an+1 = 3an. Similarly, b1 = 100, b2 = 100100 etc. What is the smallest n for which bn > a100?
Answer
99
Solution
We have 33 < 100 and hence by a trivial induction an+1 < bn (an+1 = 3an < 100an < 100bn-1 = bn).
We claim that an+1 > 6bn-1. Obviously 333 > 600 so it is true for n = 2. Suppose it is true for n. Note that 36 = 729 > 600, so an+2 > 36bn-1 > 600bn-1 = 6bn-1bn > 6bn.
Problem A3
Three chords of the circumcircle bisect the three angles of a triangle. Show that the sum of their lengths exceeds the perimeter of the triangle.
Solution
Bizarrely, this is a repeat of 82/A3.
Problem A3
In the triangle ABC, let the angle bisectors of A, B, C meet the circumcircle again at X, Y, Z. Show that AX + BY + CZ is greater than the perimeter of ABC.
Solution
The sine rule gives BC/sin A = CA/sin B = AB/sin C. Put k = BC/sin A, so perimeter = k(sin A + sin B + sin C). Applying sine rule to BCY we have BC/sin Y = BC/sin A = k = BY/sin BCY. But ∠BCY = ∠C + ∠ACY = ∠C + ∠B/2. So BY = k sin(C+B/2). Similarly for the other chords, so AX + BY + CZ = k(sin(C+B/2) + sin(A+C/2) + sin(B+A/2)).
But sin(C+B/2) = sin(90o + C/2 - A/2) = cos(A/2 - C/2) > cos(A/2 - C/2) sin(A/2 + C/2) = ½ sin A + ½ sin C. Adding the two similar relations gives the required result. Note that we cannot have equality because A + C < 180o, so sin(A/2 + C/2) < 1.
Problem B1
C is a circle of unit radius. For any line L define d(L) to be the distance between the two points of intersection if L meets C in two points, and zero otherwise. For any point P define f(P) to be the maximum value of d(L) + d(L') for two perpendicular lines through P. Which P have f(P) > 2?
Answer
points inside (but not on) the circle with the same center and radius √(3/2)
Solution
It is obvious that points inside C qualify (take L to be a diameter), and fairly obvious that points on C qualify (L and L' meet C again at the ends of a diameter). So we consider P outside C.
Let O be the center of C. Obviously if PO ≥ √2, then only one of L, L' meets C, so f(P) = 2. So suppose P is close enough for L and L' to meet C.
Let OP = r. Take the angle x as shown. Then d = d(L) + d(L') = 2√(1 - r2cos2x) + 2√(1 - r2sin2x). So d2/4 = 2 - r2 + 2√(1 - r2 + r4sin2xcos2x) = 2 - r2 + √(4 - 4r2 + r4sin2x). That is obviously maximised by taking x = π/4 giving 2 - r2 + √(4 - 4r2 + r4) = 4 - 2r2, since r2 < 2. So f(P) > 2 iff 4 - 2r2 > 1, or r < √(3/2).
Problem B2
Define the sequence a1, a2, a3, ... by a1 = 1, a2 = b, an+2 =2an+1 - an + 2, where b is a positive integer. Show that for any n we can find an m such that anan+1 = am.
Solution
A trivial induction shows that an+2 = (n+1)b + n2. So an+1an+2 = n(n+1)b2 + (2n3-n2-n+1)b + n2(n-1)2 = aN+2, where N = nb+n2-n.
Problem B3
A real polynomial of degree n > 2 has all roots real and positive. The coefficient of xn is 1, the coefficient of xn-1 is -1. The coefficient of x1 is non-zero and -n2 times the coefficient of x0. Show that all the roots are equal.
Solution
Let the roots be a1, a2, ... , an. Then a1 + a2 + ... + an = 1 and 1/a1 + 1/a2 + ... + 1/an = n2, so (a1 + ... + an)(1/a1 + ... + 1/an) = n2. But (a1 + ... + an)(1/a1 + ... + 1/an) ≥ n2 with equality iff all ai are equal. (A well-known result = prove for example by Cauchy-Schwartz on √ai and 1/√ai).
Solutions are also available in Australian Mathematical Olympiads 1979-1995
by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust
by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust