A1. Let A, B, C, D be four distinct points on a line, in that order. The circles with diameter AC and BD intersect at X and Y. The line XY meets BC at Z. Let P be a point on the line XY other than Z. The line CP intersects the circle with diameter AC at C and M, and the line BP intersects the circle with diameter BD at B and N. Prove that the lines AM, DN, XY are concurrent.
A2. Let a, b, c be positive real numbers with abc = 1. Prove that: 1/(a3(b + c)) + 1/(b3(c + a)) + 1/(c3(a + b)) ≥ 3/2.
A3. Determine all integers n > 3 for which there exist n points A1, ... , An in the plane, no three collinear, and real numbers r1, ... , rn such that for any distinct i, j, k, the area of the triangle AiAjAk is ri + rj + rk.
B1. Find the maximum value of x0 for which there exists a sequence x0, x1, ... , x1995 of positive reals with x0 = x1995 such that for i = 1, ... , 1995: xi-1 + 2/xi-1 = 2xi + 1/xi.
B2. Let ABCDEF be a convex hexagon with AB = BC = CD and DE = EF = FA, such that ∠BCD = ∠EFA = 60o. Suppose that G and H are points in the interior of the hexagon such that ∠AGB = ∠DHE = 120o. Prove that AG + GB + GH + DH + HE ≥ CF.
B3. Let p be an odd prime number. How many p-element subsets A of {1, 2, ... , 2p} are there, the sum of whose elements is divisible by p?
Solutions
Problem A1
Let A, B, C, D be four distinct points on a line, in that order. The circles with diameter AC and BD intersect at X and Y. The line XY meets BC at Z. Let P be a point on the line XY other than Z. The line CP intersects the circle with diameter AC at C and M, and the line BP intersects the circle with diameter BD at B and N. Prove that the lines AM, DN, XY are concurrent.
Solution
Let DN meet XY at Q. Angle QDZ = 90o - angle NBD = angle BPZ. So triangles QDZ and BPZ are similar. Hence QZ/DZ = BZ/PZ, or QZ = BZ·DZ/PZ. Let AM meet XY at Q'. Then the same argument shows that Q'Z = AZ·CZ/PZ. But BZ·DZ = XZ·YZ = AZ·CZ, so QZ = Q'Z. Hence Q and Q' coincide.
Problem A2
Let a, b, c be positive real numbers with abc = 1. Prove that:
1/(a3(b + c)) + 1/(b3(c + a)) + 1/(c3(a + b)) ≥ 3/2.
Solution
Put a = 1/x, b = 1/y, c = 1/z. Then 1/(a3(b+c)) = x3yz/(y+z) = x2/(y+z). Let the expression given be E. Then by Cauchy's inequality we have (y+z + z+x + x+y)E ≥ (x + y + z)2, so E ≥ (x + y + z)/2. But applying the arithmetic/geometric mean result to x, y, z gives (x + y + z) ≥ 3. Hence result.
Thanks to Gerhard Woeginger for pointing out that the original solution was wrong.
Problem A3
Determine all integers n > 3 for which there exist n points A1, ... , An in the plane, no three collinear, and real numbers r1, ... , rn such that for any distinct i, j, k, the area of the triangle AiAjAk is ri + rj + rk.
Answer
n = 4.
Solution
The first point to notice is that if no arrangement is possible for n, then no arrangement is possible for any higher integer. Clearly the four points of a square work for n = 4, so we focus on n = 5.
If the 5 points form a convex pentagon, then considering the quadrilateral A1A2A3A4 as made up of two triangles in two ways, we have that r1 + r3 = r2 + r4. Similarly, A5A1A2A3 gives r1 + r3 = r2 + r5, so r4 = r5.
We show that we cannot have two r's equal (whether or not the 4 points form a convex pentagon). For suppose r4 = r5. Then A1A2A4 and A1A2A5 have equal area. If A4 and A5 are on the same side of the line A1A2, then since they must be equal distances from it, A4A5 is parallel to A1A2. If they are on opposite sides, then the midpoint of A4A5 must lie on A1A2. The same argument can be applied to A1 and A3, and to A2 and A3. But we cannot have two of A1A2, A1A3 and A2A3 parallel to A4A5, because then A1, A2 and A3 would be collinear. We also cannot have the midpoint of A4A5 lying on two of A1A2, A1A3 and A2A3 for the same reason. So we have established a contradiction. hence no two of the r's can be equal. In particular, this shows that the 5 points cannot form a convex pentagon.
Suppose the convex hull is a quadrilateral. Without loss of generality, we may take it to be A1A2A3A4. A5 must lie inside one of A1A2A4 and A2A3A4. Again without loss of generality we may take it to be the latter, so that A1A2A5A4 is also a convex quadrilateral. Then r2 + r4 = r1 + r3 and also = r1 + r5. So r3 = r5, giving a contradiction as before.
