A game takes place on a squared 9 × 9 piece of checkered paper. Two players play in turns. The first player puts crosses in empty cells, its partner puts noughts. When all the cells are filled, the number of rows and columns in which there are more crosses than zeros is counted, and is denoted by the number K, and the number of rows and columns in which there are more zeros than crosses is denoted by the number H $($ 18 rows in total $)$. The difference B = K – H is considered the winnings of the player who goes first. Find a value of B such that
$\\$$\\$
1$)$ the first player can secure a win of no less than B, no matter how the second player played;
$\\$$\\$
2$)$ the second player can always make it so that the first player will receive no more than B, no matter how he plays.
Solution 1
$\\$$\\$$\\$$\\$
One of the possible strategies of the first player: with his first move he places a cross at the center of the board; then for every move of the second player in any cell, it responds with a move in a cell symmetric with respect to the center. This is possible, since symmetry is violated each time by the second player.
In the final position in each pair of cells, symmetrical with respect to the center, there is a cross and a nought. Therefore, in the central row and in the central column, there are more crosses than zeros; and each row $($ column $)$ that does not pass through the center corresponds to a symmetrical row $($ column $)$, with one of them having more zeros than crosses and one of them having fewer zeros than crosses. Thus, with this strategy, the first player’s win is equal to 2.
Now we indicate the strategy of the second player. If he has the opportunity to occupy a cell that is a symmetrical $($ relative to the center $)$ cell, which has just been occupied by the first player then it does so. Otherwise, he makes any move. With such a strategy, after the turn of the second player, the set of occupied cells is either symmetrical $($ until the first has occupied the center $)$, or differs from the symmetrical one by exactly one cell. In this case, in each pair of occupied symmetrical cells there will be one cross and one zero. With his last move, the first player will have to create the position described in the previous paragraph, and his winnings are again 2.
$\\$$\\$$\\$$\\$
Solution 2
$\\$$\\$$\\$$\\$
We cover the board with tiles as shown in the figure. The strategy of the first: take the first lower left corner, and then go to the same domino tile on the board, where the second one went before it.
$\\$$\\$
$\\$$\\$
The strategy of the second: if possible, take a cell in the same part of the board, where the first player went first; if it is impossible – any move.
If at least one of the players adheres to the strategy indicated for him, then in the end in the lower left corner there will be a cross, and in each of the other parts of the board there will be an equal number of noughts and crosses $($ for reasons similar to those stated in the first solution: after each move of the second player in all filled dominoes there will stand a cross and a nought, and, perhaps, in one there is just a nought $)$.
Consider the two upper rows. In total, they contain 9 crosses and 9 noughts $($ since they are divided into 9 “dominoes” $)$. Therefore, in one of them there are more crosses, in the other – more zeroes. A “Draw” is also achieved in the third and fourth, …, seventh and eighth rows from the top, and in the bottom rows the first “wins” $($ it has 5 crosses and 4 noughts $)$. The situation is similar in the columns $($ pairs of neighboring columns, starting counting from the right, also consist of 9 “dominos” $)$.
B = 2.
17 squares are marked on an 8×8 chessboard. In chess a knight can move horizontally or vertically, one space then two or two spaces then one – eg: two down and one across, or one down and two across. Prove that it is always possible to pick two of these squares so that a knight would need no less than three moves to get from one to the other.
Divide the board into 16 identical groups containing 4 squares – two of which are marked on the diagram below. We get another two by translating both groups down two squares, and the remaining groups by translating to the right to fill the empty columns.
It is easy to check that no fewer than 3 moves of a knight are required in order to get from one square in a figure to any other in that figure. By the pigeon hole principle, at least two of the 17 marked squares will be those in one of the figures.
Consider two cases:
1) Amongst the marked squares, squares of both colours are present. Then there must be no less than 9 marked squares of one of the colours, black for example. From a marked white square, a knight can only get to at most 8 different black squares in one move. The second move would then have to be to a white square, meaning that in order to get to at least one black square three moves will be required.
2) All of the marked squares are the same colour, for example black. Consider the smallest rectangle that includes all of the marked squares. Its area must be at least 33 – it includes no less than 17 black squares, and therefore at least 16 white squares. Therefore one of its sides, say the horizontal side, must be at least of length 6. In this case a knight standing on a marked square on the left edge of the rectangle cannot reach a marked square on the right edge of the rectangle in two moves. Therefore three moves are required.
The value of 17 squares given in the question is not a minimum. Let us call the distribution of marked squares ‘correct’ if it the ‘distance’ between any two marked squares on the board is no greater than two moves of a knight. We can prove that it is possible to ‘correctly’ distribute no more than 9 squares.
Note, that any two ‘correctly distributed’ squares of different colours must be exactly one move of a knight away, and any two ‘correctly distributed’ squares of the same colour must be exactly two moves of a knight away.
