Two players alternately choose the sign for one of the numbers 1, 2, ..., 20. Once a sign has been chosen it cannot be changed. The first player tries to minimize the final absolute value of the total and the second player to maximize it. What is the outcome (assuming both players play perfectly)?

Example: the players might play successively: 1, 20, - 19, 18, - 17, 16, - 15, 14, - 13, 12, - 11, 10, - 9, 8, - 7, 6, - 5, 4, - 3, 2. Then the outcome is 12. However, in this example the second player played badly!

Solution:
Answer: 30.

The second player can play the following strategy: (1) if the first player plays \(2\mathrm{n} - 1\) for \(1< = \mathrm{n}\) \(< = 9\) , then he replies \(2\mathrm{n}\) with the opposite sign; (2) if the first player plays \(2\mathrm{n}\) for \(1< = \mathrm{n}< = 9\) then he replies \(2\mathrm{n} - 1\) with the opposite sign; (3) if the first player plays 19 or 20, then he plays the other with the same sign. This secures a score of at least 39 (from (3) ) less \(9\mathrm{x}1\) (from (1) and (2) ). So he can ensure a score of at least 30.

The first player can play the following strategy: (1) he opens with 1; (2) if the second player plays \(2\mathrm{n}\) for \(1\leq \mathrm{n}< = 9\) , then he replies with \(2\mathrm{n} + 1\) with the opposite sign; (3) if the second player plays \(2\mathrm{n} + 1\) for \(1< = \mathrm{n}< = 9\) , then he replies with \(2\mathrm{n}\) with the opposite sign; (4) if any of these replies are impossible, or if the second player plays 20, then he replies with the highest

number available with the opposite sign. If the second player does not play 20 until the last move, then this strategy ensures a score of at most 1 (from (1) ) + 9 x 1 (from (2) and (3) ) + \(20 = 30\) . Now suppose that the second player plays 20, a1, a2, ..., an (where \(1 \leq n \leq 9\) ) which require a reply under (4). The reason a1 required a move under (4) was that a1- 1 or a1+1 was the 1st player's response to 20. Similarly, the reason a2 required a move under (4) was that a2- 1 or a2+1 was the 1st player's response to a2, and so on. Thus the increment to the absolute value from these moves is at most \(120 - a_1 + 11 + |a_1 - a_2 + 11 + \ldots + |a_{n - 1} - a_n + 11 + |a_n| = 20 + n\) . The increment from the moves under (2) and (3) is \((9 - n) \times 1\) , and the increment from the move under (1) is 1. Hence the maximum absolute value is 30.

Since the 1st player has a strategy to do no worse than 30 and the 2nd player has a strategy to do no worse than 30, these strategies must actually be optimal.


In the acute- angled triangle ABC, AH is the longest altitude (H lies on BC), M is the midpoint of AC, and CD is an angle bisector (with D on AB).

(a) If \(\mathrm{AH}< = \mathrm{BM}\) , prove that the angle \(\mathrm{ABC}< = 60\) .

(b) If \(\mathrm{AH} = \mathrm{BM} = \mathrm{CD}\) , prove that ABC is equilateral.

Solution:
As usual let a, b, c be the lengths of BC, CA, AB respectively and let A, B, C denote the angles BAC, ABC, BCA respectively. We use trigonometry and try to express the quantities of interest in terms of a, b and C.

(a) Since AH is the longest altitude, BC must be the shortest side (use area = side x altitude/2). So \(\mathrm{b}^2 > = \mathrm{a}^2\) , and \(\mathrm{c}^2 > = \mathrm{a}^2\) . Using the formula \(\mathrm{c}^2 = \mathrm{a}^2 + \mathrm{b}^2 - 2\mathrm{ab}\cos \mathrm{C}\) , we deduce that \(\mathrm{b}^2 > = 2\mathrm{ab}\cos \mathrm{C}\) . Hence \(2\mathrm{b}^2 > = \mathrm{a}^2 + 2\mathrm{ab}\cos \mathrm{C}\) . After a little manipulation this gives: \(\mathrm{a}^2 + \mathrm{b}^2 - 2\mathrm{ab}\cos \mathrm{C} > = 4 / 3(\mathrm{a}^2 + \mathrm{b}^2 /4 - \mathrm{ab}\cos \mathrm{C})\) or \(\mathrm{c}^2 > = 4 / 3\mathrm{BM}^2\) . But we are given that \(\mathrm{BM} > = \mathrm{AH} = \mathrm{b}\) sin C, so \((\mathrm{b}^2\sin^2\mathrm{C}) / \mathrm{c}^2 < = 3 / 4\) . But the sine formula gives \(\sin \mathrm{B} = (\mathrm{b}\sin \mathrm{C}) / \mathrm{c}\) , so \(\sin^2\mathrm{C} < = 3 / 4\) . The triangle is acute- angled, hence \(\mathrm{B}< = 60\) degrees.

(b) The angle bisector theorem gives \(\mathrm{AD / BD} = \mathrm{b / a}\) , hence \(\mathrm{AD / AB} = \mathrm{b / (a + b)}\) , so \(\mathrm{AD} = \mathrm{bc / (a + b)}\) . Hence, using the sine formula, \(\mathrm{CD / sin A} = \mathrm{AD / (sin C / 2)}\) . So \(\mathrm{CD} = \mathrm{bc}\) sin \(\mathrm{A / ((a + b)}\) sin \(\mathrm{C / 2)} = \mathrm{ba}\) sin \(\mathrm{C / ((a + b)}\) sin \(\mathrm{C / 2)}\) , using the sine formula again. But we are given that \(\mathrm{CD}\geq\) \(\mathrm{AH} = \mathrm{b}\) sin C, so \(\mathrm{a / ((a + b)}\) sin \(\mathrm{C / 2)}\geq 1\) . But a is the shortest side, so \(\mathrm{a / (a + b)}\leq 1 / 2\) and hence sin \(\mathrm{C / 2}< 1 / 2\) . The triangle is acute- angled, so \(\mathrm{C / 2}\leq 30\) degrees, and \(\mathrm{C}\leq 60\) degrees. BC is the shortest side, so A is the smallest angle and hence \(\mathrm{A}\leq 60\) degrees. Also since \(\mathrm{AH}\leq \mathrm{BM}\) , \(\mathrm{B}\leq\) 60 degrees. But the angles sum to 180 degrees, so they must all be 60 degrees and hence the triangle is equilateral.


(a) The digits of a natural number are rearranged and the resultant number is added to the original number. Prove that the answer cannot be 99 ... 9 (1999 nines).

(b) The digits of a natural number are rearranged and the resultant number is added to the original number to give \(10^{10}\) . Prove that the original number was divisible by 10.

Solution:
(a) Let the digits of the original number be \(\mathrm{a_1}\) , \(\mathrm{a_2}\) , ... and the rearranged digits be \(\mathrm{b_1}\) , \(\mathrm{b_2}\) , ... Suppose that in the addition there is a carry, in other words \(\mathrm{a_1 + b_1 > 9}\) for some i. Take the largest such i. Then the resulting digit in that position cannot be a 9. Contradiction. So there cannot be any carries. Hence each pair \(\mathrm{a_1 + b_1 = 9}\) . Let n be the total number of digits 0, 1, 2, 3, and 4 in the number. Then each of these must be paired with a digit 5, 6, 7, 8 or 9. So the total number of digits 5, 6, 7, 8 and 9 must also be n, and hence the number must have an even number of digits. But we are told that the answer and hence the original number has an odd number of digits.

(b) In the addition the carry can never be 2, because that would require the previous carry to be at least 2, and the first carry cannot be 2. So all carries are 0 or 1. If a carry is 1, then all subsequent carries must also be 1. If the first carry is 0, then the corresponding digits must be 0 and hence the original number is divisible by 10. If it is not, then all carries are 1 and hence after the first carry all the digit pairs sum to 9. But arguing as in (a), this means that there must be an even number of digits, excluding the last (where we have a digit sum 10), and hence an odd number of digits in the original number. But \(10^{10}\) has an odd number of digits and hence the original number had an even number of digits. Contradiction.


Four lighthouses are arbitarily placed in the plane. Each has a stationary lamp which illuminates an angle of 90 degrees. Prove that the lamps can be rotated so that at least one lamp is visible from every point of the plane.

Solution:
Take a north direction, arbitary except that no points are aligned north- south or east- west. Take the two most northerly points. Point the lamp for the more easterly of these two in the direction SW (so that it covers directions S to W). Point the lamp for the other in the direction SE. For the other two points, point the lamp for the more easterly in the direction NW, and the lamp for the other in the direction NE.

Clearly the lamps cover all directions, the only possible problem is uncovered strips. However, the two lamps pointing N are below the two lamps pointing S, and the two lamps pointing E are west of the two lamps pointing W, so there are no uncovered strips.


(a) Can you arrange the numbers 0, 1, ..., 9 on the circumference of a circle, so that the difference between every pair of adjacent numbers is 3, 4 or 5? For example, we can arrange the numbers 0, 1, ..., 6 thus: 0, 3, 6, 2, 5, 1, 4.

(b) What about the numbers 0, 1, ..., 13?

Solution:
No. Each of the numbers 0, 1, 8, 9 can only be adjacent to 3, 4, 5 or 6. But they can only accommodate 3 numbers, not 4.

0, 3, 7, 10, 13, 9, 12, 8, 11, 6, 2, 5, 1, 4 is a solution for 13.

In passing, there are obviously no solutions for 4 or 5. There is just the one solution for 6 (given in the question). For 7 there are 5 solutions: 0, 3, 6, 1, 5, 2, 7, 4; 0, 3, 6, 1, 4, 7, 2, 5; 0, 3, 6, 2, 7, 4, 1, 5; 0, 3, 7, 4, 1, 6, 2, 5; 0, 4, 1, 6, 3, 7, 1, 5. For 8 there is the solution 0, 3, 7, 2, 6, 1, 5, 8, 4, and maybe others.


Prove that there exists a number divisible by \(5^{1000}\) with no zero digit.

Solution:
We first find a multiple of \(5^{1000}\) which has no zeros in the last 1000 digits. Suppose that we have a multiple n.5 \(^{1000}\) whose last zero is in place r (treating the last place as place 0, the next to last as place 1 and so on). Then \(\mathrm{n}(10^{\mathrm{r}} + 1)\) has the same digits in places 0 to r- 1 and a nonzero digit in place r, and hence no zeros in places 0 to r. So repeating, we find a multiple n.5 \(^{1000}\) with no zeros in the last 1000 digits.

Now let m be the remainder when n is divided by \(2^{1000}\) , so \(\mathrm{n} = \mathrm{k}.2^{1000} + \mathrm{m}\) , and hence m.5 \(^{1000} = \mathrm{n}.5^{1000} - \mathrm{k}.10^{1000}\) . So m.5 \(^{1000}\) has the same last 1000 digits as n.5 \(^{1000}\) . But it has less than 1001 digits, and hence it has exactly 1000 digits and no zeros.


Find all integers x, y satisfying \(\mathrm{x}^{2} + \mathrm{x} = \mathrm{y}^{4} + \mathrm{y}^{3} + \mathrm{y}^{2} + \mathrm{y}\) .

Solution:
The only solutions are \(\mathrm{x},\mathrm{y} = - 1,1 - ;0, - 1,0;0,0; - 6,2;\) or 5,2.

\((\mathrm{y}^{2} + \mathrm{y} / 2 - 1 / 2)(\mathrm{y}^{2} + \mathrm{y} / 2 + 1 / 2) = \mathrm{y}^{4} + \mathrm{y}^{3} + 1 / 4\mathrm{y}^{2} - 1 / 4< \mathrm{y}^{4} + \mathrm{y}^{3} + \mathrm{y}^{2} + \mathrm{y}\) except for \(- 1< = \mathrm{y}< =\) \(- 1 / 3\) . Also \((\mathrm{y}^{2} + \mathrm{y} / 2)(\mathrm{y}^{2} + \mathrm{y} / 2 + 1) = \mathrm{y}^{4} + \mathrm{y}^{3} + 5 / 4\mathrm{y}^{2} + \mathrm{y} / 2\) which is greater than \(\mathrm{y}^{4} + \mathrm{y}^{3} + \mathrm{y}^{2}+\) y unless \(0< = \mathrm{y}< = 2\)

But no integers are greater than \(\mathrm{y}^{2} + \mathrm{y} / 2 - 1 / 2\) and less than \(\mathrm{y}^{2} + \mathrm{y} / 2\) . So the only possible solutions have y in the range - 1 to 2. Checking these 4 cases, we find the solutions listed.


What is the maximum possible length of a sequence of natural numbers \(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \ldots\) such that \(\mathrm{x}_{i} \leq 1998\) for \(i \geq = 1\) , and \(\mathrm{x}_{i} = |\mathrm{x}_{i - 1} - \mathrm{x}_{i - 2}|\) for \(i \geq 3\) .

Solution:
Answer 2998.

The sequence is completely determined by its first two elements. If the largest element of the sequence is n, then it must occur as one of the first two elements. Because \(\mathrm{x}_{3}\) and \(\mathrm{x}_{4}\) are both smaller than the largest of the first two elements and hence all subsequent elements are too.

Let \(\mathrm{f}(\mathrm{n}, \mathrm{m})\) be the length of the sequence with \(\mathrm{x}_{1} = \mathrm{n}\) , \(\mathrm{x}_{2} = \mathrm{m}\) . It is straightforward to verify by induction that \(\mathrm{f}(1,2\mathrm{n}) = \mathrm{f}(2\mathrm{n} - 1,2\mathrm{n}) = 3\mathrm{n} + 1\) , \(\mathrm{f}(2\mathrm{n},1) = \mathrm{f}(2\mathrm{n},2\mathrm{n} - 1) = 3\mathrm{n}\) , \(\mathrm{f}(2\mathrm{n},2\mathrm{n} + 1) = 3\mathrm{n} + 3\) , \(\mathrm{f}(1,2\mathrm{n} + 1) = \mathrm{f}(2\mathrm{n} + 1,1) = 3\mathrm{n} + 2\) , \(\mathrm{f}(2\mathrm{n} + 1,2\mathrm{n}) = 3\mathrm{n} + 1\) . A rather more fiddly induction then shows that these are the best possible lengths. Hence the longest sequence with no element more than 1998 is that starting 1, 1998 which has length 2998.


