International Mathematics TOURNAMENT OF THE TOWNS
Senior A-Level Paper Spring 2001.
1. Find a polynomial P(x) of degree 2001 such that P(x) +P(1−x) = 1 for all real numbersx.
2. In a class with at least 5 students, each subject taken by a student results in pass or failure.
For any arbitrarily chosen group of no less than 5 students, at least 80% of the failures received by this group are given to at most 20% of the students in the group. Prove that at least 75%
of all the failures in this class are given to one student.
3. AD, BE and CF are the altitudes of triangle ABC. K, M and N are the respective or- thocentres of triangles AEF, BF D and CDE. Prove that KM N and DEF are congruent triangles.
4. Each entry in two m×n tables A and B is either 0 or 1. The number of 1’s in A is equal to the number of 1’s in B. In both A and B, the numbers do not decrease from the left to the right in any row, nor from top to bottom in any column. For every k, 1≤ k ≤ m, the sum of the entries in the top k rows of A is not less than the sum of entries in the top k rows of B. Prove that for any `, 1≤ ` ≤n, the sum of the numbers in the leftmost ` columns of B is at least the sum of the numbers in the leftmost` columns of A.
5. In a chess tournament, every participant played with the others exactly once, getting 1 point for a win, 12 for a draw and 0 for a loss. For each player, the sum of the points earned by the players who were beaten by this player was computed, as was the sum of the points earned by the players who beat this player.
(a) Is it possible for the first sum to be greater than the second one for every player?
(b) Is it possible for the first sum to be less than the second one for every player?
6. Prove that there exist 2001 convex polyhedra such that any three of them do not have any common points, and any two of them have at least one common boundary point but no common inner points.
7. Several boxes are placed along a circle. Each box may contain any number of chips, including zero. A move consists of taking all the chips from some box and placing them in the subsequent boxes clockwise, one chip in every box, beginning from the next box in the clockwise direction.
(a) Suppose that in each move after the first one, one must take the chips from the box in which the last chip was placed on the previous move. Prove that after several moves, the initial distribution of the chips among the boxes will reappear.
(b) Suppose that in each move, one may take the chips from any box. Is it true that for every initial distribution of the chips, one can get any possible distribution by performing an appropriate sequence of moves?
Note: The problems are worth 3, 5, 5, 5, 4+4, 8 and 4+4 points respectively.
Solution to Senior A-Level Spring 2001
1. Let P(x) = x2001−(1−x)2001+12. This is a polynomial of degree 2001 since the leading term is 2x2001. NowP(x) +P(1−x) = x2001−(1−x)2001+12+ (1−x)2001+ (1−(1−x))2001+12 = 1 for all real numbers x.
2. Let a1 ≥ a2 ≥ · · · ≥ an be the respective numbers of failures given to the n students in the class. For any k such that k + 4 ≤ n, we have ak ≥ 45(ak +ak+1 +ak+2 +ak+3 +ak+4, or equivalently, ak ≥3(ak+1+ak+2+ak+3+ak+4). Supposen ≡1 (mod 4). Then
a1 ≥ 4(a2 +a3+a4+a5)
≥ 3(a2 +a3+a4+a5) +a5
≥ 3(a2 +a3+a4+a5) + 4(a6+a7+a8+a9)
≥ 3(a2 +a3+· · ·+a9) +a9
≥ · · ·
≥ 3(a2 +a3+· · ·+an).
This is equivalent toa1 ≥ 34(a1+a2+· · ·+an), which is the desired result. It does not matter ifn 6≡1 (mod 4) as we simply takean−4 as the extra term to generate the last inequality. For instance, ifn = 8, we have
a1 ≥ 3(a2 +a3+a4+a5) +a4
≥ 3(a2 +a3+a4+a5) + 4(a5+a6+a7+a8)
≥ 3(a2 +a3+· · ·+a8).
3. Let H be the orthocentre of triangle ABC. Note that F M and HD are both perpendicular toBC. Hence they are parallel to each other. Similarly, so are F H and M D. It follows that DHF M is a parallelogram, Similarly, so is DHEN. Hence F M =EN and they are parallel to each other. It follows that EF M N is also a parallelogram, so thatEF =N M. Similarly, we haveF D =KN and DE =M K. Hence triangles DEF and KM N are congruent.
