Solution

This is a next step following the Example 3. Here we have two dimensions instead of 1 and so each point is represented by a pair of numbers, for example $(a,b)$. Both of these numbers are integers in case of each of the marked points. The other point might have coordinates $(c,d)$. Then the midpoint of these points has coordinates $(\frac{a+c}2,\frac{b+d}2)$. Again, each coordinate is an integer if and only if the sum of two respective coordinates is an even number. We need both coordinates to be integers this time – so both sums need to be even. They are even if and only if x coordinate of both points have same parity AND y coordinates of these points also have same parity. For example, if $a$ and $c$ are both even and $b$ and $d$ are both odd, both the sums are even and both the coordinates of the midpoint are integers. $\\$
Each point can either have both coordinates even, both odd, the first even and the other odd or the first odd and the second even. These are $4$ possibilities, compared to $5$ points that were marked. Therefore, 2 of the points belong to the same category and so, their midpoint has both coordinates integer.

Hint

Solution

Let’s say the three marked points have integer coordinates $a,b$ and $c$. A midpoint of two points of coordinates $a$ and $b$ respectively is a point of coordinates $\frac{a+b}2$. This is an integer if and only if $a+b$ is even. There are three numbers marked on the line by Angelina, and so three midpoints. Their coordinates correspond to three sums: $a+b, b+c$ and $c+a$. Such a sum is even only if both numbers are even or both numbers are odd. $\\$
An integer can only be even or odd. These are our pigeonholes. Numbers are our pigeons this time – there are three of them. That means two of these numbers are placed in the same pigeonhole – they have the same parity (are both even or both odd). Thus, one of the three sums is even and the respective midpoint has the integer coordinate.

Hint

Solution

Solution 1

We denote the 25-gon by $A_1A_2 … A_{24}A_{25}$. We call the index of the diagonal $A_iA_j$ $($i $<$ j$)$ the number min {j - i, 25 - j + i}. The index of the diagonal $A_iA_j$ passing through the point at which nine diagonals intersect can assume the values 9, 10, 11 and 12, since there are 8 diagonal ends between the points $A_i$ and $A_j$ on the contour of the 25-gon.

It is clear that the diagonals of one index are tangent to one circle.

Let nine diagonals of a regular 25-gon intersect at one point. Then among them there are at least three single indexes. This means that from a certain point three tangents to one circle can be drawn. And so, we have a contradiction.

Solution 2

Suppose that the n diagonals $A_1B_1, …, A_nB_n$ intersect at the point M. We can assume that the points $A_1, …, A_n, B_1, …, B_n$ are arranged in this order “clockwise”. Let us describe a circle around a 25-gon. Then the following sum of arcs
$(A_1A_2 + B_1B_2) + (A_2A_3 + B_2B_3) + … + (A_{n-1}A_n + B_{n-1}B_n) + (A_nB_1 + B_nA_1)$ is equal to 2φ. The value of each arc is a multiple of φ = 2π / 25, therefore each of the sums in parentheses equals at least 2φ.

Suppose that one of these sums $(for example, A_1A_2 + B_1B_2)$ is equal to 2φ. Then the chords $A_1A_2$ and $B_1B_2$ are equal, hence the quadrilateral $A_1A_2B_1B_2$ is an isosceles trapezoid. The perpendicular dropped from the centre O of the circle to the base $A_1B_2$ of this trapezoid passes through the middle of its bases, and hence through the point M of the intersection of its diagonals. Moreover, the straight line l perpendicular to OM and passing through the point M is parallel to the bases of the trapezium and, therefore, intersects its lateral sides, and hence also the arcs $A_1A_2$ and $B_1B_2$. If one of the sums is equal to 2φ, then the same line l intersects two more pairs of the arcs that are under consideration, which is impossible. Consequently, all the sums in parentheses, except perhaps by one, are no less than 3φ.
And so, we obtain the inequality $(n – 1) \times 3φ + 2φ \leq 2φ $ = 25φ, and hence n $\leq$ 26/3 $<$ 9.

Answer

6 points

Hint

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Solution

Assume that this is not true – that is, in every circle of radius 1995 will contain at least one marked point. Consider an infinitely long horizontal strip of width 4000. Divide this strip into squares of size 4000×4000. Consider 4001 such squares that are adjacent to one another, creating a rectangle we will call $P$. Each square must contain at least one point, as a circle of radius 1995 can fit into each square. The vertical distances of all of the points from the bottom edge of the strip cannot all be different as there are only 4000 possible values for the distance and at least 4001 points. Hence there must be at least two points which are the same vertical distance above the bottom edge of the $P$. The abscissae – modulus of the $x$ co-ordinates – of these two points are some two whole numbers, so in a within $P$ there is a finite number of values they could have.

Now consider a relatively large number of these horizontal strips, which do not overlap with one another. In each of these, in the rectangles located above $P$, there will be a pair of points with the same height above the bottom edge of the rectangle or the same $y$ co-ordinate. There will be two pairs of points, such that the abscissae of one pair corresponds to the abscissae of the other pair. These 4 points thus form a rectangle, therefore all four points lie on the same circle. This is a contradiction.

Hint

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Hint

Note that in order to fulfil the conditions of the problem, the distance between adjacent plants must be no greater than 40m

Solution

In each 40m segment of the path there must be one plant. The 400m path must contain exactly 10 such segments, therefore we need 10 plants with the plant planted at the centre of each segment. We would have the first plant 20m from the start of the path, and the last 20m before the end, with the rest being at 40m intervals.

Hint

Note that in order to fulfil the conditions of the problem, the distance between adjacent plants must be no greater than 40m

Hint

Try splitting the original square into unit squares.

Solution

Divide the original square into 16 $1 \times 1$ squares. Amongst the 16 squares there will always be at least one that does not contain a point – 15 points cannot be placed into 16 different squares.

Hint

Try splitting the original square into unit squares.

Hint

Can some two polygons share more than one side?

Solution

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.

Answer

Yes.

Hint

Can some two polygons share more than one side?

Solution

Divide the carpet into 16 squares of size 1m x 1m. Since there are 15 holes, or points, it is always true that at least one square will not contain any holes.

Answer

Yes.

Hint

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Solution

Consider an arbitrary regular triangle with side l. It has two vertices of the same colour. The segment connecting these vertices is the one you are looking for.

Hint

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Solution

By drawing lines through the centre of the square, parallel to its sides, we can divide the square into 4 parts. Some two of these 4 points lie in the same one of these 4 parts, and the distance between them cannot exceed the length of the diagonal of this quarter of the square, which is equal to $ \frac{1}{2} AC$.

Hint

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Solution

We consider all segments with endpoints at the marked points. Since no three of the marked 64 points lie on a single straight line, there will be $64\times63$/2 = 2016 $>$ 1981 such segments. From problem 116573 it follows that among these segments there will be at least one pair of parallel points. A quadrilateral with vertices at the ends of these segments is the desired one $($it cannot be a parallelogram, since there is no parallelogram with vertices at the vertices of the regular 1981-gon$)$.

Hint

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Solution

Let’s mark the upper left corner of each square. By hypothesis, for different squares the marked points are different. Therefore, the number of squares does not exceed mn. It is also clear that mn cells have the required property, therefore the number of squares can be equal to mn.

Answer

mn squares.

Hint

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Solution

We prove by mathematical induction on n that the number of distinct arcs after n intersections does not exceed 3. For n = 2 this is obvious. Denote by $A_k$ the notch with the number k. Let n be notches be made and the point $A_n$ be on the arc $A_{k}A_{l}$. Then the point $A_{n-1}$ falls on the arc $A_{k-1}$$A_{l-1}$. Therefore, for k, l $\neq$1, no new arcs appear, and the required assertion is proved. Suppose now that, for example, l = 1. Let us prove that then the length of any arc $A_{p}A_{q}$ between adjacent notches is equal to the length of one of the arcs $A_{k}A_{1}$, $A_{n}A_{1}$, $A_{1}A_{s}$, where $A_s$ is the notch closest to $A_1$, and different to $A_n$. Indeed, for p, q $\neq$ 1, the length of the arc $A_{p}A_{q}$ is equal to the length of the arc $A_{p-1}A_{q-1}$, and between the notches $A_{p-1}$ and $A_{q-1}$ there are no other notches, except possibly, An. Thus, in a finite number of steps we arrive at one of the three considered arcs.

