International Mathematics TOURNAMENT OF THE TOWNS
Senior O-Level Paper Spring 2001.
1. At 12:20 pm, a bus leaves on a 100-kilometre route. It is equiped with a computer, which makes the following announcement at 1:00 pm, 2:00 pm, 3:00 pm, 4:00 pm, 5:00 pm and 6:00 pm: “Assuming that the average speed for the remaining part of the route is the same as that in the part already covered, the bus will arrive at its destination in another hour.” Can it be right, and if so, how many kilometers has the bus covered at 6 pm?
2. The cube of an n-digit number has m digits. Isn+m= 2001 possible?
3. In triangle ABC, X is a point on AB and Y is a point on BC. The segments AY and CX
intersect at Z. If AY =Y C and AB =ZC, prove that the points B, X, Z and Y lie on a
circle.
4. On a 3×100 board, two players alternately place dominoes. The first player places them as
1×2 rectangles while the second places them as 2×1 rectangles. The loser is the one who
cannot make a move. Which player can force a win, and what is the winning strategy?
5. Nine points are drawn on the surface of a regular tetrahedron of edge 1 centimetre. Prove that among these points there are two such that the distance in space between them is at most 0.5 centimetre.
Note: The problems are worth 3, 4, 4, 5 and 5 points respectively.
Solution to Senior O-Level Spring 2001
1. The computer can be right if at k : 00 pm, the distance covered is 60k60k+40−20 of the total distance
for 1≤ k ≤6. At k : 00 pm, the average speed so far is 60k+401 , and 60k+4060 of the distance is
still to be covered, requiring another hour at the same speed. Note that between (k−1) : 00
pm andk : 00 pm, the distance coverd is 60k60k+40−20−60k60k−−8020 = (60k+40)(60k80 −20) of the total distance,
and these quantites are all positive. At 6:00 pm, 340400 = 1720 of the distance has been covered,
which translates into 85 kilometres.
2. Note that 10500has 501 digits while (10500)3 has 1501 digits for a total of 2002. For any integer
a >10500, a and a3 will also have at least 2002 digits between them. Ifa <10500, then it has
at most 500 digits while a3 < 101500 has at most 1500 digits. Thus it is not possible for the
sum of the numbers of digits to be 2001.
3. First, note that CY = AY < CZ = AB. This is evident if AY is perpendicular to BC. If
not, one of 6 AY B and 6 AY C will be obtuse. In the former case, AB > AY. In the latter
case, CZ > CY.
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B Y C
X
Z
A W D
Complete the parallelogram ABCD and let W onAD be such that 6 W CY =6 AY C. Then
CY = CW, CZ =AB = CD and 6 AY C = 6 W CY = 6 DW C. Since AB > AY, triangles
ZY C and DW C are congruent, so that 6 AY C =6 W CY =6 DCZ. It follows that triangles
Y AC andDZC are similar, so that6 ZAC =6 ZDC. Hence ADCZ is a cyclic quadrilateral.
Now 6 XBY +6 XZY =6 ADC +6 AZC = 180◦, so that BXZY is also cyclic.
4. We first play the game on a 3× 4 board. The first player can force a win by placing the
first domino anywhere on the middle row. This leaves the second player with two columns on which dominoes can be placed, while the first player can place two more dominoes above
and below the first one. Now divide the 3×1000 board into 250 3×4 boards. Since the
second player can only place dominoes within such a board, the first player can force a win
by adopting the earlier strategy on each 3×4 board.
5. Connect the midpoints of two edges of the tetrahedron if and only if they share a common vertex. This divides the solid tetrahedron into four corner tetrahedra plus a central octahe- dron. If two of the nine points lie on the surface of one of those corner tetrahedra, then the distance between them is at most 0.5 centimetre. Hence we may assume that each contains at most one point, leaving five on the surface of the central octahedron. Only four of its faces are on the surface of the original tetrahedron. By the Pigeonhole Principle, two of the five points will lie on the same face, and the distance between them is at most 0.5 centimetre.