Assume that we have managed to ‘correctly’ distribute some marked squares. We will consider two cases.
1) There are no fewer than two squares of each colour marked. Then all of the white squares must be within two moves of a knight from both of the marked black squares. Therefore the two white squares must lie on the intersections of two circles of radius $sqrt{5}$, whose centres lie on the black squares. However two circles intersect at no more than two points. Hence there are no more than two white squares present. Analogously, there are no more than two black squares present meaning there must be exactly 4 marked squares.
2) There are fewer than two marked squares of one of the colours, for example white. As shown in problem 116486 there can be no more than 8 marked squares of the other colour. Therefore in total, there can be no more than 9 marked squares that are ‘correctly’ distributed.
It is possible to ‘correctly’ distribute 9 squares, by placing the knight in the middle of the board, and marking the square the knight stands on, and the 8 different squares it can reach from there.
On a plane there is a square, and invisible ink is dotted at a point P. A person with special glasses can see the spot. If we draw a straight line, then the person will answer the question of on which side of the line does P lie $($ if P lies on the line, then he says that P lies on the line $)$.
What is the smallest number of such questions you need to ask to find out if the point P is inside the square?
During the ball every young man danced the waltz with a girl, who was either more beautiful than the one he danced with during the previous dance, or more intelligent, but most of the men $($ at least 80% $)$ – with a girl who was at the same time more beautiful and more intelligent. Could this happen? $($ There was an equal number of boys and girls at the ball.$)$
A square piece of paper is cut into 6 pieces, each of which is a convex polygon. 5 of the pieces are lost, leaving only one piece in the form of a regular octagon $($see the drawing$)$. Is it possible to reconstruct the original square using just this information?
Can some two polygons share more than one side?
Since the remaining 5 polygons are convex, none of them can share more than one side with the octagon. This means that at least 3 sides of the octagon must belong to the square. This conclusion allows us to uniquely define the size of the square – its side length is equal to the distance between opposite sides of the octagon.
It is interesting to note that although we can deduce the size of the square, we cannot say exactly what polygons it consists of. We can only reconstruct 4 of the polygons – the octagon and three corner triangles. For the remaining two polygons we only know that together they must make up the fourth corner triangle. We cannot even definitively state how many sides each of these two polygons has – they could be a triangle and a quadrilateral, or two triangles, as in the diagram.
Yes.
An entire set of dominoes, except for 0-0, was laid out as shown in the figure. Different letters correspond to different numbers, the same – the same. The sum of the points in each line is 24. Try to restore the numbers.
What letter is zero encrypted as? Can the letter “a” denote an odd number $($see the last line in the figure for the task$)$? Can “a” be equal to 2 $($see the first line in the figure$)$?
Consider the drawing.
Twenty-eight dominoes can be laid out in various ways in the form of a rectangle of $8 \times 7$ cells. In Fig. 1-4 four variants of the arrangement of the figures in the rectangles are shown. Can you arrange the dominoes in the same arrangements as each of these options?
Option 1 $($see figure. 1 in the task$)$. Try to determine where the duplicate dominoes are located. Option 4 $($see figure. 4 in the task$)$. Where is the domino 4-2? Can the second row contain the dominoes 0-1 and 5-6? Can the first row of dominoes contain the dominoes 1-2 and 4-5? Can the dominoes 2-5 and 1-4 lie in the second row? Can the domino 1-1 stand in the second column? And can the domino 5-5 stand in the seventh column? Where is the domino 0-6? Can the dominoes 6-3 and 3-0 be in the third or fourth rows?
All similar problems are solved in the same way. The usual properties of the dominoes are used. For example, in a domino set, a domino with a pair of any numbers from 0 to 6 is necessarily encountered, and no such pair is repeated twice. Dominoes cannot be arranged on an odd number of cells, etc. Since the variations are proposed in order of increasing difficulty, we will consider detailed solutions only for the first – because of its clarity and for the latter – because of its complexity.
Option 1 $($Figure 1$)$:
1. The places of all duplicates in this layout are uniquely determined. Let us note them. Hence, the locations of the dominoes 5-0, 5-3, 0-2, 3-4, 0-6, 2-5 are uniquely determined. Let us note now that if the above-mentioned pairs of digits stand side by side, then they cannot form dominoes. The resulting position is shown in Fig. 2.
2. Hence the locations of the dominoes 2-4 and 0-3 are also uniquely determined. Noting that these dominoes cannot be in other places, we get the locations of these dominoes $($the position in Figure 3$)$.
3. And again, we pay attention to the fact that, if the above-mentioned pairs of figures are nearby, they cannot form dominoes. From here it is already possible to unambiguously restore the entire layout $($Figure 4$)$.
