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International Mathematics and Science Olympiad 2016 EXPLORATION PROBLEMS (1) The diagram below shows the V-, L- and P-pentominoes. Use two copies of each to construct each of the given figures. The pieces may be rotated or reflected. (To solve this problem, you may use scissors to cut the puzzle pieces in the attached colour page.)
(2) Find all ways of filling in the nine boxes with the digits 1 to 9, using each once, in order to make the addition and the multiplication correct. Submitted by Philippines = + = The multiplication must be one of 3 16= 48, 3 18= 54, 3 19= 57, 4 17= 68, 4 19= 76, 4 13= 52, 4 18= 72, 6 13= 78, 7 12= 84, 7 14= 98 and 8 12= 96. In the last six cases, the sum will exceed 100. For 4 19= 76, we must add 23 to stay under 100, but 99 is not an acceptable sum. For 3 19= 57, only the even digits are left, and parity will fail in the addition. For 3 16= 48, the units digits in the sum must be 8+ 7 = (1)5 but 4+ 2 + (1) < 9. For 3 18= 54, the tens digits in the sum must be 5+ 2= 7, but now the units digits do not work. Hence there is only one possible way, as shown in the diagram below. 1 7 4 = 6 8 + 2 5 = 9 3 (3) Two of 13 coins are counterfeit with equal weight. The other 11 coins are genuine coins also with equal weight, but a different weight from that of the counterfeit coins. We are not trying to identify the counterfeit coins but to determine whether a counterfeit coin is heavier or lighter than a genuine coin. How can this be done in three weighings on a balance? Submitted by Jury Label the coins A, B, C, D, E, F, G, H, I, J, K, L and M. In the first weighing, put (A, B, C, D, E, F) against (G, H, I, J, K, L). There are two cases. Case 1. We have equilibrium. Then each side contains a counterfeit coin. Hence one of (A, B), (C, D) and (E, F) contains a counterfeit coin. In the next two weighing, put A and B against C and D, and then against E and F. We cannot have equilibrium both times. If we do not have equilibrium at all, then (A, B) contains the counterfeit coin. If we have equilibrium once, say between (A, B) and (C, D), then (E, F) contains the counterfeit coin. In either case, we can tell if the counterfeit coin is heavier or lighter. Case 2. One side, say (A, B, C, D, E, F), is heavier. The next two weighings are the same as in Case 1. If we have equilibrium both times, the counterfeit coin is lighter. Otherwise, it is heavier.
(4) (a) Cut each figure in the diagram below into two pieces such that each piece has an axis of symmetry. (b) Cut each of the figures in the diagram below into two pieces so that the six pieces consist of three identical pairs. The pieces may be rotated and reflected. (a) The cuts are shown in the diagram below. or There is an alternative solution to the first figure. Just cut along a diagonal of the complete square at the top left corner. (b) The cuts are shown in the diagram below.
(5) We introduce a new chess piece called a Catapult. Its range of attack is shown below. X X X X X X X X X X C X X X X X X X X X X In the 8 8 chessboard below, place as many catapults as possible, so that no two them can attack each other. (Mark a C on the chessboard for a catapult) Submitted by Singapore C C C C C C C C C C C C C The diagram above shows that as many as 13 non-attacking catapults, represented by C s, can be placed on the 8 8 chessboard. This formation is unique up to symmetry. We now prove that 13 is maximum. Divide the chessboard into four quadrants. In each quadrant, at most 4 catapults can be placed, and if 4 are placed, they must be in either formations shown in the diagram below, up to symmetry. Note that 2 of the catapults must occupy opposite corners of such a quadrant. C C C C C C C C Suppose there are more than 13 catapults. Then there must be 4 of them in each of at least two quadrants. If two such quadrants are adjacent, then the catapults must be in the formation shown in the diagram below, up to symmetry. The X s represent spaces on which catapults may not be placed. Now there are room for at most 2 more
catapults in the northeastern quadrant and at most 3 more in the southeastern one. Hence the maximum in this case is only 13. C C X X X X X C X C X X C X X X C X X X X X C C X if no two adjacent quadrants hold 4 catapults, then we must have 4 catapults in each of two opposite quadrants and 3 in each of the other two. The corner catapults in the two quadrants with 4 catapults must be in either formation shown in the diagram below, up to symmetry. Now there is room for at most 2 catapults in each of the other two quadrants. Thus the maximum in this case is only 12. X X X C X X X C X X X X X X X X X X X X X C X X X C C X X X X C X X X X X X X X X X X X X X X C C X X X Marking Scheme Place 10 or 11 catapults with correct diagram, 1 mark. Place 12 catapults with correct diagram, 3 mark. Place 13 catapults with correct diagram, 6 mark.
(6) The diagram below shows seven pieces on the left and a board on the right. On the pieces are black circles which represent the binary digit 1, and white circles which represent the binary digit 0. Use the seven pieces to cover the board so that the resulting configuration is a correct multiplication in binary numbers. The pieces may be rotated or reflected. Submitted by Jury We first determine what the multiplication should be. Since there are three rows between the two long horizontal lines in the board, the three-digit multiplier must be 111. The four-digit multiplicand cannot be 1000 or 1001 as otherwise the product will only have six digits. Hence the multiplication is one of the following. 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 1 0 1 0 1 1 1 0 1 0 1 0 1 1 1 0 1 0 1 0 1 1 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 0 0 1 0 1 1 0 1 0 0 1
There are 7 white circles among the seven pieces, and there is only one multiplication with the correct number of 0s, namely, 1011 111 = 1001101. It follows that the target diagram is as shown below. The O-tetromino has a checkered pattern, and there are only two possible positions for it. In the upper position, as shown in the diagram below on the left, it cuts the board into two pieces, consisting of ten and twelve squares respectively. Hence the smaller piece must use both trominoes and one tetramino. Since it has only one white circle, we must use the S-tetromino. The white circle forces the position of the V-tromino, but now we cannot fit the other two pieces in the remaining part. So the O-tetromino must go into the lower position. This forces the positions of the T-tetromino, I-tetromino and the S-tetromino. It is easy to complete the reconstruction of the multiplication configuration as shown in the diagram above on the right. Marking Scheme Find the correct target diagram only, 3 mark. Use the seven pieces to make a correct multiplication, 6 mark.