The final case is the convex hull a triangle, which we may suppose to be A1A2A3. Each of the other two points divides its area into three triangles, so we have: (r1 + r2 + r4) + (r2 + r3 + r4) + (r3 + r1 + r4) = (r1 + r2 + r5) + (r2 + r3 + r5) + (r3 + r1 + r5) and hence r4 = r5, giving a contradiction.
So the arrangement is not possible for 5 and hence not for any n > 5.
Problem B1
Find the maximum value of x0 for which there exists a sequence x0, x1, ... , x1995 of positive reals with x0 = x1995 such that for i = 1, ... , 1995:
xi-1 + 2/xi-1 = 2xi + 1/xi.
Answer
2997.
Solution
The relation given is a quadratic in xi, so it has two solutions, and by inspection these are xi = 1/xi-1 and xi-1/2. For an even number of moves we can start with an arbitrary x0 and get back to it. Use n-1 halvings, then take the inverse, that gets to 2n-1/x0 after n moves. Repeating brings you back to x0 after 2n moves. However, 1995 is odd!
The sequence given above brings us back to x0 after n moves, provided that x0 = 2(n-1)/2. We show that this is the largest possible x0. Suppose we have a halvings followed by an inverse followed by b halvings followed by an inverse. Then if the number of inverses is odd we end up with 2a-b+c- .../x0, and if it is even we end up with x0/2a-b+c- .... In the first case, since the final number is x0 we must have x0 = 2(a-b+-...)/2. All the a, b, ... are non-negative and sum to the number of moves less the number of inverses, so we clearly maximise x0 by taking a single inverse and a = n-1. In the second case, we must have 2a-b+c- ... = 1 and hence a - b + c - ... = 0. But that implies that a + b + c + ... is even and hence the total number of moves is even, which it is not. So we must have an odd number of inverses and the maximum value of x0 is 2(n-1)/2.
Problem B2
Let ABCDEF be convex hexagon with AB = BC = CD and DE = EF = FA, such that ∠BCD = ∠EFA = 60o. Suppose that G and H are points in the interior of the hexagon such that ∠AGB = ∠DHE = 120o. Prove that AG + GB + GH + DH + HE ≥ CF.
Solution
BCD is an equilateral triangle and AEF is an equilateral triangle. The presence of equilateral triangles and quadrilaterals suggests using Ptolemy's inequality. [If this is unfamiliar, see ASU 61/6 solution.]. From CBGD, we get CG·BD ≤ BG·CD + GD·CB, so CG ≤ BG + GD. Similarly from HAFE we get HF ≤ HA + HE. Also CF is shorter than the indirect path C to G to H to F, so CF ≤ CG + GH + HF. But we do not get quite what we want.
However, a slight modification of the argument does work. BAED is symmetrical about BE (because BA = BD and EA = ED). So we may take C' the reflection of C in the line BE and F' the reflection of F. Now C'AB and F'ED are still equilateral, so the same argument gives C'G ≥ AG + GB and HF' ≤ DH + HE. So CF = C'F' ≤ C'G + GH + HF' ≤ AG + GB + GH + DH + HE.
Problem B3
Let p be an odd prime number. How many p-element subsets A of {1, 2, ... , 2p} are there, the sum of whose elements is divisible by p?
Answer
2 + (2pCp - 2)/p, where 2pCp is the binomial coefficient (2p)!/(p!p!).
Solution
Let A be a subset other than {1, 2, ... , p} and {p+1, p+2, ... , 2p}. Consider the elements of A in {1, 2, ... , p}. The number r satisfies 0 < r < p. We can change these elements to another set of r elements of {1, 2, ... , p} by adding 1 to each element (and reducing mod p if necessary). We can repeat this process and get p sets in all. For example, if p = 7 and the original subset of {1, 2, ... , 7} was {3 , 5}, we get:
{3 , 5}, {4, 6}, {5, 7}, {6, 1}, {7, 2}, {1, 3}, {2, 4}.
The sum of the elements in the set is increased by r each time. So, since p is prime, the sums must form a complete set of residues mod p. In particular, they must all be distinct and hence all the subsets must be different.
Now consider the sets A which have a given intersection with {p+1, ... , n}. Suppose the elements in this intersection sum to k mod p. The sets can be partitioned into groups of p by the process described above, so that exactly one member of each group will have the sum -k mod p for its elements in {1, 2, ... , p}. In other words, exactly one member of each group will have the sum of all its elements divisible by p.
There are 2pCp subsets of {1, 2, ... , 2p} of size p. Excluding {1, 2, ... , p} and {p+1, ... , 2p} leaves (2pCp - 2). We have just shown that (2pCp - 2)/p of these have sum divisible by p. The two excluded subsets also have sum divisible by p, so there are 2 + (2pCp - 2)/p subsets in all having sum divisible by p.
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