499 white rooks and a black king are placed on a \(1000 \times 1000\) chess board. The rook and king moves are the same as in ordinary chess, except that taking is not allowed and the king is allowed to remain in check. No matter what the initial situation and no matter how white

moves, the black king can always:

(a) get into check (after some finite number of moves);
(b) move so that apart from some initial moves, it is always in check after its move;
(c) move so that apart from some initial moves, it is always in check (even just after white has moved).

Prove or disprove each of (a) - (c).

Solution:
(a) True. Black moves to one end of a main diagonal and then moves along the diagonal to the opposite end. Each of the 499 rooks is in some row. Since black moves through each row, every rook must change row. But each of the rooks is also in some column and so every rook must also change column. A rook cannot change row and column in the same move, so white must make at least 998 moves before black reaches the opposite end of the diagonal. But it cannot start until black is two moves from its starting position, because if it moves a rook into row (or column) one or two earlier, then black is checked or can move into check. So it has only 997 moves available, which is one too few.

(b) False. Suppose the contrary, that after move n, the king is always in check after its move. Let the corners of the board be A, B, C, D. After move n, white moves all its rooks inside a square side 23 at corner A. The king must now be in the 23 rows between A and B or in the 23 columns between A and D. Suppose the latter. Then white moves all its rooks inside a square side 23 at corner B. This should take 499 moves. However, it could take longer if black used his king to obstruct the move. The worst case would be 3 x 23 additional moves (the king can only obstruct one row of 23 rooks, and each rook in the obstructed row could take 4 moves instead of one to reach its destination.). During this period the king must remain in the 23 rows from A to B or the 23 columns from A to D, since it must remain in check. Thus it cannot get to B by the completion of the process. In fact, it must be at least 999 - 46 (the total number of moves required) - (499 + 69) (the number of moves available) = 385 moves behind.

White now moves all the rooks inside a square side 23 at corner C. The king cannot cut across (or it will be unchecked). It must keep within 23 squares of the edge. So it ends up 770 moves behind (more in fact, since it cannot obstruct the move as effectively). Finally, white moves all the rooks inside a square side 23 at corner D. The king cannot get to the side CD by the time this process is completed. So there is then a lag of over two hundred moves before it can get back into check. Note that it does not help black to change direction. Whatever black does, white ends up with all the rooks at a corner and the king a long way from the two checked sides.

(c) False. This follows from (b). But we may also use a simpler argument. Take coordinates \(x = 1\) to 1000, \(y = 1\) to 1000. White gets its pieces onto (2,0), (4,0), ..., (998,0). If the king moves onto \((2n,*)\) , then white moves its rook from \((2n,0)\) to \((2n-1,0)\) , leaving the king unchecked. If the king moves to \((2n-1,*)\) or \((2n+1,*)\) , then white moves its rook back to \((2n,0)\) , leaving the king unchecked. If the king stays on the line \((2n,*)\) , then white fills in time by toggling one of its endmost rooks to an adjacent square (and the king remains unchecked). The only way the black king can escape this repeated unchecking is by moving up to the line \(y = 0\) . If it does so, then white transfers all its rooks to the line \(y = 1000\) and repeats the process. The transfer takes 499 moves. It takes black 1000 moves to follow, so during the 501 moves before black catches up, the king is subject to repeated unchecking.


ABCD is a unit square. One vertex of a rhombus lies on side AB, another on side BC, and a third on side AD. Find the area of the set of all possible locations for the fourth vertex of the rhombus.

Solution:
Answer: \(2 1 / 3\)

Let the square be ABCD. Let the vertices of the rhombus be P on AB, Q on AD, and R on BC. We require the locus of the fourth vertex S of the rhombus. Suppose P is a distance x from B. We may take \(\mathrm{x}< = 1 / 2\) , since the locus for \(\mathrm{x} > 1 / 2\) is just the reflection of the locus for \(\mathrm{x}< 1 / 2\) . Then since PR is parallel to QS, S is a distance x from the line AD. Also, by continuity, as Q varies over AD (with P fixed a distance x from B), the locus of S is a line segment.

The two extreme positions for S occur when Q coincides with A and when R coincides with C. When Q coincides with A the rhombus has side \(1 - \mathrm{x}\) . Hence \(\mathrm{BR}^2 = (1 - \mathrm{x})^2 - \mathrm{x}^2 = 1 - 2\mathrm{x}\) . In this case SR is parallel to AB, so the distance of S from AB is \(\sqrt{(1 - 2\mathrm{x})}\) . When R coincides with C, the rhombus has side \(\sqrt{(1 + \mathrm{x}^2)}\) , so \(\mathrm{AQ}^2 = 1 + \mathrm{x}^2 - (1 - \mathrm{x})^2 = 2\mathrm{x}\) . Hence the distance of S from AB is \(1 + \sqrt{(2\mathrm{x})}\) .

Thus the locus of S over all possible rhombi is the interior of a curvilinear quadrilateral with vertices MDNC, where M is the midpoint of AB and N is the reflection of M in CD. Moreover the curve from M to C is just the translate of the curve from D to N, for if we put y \(= 1 / 2 - \mathrm{x}\) , then \(\sqrt{(1 - 2\mathrm{x})}\) becomes \(\sqrt{(2\mathrm{y})}\) . Thus if L is the midpoint of CD, then the area in the MLC plus the area in DLN is just \(1 / 2\) , and the total area of the curvilinear quadrilateral is 1.

However, the arrangement of the vertices discussed above is not the only one. The order of vertices above is PQSR. We could also have PQRS or PSQR. In either case QR is a side rather than a diagonal of the rhombus. We consider the case PQRS (the case PSQR is just the reflection in the line MN). As before it is convenient to keep P fixed, but this time we take x to be the distance AP. Take y to be the distance AQ.

As before we find that S must lie on a line parallel to BC a distance x from it (on the other side to AD). Again we find that for fixed P, the locus of S is a segment of this line. If we assume that \(\mathrm{AQ} > \mathrm{BR}\) , then the two extreme positions are (1) QR parallel to AB, giving S on the line AB, (2) Q at D, giving S a distance x from the line AB. So as x varies from 0 to 1 we get a right- angled triangle sides 1, 1 and \(\sqrt{2}\) and area \(1 / 2\) . However, we can also have \(\mathrm{BR} > \mathrm{AQ}\) . This gives points below the line AB. The extreme position is with R at C. Suppose \(\mathrm{QD} = \mathrm{y}\) . Then \(1 + \mathrm{y}^2 = \mathrm{x}^2 + (1 - \mathrm{y})^2\) , so \(\mathrm{y} = \mathrm{x}^2 /2\) . This gives S a distance y below the line AB. This gives an additional area of \(1 / 6\) (by calculus - integrate \(\mathrm{x}^2 /2\) from 0 to 1; I do not see how to do it without).

The triangle and the curvilinear triangle together form a curvilinear triangle area \(1 / 2 + 1 / 6 = 2 / 3\) . There is an identical triangle formed by reflection in MN. Thus the total area is \(1 + 2 / 3 + 2 / 3 = 2 1 / 3\) .

Thanks to Robert Hill and John Jones for pointing out that the original solution missed out the two triangles.


A natural number k has the property that if k divides n, then the number obtained from n by reversing the order of its digits is also divisible by k. Prove that k is a divisor of 99.

Solution:
Let \(\mathrm{r(m)}\) denote the number obtained from m by reversing the digits.

We show first that k cannot be divisible by 2 or 5. It cannot be divisible by both, for then it ends in a zero and hence \(\mathrm{r(k)}< \mathrm{k}\) and so is not divisible by k (contradiction). So if 5 divides k, then the last digit of k must be 5. Since \(\mathrm{r(k)}\) is divisible by 5 its last digit must also be 5, so the first digit of k is 5. But now 3k has first digit 1 \((3.5 > 10\) and \(3.6< 20\) ), so \(\mathrm{r(3k)}\) has last digit 1 and cannot be divisible by 5. Contradiction. If 2 divides k, then every multiple of k must be even. So the last digit of \(\mathrm{r(k)}\) must be even and hence the first digit of k must be 2, 4, 6, or 8. If 2, then 5k has first digit 1, so \(\mathrm{r(2k)}\) is odd. Contradiction. Similarly, if the first digit is 4, 3k has first digit 1; if 6, then 5k has first digit 3; if 8, then 2k has first digit 1. Contradiction. So k is not divisible by 2 or 5.

Suppose \(\mathrm{k} = 10^{\mathrm{n}}\mathrm{a}_{\mathrm{n}} + \ldots +\mathrm{a}_{0}\) k divides \(\mathrm{r(k)}\) , so \(\mathrm{a}_{0} > = 1\) . Hence \((10^{\mathrm{n} + 1} - 1)\mathrm{k} = 10^{2\mathrm{n} + 1}\mathrm{a}_{\mathrm{n}} + \ldots +\) \(10^{\mathrm{n} + 1}\mathrm{a}_{0} - (10^{\mathrm{n}}\mathrm{a}_{\mathrm{n}} + \ldots +\mathrm{a}_{0}) = 10^{2\mathrm{n} + 1}\mathrm{a}_{\mathrm{n}} + \ldots +10^{\mathrm{n} + 1}(\mathrm{a}_{0} - 1) + 10^{\mathrm{n}}\mathrm{c}_{\mathrm{n}} + \ldots +10\mathrm{c}_{1} + (\mathrm{c}_{0} + 1)\) , where \(\mathrm{c_{i}} =\) 9 - a. The reverse of this, \(10^{2\mathrm{n} + 1}(\mathrm{c}_{0} + 1) + 10^{2\mathrm{n}}\mathrm{c}_{1} + \ldots +10^{\mathrm{n} + 1}\mathrm{c}_{\mathrm{n}} + 10^{\mathrm{n}}(\mathrm{a}_{0} - 1) + \ldots +\mathrm{a}_{\mathrm{n}}\) , is also divisible by k. So is the reverse of k, \(10^{\mathrm{n}}\mathrm{a}_{0} + \ldots +\mathrm{a}_{\mathrm{n}}\) and hence also their difference: \(10^{\mathrm{n}}(10^{\mathrm{n} + 1}(\mathrm{c}_{0} + 1) + 10^{\mathrm{n}}\mathrm{c}_{1} + \ldots +10\mathrm{c}_{\mathrm{n}} - 1)\) . k has no factors 2 or 5, so k must divide \(10^{\mathrm{n} + 1}(\mathrm{c}_{0} + 1)\) \(+10^{\mathrm{n}}\mathrm{c}_{1} + \ldots +10\mathrm{c}_{\mathrm{n}} - 1\) . Adding 10k, we find that k also divides \(10^{\mathrm{n} + 2} + 10^{\mathrm{n}}9 + \ldots +10.9 - 1 =\) 1099. .989 (n - 2 consecutive 9s) \(= 11(10^{\mathrm{n} + 1} - 1)\)

We can now carry out exactly the same argument starting with \((10^{\mathrm{n} + 2} - 1)\mathrm{k}\) . This leads to k dividing \(10^{\mathrm{n} + 2}(\mathrm{c}_{0} + 1) + \ldots +10^{2}\mathrm{c}_{0} + 10.9 - 1\) and hence also \(10^{\mathrm{n} + 3} + 10^{\mathrm{n} + 1}9 + \ldots +10^{2}9 + 10.8+\) \(9 = 11(10^{\mathrm{n} + 2} - 1)\) . Subtracting 10 times this from the previous number we conclude that k must divide \(11(10^{\mathrm{n} + 1} - 1) - 11(10^{\mathrm{n} + 1} - 10) = 99\)

Finally, we note that any factor of 99 has the required property. For 3 and 9 divide a number if and only if they divide its digit sum. So if m is divisible by 3 or 9, then the number formed by any rearrangement of its digits is also divisible by 3 or 9. m is divisible by 11 if and only if the difference between the sums of alternate digits is divisible by 11, so if m is divisible by 11, then so is its reverse.


An octagon has equal angles. The lengths of the sides are all integers. Prove that the opposite sides are equal in pairs.

Solution: Extend the sides to form two rectangles. Let the sides of the octagon have length a, b, c, d, e, f, g, h. Then we can find the rectangle sides. For example, one of the rectangles has opposite sides \(\mathrm{a + (b + h) / \sqrt{2}}\) and \(\mathrm{e + (d + f) / \sqrt{2}}\) . Hence either \(\mathrm{a = e}\) or \(\sqrt{2} = (\mathrm{b + h - d - f}) / (\mathrm{a - e})\) . The root is irrational, so we must have \(\mathrm{a = e}\) . Similarly for the other pairs of opposite sides.


Which is greater: \(31^{11}\) or \(17^{14}\) ? [No calculators allowed!]

Solution: \[17^{2} = 289 > 9.31. \mathrm{So} 17^{14} > 9^{7} 31^{7}. \mathrm{But} 3^{7} = 2187 > 31^{2}. \mathrm{Hence} 17^{14} > 31^{11}.\]


A circle radius 100 is drawn on squared paper with unit squares. It does not touch any of the grid lines or pass through any of the lattice points. What is the maximum number of squares can it pass through?

Solution: Take compass directions aligned with the grid. Let N, E, S, W be the most northerly, easterly, southerly and westerly points on the circle. The arc from N to E must cross 100 north- south grid lines and 100 east- west grid lines. Each time it crosses a grid line it changes square (and it never crosses two grid lines at once, because it does not pass through any lattice points), so the arc N to E must pass through 200 in addition to the starting square. Similarly for the other 4 arcs. So the circle passes through a total of 800 squares (we count the starting square in the last 200).