Option 4 $($Figure 5$)$:
1. The only domino, the location of which can be determined unambiguously, is 4-2 $($nowhere else do the numbers 4 and 2 stand side by side$)$.
2. In the second row there cannot be a domino 0-1 $($otherwise we get two identical dominoes 0-1 in the first and second rows$)$. For the same reasons, in the second row there cannot be a domino 5-6, and in the sixth – the dominoes 4-6 and 0-2. Note this $($Figure 6$)$.
3. The domino 1-2 cannot be placed in the first row $($otherwise the number 1 at the intersection of the second row and the second column will form one more domino 1-2$)$. By analogy, in the first row there cannot be a domino 4-5.
4. Similarly, as was done in step 2, we will determine that dominoes 2-5 and 1-4 cannot lie in the second row. Note this $($Figure 7$)$.
5. Now it is obvious that the domino 1-1 cannot stand in the second column $($otherwise there is nowhere to place the domino 1-2$)$. By analogy, 5-5 cannot be in the seventh column $($otherwise there is no room for the domino 4-5$)$. From here you can unambiguously restore the locations of dominoes 0-1, 0-5, 5-6 and 6-1. Noting that in other places these dominoes cannot be located, we will get an unequivocal opportunity to place the dominoes 1-1 and 5-5. Note that any other location for them is impossible. Thus, we obtain the position of the dominoes in Fig. 8.
6. 0 and 6 are found three times side by side, and all three times – in the sixth row. However, only the pair that lies exactly under the domino 4-2 can form a domino 0-6 $($otherwise we will get two dominoes 0-6$)$. Hence the position of the dominoes 4-0 and 2-6 is uniquely determined. Note the impossibility of placing these dominoes in other places $($Figure 9$)$.
7. Let’s pay attention to the fact that in the third and fourth rows the dominoes 6-3 and 3-0 cannot be placed; hence the location of the dominoes 6-6 and 0-0 can be determined. Furthermore, the domino 3-3 cannot be in the third row, nor in the fourth or fifth columns. From this we determine the location of the dominoes 3-3, 5-3 and 1-3. Noting that these dominoes cannot be located elsewhere, we will get an unambiguous arrangement of all the dominoes $($Fig. 10$)$.
See the figure.
Is it possible to cut a square into four parts so that each part touches each of the other three $($ie has common parts of a border$)$?
Notice, in both cases, we are not given any restrictions on the shape of the pieces, only their number is defined.
Yes, it is possible. See the picture below.
Yes you can. See the figure.
Decipher the following rebus $\\$
$\\$
All the digits indicated by the letter “E” are even $($ not necessarily equal $)$; all the numbers indicated by the letter O are odd $($ also not necessarily equal $)$.
For the convenience of further reasoning, we replace all even numbers with vowels, and odd ones with consonants, bearing in mind that the same number can correspond to different letters. Therefore the rebus will take the following form:$\\$
$\\$
Note that the numbers AEB and YHK differ by the second digit, so C $>$ 1. For C $>$ 3 or A $>$ 2, the product $AEB \times C$ will be four-digit, therefore C = 3, A = 2. Hence, I is at most 2, that is it is equal to 2. It follows that D = 9 $($ for a smaller value of D, the expression $AEB \times D$ will be less than 2100, and IFUG $>$ 2100 $)$ Y = 8 $($ otherwise the entire product will not be a five-digit number $)$. E = 8 $($ if E $<$ 8, then YHK $<$ 810 $)$. If B varies from 1 to 9, then the number represented by IFUG will vary from 2529 to 2601, therefore F = 5, and B $<$ 9. H = 5 $($ with B changing from 1 to 7, YHK will vary from 843 to 861 $)$. K = 5 $($ otherwise the number 85K will not be divisible by 3 $)$.
285 × 39 = 11115.
In Neverland, there are magic laws of nature, one of which reads: “A magic carpet will fly only when it has a rectangular shape.” Frosty the Snowman had a magic carpet measuring $9 \times 12$. One day, the Grinch crept up and cut off a small rug of size $1 \times 8$ from this carpet. Frosty was very upset and wanted to cut off another $1 \times 4$ piece to make a rectangle of $8 \times 12$, but the Wise Owl suggested that he act differently. Instead he cut the carpet into three parts, of which a square magic carpet with a size of $10 \times 10$ could be sown with magic threads. Can you guess how the Wise Owl restructured the ruined carpet?
Think about how the magic carpet looked after the Grinch cut off a piece of it.
After the Grinch messed up the carpet, Frosty could cut off a piece of size $1 \times 4$ from this carpet and turn it into a carpet measuring $8 \times 12$. This means that after the actions of the Grinch the carpet looked as shown in Fig. 1. The Wise Owl cut this carpet as shown in Fig. 2, and sewed it as shown in Fig. 3.
