Solution

Example: Peter can mark the cells shown in the picture on the left. Then Richard cannot cover more than one marked cell with one corner. But nine corners that do not overlaps do not fit on the board, since 27$>$25.

Evaluation: if Peter marks fewer than nine cells, then at least one of the black cells specified in the example will not be marked. Then Richard can cover all of the cells of the board, except this one. Indeed, in the figure on the right, the shaded cells can be added to the corners so that only one of the cells 1, 2 or 3 is free. If you want to exclude another black cell, the pattern needs to be rotated.

Answer

9 cells.

Hint

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Solution

The required tilling is shown in the figure below.

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Solution

Let A be the largest angle of the triangle ABC. Then the angles B and C are acute. After cutting through the midpoints of the sides AB and AC perpendicular to BC, we rearrange the resulting parts, as shown in the figure.

Hint

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Solution

Divide the given unit square into 25 identical smaller squares of side 0.2. One of these must contain at least three points. A circle that completely encloses this smaller square will have a radius of $\frac{1}{5} \sqrt{2} < \frac{1}{7}$, therefore the smaller square can be completely enclosed in a circle of radius $\frac{1}{7}$.

Hint

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Hint

What is the minimum length of the longest rug?

Solution

We will prove that no more than 50m can remain uncovered. In fact, of the 20 rugs with a total length of $20 \times 50=1000$ at least one must have a length of no less than 50m – if every rug was shorter than 50m, then the total length of the rugs would be less than 1km. In this way the rugs will cover at least 50m of the corridor.

On the other hand if we take 20 rugs each of length 50m and place them on top of one another, then the total length of the rugs will be 1km, but 50m of the corridor will not be covered by a rug.

Answer

50m.

Hint

What is the minimum length of the longest rug?

Hint

Number the rational numbers and cover each successive rational number by an interval of length two times smaller than the length of the interval covering the previous rational number.

Solution

Consider the finite sets $A_n$, n = 1, 2, …, where $A_n$ denotes the set of rational numbers in the interval [0; 1] that can be represented as a fraction m / n for some integer m. A finite set of points can be covered by a system of intervals of arbitrarily small length. Therefore, we can cover the set $A_1$ by intervals of total length $(1/2)$ * $(1/1000)$, the set $A_2$ – by intervals of total length $(1/4)$ * $(1/1000)$, etc. , the set $A_k$ by intervals of the total length $(1 / 2k)$ * $(1/1000)$. Thus, all rational numbers turn out to be covered by a system of intervals whose total length is not greater than $(1/2)$ * $(1/1000)$ + $(1/4)$ * $(1/1000)$ + … + $(1 / 2k )$ * $(1/1000)$ + … = 1/1000. $($Here we used the formula for the sum of an infinite geometric progression.$)$

Hint

Number the rational numbers and cover each successive rational number by an interval of length two times smaller than the length of the interval covering the previous rational number.

Hint

A straight line passing through both points crosses two sides of the pentagon.

Solution

Draw a straight line through the two given points. In one of the half-planes created by the straight line lie at least three vertices $A, B, C$ of the pentagon. If we cut off a triangle with vertices at $A, B, C$ from the pentagon we are left with a quadrilateral that contains both points, as required.

Hint

A straight line passing through both points crosses two sides of the pentagon.

Hint

Compare the problem to the case where the magazines do not overlap.

Solution

Let’s number the magazines from 1 to 10. From the second magazine, let’s cut off that part $($if there is one$)$, which is already covered by the first magazine. From the third magazine, let’s cut off the part that is already covered by the first and second magazines. We continue in this way and in the end, from the tenth magazine, let’s cut off the part that is already covered by all of the magazines from the first to the ninth. After the described procedure, 10 magazines $($some of them may have some part cut off and it could be the case that an entire magazine has been removed$)$ will cover the table in one layer. Consequently, 5 magazines, that are the largest by area, cover at least half of the table. It is clear that the same 5 magazines covered at least half of the table area before we cut them. Therefore, the remaining 5 magazines can be removed from the table so that the condition is met.

Hint

Compare the problem to the case where the magazines do not overlap.