In a group of students, 50 speak English, 50 speak French and 50 speak Spanish. Some students speak more than one language. Prove it is possible to divide the students into 5 groups (not necessarily equal), so that in each group 10 speak English, 10 speak French and 10 speak Spanish.

Solution:
Let EF denote the number of students speaking English and French. Similarly define ES, FS, E, F, S, EFS. Then ES + EF + E + EFS = 50, EF + FS + F + EFS = 50. Subtracting: ES - F = FS - E. Similarly, ES - F = EF - S.

Pair off members of FS with members of E. Similarly, members of ES with F, and members of EF with S. The resulting pairs have one person speaking each language. If \(\mathrm{ES} = \mathrm{F}\) , then the only remaining students are those in EFS, who speak all three languages. We thus have a collection of units (pairs or individuals) each containing one speaker of each language.

If \(\mathrm{ES}< \mathrm{F}\) , then after the pairing off we are left with equal numbers of members of E, F, and S. These may be formed into triplets, with each triplet containing one speaker of each language. As before we also have the students in EFS. Again, we have partitioned the student body into units with each unit containing one speaker of each language.

If \(\mathrm{ES} > \mathrm{F}\) , then after the pairing off, we are left with an equal number of members of ES, FS and EF. These may be formed into triplets, with each triplet containing two speakers of each language. So, in this case we partition the student body into units with each unit containing either one speaker of each language, or two speakers of each language.

Finally, we may divide the units into 5 groups with 10 speakers of each language in each group.


Solution: \[\mathrm{lhs} = 1 / (\mathrm{x} - 1) - 1 / (\mathrm{x} + 1) + 1 / (\mathrm{x} - 2) - 1 / (\mathrm{x} + 2) + \ldots +1 / (\mathrm{x} + 10) - 1 / (\mathrm{x} - 10) =\] \[1 / (\mathrm{x} - 1) - 1 / (\mathrm{x} + 10) + 1 / (\mathrm{x} - 2) - 1 / (\mathrm{x} + 9) + \ldots +1 / (\mathrm{x} - 10) - 1 / (\mathrm{x} + 1) = \mathrm{rhs}.\]


The difference between the longest and shortest diagonals of the regular n- gon equals its side. Find all possible n.

Solution:
Answer: \(\mathrm{n} = 9\)

For \(\mathrm{n}< 6\) , there is at most one length of diagonal. For \(\mathrm{n} = 6\) , 7 the longest and shortest, and a side of the n- gon form a triangle, so the difference between the longest and shortest is less than the side.

For \(\mathrm{n} > 7\) the side has length 2R sin \(\pi /\mathrm{n}\) , the shortest diagonal has length 2R sin \(2\pi /\mathrm{n}\) , and the longest diagonal has length 2R for n even and 2R cos \(\pi /2\mathrm{n}\) for n odd (where R is the radius of the circumcircle). Thus we require:

\(\sin 2\pi /\mathrm{n} + \sin \pi /\mathrm{n} = 1\) and n even, or

\(\sin 2\pi /\mathrm{n} + \sin \pi /\mathrm{n} = \cos \pi /2\mathrm{n}\) and n odd.

Evidently the lhs is a strictly decreasing function of n and the rhs is an increasing function of n, so there can be at most one solution of each equation. The second equation is satisfied by n \(= 9\) , although it is easier to see that there is a quadrilateral with the longest diagonal and shortest diagonals as one pair of opposite sides, and 9- gon sides as the other pair of opposite sides. The angle between the longest side and an adjacent side is 60, so that its length is the length of the shortest diagonal plus \(2 \times 9\) - gon side x cos 60. Hence that is the only solution for n odd.

For \(\mathrm{n} = 8\) we have the same quadrilateral as for the 9- gon except that the angle is 67.5 and hence the difference is less than 1. For \(\mathrm{n} = 10\) , \(\sin 2\pi /10 + \sin \pi /10 = \sin \pi /10\) (2 cos \(\pi /10 + 1\) ) \(< 3\) sin \(\pi /10< 3\pi /10< 1\) . So there are no solutions for n even \(\geq 10\) , and hence no solutions for n even.


The sequence \(\mathrm{a}_{\mathrm{n}}\) is defined as follows: \(\mathrm{a}_{1} = 1\) , \(\mathrm{a}_{\mathrm{n} + 1} = \mathrm{a}_{\mathrm{n}} + 1 / \mathrm{a}_{\mathrm{n}}\) for \(\mathrm{n} \geq 1\) . Prove that \(\mathrm{a}_{100} > 14\) .

Solution:
First we must notice that for \(1 \leq \mathrm{a}\) , b we have \(\mathrm{a} < \mathrm{b}\) , then \(\mathrm{a} + 1 / \mathrm{a} < \mathrm{b} + 1 / \mathrm{b}\) . This is basic to any estimation.

The obvious approach is to notice that if \(\mathrm{a}_{\mathrm{i}} \leq \mathrm{n}\) , then \(\mathrm{a}_{\mathrm{i} + 1} \geq \mathrm{a}_{\mathrm{i}} + 1 / \mathrm{n}\) . Hence it takes at most n steps to get from \(\mathrm{n} - 1\) to \(\mathrm{n}\) . Unfortunately, this does not quite work: we need \(2 + 3 + \ldots + 14 = 104\) steps to get from 1 to 14.

The trick is to notice that \(\mathrm{a}_{\mathrm{n} + 1}^{2} > \mathrm{a}_{\mathrm{n}}^{2} + 2\) . But \(\mathrm{a}_{2} = 2\) , so \(\mathrm{a}_{\mathrm{n}}^{2} > 2\mathrm{n}\) . That gives \(\mathrm{a}_{100}^{2} > 200 > 14^{2}\) .


Given point O inside the acute- angled triangle ABC, and point O' inside the acute- angled triangle A'B'C'. D, E, F are the feet of the perpendiculars from O to BC, CA, AB respectively, and D', E', F' are the feet of the perpendiculars from O' to B'C', C'A', A'B' respectively. OD is parallel to O'A', OE is parallel to O'B' and OF is parallel to O'C'. Also OD- O'A' = OE- O'B' = OF- O'C'. Prove that O'D' is parallel to OA, O'E' to OB and O'F' to OC, and that O'D' OA = O'E' OB = O'F' OC.

Solution:
![md5:ca87b07b2609f5391dc57bee5f28e69d](ca87b07b2609f5391dc57bee5f28e69d.jpeg)

Let \(\Gamma\) be the circumcircle of DEF. Let OD, OE, OF meet it again at A", B", C" respectively. Then the figure O'A'B'C' must be similar to OA"B"C". So to prove that OD is parallel to O'A', we have to prove that AO is perpendicular to B"C".

So AO meets B"C" at D". Now since \(\square \mathrm{AFC}'' = 90^{0}\) and \(\square \mathrm{AD}''\mathrm{C}'' = 90^{0}\) , both F and D' lie on the circle diameter AC". Hence AO- OD" = OF- OC". Similarly, BO meets C"A" at E", and CO meets A"B" at F", and BO- OE" = OD- OA" and CO- OF" = OE- OB". Hence OD". OA = OE". OB = OF". OC. So using the similarity, OD'- OA = OE'- OB = OF'. OC.


Prove that any positive integer not exceeding n! can be written as a sum of at most n distinct factors of n!.

Solution: Given \(\mathrm{m}\leq \mathrm{n}!\) write \(\mathrm{m} = \mathrm{nq} + \mathrm{r}\) (\*), with \(0\leq \mathrm{q},0\leq \mathrm{r}< \mathrm{m}\) . Then \(\mathrm{q}\leq (\mathrm{n} - 1)!\) , so \(\mathrm{q}\) is a sum of at most n- 1 distinct factors of (n- 1)!. r is itself a factor of n! and is not divisible by n, so (\*) expresses m as a sum of at most n distinct factors of n!.


Given a triangle ABC, and D on the segment AB, E on the segment AC, such that AD = DE = AC, BD = AE, and DE is parallel to BC. Prove that BD equals the side of a regular 10- gon inscribed in a circle with radius AC.

Solution:
\(\mathrm{DA} = \mathrm{DE}\) , so DAE is isosceles. DE is parallel to BC, so ABC is isosceles, so \(\mathrm{BA} = \mathrm{AC} / (2 \cos \mathrm{A})\) . Hence \(\mathrm{BD} = \mathrm{AC} / (2 \cos \mathrm{A}) - \mathrm{AC}\) . But \(\mathrm{AE} = 2 \mathrm{AC} \cos \mathrm{A}\) , so we have an equation for \(\mathrm{c} = \mathrm{cos} \mathrm{A}: 4 \mathrm{c}^{2} + 2 \mathrm{c} - 1 = 0\) .

\(2\pi /5,4\pi /5,6\pi /5,8\pi /5\) and \(10\pi /5\) are the roots of: real part of \((\cos \theta + \mathrm{i}\sin \theta)^{5} = 1\) . Expanding this gives that \(\cos 2\pi /5\) , \(\cos 4\pi /5\) , \(\cos 6\pi /5\) , \(\cos 8\pi /5\) and 1 are the roots of \(16\mathrm{c}^{5} - 20\mathrm{c}^{3} + 5\mathrm{c} - 1 = 0\) . Dividing by \((\mathrm{c} - 1)\) gives \(16\mathrm{c}^{4} + 16\mathrm{c}^{3} - 4\mathrm{c}^{2} - 4\mathrm{c} + 1 = (4\mathrm{c}^{2} + 2\mathrm{c} - 1)^{2}\) . So \(\cos 2\pi /5\) ( \(=\) cos \(8\pi /5\) ) and \(\cos 4\pi /5\) ( \(=\) cos \(6\pi /5\) ) are the roots of \(4\mathrm{c}^{2} + 2\mathrm{c} - 1 = 0\) .

We know that \(\mathrm{A}< 90^{\circ}\) (since \(\mathrm{A} = \mathrm{C}\) and their sum is less than \(180^{\circ}\) ). Hence \(\mathrm{A} = 2\pi /5\) . So BD \(= 2\) AC cos \(2\pi /5 = 2\) AC sin \(\pi /10\) , which is the side length for a regular 10- gon inscribed in a circle radius AC.


Given a regular tetrahedron ABCD, prove that it is contained in the three spheres on diameters AB, BC and AD. Is this true for any tetrahedron?

Solution:
Let the tetrahedron have side 1. Then the center O is a distance \(1 / \sqrt{8}\) from the center of each of the spheres, so it is contained in each of the spheres. We now use convexity.

Two circles with diameters two of the sides of a triangle cover the triangle (consider the foot of the altitude to the third side), so faces ABC and ABD are certainly contained in the spheres. Consider face ACD. The sphere on BC passes through the midpoints of AC and CD, and through C, so it contains the triangle formed by these three points (by convexity). But the rest of ACD is contained in the sphere on AD. Similarly for the face BCD. Hence all the faces are contained in the spheres. But now take any point P inside the tetrahedron. Extend OP to meet a face at X. X lies in one of the spheres, but O also lies in the sphere and hence all points on OX, including P (by convexity).

False in general. Take ABCD to be a plane square, then no points on CD are in the spheres except C and D (and we can obviously distort this slightly to make it less degenerate).


(a) Given a \(4 \times 4\) array with \(+\) signs in each place except for one non-corner square on the perimeter which has a - sign. You can change all the signs in any row, column or diagonal. A diagonal can be of any length down to 1. Prove that it is not possible by repeated changes to arrive at all \(+\) signs.

(b) What about an \(8 \times 8\) array?

Solution:
(a) Let S be the set of the 8 positions on the perimeter not at a corner. Any move changes the sign of either 2 or 0 of the members of S. We start with an odd number of members of S with a minus sign, so we must always have an odd number of members of S with a minus sign and hence cannot get all plus signs.

(b) Also impossible. The same argument works.


The medians divide a triangle into 6 smaller triangles. 4 of the circles inscribed in the smaller triangles have equal radii. Prove that the original triangle is equilateral.

Solution:
Denote the side lengths by a, b, c and the corresponding median lengths by \(\mathrm{m}_{\mathrm{a}}\) , \(\mathrm{m}_{\mathrm{b}}\) , \(\mathrm{m}_{\mathrm{c}}\) . The six small triangles all have equal area. [Let the areas be \(\mathrm{t}_1\) , ..., \(\mathrm{t}_6\) . It is obvious that the adjacent pairs have equal height and equal base, so we have \(\mathrm{t}_1 = \mathrm{t}_2\) , \(\mathrm{t}_3 = \mathrm{t}_4\) , \(\mathrm{t}_5 = \mathrm{t}_6\) . The three on each side of a median sum to the same area, so \(\mathrm{t}_1 + \mathrm{t}_2 + \mathrm{t}_3 = \mathrm{t}_4 + \mathrm{t}_5 + \mathrm{t}_6\) , \(\mathrm{t}_1 + \mathrm{t}_5 + \mathrm{t}_6 = \mathrm{t}_2 + \mathrm{t}_3 + \mathrm{t}_4\) . Subtracting gives \(\mathrm{t}_1 = \mathrm{t}_4\) . Similarly, \(\mathrm{t}_2 = \mathrm{t}_5\) and we are home.] So by the usual result that the twice area of a triangle equals its perimeter times its in- radius, we conclude that the perimeters of four of the small triangles are equal.

Two of them must share a side of the original triangle. Suppose it is a. Then we have: \(a / 2+\) \(\mathrm{m}_{\mathrm{a}} / 3 + 2\mathrm{m}_{\mathrm{b}} / 3 = a / 2 + \mathrm{m}_{\mathrm{a}} / 3 + 2\mathrm{m}_{\mathrm{c}} / 3\) . So \(\mathrm{m}_{\mathrm{b}} = \mathrm{m}_{\mathrm{c}}\) . That implies that \(\mathrm{b} = \mathrm{c}\) . [Because the triangle formed by the centroid and side a is isosceles, so the median is perpendicular to the side, so the main triangle is isosceles.]

