khdavid June 25, 2013 at 01:25

Tasks with beautiful solutions

There is a class of problems that are mostly passed on by word of mouth, one might say they are part of mathematical folklore. Sometimes there are problems with very beautiful solutions. You look at the decision, you seem to understand every step in the reasoning, but you feel as if you were deceived. You understand everything and at the same time understand nothing. You can probably draw an analogy, for example, with this optical illusion:

Here you see either a large cube with a cut piece, or a small cube standing in the corner.

In this post, I collected some of my favorite tasks, the solutions of which, it seems to me, cause this elusive dualism of feelings: "I understand - I don't understand."

Circle and Ruler

Prove that using only one ruler, you cannot find the center of a circle drawn on the plane (it is believed that the ruler has infinite length; it can connect any given points on the plane; there is no scale on the ruler and nothing can be marked on it).

Decision

Consider an inclined cone, the base of which is a certain circle S_1

. Since the cone is inclined, there exists a plane that is not parallel to the base and which cuts out the second circle from the cone; we call it S_2

.
Place a light bulb on top of the cone. This light will cast a shadow on each line lying on the plane of the “upper” circle S_2

onto the plane of the “lower” circle S_1

. Moreover, the shadow of any line will also be a line. Note that this shadow is so tricky that despite the fact that it maps the “upper” circle S_2

to the “lower” one S_1

, the shadow of the center of the “upper” circle does not fall into the center of the “lower” one.

Now let's imagine for a moment that there is such a wonderful algorithm that says how to find the center of any circle using one ruler. This algorithm, read a computer program, should consist of a sequence of actions such as: draw an arbitrary straight line, draw a second arbitrary straight line, connect such and such intersection point with such and such point, then connect this intersection point of this straight line and circle with some other point , and so on ... Note that if we use this miracle algorithm on the “upper” plane to find the center of the circle S_2

, then the “shadow” of this algorithm will execute exactly the same commands on the “lower” plane. And since we assumed that our algorithm (a set of commands) finds the center of any circle, the "shadow" of the algorithm that performs exactly the same commands must find the center of the lower circle. We immediately come to a contradiction, because, as we noted earlier, the shadow of the found center of the “upper” circle does not fall into the center of the “lower” one.

The task of the Moscow metro

In the Moscow metro there is a rule that prohibits carrying items whose sum of heights, widths and depths are greater than 150

cm. Let's agree that we are talking about rectangular boxes. To prove that it is impossible to deceive the system and completely put the box, the sum of measurements of which is more than 150

cm, into the box with the sum of measurements of less than 150

cm. You can try to stack the box as you want, but it can not be wrinkled.

Decision

The decision was told to me by Hovhannes Khudaverdyan .

For our proof, we need the concept of $\ varepsilon$ -blossing over the body. Let us take an arbitrary body in space, by its $\ varepsilon$ inflation we mean the set of points that are on the body or at a distance less than $\ varepsilon$ from it. Say, the $\ varepsilon$ bloating point in space is a ball of radius $\ varepsilon$ , and the $\ varepsilon$ bloating segment is a body that looks like a sausage.

Now take our box (box plus its inside) with dimensions

and

, and volume, respectively V = abc

. We will try to calculate the volume of $V_ \ varepsilon$ its- $\ varepsilon$ inflation. This $\ varepsilon$ bloating includes:

volume box itself ;
growths over the edges of the box. If we denote the total surface area of the box for , then the volume of these growths will be $S \ varepsilon$ .
growths over the edges of the box. Each such growth is a quarter of a cylinder with a radius of the base $\ varepsilon$ . Since in a drawer for four ribs lengths , and then each growths with four identical ribs can be combined into one integral cylinder. The total volume of the resulting three cylinders will be $(a + b + c) \ pi \ varepsilon ^ 2$ ;
growths over the tops of the box. Each such growth is an eighth of a ball of radius $\ varepsilon$ . Therefore, from the growths above all eight vertices of the box, you can collect one whole ball of radius $\ varepsilon$ , that is, volume $\ frac {4} {3} \ pi \ varepsilon ^ 3$ .

We get that the volume of the $\ varepsilon$ -blown-up box will be

$\ frac {4} {3} \ pi \ varepsilon ^ 3 + (a + b + c) \ pi \ varepsilon ^ 2 + S \ varepsilon + abc$

Let now in the box with the dimensions of the sides

and

there is a second box a '

and

. It is clear that no matter what number $\ varepsilon$ we take, the $\ varepsilon$ inflation of the inner box will lie in the $\ varepsilon$ inflation of the outer box, therefore its volume will be less:

$V '_ \ varepsilon <V_ \ varepsilon$

We substitute the expressions for volumes into the inequality, reduce the same terms and divide everything by $\ varepsilon ^ 2$ :

$(a '+ b' + c ') \ pi + \ frac {S'} {\ varepsilon} + \ frac {a'b'c '} {\ varepsilon ^ 2} <(a + b + c) \ pi + \ frac {S} {\ varepsilon} + \ frac {abc} {\ varepsilon ^ 2}$

Note that the last inequality must be fulfilled for anyone $\ varepsilon$ , both small and large. Therefore, we can always go to the limit $\ varepsilon \ to \ infty$ , we get:

$(a '+ b' + c ') \ pi \ leqslant (a + b + c) \ pi$

So we proved that if one box is in the second, then the sum of its dimensions cannot be more.