AA AA AA AA AA AA
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@
B D C
M
N H
F K
E A
4. Suppose the result is false. Let j be the smallest integer such that the sum of the numbers in the leftmost j columns of B is less than the sum of the numbers in the leftmostj columns of A. Then there are more 1’s in column j of A than in column j of B. Let the number of 1’s
in column j of A be i. Divide each of A and B into four quadrants, the dividing lines being between the (m−i)-th row and the (m−i+ 1)-st row and between thej-th column and the (j+ 1)-st column.
j columns n−j columns
m−i II I
rows quadrant quadrant
i III IV
rows quadrant quadrant
Let the numbers of 1’s in the four quadrants of A be a1, a2, a3 and a4, and those of B be b1, b2, b3 and b4, respectively. Then a1+a2+a3+a4 =b1+b2+b3+b4 and a2+a3 > b2+b3, so thata1+a4 < b1+b4. Since the lastinumbers in columnj of Aare 1’s, all of the numbers in the fourth quadrant of A are 1’s. Hence b4 ≤ a4, so that a1 < b1. Since the first m−i numbers in column j of A and B are 0’s, a2 =b2 = 0 =. Hence a1+a2 < b1+b2, which is a contradiction.
5. Let there ben players. For playerk, denote the first sum bywk, the second sum by`kand the player’s own score by sk. Let T =
Xn k=1
sk(wk−`k). We claim that T = 0. Consider the game between players i and j. If it is a tie, it contributes nothing to T. Suppose i beats j. Then it contributes sisj to the term si(wi −`i) while it contributes −sjsi to the term sj(wj −`j).
The net contribution toT is still nothing, and the same holds by symmetry ifj beatsi. This justifies the claim.
(a) Since sk≥0 for all k, we cannot havewk > `k for all k as otherwise T >0.
(b) Since sk≥0 for all k, we cannot havewk < `k for all k as otherwise T <0.
6. We construct the polyhedra as follows. Start with an infinite inverted cone with vertex at the originO and axis along the positivez-axis. On the surface of the cone are equally spaced lines `1, `2, . . . , `2001. For any positive number t, the plane z = t intersects these lines at At1, At2, . . . , At2001. Let t1 be an arbitrary positive number. Let D1 be the disc which is the part of the plane z = t1 inside the cone. Let B1 be the midpoint of the minor arc At11At21. Let M1 be the convex polygon At11At21. . . At20011 . Consider the infinite oblique prism with base M1 and lateral edges parallel to OB1. This will eventually be truncated to yield a convex polyhedron P1. For sufficiently large t, the distance between At1At2 and the midpoint of the minor arc At1At2 will be larger than the diameter of D1 so that P1 will not intersect the polygonAt1At2. . . At2001. Lett2be such a value and letD2 be the part of the planez =t2 inside the cone. Let B2 be the midpoint of the minor arc At22At32. Let M2 be the convex polygon At12At2−2. . . At20012 if it already touches P1. If not, insert between At12At22 an additional vertex which lies on D2 ∩P1. Let P2 be the convex polyhedron obtained by eventually truncating
the oblique infinite prism with base M2 and lateral edges parallel to OB2. As before, for sufficiently larget,P2will not intersect the polygonAt1At2. . . At2001. LetT3be such a value, and we can continue the construction as before withM3 being the convex polygonAt13At23. . . At20013 , expanded if necessary, so that it touches both P1 and P2. For 1 ≤ i < j < k≤ 2001, Pi and Pj have only common boundary points on the plane z =ti. Since Pk does not intersect this plane, Pi, Pj and Pk have no common points.
7. (a) We construct a directed graph as follows. Each vertex represents a distribution of chips coupled with the box from which chips must be taken. An arc goes from vertex A to vertex B if the state represented fromA leads to the state represented by B. The out- degree of each vertex is clearly 1. We claim that so is its in-degree. From any given state, we can reverse the process by taking a chip from the box which is the last one to receive a chip, collecting 1 chip from each subsequent box counterclockwise, and stopping when we reach an empty box. We can only deposit all the chips into this box, and mark it as the one from which chips must be taken. This is the unique state which leads to the given one, and the claim is justified. Since the number of vertices is finite, the directed graph is a union of disjoint directed cycles. It does not matter which cycle contains the vertex which represents the initial state, as all states represented by vertices in this cycle will reappear.
(b) We construct a directed graph as follows. Each vertex represents a distribution of the chips. An arc goes from vertexA to vertex B if it is possible to change the distribution represented by A directly into the distribution represented by B. The out-degree and the in-degree of each vertex are both equal to the number of non-empty boxes in the distribution represented by that vertex. Hence the directed graph consists of strongly connected components. It has only one component because the distribution in which all chips are in a specific box is reachable from any other distribution. We simply do not take chips from that box. Hence we can change any distribution into any other distribution.