Using the facts that \(\mathrm{b} = \mathrm{c}\) and \(\mathrm{m}_{\mathrm{b}} = \mathrm{m}_{\mathrm{c}}\) , we see that two of the remaining small triangles have perimeter \(\mathrm{b} / 2 + \mathrm{m}_{\mathrm{b}}\) and two have perimeter \(\mathrm{b} / 2 + \mathrm{m}_{\mathrm{b}} / 3 + 2\mathrm{m}_{\mathrm{a}} / 3\) . So there are two cases to consider. In the first case \(a / 2 - \mathrm{m}_{\mathrm{a}} / 3 = \mathrm{b} / 2 - \mathrm{m}_{\mathrm{b}} / 3\) . That implies \(\mathrm{a} = \mathrm{b}\) , since if \(\mathrm{a}< \mathrm{b}\) , \(\mathrm{m}_{\mathrm{a}} > \mathrm{m}_{\mathrm{b}}\) (consider the triangle formed by the centroid and the side c). So the triangle is equilateral.

The second case is harder. We have: \(a / 2 + \mathrm{m}_{\mathrm{a}} / 3 + 2\mathrm{m}_{\mathrm{b}} / 3 = \mathrm{b} / 2 + \mathrm{m}_{\mathrm{b}}\) , and hence \(a / 2 + \mathrm{m}_{\mathrm{a}} / 3 = \mathrm{b} / 2 + \mathrm{m}_{\mathrm{b}} / 3\) (\*). Take the angle between a and b to be \(\theta\) . Then \(\mathrm{m}_{\mathrm{a}} = \mathrm{b}\) sin \(\theta\) , \(\mathrm{a} = 2\mathrm{b}\cos \theta\) , and \(\mathrm{m}_{\mathrm{b}}^{2} = \mathrm{b}^{2} / 4 + \mathrm{a}^{2} - \mathrm{ab}\cos \theta = \mathrm{b}^{2} / 4 + 2\mathrm{b}^{2}\cos^{2}\theta\) . We can now use (\*) to get an equation for \(\theta\) . First we square (\*) to get: \(\mathrm{m}_{\mathrm{b}}^{2} = (3a / 2 - 3b / 2 + \mathrm{m}_{\mathrm{a}})^{2}\) . We divide out the factor \(\mathrm{b}^{2}\) to get: \(1 / 4 + 2\cos^{2}\theta = 3 1 / 4 + 8\cos^{2}\theta - 9\cos \theta + 3\sin \theta (2\cos \theta - 1)\) . Squaring, so that we can use \(\sin^{2}\theta = 1 - \cos^{2}\theta\) , and writing \(\mathrm{c} = \cos \theta\) , we get: \((1 - \mathrm{c}^{2})(4\mathrm{c}^{2} - 4\mathrm{c} + 1) = 4\mathrm{c}^{4} - 12\mathrm{c}^{3} + 13\mathrm{c}^{2} - 6\mathrm{c} + 1\) . Hence \(8\mathrm{c}^{4} - 16\mathrm{c}^{3} + 10\mathrm{c}^{2} - 2\mathrm{c} = 0\) . Factorizing: \(\mathrm{c}(\mathrm{c} - 1)(2\mathrm{c} - 1)^{2} = 0\) . \(\mathrm{c} = 0\) and \(\mathrm{c} = 1\) give degenerate triangles, so we must have \(\mathrm{c} = 1 / 2\) and hence the triangle is equilateral.


Prove that we can find positive integers \(\mathrm{x}\) , y satisfying \(\mathrm{x}^{2} + \mathrm{x} + 1 = \mathrm{py}\) for an infinite number of primes p.

Solution: This is a trivial variant on the proof that there are an infinite number of primes. Suppose that we can only find \(\mathrm{x}\) , y for a finite number of primes \(\mathrm{p}_1\) , \(\mathrm{p}_2\) , ..., \(\mathrm{p}_n\) . Set \(\mathrm{x} = \mathrm{p}_1\mathrm{p}_2\dots \mathrm{p}_n\) . Then none of the \(\mathrm{p}_i\) can divide \(\mathrm{x}(\mathrm{x} + 1) + 1\) . But it must have prime factors. Contradiction.


9 judges each award 20 competitors a rank from 1 to 20. The competitor's score is the sum of the ranks from the 9 judges, and the winner is the competitor with the lowest score. For each competitor the difference between the highest and lowest ranking (from different judges) is at most 3. What is the highest score the winner could have obtained?

Solution:
Answer: 24.

At most 4 competitors can receive a rank 1. For a competitor with a rank 1 can only receive ranks 1, 2, 3 or 4. There are only 36 such ranks available and each competitor with a rank 1 needs 9 of them.

If only one competitor receives a rank 1, then his score is 9. If only 2 competitors receive a rank 1, then one of them must receive at least five rank 1s. His maximum score is then \(5.1 + 4.4 = 21\) . If 4 competitors receive a rank 1, then they must use all the 36 ranks 1, 2, 3, and 4. The total score available is thus \(9(1 + 2 + 3 + 4) = 90\) , so at least one competitor must receive 22 or less. Thus the winner's maximum score is at most 22. If 3 competitors receive a rank 1, then the winner's score is maximised by giving all three competitors the same score and letting them share the 27 ranks 1, 3 and 4. That gives a winner's score of \(9(1 + 3 + 4) / 3 = 24\) . That can be achieved in several ways, for example: each competitor gets 3 1s, 3 3s and 3 4s, or one competitor gets 4 1s and 5 4s, another gets 3 1s, 3 3s and 3 4s, another gets 2 1s 6 3s and one 4. Note that it is trivial to arrange ranks for the remaining 17 competitors. For example: give one 5 2s and 4 5s total 30, one 4 2s and 5 5s total 33, and then one 9 6s, one 9 7s and so on.

Thus the answer is 24, with three joint winners. If there is required to be a single winner, then the answer is 23.


\(\{\mathrm{a_i}\}\) and \(\{\mathrm{b_i}\}\) are permutations of \(\{1 / 1,1 / 2,\dots ,1 / \mathrm{n}\}\) . \(\mathrm{a_1 + b_1\geq a_2 + b_2\geq \ldots \geq a_n + b_n}\) . Prove that for every m \((1\leq \mathrm{m}\leq \mathrm{n})\) \(\mathrm{a_m + a_n\geq 4 / m}\)


There is a set of scales on the table and a collection of weights. Each weight is on one of the two pans. Each weight has the name of one or more pupils written on it. All the pupils are outside the room. If a pupil enters the room then he moves the weights with his name on them to the other pan. Show that you can let in a subset of pupils one at a time, so that the scales change position after the last pupil has moved his weights.


The streets in a city are on a rectangular grid with m east- west streets and n north- south streets. It is known that a car will leave some (unknown) junction and move along the streets at an unknown and possibly variable speed, eventually returning to its starting point without ever moving along the same block twice. Detectors can be positioned anywhere except at a junction to record the time at which the car passes and it direction of travel. What is the minimum number of detectors needed to ensure that the car's route can be reconstructed?


The circle inscribed in the triangle ABC touches the side AC at K. Prove that the line joining the midpoint of AC with the center of the circle bisects the segment BK.


The sequence \(\mathrm{a_1,a_2,\ldots,a_n}\) satisfies the following conditions: \(\mathrm{a_1 = 0}\) \(\mathrm{|a_i| = |a_{i - 1} + 1|}\) for \(\mathrm{i} = 2\) , 3, ... , n. Prove that \((\mathrm{a_1 + a_2 + \ldots + a_n}) / \mathrm{n}\geq - 1 / 2\) .


The sides and diagonals of ABCD have rational lengths. The diagonals meet at O. Prove that the length AO is also rational.

Solution: \(\mathrm{AB} = \mathrm{AO}\cos \mathrm{OAB} + \mathrm{BO}\cos \mathrm{OBA}\) . We can derive a rational expression for cos OAB using the cosine rule for triangle ABC. Similarly for cos OBA using the cosine rule for triangle DAB. So \(\mathrm{OA} = \mathrm{r}_1 + \mathrm{r}_2\mathrm{OB}\) , where \(\mathrm{r}_1\) denotes a rational number. Similarly, \(\mathrm{OB} = \mathrm{r}_3 + \mathrm{r}_4\mathrm{OC}\) , so \(\mathrm{OA} = \mathrm{r}_5 + \mathrm{r}_6\mathrm{OC}\) . But \(\mathrm{OA} + \mathrm{OC} = \mathrm{AC} = \mathrm{r}_7\) . Hence OA is rational.


In the quadrilateral ABCD, BC is parallel to AD. The point E lies on the segment AD and the perimeters of ABE, BCE and CDE are equal. Prove that \(\mathrm{BC} = \mathrm{AD} / 2\) .

Solution:
Take \(\mathrm{E_1}\) on the line AD so that \(\mathrm{AE_1CB}\) is a parallelogram. Then \(\mathrm{AE_1 = BC}\) \(\mathrm{AB = CE_1}\) , so triangles \(\mathrm{ABE_1}\) and \(\mathrm{BCE_1}\) have equal perimeters. Moreover, \(\mathrm{E_1}\) is the only point on the line for which this is true. For if we move E a distance \(\mathbf{x}\) from \(\mathrm{E_1}\) , then we change \(\mathrm{AE_1}\) by \(\mathbf{x}\) , and \(\mathrm{CE_1}\) by less than \(\mathbf{x}\) . AB and BC are unchanged. So \(\mathrm{AB + BE_1}\) and \(\mathrm{BC + CE_1}\) are changed by different amounts. Hence the perimeters of \(\mathrm{ABE_1}\) and \(\mathrm{BCE_1}\) are no longer equal.

Similarly, let \(\mathrm{E_2}\) be the point on the line AD so that \(\mathrm{BCDE_2}\) is a parallelogram. Then \(\mathrm{E_2}\) is the unique point such that \(\mathrm{BCE_2}\) and \(\mathrm{CDE_2}\) have equal perimeters. So if all three triangles have equal perimeters, then \(\mathrm{E_1}\) and \(\mathrm{E_2}\) must coincide and hence \(\mathrm{BC} = \mathrm{AE} = \mathrm{DE}\) , so \(\mathrm{BC} = \mathrm{AD} / 2\) .


A wolf is in the center of a square field and there is a dog at each corner. The wolf can run anywhere in the field, but the dogs can only run along the sides. The dogs' speed is 3/2 times the wolf's speed. The wolf can kill a single dog, but two dogs together can kill the wolf. Prove that the dogs can prevent the wolf escaping.


A finite sequence of 0s and 1s has the following properties: (1) for any \(\mathrm{i}< \mathrm{j}\) , the sequences of length 5 beginning at position i and position j are different; (2) if you add an additional digit at either the start or end of the sequence, then (1) no longer holds. Prove that the first 4 digits of the sequence are the same as the last 4 digits.

Solution: Let the last 4 digits be abcd. Then the 5 digit sequences abcd0 and abcd1 must occur somewhere. If neither of them are at the beginning then there are three 5 digit sequences xabcd, two of which must therefore be the same, contradicting (1). Hence abcd are the first 4 digits.


Given positive numbers a, b, c, d prove that at least one of the inequalities does not hold: \(\mathrm{a + }\) \(\mathrm{b< c + d}\) ; \((\mathrm{a + b})(\mathrm{c + d})< \mathrm{ab + cd}\) ; \((\mathrm{a + b})\mathrm{cd}< \mathrm{ab}(\mathrm{c + d})\) .

Solution: From the first and second inequalities we have \(\mathrm{ab + cd > a(c + d) + b(a + b)}\) , so \(\mathrm{cd > ad}\) , and hence \(\mathrm{c > a}\) . We also have \(\mathrm{ab + cd > a(a + b) + b(c + d)}\) , so \(\mathrm{cd > bc}\) , and hence \(\mathrm{d > b}\) . So \(1 / \mathrm{a} + 1 / \mathrm{b} > 1 / \mathrm{c} + 1 / \mathrm{d}\) , which contradicts the third inequality.


What is the smallest positive integer a such that we can find integers b and c so that \(\mathrm{ax}^2 +\mathrm{bx} + \mathrm{c}\) has two distinct positive roots less than 1?

Solution:
\(4\mathrm{x}^2 - 4\mathrm{x} + 1 = (2\mathrm{x} - 1)^2\) , which has the double root 1/2. So it remains to consider \(\mathrm{a} = 1,2,3\)

- b/a is the sum of the roots, so b is negative. c/a is the product of the roots, so c is positive. If \(\mathrm{a} = 1\) , then the product of the roots is c, which is at least 1, so both roots cannot lie strictly between 0 and 1. If \(\mathrm{a} = 2\) , then the sum of the roots is less than 2, so b must be -1, -2, or -3. The roots are real so \(\mathrm{b}^2 > 4\mathrm{ac} = 8\mathrm{c}\) . Hence \(\mathrm{b} = -3\) and \(\mathrm{c} = 1\) . But \(2\mathrm{x}^2 - 3\mathrm{x} + 1 = (2\mathrm{x} - 1)(\mathrm{x} - 1)\) and one root is not less than 1. If \(\mathrm{a} = 3\) , then b must be -1, -2, ..., or -5. But \(\mathrm{b}^2 > 4\mathrm{ac} = 12\mathrm{c}\) , so \((\mathrm{b},\mathrm{c}) = (-4,1),(-5,1)\) or (-5, 2). In the first and last case, the equation has a root 1. In the middle case it has a root \(5 / 6 + \sqrt{13 / 6} = 1.434 > 1\) . Thus there are no solutions for \(\mathrm{a} = 1,2,3\) and so the smallest value of a is 4.


n is an integer. Prove that the sum of all fractions 1/rs, where r and s are relatively prime integers satisfying \(0< \mathrm{r}< \mathrm{s}\leq \mathrm{n}\) , \(\mathrm{r} + \mathrm{s} > \mathrm{n}\) , is 1/2.