Brick wall

Imagine that we have so many different rectangles (two-dimensional bricks) such that each brick has at least one side that has an integer length. From such bricks they built an even rectangular wall, without overlays and holes, the bricks are not inclined. Prove that at least one side of the resulting wall has an integer length.

Decision

Before solving the problem, let's recall one remarkable property of a function $\ sin (x)$ : its integral over any interval whose length is a multiple of a number $2 \ pi$ is zero. Indeed,

$\ int \ limits_a ^ {a + 2 \ pi n} \ sin (x) dx = - \ cos (a + 2 \ pi n) + \ cos (a) = - cos (a) + cos (a) = 0$

Moreover, if the integral of a function $\ sin (x)$ from zero to some number

is equal to zero, then we can safely assume that the number is a

multiple $2 \ pi$ .

It is similarly shown that for a function “horizontally compressed” $\ sin (2 \ pi x)$ , the integral over any segment whose length is a multiple of unity (an integer) is equal to zero:

$\ int \ limits_a ^ {a + n} \ sin (2 \ pi x) dx = 0$

Now we consider the function $\ sin (2 \ pi x) \ cdot \ sin (2 \ pi y)$ (we place the origin in the lower left corner of the wall). This function has such a wonderful property that its integral over any brick on the wall is zero:

$\ int \ limits_ {kirpich} \ sin (2 \ pi x) \ sin (2 \ pi y) dxdy = \ int \ limits_ {goriz. \ storona} \ hspace {-7mm} \ sin (2 \ pi x) dx \ cdot \ hspace {-7mm} \ int \ limits_ {vert. \ storona} \ hspace {-7mm} \ sin (2 \ pi y) dy$

Indeed, after all, one of the integrals on the right is taken over a segment of length an integer, and therefore equal to zero.

We see that the integral of our wonderful function $\ sin (2 \ pi x) \ cdot \ sin (2 \ pi y)$ for any of the bricks on the wall is zero, therefore this integral is zero on the entire wall built by these bricks, as it is simply the sum of the integrals over each of the bricks. We get :

$\ int \ limits_ {stena} \ sin (2 \ pi x) \ sin (2 \ pi y) dxdy = \ int \ limits_ {goriz. \ storona} \ hspace {-7mm} \ sin (2 \ pi x) dx \ cdot \ hspace {-7mm} \ int \ limits_ {vert. \ storona} \ hspace {-7mm} \ sin (2 \ pi y) dy$

So either $\ int \ limits_ {goriz. \ storona} \ hspace {-7mm} \ sin (2 \ pi x) dx$ , or $\ int \ limits_ {vert. \ storona} \ hspace {-7mm} \ sin (2 \ pi y) dy$ should be equal to zero. From which it immediately follows that either the horizontal or vertical side of the wall has a whole length.

( upd: as my readers suggest, the problem has at least 14 solutions )

The task of the well

A round well was dug in the field. We have a lot of different endlessly long boards. Each board has its own width. And we completely closed the well with these boards so that there are no gaps (the boards are not necessarily all parallel to each other). Prove that the sum of the widths of the boards will always be no less than the diameter of the well.

Decision

The decision, if I am not mistaken, belongs to Alexander Karabegov .

Let's cover the well with a hemisphere, as shown in the figure, and install a huge spotlight in the well, which shines with parallel vertical rays up. And consider a very, very thin board, the width

that lies on the well.

Note that the farther the distance of the board from the center of the well, the less becomes the length

that the board takes directly above the well, but at the same time, the angle of the shadow from this board in the hemisphere becomes steeper. It turns out that these two processes cancel each other out, and the shadow area

does not depend on the distance of the board from the center of the well. Indeed, the length of the board above the well $l = 2 \ sqrt {R ^ 2-x ^ 2}$ , and the tangent of the angle of the shadow is $\ tan {\ varphi} = \ sqrt {R ^ 2-x ^ 2} ^ {'} = - \ frac {x} {\ sqrt {R ^ 2-x ^ 2}}$ . We get the formula for the

shadow area from the board, which is equal to the length of the shadow multiplied by its width:

$dS = \ frac {\ pi l} {2} \ cdot dx \ big / \ cos {\ varphi} = \ frac {\ pi l} {2} \ cdot dx \ cdot \ sqrt {\ tan ^ 2 {\ varphi } +1} = \ pi Rdx$

We see that, indeed, wherever a very thin board is wide above the well

, the area of

its shadow on the hemisphere will always be equal $\ pi R dx$ , that is, it will depend only on the width of the board

. This property of “independence” is also fulfilled for boards of any width, because they can be imagined as many thin boards bound together. As a result, we get a wonderful result: if the width of the board above the well is equal

, then its shadow area

is equal $\ pi R d$ .

Now let many planks wide d_1, d_2, ..., d_n

completely cover our well. Some of the planks may, of course, not be located above the well with their entire width. Therefore, the shadow area of each of the boards $S_n \ leqslant \ pi Rd_n$ . Different boards can overlap each other, so the area of the common shadow

$S \ leqslant S \ leqslantS_1 + ... + S_n \ leqslant \ pi R (d_1 + ... + d_n)$

But since the boards cover the well without gaps, their common shadow fills the entire hemisphere, which means it has an area $S = 2 \ pi R ^ 2$ . As a result, we get that which was

$2 \ pi R ^ 2 \ leqslant \ pi R (d_1 + ... + d_n)$

$2R \ leqslant d_1 + ... + d_n$

required to prove.

Tags:

Tasks with beautiful solutions

Circle and Ruler

The task of the Moscow metro

Brick wall

The task of the well

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