Answer

No more than three.

Hint

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Solution

Answer: at the height of $\frac{a}{2 \sqrt2}$ , where a is the side of the square. Let’s consider the following 5 points: the vertices of the square and its centre. If the diameter of the circle is less than $\frac{a}{\sqrt2}$, then it can contain no more than one of these points. Therefore, 4 circles of radius less than $\frac{a}{\sqrt2}$ cannot completely cover a square with side a. It is easy to see that 4 circles, whose diameters are the segments connecting the centre of the square with its vertices, completely cover the square.

Hint

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Solution

Consider a particular arrangement of kings that satisfies these conditions. A king standing on the edge squares of the board can never be attacked by another king, therefore all of the kings must lie in the ‘inner’ square of size 6×6. Notice that in any square of size 3×3 there can be no more than 4 kings. If such a square contains 5 kings then the centre square must definitively be a black one, and the kings must stand on all of the black squares in that 3×3 square. However in this situation the king standing in the centre cannot be attacked by any other king. Hence a 6×6 square can contain no more than $4 \times 4 = 16$ kings.

One such arrangement is shown in the drawing below.

Answer

16 kings.

Hint

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Solution

Answer: Yes, it is possible. Let’s cut this cube to $13^3$ = 2197 cubes with edge 1. If there was a selected point inside each of these cubes, then the number of selected points would be not less than 2197, which contradicts the condition. Consequently, at least one selected point does not lie inside at least one of these cubes.

Hint

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Solution

Let’s think of situation where 12 fuses are not enough. We can assume that Ahmed is allowed to install fuses only at the ends of the cords. After all, if you transfer the fuse from the inner point of the cord to the end from which it burns, the situation will not change.

On each cord, there is an arrow in the direction in which it will burn. Let’s paint the knots of the cords in a chequered pattern. There will be 13 knots of one colour, let’s say, black. Let’s assume that all of the arrows come from these black knots $($see the figure$)$. Then each of the cords originating from a black point can only burn if a fuse is placed at this point. But we only have 12 fuses and there are 13 points.

Answer

Yes, Ahmed will be able to arrange the cords in such a way that Helen will not be able to burn all of them.

Hint

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Solution

To draw the perimeter of the 25×25 square we have no choice but to draw all 96 edge and corner squares. Divide the inner remaining 23×23 square into 264 ‘dominoes’ of size 2×1 and one 1×1 square. We need to draw one of the 1×1 squares in each ‘domino’ otherwise the line dividing the 2×1 square into two 1×1 squares will not be visible. Therefore in total we need to draw $96 + 264 = 360$ 1×1 squares.

From another perspective – 350 squares is sufficient. Consider a ‘chessboard’ shading of the 25×25 square, in which all of the corner 1×1 squares are black. Each line segment in the ‘grid’ we are trying to create is an edge of either a ‘border’ square – edge or corner squares – or a white square. Therefore it is sufficient to draw all of the border squares, of which there are 96, and all of the inside white squares, of which there are 264.

Answer

360 squares.

Hint

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Solution

Imagine that the knights stand on 18 squares, located on two circular platforms, and which each hour they turn in opposite directions by 40$^{\circ}$. According to the question, at some point there was no one on five of the sites, and on the corresponding sites $($in the same towers$)$ there was one knight. Let us examine two cases.

1$)$ There are empty squares on both platforms. Then at some point they will be in the same tower, that is, there will be no knights on this tower.

2$)$ All empty platforms are on the same platform, and all the lone knights are on the other. Since there are only 9 $<$ 5 + 5 towers, at each moment there is a tower, where there will be both an empty square and a lone knight. But this contradicts the question.

Answer

5 points

Hint

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Solution

Since the diagonals going from one vertex divide the $n$-sided polygon into $n-2$ triangles, then $n-2$ points are necessary.

From the diagram we can see how $n-2$ points can be placed to fulfil the requirements; it is sufficient to place one marked point in each of the black triangles. In fact, inside any triangle $A_pA_qA_r$ where $p < q < r$ there will always be a black triangle at vertex $A_q$

Answer

$n-2$

Hint

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