Solution: We use induction on n. If \(\mathrm{n} = 2\) , then the only such fraction is \(\mathrm{r} = 1\) , \(\mathrm{s} = 2\) , giving \(1 / \mathrm{rs} = 1 / 2\) , so the result holds. Suppose it holds for \(\mathrm{n} - 1\) . As we move to n, we lose the fractions with \(\mathrm{r} + \mathrm{s} = \mathrm{n}\) . The other fractions \(1 / \mathrm{rs}\) which satisfy the conditions for \(\mathrm{n} - 1\) also satisfy the conditions for n. We also gain the fractions with \(\mathrm{s} = \mathrm{n}\) . These have sum \(= 1 / \mathrm{n}\) (sum \(1 / \mathrm{r}\) for all r satisfying \(0< \mathrm{r}< \mathrm{n}\) and r relatively prime to n). But if r is relatively prime to n, then so is n - r, and n - r does not equal r (otherwise r divides n). The pair \(1 / \mathrm{r}\) , \(1 / (\mathrm{n} - \mathrm{r})\) has sum \(\mathrm{n} / (\mathrm{r}(\mathrm{n} - \mathrm{r}))\) . So the fractions with \(\mathrm{s} = \mathrm{n}\) have sum equal to the sum of all \(1 / (\mathrm{r}(\mathrm{n} - \mathrm{r}))\) with \(0< \mathrm{r}< \mathrm{n}\) and r relatively prime to n. But that is exactly the sum of the fractions lost. Thus the total is unchanged as we move from n- 1 to n.


Given n points in space such that the triangle formed from any three of the points has an angle greater than 120 degrees. Prove that the points can be labeled 1, 2, 3, ..., n so that the angle defined by i, \(\mathrm{i} + 1\) , \(\mathrm{i} + 2\) is greater than 120 degrees for \(\mathrm{i} = 1,2,\ldots ,\mathrm{n} - 2\) .


Find four different three- digit numbers (in base 10) starting with the same digit, such that their sum is divisible by three of the numbers.

Solution:
Answer: 108, 117, 135, 180. Sum \(540 = 108\) . \(5 = 135\) . \(4 = 180\) . 3.

Try looking for a number of the form 3. 4. 5. n. We want 12n, 15n and 20n to have the same first digit. If the first digit is 1, this requires \(\mathrm{n} = 9\) . We must now check that the fourth number which must be \(60\mathrm{n} - 12\mathrm{n} - 15\mathrm{n} - 20\mathrm{n} = 13\mathrm{n}\) also has three digits starting with 1. It does, so we are home. [In fact, in this case the first digit must be 1, since \(20\mathrm{n} > 3 / 2 12\mathrm{n}\) .]


Every city in a certain state is directly connected by air with at most three other cities in the state, but one can get from any city to any other city with at most one change of plane. What is the maximum possible number of cities?

Solution:
Answer: 10.

Take a particular city X. At most 3 cities are directly connected to X. Each of those is directly connected to at most 2 other cities (apart from X). So X is connected with at most one change to at most 9 other cities. Thus the maximum number is at most 10.

We can achieve 10 as follows. Label the cities 1, ..., 10. Make direct connections as follows: 1: 2, 3, 4; 2: 1, 5, 6; 3: 1, 7, 8; 4: 1, 9, 10; 5: 2, 7, 9; 6: 2, 8, 10; 7: 3, 5, 10; 8: 3, 6, 9; 9: 4, 5, 8; 10: 4, 6, 7.


Given a pentagon with equal sides.

(a) Prove that there is a point X on the longest diagonal such that every side subtends an angle at most 90 degrees at X.

(b) Prove that the five circles with diameter one of the pentagon's sides do not cover the pentagon.


Given the equation \(\mathrm{x}^{3} + \mathrm{ax}^{2} + \mathrm{bx} + \mathrm{c} = 0\) , the first player gives one of a, b, c an integral value. Then the second player gives one of the remaining coefficients an integral value, and finally the first player gives the remaining coefficient an integral value. The first player's objective is to ensure that the equation has three integral roots (not necessarily distinct). The second player's objective is to prevent this. Who wins?

Solution:
Answer: the first player.

The first player starts by choosing \(\mathrm{c} = 0\) . Now if the second player selects a, then he can take b = a - 1. Then the polynomial factorizes as: \(\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + \mathrm{a} - 1)\) with integral roots, 0, - 1, 1- a. If the second player selects b, then he can take a = b + 1. Then the polynomial factorizes as \(\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + \mathrm{b})\) with integral roots 0, - 1, - b.


20 teams compete in a competition. What is the smallest number of games that must be played to ensure that given any three teams at least two play each other?


A regular n- gon is inscribed in a circle radius R. The distance from the center of the circle to the center of a side is \(\mathbf{h}_{\mathrm{n}}\) . Prove that \((n + 1) \mathbf{h}_{\mathrm{n} + 1} - \mathbf{n} \mathbf{h}_{\mathrm{n}} > \mathbf{R}\) .


Prove that for any positive numbers \(\mathbf{a}_{1}, \mathbf{a}_{2}, \ldots , \mathbf{a}_{\mathrm{n}}\) we have:

\[\mathrm{a}_{1} / (\mathrm{a}_{2} + \mathrm{a}_{3}) + \mathrm{a}_{2} / (\mathrm{a}_{3} + \mathrm{a}_{4}) + \ldots +\mathrm{a}_{\mathrm{n} - 1} / (\mathrm{a}_{\mathrm{n}} + \mathrm{a}_{1}) + \mathrm{a}_{\mathrm{n}} / (\mathrm{a}_{1} + \mathrm{a}_{2}) > \mathrm{n} / 4.\]


Given a circle, diameter AB and a point C on AB, show how to construct two points X and Y on the circle such that (1) Y is the reflection of X in the line AB, (2) YC is perpendicular to XA.

Solution: AB is a diameter, so BX is perpendicular to AX and hence parallel to YC. YX is perpendicular to BC, so \(\mathrm{YC} = \mathrm{YB}\) . Hence X and Y lie on the perpendicular bisector of BC.


The product of three positive numbers is 1, their sum is greater than the sum of their inverses. Prove that just one of the numbers is greater than 1.

Solution:
The product of the numbers is 1, so they cannot all be greater than 1 or all less than 1. If all equalled 1, then the sum would not be greater than the sum of the inverses. So we must have either one or two greater than 1. Thus it is sufficient to show that we cannot have two of the numbers greater than 1.

Suppose that a, \(\mathrm{b} > 1\) . Then since \(\mathrm{a} + \mathrm{b} + \mathrm{c} > 1 / \mathrm{a} + 1 / \mathrm{b} + 1 / \mathrm{c}\) , we have \(\mathrm{a} + \mathrm{b} + 1 / \mathrm{ab} > 1 / \mathrm{a} + 1 / \mathrm{b} + \mathrm{ab}\) , and hence \((1 - 1 / \mathrm{a})(1 - 1 / \mathrm{b}) > (\mathrm{a} - 1)(\mathrm{b} - 1)\) . Dividing by \((\mathrm{a} - 1)(\mathrm{b} - 1)\) gives \(\mathrm{ab}< 1\) . Contradiction.


What is the greatest number of sides of a convex polygon that can equal its longest diagonal?

Solution:
Answer: 2, except for the equilateral triangle.

It is easy to find two. Take the two sides to be AB and AC with angle \(\mathrm{BAC} = 60\) deg, and take the other vertices on the minor arc of the circle center A radius AB between B and C.

Let the longest diagonal have length k. Suppose there are three sides with length k. Extend them (if necessary) so they meet at A, B, C. Suppose \(\square \mathrm{A} > 60^{\circ}\) . Take the vertices on side AB to be P, Q (where we may have \(\mathrm{P} = \mathrm{A}\) , or \(\mathrm{Q} = \mathrm{B}\) , or both). Take the vertices on side AC to be R, S (where we may have \(\mathrm{R} = \mathrm{A}\) , or \(\mathrm{S} = \mathrm{C}\) , or both). Then \(\mathrm{AQ}\geq \mathrm{k}\) , \(\mathrm{AS}\geq \mathrm{k}\) , so \(\mathrm{QS} > \mathrm{k}\) . Contradiction. Hence angle \(\mathrm{A}\leq 60^{\circ}\) . The same is true for \(\square \mathrm{B}\) and \(\square \mathrm{C}\) . Hence \(\square \mathrm{A} = \square \mathrm{B} = \square \mathrm{C} = 60^{\circ}\) . But now \(\mathrm{QS} > \mathrm{k}\) unless \(\mathrm{A} = \mathrm{P} = \mathrm{R}\) . Similarly, B and C must be vertices of the convex polygon, so that it is just an equilateral triangle.


n is a 17 digit number. m is derived from n by taking its decimal digits in the reverse order. Show that at least one digit of n + m is even.

Solution:
Let the number be n with digits \(\mathrm{d_1d_2\ldots d_{17}}\) , so that the reversed number n' has digits \(\mathrm{d_{17}d_{16}\ldots}\) \(\mathrm{d_1}\) . Let the digits of \(\mathrm{n + n'}\) be \(\mathrm{a_0a_1\ldots a_{17}}\) , where \(\mathrm{a_0}\) may be zero. Let the carry forward when adding digits to get \(\mathrm{a_i}\) be \(\mathrm{c_{i - 1}}\) , so that, in general, \(\mathrm{c_i + d_i + d_{18 - i} = a_i + 10 c_{i - 1}}\) . Obviously \(\mathrm{c_i}\) is 0 or 1.

Suppose all the digits \(\mathrm{a_i}\) are odd (except that \(\mathrm{a_0}\) may be zero). Now \(\mathrm{c_9 + 2d_9 = a_9 + 10 c_8}\) . Since \(\mathrm{a_9}\) is odd, \(\mathrm{c_9}\) must be 1. But if we consider \(\mathrm{c_{10} + d_{10} + d_8 = a_{10} + 10 c_9}\) , we see that since \(\mathrm{a_{10}}\) is odd it is at least 1 and hence \(\mathrm{d_8 + d_{10}}\) is at least 10. Hence there must be a non- zero carry \(\mathrm{c_9}\) in \(\mathrm{c_{10} + d_{10} + d_8 = a_{10} + 10 c_9}\) irrespective of the value of \(\mathrm{c_{10}}\) .

We can now iterate and conclude successively that \(\mathrm{c_{12}}\) , \(\mathrm{c_{14}}\) , \(\mathrm{c_{16}}\) must be non- zero.


A room is an equilateral triangle side 100 meters. It is subdivided into 100 rooms, all equilateral triangles with side 10 meters. Each interior wall between two rooms has a door. If you start inside one of the rooms and can only pass through each door once, show that you cannot visit more than 91 rooms. Suppose now the large triangle has side k and is divided into \(\mathrm{k^2}\) small triangles by lines parallel to its sides. A chain is a sequence of triangles, such that a triangle can only be included once and consecutive triangles have a common side. What is the largest possible number of triangles in a chain?


Given 5 segments such that any 3 can be used to form a triangle. Show that at least one of the triangles is acute- angled.

Solution: Let the segments have length \(\mathrm{a\leq b\leq c\leq d\leq e}\) . Then if all triangles are obtuse we have \(\mathrm{e^2 > c^2}\) \(+\mathrm{d}^2\) \(\mathrm{d}^2 >\mathrm{b}^2 +\mathrm{c}^2\) \(\mathrm{c}^2 >\mathrm{a}^2 +\mathrm{b}^2\) . Adding \(\mathrm{e^2 > a^2 + 2b^2 + c^2 > a^2 + 3b^2}\) . But \(\mathrm{e\leq a + b}\) , so \(\mathrm{e^2\leq a^2 + }\) \(2\mathrm{ab + b^2\leq a^2 + 3b^2}\) . Contradiction.


ABC is an acute- angled triangle. The angle bisector AD, the median BM and the altitude CH are concurrent. Prove that angle A is more than 45 degrees.

Solution: We use Ceva's theorem. Since AD, BM, CH are concurrent, we have (BD/DC).(CM/MA).(AH/BH) = 1. But CM = MA and since AD is the angle bisector BD/DC = AB/AC, so (AB/AC).(AH/BH) = 1. Hence AH/AC = BH/AB < 1. So angle HAC > angle HCA. But angle AHC = 90 deg, so angle A > 45 deg.


Five n- digit binary numbers have the property that every two numbers have the same digits in just m places, but no place has the same digit in all five numbers. Show that \(2 / 5 \leq \mathrm{m} / \mathrm{n} \leq 3 / 5\) .


Show that given 200 integers you can always choose 100 with sum a multiple of 100.


ABC is a triangle with incenter I. M is the midpoint of BC. IM meets the altitude AH at E. Show that \(\mathrm{AE} = \mathrm{r}\) , the radius of the inscribed circle.


Given any positive integer n, show that we can find infinitely many integers m such that m has no zeros (when written as a decimal number) and the sum of the digits of m and mn is the same.


Two congruent rectangles of area A intersect in eight points. Show that the area of the intersection is more than A/2.


If the numbers from 11111 to 99999 are arranged in an arbitrary order show that the resulting 444445 digit number is not a power of 2.

Solution: Let the set of numbers be S. Define the function f on S as follows. Replace each digit i in n by 9- i for \(0 < \mathrm{i} < 9\) . This gives f(n). Then f(f(n)) = n, so f is a bijection. The fixed points have only the digits 0 and 9 and so are all divisible by 9. The other points divide into pairs (n, f(n)) and the sum of each pair is divisible by 9. Hence the sum of all the numbers in S is divisible by 9.


S is the set of all positive integers with n decimal digits or less and with an even digit sum. T is the set of all positive integers with n decimal digits or less and an odd digit sum. Show that the sum of the kth powers of the members of S equals the sum for T if \(1 \leq \mathrm{k} < \mathrm{n}\) .