Hint

If all the sectors are centred at a point $O$, then if the total angle is less than $360^\circ$ there will be a ray from $O$ that is not covered by an angle.

Solution

Assume that the total angle is less than $360^\circ$. Consider a point $O$ in the plane. Translate every open sector so that it is centred on $O$. Since the angles total less than $360^\circ$, there will be a ray $m$ we can draw from $O$ that does not lie within any open sector. Now translate all of the sectors back to their original positions. At most only a line segment of ray $m$ can be covered by the sectors, that is it is possible for the open sector to not intersect with the ray $m$ at all, intersect at a point, or intersect twice with a line segment of $m$ between the two intersections. Therefore the given open sectors carve out only a finite number of line segments from the ray $m$. Therefore the ray $m$ is not completely covered by the open sectors, and therefore neither is the entire plane. This is a contradiction.

Hint

If all the sectors are centred at a point $O$, then if the total angle is less than $360^\circ$ there will be a ray from $O$ that is not covered by an angle.

Solution

The figure shows how to cut the given figure into 18 equal-sized triangles. Let us prove that this number is the maximum possible. We take the area of the cell as equal to one unit. The figure shown is a nonconvex hexagon ABCDEF of area 63 with an angle of 270 at the vertex D $($see figures$)$. If we divide the figure into triangles, then it is obvious that the point D must belong to at least two triangles, one of which lies on the line DE, and the other side lies on the DC. Moreover, for at least one of them, it lies on the corresponding segment. For certainty, suppose that this is a triangle DKL, where K ∈ [DC]. Then the base DK of this triangle is no greater than DC = 1, and the height is no greater than BC = 7. Therefore, the area of the triangle DKL is no greater than 7/2. By assumption, we can divide this figure into equal triangles. Since the area of one triangle is no greater than 7/2, then all triangles are no less than 63/[7/2] = 18.

Hint

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Solution

The total number of cells $($25$)$ is not divisible by 2, and each domino covers two cells.

Answer

You cannot.

Hint

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Solution

In any given 2×2 square on the grid at least two squares must be occupied by a ‘corner’, otherwise an extra corner would fit on that square.

An 8×8 grid can be divided into 16 2×2 squares, meaning at least 32 grid squares must be occupied by ‘corners’ which would require no fewer than 11 corners.

One such possible arrangement is shown below.

Hint

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Solution

We will show that the sniper can always hit 25 triangles 5 times. Consider the target divided into 25 triangular sections of side length 2 – that is consisting of 4 triangles – see the diagram. Then, by shooting at the centre triangle in each section until one of the equilateral triangles is hit exactly 5 times, he will get a total of 25 such triangles.

We will prove that the sniper cannot guarantee more than 25 triangles will be hit in this way. Indeed by aiming at a random triangle in a given section a shot can always hit the centre triangle of that section. Therefore the number of triangles hit 5 times cannot be guaranteed to exceed 25.

Answer

25

Hint

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Solution

If three arcs cover some part of the circle, so that they all have a common part, then from these one can always choose two arcs such that the same part of the circle is still covered. Therefore, extra arcs can be discarded so that each part of the circle will be covered no more than twice, and so the sum of the covering arcs will be no more than 720$^{\circ}$.

Hint

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Hint

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Solution

The diagram below shows how 7 squares on the three visible faces of the cube can be marked. On the rear three faces we can mark squares symmetrically to those shown.

We will now prove, that it is not possible to mark more than 14 squares in the required way.

Divide the surface of the cube as shown on the diagram below, again dividing the hidden faces symmetrically. It is easy to see that any two squares marked in the same part of the diagram will share a vertex or be adjacent to another marked piece. Therefore no more than 1 square in each part of the diagram can be marked, so no more than 14 squares can be marked in total.

Answer

14 squares.

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Solution

It is necessary to cut this corner into four small corners as shown in the picture:

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Solution

Yes, it is possible $($see the figure$)$.

Hint

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Hint

1/3 + 2/3 = 1.

Solution

For example, a square with a side of 2/3 m can be cut along the midpoint of its side into two rectangles, the perimeters of which are equal to 2 m.

Answer

No, it is not true.

Hint

1/3 + 2/3 = 1.

Solution

See the figure.

Hint

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