See the picture on the right.
What is the maximum number of pieces that a round pancake can be divided into with three straight cuts?
Remember task 11.
Since we are cutting a cylindrical cake and not a flat tart we should think of cuts in various planes. On the picture shown the way to cut the cake into 8 pieces with 2 vertical cuts and one horizontal.
8 pieces.
Is it possible to bake a cake that can be divided by one straight cut into 4 pieces?
Notice that the cake does not have to be a convex figure.
If the cake were a convex figure, it would not be possible to do this, but it is not said anywhere that it should be like that. You can, for example, bake a cake in the form of the letter “E” and cut it as shown in the picture.
Yes, see the figure.
How can you divide a pancake with three straight sections into 4, 5, 6, 7 parts?
The number of parts depends on whether the cuts cross each other inside the pancake.
We draw in the pancake three straight lines and consider the points of their intersection. Depending on where these points are located, a specific number of parts will be obtained. To get 4 pieces, you need to place all three points outside the pancake $($Figure 1$)$. The transfer of one of these points from outside of the border of the pancake to the inside adds one part. So, to get 5 pieces, you have to transfer one point to the inside of the pancake $($Figure 2$)$, 6 – one more point needs to be moved inside the pancake $($Figure 3$)$, 7 – place all the three intersection points inside the pancake $($Figure 4$)$.
See the figure.
Two people had two square cakes. Each person made 2 straight cuts from edge to edge on their cake. After doing this, one person ended up with three pieces, and the other with four. How could this be?
Is it possible to cut out such a hole in a sheet of paper through which a person could climb through?
Try to fold the sheet in half and cut a narrow hole along the crease. You will get a narrow hole with wide edges. Try to increase the “length” of the edges by reducing their “width”.
It is enough to fold the sheet in half, cut a narrow hole along the crease, and then make a lot of rectilinear cuts as shown in the figure. The first cut makes a “hole”, and the others increase the length of the “edges” of this hole.
Yes.
Four lamps need to be hung over a square ice-rink so that they fully illuminate it. What is the minimum height needed at which to hang the lamps if each lamp illuminates a circle of radius equal to the height at which it hangs?
Several pieces of carpet are laid along a corridor. Pieces cover the entire corridor from end to end without omissions and even overlap one another, so that over some parts of the floor lie several layers of carpet. Prove that you can remove a few pieces, perhaps by taking them out from under others and leaving the rest exactly in the same places they used to be, so that the corridor will still be completely covered and the total length of the pieces left will be less than twice the length corridor.
In a $10 \times 10$ square, all of the cells of the upper left $5 \times 5$ square are painted black and the rest of the cells are painted white. What is the largest number of polygons that can be cut from this square $($on the boundaries of the cells$)$ so that in every polygon there would be three times as many white cells than black cells? $($Polygons do not have to be equal in shape or size.$)$
In each cut-out polygon, there must be cells of both colours. So, there must be a black cell that borders a white one. But there are only 9 such cells. $($See the figure for an example of how to cut the square into 9 polygons$)$.
9 polygons.
Every day, James bakes a square cake size $3\times3$. Jack immediately cuts out for himself four square pieces of size $1\times1$ with sides parallel to the sides of the cake $($not necessarily along the $3\times3$ grid lines$)$. After that, Sarah cuts out from the rest of the cake a square piece with sides, also parallel to the sides of the cake. What is the largest piece of cake that Sarah can count on, regardless of Jack’s actions?
In the figure on the left, the locations of Jack’s squares are indicated, from which Sarah, obviously, cannot cut a square with a side of more than 1/3. Let us prove that a ${1/3}\times{1/3}$ box can always be cut by Sarah.
The first method:
We divide the cake into 81 boxes ${1/3}\times{1/3}$. Each of the squares Jack cuts will touch no more than 16 of these squares. Therefore, at least 81 – 4 x 16 = 17 squares remain intact. One of them can be cut Sarah.
The second method:
In the figure on the right, five pieces of a cake of size ${1/3}\times{1/3}$ are arranged so that each of the four squares cut by Jack will touch no more than one of them. Therefore, Sarah can always take one of these pieces.
${1/3}\times{1/3}$
In the isosceles triangle $ABC,$ the angle $B$ is equal to $30^{\circ}$, and $AB = BC = 6.$ The height $CD$ of the triangle $ABC$ and the height $DE$ of the triangle $BDC$ are drawn. Find the length $BE.$
$DC = ½ BC = 3$ (see the figure). In addition angle$DCB = 90^{\circ} $- \angle $ DBC = 60^{\circ}$ therefore $CE = ½,\, DC = 1.5.$ Thus $BE = BC – CE = 4.5.$
4.5.