The vertices of a regular n- gon are colored (each vertex has only one color). Each color is applied to at least three vertices. The vertices of any given color form a regular polygon. Show that there are two colors which are applied to the same number of vertices.


Prove that we can find a number divisible by \(2^{n}\) whose decimal representation uses only the digits 1 and 2.

Solution: Induction on n. We claim that we can find N with n digits, all 1 or 2, so that N is divisible by \(2^{n}\) . True for \(\mathrm{n} = 1\) : take \(\mathrm{N} = 2\) . Suppose it is true for n. If \(2^{n + 1}\) divides N, then since \(2^{n + 1}\) divides \(2 \times 10^{n}\) it also divides N' obtained from N by placing a 2 in front of it. If \(2^{n + 1}\) does not divide N, then \(\mathrm{N} = 2^{n} \times \mathrm{odd}\) and \(10^{n} = 2^{n} \times \mathrm{odd}\) , so \(\mathrm{N} + 10^{n}\) (in other words the \(\mathrm{n} + 1\) digit number obtained by placing a 1 in front of N) is divisible by \(2^{n + 1}\) .


(1) \(\mathrm{A_1A_2A_3}\) is a triangle. Points \(\mathrm{B_1}\) , \(\mathrm{B_2}\) , \(\mathrm{B_3}\) are chosen on \(\mathrm{A_1A_2}\) , \(\mathrm{A_2A_3}\) , \(\mathrm{A_3A_1}\) respectively and points \(\mathrm{D_1}\) , \(\mathrm{D_2}\) \(\mathrm{D_3}\) on \(\mathrm{A_3A_1}\) , \(\mathrm{A_1A_2}\) , \(\mathrm{A_2A_3}\) respectively, so that if parallelograms \(\mathrm{A_iB_iC_iD_i}\) are formed, then the lines \(\mathrm{A_iC_i}\) concur. Show that \(\mathrm{A_1B_1}\) : \(\mathrm{A_2B_2}\) : \(\mathrm{A_3B_3} = \mathrm{A_1D_1}\) : \(\mathrm{A_2D_2}\) : \(\mathrm{A_3D_3}\) .

(2) \(\mathrm{A_1A_2\ldots A_n}\) is a convex polygon. Points \(\mathrm{B_i}\) are chosen on \(\mathrm{A_iA_{i + 1}}\) (where we take \(\mathrm{A_{n + 1}}\) to mean \(\mathrm{A_1}\) ), and points \(\mathrm{D_i}\) on \(\mathrm{A_{i - 1}A_i}\) (where we take \(\mathrm{A_0}\) to mean \(\mathrm{A_n}\) ) such that if parallelograms \(\mathrm{A_iB_iC_iD_i}\) are formed, then the n lines \(\mathrm{A_iC_i}\) concur. Show that \(\prod \mathrm{A_iB_i} = \prod \mathrm{A_iD_i}\) .


(1) Player A writes down two rows of 10 positive integers, one under the other. The numbers must be chosen so that if a is under b and c is under d, then a + d = b + c. Player B is allowed to ask for the identity of the number in row i, column j. How many questions must he ask to be sure of determining all the numbers?

(2) An m x n array of positive integers is written on the blackboard. It has the property that for any four numbers a, b, c, d with a and b in the same row, c and d in the same row, a above c (in the same column) and b above d (in the same column) we have a + d = b + c. If some numbers are wiped off, how many must be left for the table to be accurately restored?

Solution:
(1) is trivial. We can write the condition as \(\mathrm{b - a = d - c}\) , so the 10 numbers in the first row and 1 in the second row can all be chosen arbitrarily. Hence at least 11 questions are needed. But they are also sufficient. Having determined those numbers, the others immediately follow.

(2). The m+n-1 numbers in the first row and first column can all be chosen arbitrarily, but are sufficient to determine all the numbers. Hence at least m+n-1 numbers must survive.


Circles, each with radius less than R, are drawn inside a square side 1000R. There are no points on different circles a distance R apart. Show that the total area covered by the circles does not exceed \(340,000 \mathrm{R}^2\) .


You are given three positive integers. A move consists of replacing \(\mathrm{m}\leq \mathrm{n}\) by \(2\mathrm{m}\) , n- m. Show that you can always make a series of moves which results in one of the integers becoming zero. [For example, if you start with 4, 5, 10, then you could get 8, 5, 6, then 3, 10, 6, then 6, 7, 6, then 0, 7, 12. ]


The real numbers a, b, A, B satisfy \((\mathrm{B} - \mathrm{b})^{2}< (\mathrm{A} - \mathrm{a})(\mathrm{Ba} - \mathrm{Ab})\) . Show that the quadratics \(\mathrm{x}^{2}+\) \(\mathrm{ax} + \mathrm{b} = 0\) and \(\mathrm{x}^{2} + \mathrm{Ax} + \mathrm{B} = 0\) have real roots and between the roots of each there is a root of the other.


The projections of a body on two planes are circles. Show that the circles have the same radius.


An integer is written at each vertex of a regular n- gon. A move is to find four adjacent vertices with numbers a, b, c, d (in that order), so that \((\mathrm{a} - \mathrm{d})(\mathrm{b} - \mathrm{c})< 0\) , and then to interchange b and c. Show that only finitely many moves are possible. For example, a possible sequence of moves is shown below:

1	7	2	3	5	4

1	2	7	3	5	4

1	2	3	7	5	4

1	2	3	5	7	4

2	1	3	5	7	4


A polygon P has an inscribed circle center O. If a line divides P into two polygons with equal areas and equal perimeters, show that it must pass through O.


Given any set S of 25 positive integers, show that you can always find two such that none of the other numbers equals their sum or difference.


A and B are adjacent vertices of a 12- gon. Vertex A is marked - and the other vertices are marked +. You are allowed to change the sign of any n adjacent vertices. Show that by a succession of moves of this type with \(\mathrm{n} = 6\) you cannot get B marked - and the other vertices marked +. Show that the same is true if all moves have \(\mathrm{n} = 3\) or if all moves have \(\mathrm{n} = 4\) .


Equally spaced perpendicular lines divide a large piece of paper into unit squares. N squares are colored black. Show that you can always cut out a set of disjoint square pieces of paper, so

that all the black squares are removed and the black area of each piece is between 1/5 and 4/5 of its total area.


n is a positive integer. S is the set of all triples (a, b, c) such that \(1 \leq a, b, c, \leq n\) . What is the smallest subset X of triples such that for every member of S one can find a member of X which differs in only one position. [For example, for \(n = 2\) , one could take \(X = \{ (1, 1, 1), (2, 2, 2) \} .\) ]


Let \(\operatorname {f}(\mathbf{x}, \mathbf{y}) = \mathbf{x}^{2} + \mathbf{x}\mathbf{y} + \mathbf{y}^{2}\) . Show that given any real \(\mathbf{x}\) , \(\mathbf{y}\) one can always find integers \(\mathbf{m}, \mathbf{n}\) such that \(\operatorname {f}(\mathbf{x} - \mathbf{m}, \mathbf{y} - \mathbf{n}) \leq = 1 / 3\) . What is the corresponding result if \(\operatorname {f}(\mathbf{x}, \mathbf{y}) = \mathbf{x}^{2} + \mathbf{a}\mathbf{x}\mathbf{y} + \mathbf{y}^{2}\) with 0 \(\leq \mathbf{a} \leq 2\) ?


A switch has two inputs 1, 2 and two outputs 1, 2. It either connects 1 to 1 and 2 to 2, or 1 to 2 and 2 to 1. If you have three inputs 1, 2, 3 and three outputs 1, 2, 3, then you can use three switches, the first across 1 and 2, then the second across 2 and 3, and finally the third across 1 and 2. It is easy to check that this allows the output to be any permutation of the inputs and that at least three switches are required to achieve this. What is the minimum number of switches required for 4 inputs, so that by suitably setting the switches the output can be any permutation of the inputs?


ABCD is a rectangle. M is the midpoint of AD and N is the midpoint of BC. P is a point on the ray CD on the opposite side of D to C. The ray PM intersects AC at Q. Show that MN bisects the angle PNQ.


Given 50 segments on a line show that you can always find either 8 segments which are disjoint or 8 segments with a common point.


Find the largest integer n such that \(4^{27} + 4^{1000} + 4^{n}\) is a square.


a, m, n are positive integers and a \(>1\) . Show that if \(\mathrm{a^{m} + 1}\) divides \(\mathrm{a^{n} + 1}\) , then m divides n. The positive integer b is relatively prime to a, show that if \(\mathrm{a^{m} + b^{m}}\) divides \(\mathrm{a^{n} + b^{n}}\) then m divides n.


A sequence of finite sets of positive integers is defined as follows. \(\mathrm{S_0 = \{m\}}\) , where \(\mathrm{m > 1}\) . Then given \(\mathrm{S_n}\) you derive \(\mathrm{S_{n + 1}}\) by taking \(\mathrm{k^2}\) and \(\mathrm{k + 1}\) for each element k of \(\mathrm{S_n}\) . For example, if \(\mathrm{S_0 = \{5\}}\) , then \(\mathrm{S_2 = \{7, 26, 36, 625\}}\) . Show that \(\mathrm{S_n}\) always has \(2^{\mathrm{n}}\) distinct elements.


Prove that a collection of squares with total area 1 can always be arranged inside a square of area 2 without overlapping.


O is the point of intersection of the diagonals of the convex quadrilateral ABCD. Prove that the line joining the centroids of ABO and CDO is perpendicular to the line joining the orthocenters of BCO and ADO.


9 lines each divide a square into two quadrilaterals with areas 2/5 and 3/5 that of the square. Show that 3 of the lines meet in a point.


A 7- gon is inscribed in a circle. The center of the circle lies inside the 7- gon. A, B, C are adjacent vertices of the 7- gon show that the sum of the angles at A, B, C is less than 450 degrees.


Two players play the following game. At each turn the first player chooses a decimal digit, then the second player substitutes it for one of the stars in the subtraction | **** - **** |. The first player tries to end up with the largest possible result, the second player tries to end up with the smallest possible result. Show that the first player can always play so that the result is at least 4000 and that the second player can always play so that the result is at most 4000.


For positive reals x, y let f(x, y) be the smallest of x, 1/y, y + 1/x. What is the maximum value of f(x, y)? What are the corresponding x, y?


P is a convex polygon and X is an interior point such that for every pair of vertices A, B, the triangle XAB is isosceles. Prove that all the vertices of P lie on some circle center X.


Is it possible to place the digits 0, 1, 2 into unit squares of 100 x 100 cross- lined paper such that every 3 x 4 (and every 4 x 3) rectangle contains three 0s, four 1s and five 2s?


x₁, x₂, ..., xₙ are positive reals with sum 1. Let s be the largest of x₁/(1 + x₁), x₂/(1 + x₁ + x₂), ..., xₙ/(1 + x₁ + ... + xₙ). What is the smallest possible value of s? What are the corresponding xᵢ?


n teams compete in a tournament. Each team plays every other team once. In each game a team gets 2 points for a win, 1 for a draw and 0 for a loss. Given any subset S of teams, one can find a team (possibly in S) whose total score in the games with teams in S was odd. Prove that n is even.


You are given 14 coins. It is known that genuine coins all have the same weight and that fake coins all have the same weight, but weigh less than genuine coins. You suspect that 7 particular coins are genuine and the other 7 fake. Given a balance, how can you prove this in three weighings (assuming that you turn out to be correct)?

Solution: Let the coins you suspect to be genuine be \(\mathrm{G}_1\) , \(\mathrm{G}_2\) , ..., \(\mathrm{G}_7\) , and the suspected fakes by \(\mathrm{F}_1\) , \(\mathrm{F}_2\) , ..., \(\mathrm{F}_7\) . First weigh \(\mathrm{F}_1\) against \(\mathrm{G}_1\) . Assuming \(\mathrm{F}_1\) weighs less, you have proved that \(\mathrm{F}_1\) is fake and \(\mathrm{G}_1\) genuine. Second, weigh \(\mathrm{F}_1\) , \(\mathrm{G}_2\) , \(\mathrm{G}_3\) against \(\mathrm{G}_1\) , \(\mathrm{F}_2\) , \(\mathrm{F}_3\) . Assuming the first three weigh more you have proved that they include more genuine coins than the second three. But the second three includes one genuine coin \((\mathrm{G}_1)\) and the first three includes one fake \((\mathrm{F}_1)\) , so you have proved that \(\mathrm{G}_2\) and \(\mathrm{G}_3\) are genuine and \(\mathrm{F}_2\) and \(\mathrm{F}_3\) fake. Finally, weigh \(\mathrm{F}_1\) , \(\mathrm{F}_2\) , \(\mathrm{F}_3\) , \(\mathrm{G}_4\) , \(\mathrm{G}_5\) , \(\mathrm{G}_6\) , \(\mathrm{G}_7\) against \(\mathrm{F}_4\) , \(\mathrm{F}_5\) , \(\mathrm{F}_6\) , \(\mathrm{F}_7\) , \(\mathrm{G}_1\) , \(\mathrm{G}_2\) , \(\mathrm{G}_3\) . Assuming the first group weighs more it must include more genuine coins and hence just four genuine coins. Similarly, the second group must include four fakes. So you have proved the identity of the remaining coins.


Prove that a 9 digit decimal number whose digits are all different, which does not end with 5 and or contain a 0, cannot be a square.


Given \(\mathrm{n} > 4\) points, show that you can place an arrow between each pair of points, so that given any point you can reach any other point by travelling along either one or two arrows in the direction of the arrow.


OA and OB are tangent to a circle at A and B. The line parallel to OB through A meets the circle again at C. The line OC meets the circle again at E. The ray AE meets the line OB at K. Prove that K is the midpoint of OB.


\(\mathrm{p}(\mathrm{x}) = \mathrm{ax}^2 +\mathrm{bx} + \mathrm{c}\) is a real quadratic such that \(\mathrm{lp}(\mathrm{x})\leq 1\) for all \(|\mathrm{x}|\leq 1\) . Prove that \(|\mathrm{cx}^2 +\mathrm{bx} + \mathrm{al}\leq 2\) for \(|\mathrm{x}|\leq 1\) .


