A1. a1, a2, ... , an are reals. Show that if the sum of any two is non-negative, then for any non-negative real x1, x2, ... , xn with sum 1, we have a1x1 + a2x2 + ... + anxn ≥ a1x12 + a2x22 + ... + anxn2. Show that the converse is also true.
A2. ABC is a triangle. The altitude from A has length 12, the angle bisector from A has length 13. What is are the possible lengths for the median from A if the angle A is (1) acute, (2) obtuse, (3) a right-angle?
A3. The n complex numbers zi satisfy |z1| + |z2| + ... + |zn| = 1. Show that we can find a subset whose sum has modulus 1/6 or more.
B1. PQRS is a convex quadrilateral which lies inside the triangle ABC of area 1. Show that three of its vertices form a triangle of area 1/4 or less.
B2. The sequence a1, a2, .... , a3972 includes each of the numbers from 1 to 1986 twice. Can the terms be rearranged so that there are just n numbers between the two n's?
B3. Each point in the plane is colored black or red. Show that we can find three points of the same color with each pair a distance 1 apart or three points of the same color with each pair a distance √3 apart.
Solutions
Problem A1
a1, a2, ... , an are reals. Show that if the sum of any two is non-negative, then for any non-negative real x1, x2, ... , xn with sum 1, we have a1x1 + a2x2 + ... + anxn ≥ a1x12 + a2x22 + ... + anxn2. Show that the converse is also true.
Solution
We have ∑ aixi = ∑ aixi ∑ xi = ∑ aixi2 + ∑ (ai + aj)xixj ≥ ∑ aixi2.
The converse is obvious: take xi = xj = 1/2, others 0.
Problem A2
ABC is a triangle. The altitude from A has length 12, the angle bisector from A has length 13. What is are the possible lengths for the median from A if the angle A is (1) acute, (2) obtuse, (3) a right-angle?
Solution
Let M be the midpoint of BC. Let AD be the altitude from A and AE the angle bisector. Let the median AM have length k. Take MN to be the perpendicular bisector of BC with N on the opposite side of BC to A and lying on the circumcircle of ABC. If ∠A is 90o, then the circumcenter is at M. Then 12/5 = AD/DE = EM/MN = EM/AM = √(k2 - 144) - 5)/k. Solving the quadratic, k = 2028/119 = 17.04.
If A is obtuse, then the circumcenter lies on MN, so the same argument shows that k > 2028/119. Evidently any such value is achievable, because we can make DM (and hence AM) arbitrarily large. Then construct N, then pick O on MN such that AO = ON. Then construct B and C on the line DM so that OB = OC = OA.
Similarly, if A is acute, the circumcenter lies on the opposite side of BC to N, so k < 2028/119. On the other hand E must lie between D and M, so k > 13 (or equal if we allow the degenerate triangle with B = C = E = M).
Problem A3
The n complex numbers zi satisfy |z1| + |z2| + ... + |zn| = 1. Show that we can find a subset whose sum has modulus 1/6 or more.
Solution
Divide the complex plane into 120o sectors with center the origin. At least one sector must contain a subset with modulus sum at least 1/3. The sum of the projections of this subset onto the center line bisecting the sector must must be at least 1/6 (because cos k ≥ 1/2 for k ≤ 60o). So the sum has a projection length ≥ 1/6 and hence must have modulus at least 1/6.
Problem B1
PQRS is a convex quadrilateral which lies inside the triangle ABC of area 1. Show that three of its vertices form a triangle of area 1/4 or less.
Solution
We prove first that a parallelogram PQRS inside ABC has area at most 1/2. Assume the parallelogram has maximal area. If the parallelogram has less than 2 vertices on the sides of the triangle then we can expand it (about the point in contact, if any, or about its center if none) until it has two points in contact. That would increase its area. So it must have at least 2 vertices on the sides. If it had only 2, then we could translate it a small distance away from the vertex where those two sides meet. That would give it no vertices on the sides. Contradiction. So it must have at least 3 vertices on the sides. Suppose it has only 3. Let vertex P not be on a side. Translate P and Q along the line PQ preserving their distance apart until P is on a side (whilst Q remains in the triangle). That does not change the area. Now translate PR slightly along the side. Then neither Q nor S will be on a side and we could increase the distance between PR and QS. Contradiction. So we may assume that P and Q lie on BC, R on CA and S on AB. Now translate PQ along BC until P is at B. This does not change the area.
Now triangles ASR and CQR are similar to ABC and their areas are AR2/AC2 and CR2/AC2. Since PQRS has maximal area, the position or R on AC must be such as to minimise AR2 + CR2 = AR2 + (AC - AR)2 = 2(AR - AC/2)2 + AC2/2. Hence R must be the midpoint of AC and the area of PQRS must be 1/2.
Now if PQRS is any convex quadrilateral it contains a point X such that three of the vertices and X form a parallelogram. For take the vertex P such that the ray PQ does not meet the line RS and the ray PS does not meet the line QR. Then X to be the intersection of lines parallel to PQ through S and parallel to PS through Q. Now the triangle PQS has half the area of the parallelogram PQXS, which is at most half the area of the triangle.
Problem B2
The sequence a1, a2, .... , a3972 includes each of the numbers from 1 to 1986 twice. Can the terms be rearranged so that there are just n numbers between the two n's?
Solution
Generalising from 1986 to n, no such arrangement is possible for n = 1 or 2 mod 4. Hence, in particular, it is not possible for 1986.
Color the positions alternately black and white. Odd numbers must occupy positions of the same color, even numbers occupy positions of the same color. But the total number of positions is even, so the number of each color is equal and hence there must be an even number of odd numbers.
Comment. This is known as Langford's problem. The fact that sequences exist for n = 0 or 3 mod 4 can be demonstrated explicitly. For n = 4m, we have 4m-4 ... even ... 2m 4m-2 2m-3 ... odd ... 1 4m-1 1 ... odd ... 2m-3 2m ... even ... 4m-4 4m 4m-3 ... odd ... 2m+1 4m-2 2m-2 ... even ... 2 2m-1 4m-1 2 ... even 2m-2 2m+1 ... odd ... 4m-3 2m-1 4m. For n = 4m-1 we have 4m-4 ... even ... 2m 4m-2 2m-3 ... odd 1 4m-1 1 ... odd 2m-3 2m ... even ... 4m-4 2m-1 4m-3 ... odd 2m+1 4m-2 2m-2 ... even ... 2 2m-1 4m-1 2 ... even ... 2m-2 2m+1 ... odd 4m-3.
Problem B3
Each point in the plane is colored black or red. Show that we can find three points of the same color with each pair a distance 1 apart or three points of the same color with each pair a distance √3 apart.
Solution
Assume we cannot find an equilateral triangle side 1 with all vertices the same color. Take an equilateral triangle side 1. Two of the vertices must have different colors. Take a point a distance 2 from each. Then it must have a different color from one of them, so we have points AB a distance 2 apart, with A black and B white. wlog the midpoint M is white. The two points a distance 1 from B and M must both be black. But then they form an equilateral triangle side √3 with A.