Players numbered 1 to 1024 play in a knock- out tournament. There are no draws, the winner of a match goes through to the next round and the loser is knocked- out, so that there are 512 matches in the first round, 256 in the second and so on. If m plays n and \(\mathrm{m}< \mathrm{n} - 2\) then m always wins. What is the largest possible number for the winner?


Define \(\mathrm{p(x) = ax^2 + bx + c}\) . If \(\mathrm{p(x) = x}\) has no real roots, prove that \(\mathrm{p(p(x)) = 0}\) has no real roots.


At time 1, n unit squares of an infinite sheet of paper ruled in squares are painted black, the rest remain white. At time \(\mathrm{k + 1}\) , the color of each square is changed to the color held at time k by a majority of the following three squares: the square itself, its northern neighbour and its eastern neighbour. Prove that all the squares are white at time \(\mathrm{n + 1}\) .


ABC is an acute- angled triangle. D is the reflection of A in BC, E is the reflection of B in AC, and F is the reflection of C in AB. Show that the circumcircles of DBC, ECA, FAB meet at a point and that the lines AD, BE, CF meet at a point.


n people are all strangers. Show that you can always introduce some of them to each other, so that afterwards each person has met a different number of the others. [problem: this is false as stated. Each person must have 0, 1, ... or n- 1 meetings, so all these numbers must be used. But if one person has met no one, then another cannot have met everyone.]


A king moves on an \(8 \times 8\) chessboard. He can move one square at a time, diagonally or orthogonally (so away from the borders he can move to any of eight squares). He makes a complete circuit of the board, starting and finishing on the same square and visiting every other square just once. His trajectory is drawn by joining the center of the squares he moves to and from for each move. The trajectory does not intersect itself. Show that he makes at least 28 moves parallel to the sides of the board (the others being diagonal) and that a circuit is possible with exactly 28 moves parallel to the sides of the board. If the board has side length 8, what is the maximum and minimum possible length for such a trajectory.


A triangle has area 1, and sides \(\mathrm{a}\geq \mathrm{b}\geq \mathrm{c}\) . Prove that \(\mathrm{b}^2\geq 2\)


A convex n- gon has no two sides parallel. Given a point P inside the n- gon show that there are at most n lines through P which bisect the area of the n- gon.


a, b, c, d, e are positive reals. Show that \((\mathrm{a} + \mathrm{b} + \mathrm{c} + \mathrm{d} + \mathrm{e})^2\geq 4(\mathrm{ab} + \mathrm{bc} + \mathrm{cd} + \mathrm{de} + \mathrm{ea})\)


Given 4 points which do not lie in a plane, how many parallelepipeds have all 4 points as vertices?


A collection of n cards is numbered from 1 to n. Each card has either 1 or - 1 on the back. You are allowed to ask for the product of the numbers on the back of any three cards. What is the smallest number of questions which will allow you to determine the numbers on the backs of all the cards if n is (1) 30, (2) 31, (3) 32? If 50 cards are arranged in a circle and you are only allowed to ask for the product of the numbers on the backs of three adjacent cards, how many questions are needed to determine the product of the numbers on the backs of all 50 cards?


Find the smallest positive integer which can be represented as \(36^{\mathrm{m}} - 5^{\mathrm{n}}\) .

## Answer

11

Solution:
Obviously \(11 = 36^{1} - 5^{2}\) , and we guess that this is the best possible. We cannot have \(36^{\mathrm{m}} - 5^{\mathrm{n}} = \mathrm{k}\) , where 2, 3 or 5 divides k (because then 2 would divide \(5^{\mathrm{n}}\) and similarly in the other cases). So the only possible values of \(\mathrm{k}< 11\) are 1, 7.

We have \(36^{\mathrm{m}} - 5^{\mathrm{n}} = 1 \mathrm{mod} 5\) , so \(\mathrm{k} \neq 7\) . Similarly, \(36^{\mathrm{m}} - 5^{\mathrm{n}} = 3 \mathrm{mod} 4\) , so \(\mathrm{k} \neq 1\) .


Each side of a convex hexagon is longer than 1. Is there always a diagonal longer than 2? If each of the main diagonals of a hexagon is longer than 2, is there always a side longer than 1?


Circles radius r and R touch externally. AD is parallel to BC. AB and CD touch both circles. AD touches the circle radius r, but not the circle radius R, and BC touches the circle radius R, but not the circle radius r. What is the smallest possible length for AB?


Given n unit vectors in the plane whose sum has length less than one. Show that you can arrange them so that the sum of the first k has length less than 2 for every \(1< \mathrm{k}< \mathrm{n}\) .


Find all real a, b, c such that \(\mathrm{la x} + \mathrm{by} + \mathrm{czl} + \mathrm{lbx} + \mathrm{cy} + \mathrm{azl} + \mathrm{lcx} + \mathrm{ay} + \mathrm{bzl} = \mathrm{lx} + \mathrm{y} + \mathrm{zl}\) for all real x, y, z.


ABCD is a square. P is on the segment AB and Q is on the segment BC such that \(\mathrm{BP} = \mathrm{BQ}\) . H lies on PC such that BHC is a right angle. Show that DHQ is a right angle.


The n points of a graph are each colored red or blue. At each move we select a point which differs in color from more than half of the points to which it it is joined and we change its color. Prove that this process must finish after a finite number of moves.


Find all positive integers m, n such that \(\mathrm{n}^{\mathrm{n}}\) has m decimal digits and \(\mathrm{m}^{\mathrm{m}}\) has n decimal digits.


In the triangle ABC, angle C is 90 deg and \(\mathrm{AC} = \mathrm{BC}\) . Take points D on CA and E on CB such that \(\mathrm{CD} = \mathrm{CE}\) . Let the perpendiculars from D and C to AE meet AB at K and L respectively. Show that \(\mathrm{KL} = \mathrm{LB}\) .


One rat and two cats are placed on a chess- board. The rat is placed first and then the two cats choose positions on the border squares. The rat moves first. Then the cats and the rat move alternately. The rat can move one square to an adjacent square (but not diagonally). If it is on a border square, then it can also move off the board. On a cat move, both cats move one square. Each must move to an adjacent square, and not diagonally. The cats win if one of them moves onto the same square as the rat. The rat wins if it moves off the board. Who wins? Suppose there are three cats (and all three cats move when it is the cats' turn), but that the rat gets an extra initial turn. Prove that the rat wins.


Arrange the numbers 1, 2, ..., 32 in a sequence such that the arithmetic mean of two numbers does not lie between them. (For example, ... 3, 4, 5, 2, 1, ... is invalid, because 2 lies between 1 and 3.) Can you arrange the numbers 1, 2, ..., 100 in the same way?


Find all three digit decimal numbers \(\mathrm{a_1a_2a_3}\) which equal the mean of the six numbers \(\mathrm{a_1a_2a_3}\) , \(\mathrm{a_1a_3a_2}\) , \(\mathrm{a_2a_1a_3}\) , \(\mathrm{a_2a_3a_1}\) , \(\mathrm{a_3a_1a_2}\) , \(\mathrm{a_3a_2a_1}\) .


No triangle of area 1 can be fitted inside a convex polygon. Show that the polygon can be fitted inside a triangle of area 4.


f is a function on the closed interval [0, 1] with non- negative real values. \(\mathrm{f}(1) = 1\) and \(\mathrm{f}(\mathrm{x} + \mathrm{y}) \geq \mathrm{f}(\mathrm{x}) + \mathrm{f}(\mathrm{y})\) for all \(\mathrm{x}, \mathrm{y}\) . Show that \(\mathrm{f}(\mathrm{x}) \leq 2\mathrm{x}\) for all \(\mathrm{x}\) . Is it necessarily true that \(\mathrm{f}(\mathrm{x}) \leq 1.9\mathrm{x}\) for all \(\mathrm{x}\) .

## Answer

no

Solution:
We have \(\mathrm{f(x) = f(1) - f(1 - x)\leq f(1) = 1}\) . So for \(\mathrm{x}\geq 1 / 2\) \(\mathrm{f(x)}\leq 1\leq 2\mathrm{x}\) . If \(\mathrm{x}< 1 / 2\) , then for some n we have \(1 / 2^{n + 1}\leq \mathrm{x}< 1 / 2^{\mathrm{n}}\) . Hence by a trivial induction \(\mathrm{f}(2^{\mathrm{n}}\mathrm{x})\geq 2^{\mathrm{n}}\mathrm{f}(\mathrm{x})\) . But \(\mathrm{f}(2^{\mathrm{n}}\mathrm{x})\leq 1\) , so \(\mathrm{f(x)}\leq\) \(1 / 2^{\mathrm{n}}\leq 2\mathrm{x}\)

Note that \(\mathrm{f(x) = 0}\) for \(\mathrm{x}\leq 1 / 2\) and 1 for \(\mathrm{x} > 1 / 2\) satisfies the conditions. But \(\mathrm{f}(0.51) = 1>\) \((1.9)(0.51)\)


The triangle ABC has area 1. D, E, F are the midpoints of the sides BC, CA, AB. P lies in the segment BF, Q lies in the segment CD, R lies in the segment AE. What is the smallest possible area for the intersection of triangles DEF and PQR?


(1) O is the circumcenter of the triangle ABC. The triangle is rotated about O to give a new triangle A'B'C'. The lines AB and A'B' intersect at C", BC and B'C' intersect at A", and CA and C'A' intersect at B". Show that A"B"C" is similar to ABC.

(2) O is the center of the circle through ABCD. ABCD is rotated about O to give the quadrilateral A'B'C'D'. Prove that the intersection points of corresponding sides form a parallelogram.


A triangle ABC has unit area. The first player chooses a point X on side AB, then the second player chooses a point Y on side BC, and finally the first player chooses a point Z on side CA. The first player tries to arrange for the area of XYZ to be as large as possible, the second player tries to arrange for the area to be as small as possible. What is the optimum strategy for the first player and what is the best he can do (assuming the second player plays optimally)?


What is the smallest perimeter for a convex 32- gon whose vertices are all lattice points?


Given a \(7 \times 7\) square subdivided into 49 unit squares, mark the center of n unit squares, so that no four marks form a rectangle with sides parallel to the square. What is the largest n for which this is possible? What about a \(13 \times 13\) square?


Given a convex hexagon, take the midpoint of each of the six diagonals joining vertices which are separated by a single vertex (so if the vertices are in order A, B, C, D, E, F, then the diagonals are AC, BD, CE, DF, EA, FB). Show that the midpoints form a convex hexagon with a quarter the area of the original.


Show that there are \(2^{n + 1}\) numbers each with \(2^{n}\) digits, all 1 or 2, so that every two numbers differ in at least half their digits.


There are finitely many polygons in the plane. Every two have a common point. Prove that there is a straight line intersecting all the polygons.


a, b, c are positive reals. Show that a \(^{3} + \mathrm{b}^{3} + \mathrm{c}^{3} + 3\mathrm{abc}\geq \mathrm{ab}(\mathrm{a} + \mathrm{b}) + \mathrm{bc}(\mathrm{b} + \mathrm{c}) + \mathrm{ca}(\mathrm{c} + \mathrm{a})\) .

Solution: The inequality is homogeneous, so we can take \(\mathrm{a} = 1\) and put \(\mathrm{b} = 1 + \mathrm{x}\) , \(\mathrm{c} = 1 + \mathrm{y}\) , where \(\mathrm{x}\) , \(\mathrm{y} \geq 0\) . Then after some reduction the inequality is equivalent to \(\mathrm{x}^{3} + \mathrm{y}^{3} + \mathrm{x}^{2} + \mathrm{y}^{2} - \mathrm{x}^{2} - \mathrm{y} - \mathrm{xy}^{2} - \mathrm{xy} \geq 0\) , or (after factorising \(\mathrm{x}^{3} + \mathrm{y}^{3}\) ) to \((\mathrm{x} + \mathrm{y} + 1)(\mathrm{x} - \mathrm{y})^{2} + \mathrm{xy} \geq 0\) , which is obviously true.


Three flies crawl along the perimeter of a triangle. At least one fly makes a complete circuit of the perimeter. For the entire period the center of mass of the flies remains fixed. Show that it must be at the centroid of the triangle. [You may not assume, without proof, that the flies have the same mass, or that they crawl at the same speed, or that any fly crawls at a constant speed.]


The finite sequence \(\mathrm{a_n}\) has each member 0, 1 or 2. A move involves replacing any two unequal members of the sequence by a single member different from either. A series of moves results in a single number. Prove that no series of moves can terminate in a (single) different number.

Solution: Suppose we start with a 0s, b 1s and c 2s. Each move changes the parity of all of a, b, c. Each moves reduces the length of the sequence by 1, so there must be a+b+c- 1 moves in all. Hence the total number of 0s, the total number of 1s and the total number of 2s all have their parity changed a+b+c- 1 times. So if we end up with just one 0, then a must have the opposite parity to b and c. In that case, we cannot end up with just one 1, or just one 2. Similarly in the other cases.


S is a horizontal strip in the plane. n lines are drawn so that no three are collinear and every pair intersects within the strip. A path starts at the bottom border of the strip and consists of a sequence of segments from the n lines. The path must change line at each intersection and must always move upwards. Show that: (1) there are at least n/2 disjoint paths; (2) there is a path of at least n segments; (3) there is a path involving not more than n/2 + 1 of the lines; and (4) there is a path that involves segments from all n lines.


For what n can we color the unit cubes in an n x n x n cube red or green so that every red unit cube has just two red neighbouring cubes (sharing a face) and every green unit cube has just two green neighbouring cubes.


\(\mathrm{p(x)}\) is a polynomial with integral coefficients. \(\mathrm{f(n)} =\) the sum of the (decimal) digits in the value \(\mathrm{p(n)}\) . Show that \(\mathrm{f(n)}\) some value m infinitely many times.


20 teams each play one game with every other team. Each game results in a win or loss (no draws). k of the teams are European. A separate trophy is awarded for the best European team on the basis of the k(k- 1)/2 games in which both teams are European. This trophy is won by a single team. The same team comes last in the overall competition (winning fewer games than any other team). What is the largest possible value of k? If draws are allowed and a team scores 2 for a win and 1 for a draw, what is the largest possible value of k?


Given real numbers \(\mathbf{a}_{\mathrm{i}}\) , \(\mathbf{b}_{\mathrm{i}}\) and positive reals \(\mathbf{c}_{\mathrm{i}}\) , \(\mathbf{d}_{\mathrm{i}}\) , let \(\mathbf{e}_{\mathrm{ij}} = (\mathbf{a}_{\mathrm{i}} + \mathbf{b}_{\mathrm{j}}) / (\mathbf{c}_{\mathrm{i}} + \mathbf{d}_{\mathrm{j}})\) . Let \(\mathbf{M}_{\mathrm{i}} = \max_{0\leq \mathrm{j}\leq \mathrm{n}}\mathbf{e}_{\mathrm{ij}}\) , \(\mathbf{m}_{\mathrm{j}} = \min_{1\leq \mathrm{i}\leq \mathrm{n}}\mathbf{e}_{\mathrm{ij}}\) . Show that we can find an \(\mathbf{e}_{\mathrm{ij}}\) with \(1\leq \mathrm{i},\mathrm{j}\leq \mathrm{n}\) such that \(\mathbf{e}_{\mathrm{ij}} = \mathbf{M}_{\mathrm{i}} = \mathbf{m}_{\mathrm{j}}\) .


50 watches, all keeping perfect time, lie on a table. Show that there is a moment when the sum of the distances from the center of the table to the center of each dial equals the sum of the distances from the center of the table to the tip of each minute hand.


1000 numbers are written in line 1, then further lines are constructed as follows. If the number m occurs in line n, then we write under it in line \(\mathrm{n + 1}\) , each time it occurs, the number of times that m occurs in line n. Show that lines 11 and 12 are identical. Show that we can choose numbers in line 1, so that lines 10 and 11 are not identical.


(1) The circles \(\mathrm{C_1}\) , \(\mathrm{C_2}\) , \(\mathrm{C_3}\) with equal radius all pass through the point X. \(\mathrm{C_i}\) and \(\mathrm{C_j}\) also intersect at the point \(\mathrm{Y_{ij}}\) . Show that angle \(\mathrm{XO_1Y_{12} + angle XO_2Y_{23} + angle XO_3Y_{31} = 180 deg}\) where \(\mathrm{O_i}\) is the center of circle \(\mathrm{C_i}\) .


\(\mathrm{a_1}\) and \(\mathrm{a_2}\) are positive integers less than 1000. Define \(\mathrm{a_n = min\{|a_i - a_j|:0< i< j< n\}}\) . Show that \(\mathrm{a_{21} = 0}\) .


Can you label each vertex of a cube with a different three digit binary number so that the numbers at any two adjacent vertices differ in at least two digits?


a, b, c, d are vectors in the plane such that \(\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = 0\) . Show that \(\left|\mathbf{a}\right| + \left|\mathbf{b}\right| + \left|\mathbf{c}\right| + \left|\mathbf{d}\right|\geq \left|\mathbf{a}\right|\) \(+\mathbf{d}\mathbf{i} + \mathbf{i}\mathbf{b} + \mathbf{d}\mathbf{i} + \mathbf{i}\mathbf{c} + \mathbf{d}\mathbf{i}\) .


S is a set of 1976 points which form a regular 1976- gon. T is the set of all points which are the midpoint of at least one pair of points in S. What is the greatest number of points of T which lie on a single circle?


n rectangles are drawn on a rectangular sheet of paper. Each rectangle has its sides parallel to the sides of the paper. No pair of rectangles has an interior point in common. If the rectangles were removed show that the rest of the sheet would be in at most \(\mathrm{n + 1}\) parts.


There are three straight roads. On each road a man is walking at constant speed. At time \(t = 0\) , the three men are not collinear. Prove that they will be collinear for \(t > 0\) at most twice.


Initially, there is one beetle on each square in the set S. Suddenly each beetle flies to a new square, subject to the following conditions: (1) the new square may be the same as the old or different; (2) more than one beetle may choose the same new square; (3) if two beetles are initially in squares with a common vertex, then after the flight they are either in the same square or in squares with a common vertex. Suppose S is the set of all squares in the middle row and column of a \(99 \times 99\) chess board, is it true that there must always be a beetle whose new square shares a vertex with its old square (or is identical with it)? What if S also includes all the border squares (so S is rows 1, 50 and 99 and columns 1, 50 and 99)? What if S is all squares of the board?


Call a triangle big if each side is longer than 1. Show that we can draw 100 big triangles inside an equilateral triangle with side length 5 so that all the triangles are disjoint. Show that you can draw 100 big triangles with every vertex inside or on an equilateral triangle with side 3, so that they cover the equilateral triangle, and any two big triangles either (1) are disjoint, or (2) have as intersection a common vertex, or (3) have as intersection a common side.


n is a positive integer. A universal sequence of length m is a sequence of m integers each between 1 and n such that one can obtain any permutation of 1, 2, ..., n by deleting suitable members of the sequence. For example, 1, 2, 3, 1, 2, 1, 3 is a universal sequence of length 7 for \(n = 3\) . But 1, 2, 3, 2, 1, 3, 1 is not universal, because one cannot obtain the permutation 3, 1, 2. Show that one can always obtain a universal sequence for n of length \(n^2 - n + 1\) . Show that a universal sequence for n must have length at least \(n(n + 1) / 2\) . Show that the shortest sequence for \(n = 4\) has 12 members. [You are told, but do not have to prove, that there is a universal sequence for n of length \(n^2 - 2n + 4\) .]


n real numbers are written around a circle. One of the numbers is 1 and the sum of the numbers is 0. Show that there are two adjacent numbers whose difference is at least \(n / 4\) . Show that there is a number which differs from the arithmetic mean of its two neighbours by at least \(8 / n^2\) . Improve this result to some \(k / n^2\) with \(k > 8\) . Show that for \(n = 30\) , we can take \(k = 1800 / 113\) . Give an example of 30 numbers such that no number differs from the arithmetic mean of its two neighbours by more than \(2 / 113\) .


You are given a regular n- gon. Each vertex is marked \(+1\) or - 1. A move consists of changing the sign of all the vertices which form a regular k- gon for some \(1 < k \ll = n\) . [A regular 2- gon means two vertices which have the center of the n- gon as their midpoint.]. For example, if we label the vertices of a regular 6- gon 1, 2, 3, 4, 5, 6, then you can change the sign of {1, 4}, {2, 5}, {3, 6}, {1, 3, 5}, {2, 4, 6} or {1, 2, 3, 4, 5, 6}. Show that for (1) \(n = 15\) , (2) \(n = 30\) , (3) any \(n > 2\) , we can find some initial marking which cannot be changed to all \(+1\) by any series of

moves. Let f(n) be the largest number of markings, so that no one can be obtained from any other by any series of moves. Show that \(\mathrm{f}(200) = 2^{80}\) .


S is a sphere with unit radius. P is a plane through the center. For any point x on the sphere \(\mathrm{f}(\mathbf{x})\) is the perpendicular distance from \(\mathbf{x}\) to P. Show that if \(\mathbf{x}\) , \(\mathbf{y}\) , \(\mathbf{z}\) are the ends of three mutually perpendicular radii, then \(\mathrm{f}(\mathbf{x})^2 + \mathrm{f}(\mathbf{y})^2 + \mathrm{f}(\mathbf{z})^2 = 1\) (\\*). Now let \(\mathrm{g}(\mathbf{x})\) be any function on the points of S taking non- negative real values and satisfying (\\*). Regard the intersection of P and S as the equator, the poles as the points with \(\mathrm{f}(\mathbf{x}) = 1\) and lines of longitude as semicircles through both poles. (1) If \(\mathbf{x}\) and \(\mathbf{y}\) have the same longitude and both lie on the same side of the equator with \(\mathbf{x}\) closer to the pole, show that \(\mathrm{g}(\mathbf{x}) > \mathrm{g}(\mathbf{y})\) . (2) Show that for any \(\mathbf{x}\) , \(\mathbf{y}\) on the same side of the equator with \(\mathbf{x}\) closer to the pole than \(\mathbf{y}\) we have \(\mathrm{g}(\mathbf{x}) > \mathrm{g}(\mathbf{y})\) . (3) Show that if \(\mathbf{x}\) and \(\mathbf{y}\) are the same distance from the pole then \(\mathrm{g}(\mathbf{x}) = \mathrm{g}(\mathbf{y})\) . (4) Show that \(\mathrm{g}(\mathbf{x}) = \mathrm{f}(\mathbf{x})\) for all \(\mathbf{x}\) .


P is a polygon. Its sides do not intersect except at its vertices, and no three vertices lie on a line. The pair of sides AB, PQ is called special if (1) AB and PQ do not share a vertex and (2) either the line AB intersects the segment PQ or the line PQ intersects the segment AB. Show that the number of special pairs is even.


n points lie in the plane, not all on a single line. A real number is assigned to each point. The sum of the numbers is zero for all the points lying on any line. Show that all the assigned numbers must be zero.


(1) The triangle ABC is inscribed in a circle. D is the midpoint of the arc BC (not containing A), similarly E and F. Show that the hexagon formed by the intersection of ABC and DEF has its main diagonals parallel to the sides of ABC and intersecting in a single point.

(2) EF meets AB at X and AC at Y. Prove that AXIY is a rhombus, where I is the center of the circle inscribed in ABC.


Black and white tokens are placed around a circle. First all the black tokens with one or two white neighbors are removed. Then all white tokens with one or two black neighbors are removed. Then all black tokens with one or two white neighbors and so on until all the tokens have the same color. Is it possible to arrange 40 tokens so that only one remains after 4 moves? What is the minimum possible number of moves to go from 1000 tokens to one?


\(\mathbf{a}_{\mathrm{n}}\) is an infinite sequence such that \((\mathbf{a}_{\mathrm{n} + 1} - \mathbf{a}_{\mathrm{n}}) / 2\) tends to zero. Show that \(\mathbf{a}_{\mathrm{n}}\) tends to zero.


There are direct routes between every two cities in a country. The fare between each pair of cities is the same in both directions. Two travellers decide to visit all the cities. The first traveller starts at a city and travels to the city with the most expensive fare (or if there are several such, any one of them). He then repeats this process, never visiting a city twice, until he has been to all the cities (so he ends up in a different city from the one he starts from). The second traveller has a similar plan, except that he always chooses the cheapest fare, and does not necessarily start at the same city. Show that the first traveller spends at least as much on fares as the second.


Each vertex of a convex polyhedron has three edges. Each face is a cyclic polygon. Show that its vertices all lie on a sphere.


Given a polynomial \(\mathrm{x}_{10} + \mathrm{a}_{9}\mathrm{x}^{9} + \ldots +\mathrm{a}_{1}\mathrm{x} + 1\) . Two players alternately choose one of the coefficients \(\mathrm{a}_{1}\) to \(\mathrm{a}_{9}\) (which has not been chosen before) and assign a real value to it. The first player wins iff the resulting polynomial has no real roots. Who wins?


Seven elves sit at a table. Each elf has a cup. In total the cups contain 3 liters of milk. Each elf in turn gives all his milk to the others in equal shares. At the end of the process each elf has the same amount of milk as at the start. What was that?


We call a number doubly square if (1) it is a square with an even number 2n of (decimal) digits, (2) its first n digits form a square, (3) its last n digits form a non- zero square. For example, 1681 is doubly square, but 2500 is not. (1) find all 2- digit and 4- digit doubly square numbers. (2) Is there a 6- digit doubly square number? (3) Show that there is a 20- digit doubly square number. (4) Show that there are at least ten 100- digit doubly square numbers. (5) Show that there is a 30- digit doubly square number.


Given a sequence \(\mathrm{a}_{1}\) , \(\mathrm{a}_{2}\) , ..., \(\mathrm{a}_{\mathrm{n}}\) of positive integers. Let S be the set of all sums of one or more members of the sequence. Show that S can be divided into n subsets such that the smallest member of each subset is at least half the largest member.


You have 1000 tickets numbered 000, 001, ..., 999 and 100 boxes numbered 00, 01, ..., 99. You may put each ticket into any box whose number can be obtained from the ticket number by deleting one digit. Show that you can put every ticket into 50 boxes, but not into less than 50. Show that if you have 10000 4- digit tickets and you are allowed to delete two digits, then you can put every ticket into 34 boxes. For \(\mathrm{n} + 2\) digit tickets, where you delete n digits, what is the minimum number of boxes required?


Given a \(100\mathrm{x}100\) square divided into unit squares. Several paths are drawn. Each path is drawn along the sides of the unit squares. Each path has its endpoints on the sides of the big square, but does not contain any other points which are vertices of unit squares and lie on the big square sides. No path intersects itself or any other path. Show that there is a vertex apart from the four corners of the big square that is not on any path.


The positive integers \(\mathbf{a}_1\) , \(\mathbf{a}_2\) , ..., \(\mathbf{a}_{\mathrm{m}}\) , \(\mathbf{b}_1\) , \(\mathbf{b}_2\) , ..., \(\mathbf{b}_{\mathrm{n}}\) satisfy: \((\mathbf{a}_1 + \mathbf{a}_2 + \ldots + \mathbf{a}_{\mathrm{m}}) = (\mathbf{b}_1 + \mathbf{b}_2 + \ldots + \mathbf{b}_{\mathrm{n}}) < \mathrm{mn}\) . Show that we can delete some (but not all) of the numbers so that the sum of the remaining a's equals to the sum of the remaining b's.