Solutions to problems M1116—M1120; F1128—F1132

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Fixed motionless, and the other can move, remaining parallel to the first plate. The movable plate is attached to the wall using a spring with a stiffness of $k=10$ N/m. The initial distance between the plates is $d=3$ cm, the area of ​​each plate is $S=0,5$ m². Calculate the voltmeter scale. What is the maximum voltage that can be changed with it? Consider separately the case when the high training is very small and quite large.

A. R. Zilberman

F1152. Behind the lens at a distance of $l=4$ cm (large focal length), a flat mirror is located perpendicular to the main optical axis. In front of the lens, also perpendicular to the main optical axis, there is a sheet of checkered paper (Fig. 2). On this sheet, images of its cells are obtained at two positions of the sheet relative to the lens. These positions differ by $l=9$ cm. Determine the focal length of the lens.

E. P. Kuznetsov

Problem solutions
M1116—M1120, F1128—F1132

Answer: a) 74; b) 6) $2S+2$.
A tract of size $m \times n$ cells contains $f=(m+1)(n+1)$ nodes. For a given area $mn=S$ the value

$$f=S+m+n+1=S+m+\frac{S}{m}+1$$

takes the most value $2S+2$ for $m=1$ or $m=S$ (for a rectangle of size $1 \times S$). For other $m$ ($1<m<S$), the value of $f$ is less than $2S+2$. This follows from inequality $m+\frac{S}{m}<S+1$, which is equivalent to the next inequality $(m-S)(m-1)<0$. Note that a more general statement is true: given $S$ (S is an integer or the resulting number), of all polygons with vertices at checkered paper nodes, the polygon that has no internal nodes contains the largest number of nodes. This follows from Pick’s formula for cell $S$ of such polygons*: $S=i+\frac{r}{2}-1$, where $i$ is the number of nodes lying inside the polygon, and $r$ is the number of nodes on its boundary: the total number of nodes is $i+r \leq 2i+r=2S+2$. (Inequality turns into equality only at $i=0$.)

V. L. Gugenacher

M1117. An arbitrary triangle is given. Prove that: a) it is possible to construct three circles with centers at its vertices, by

Points $K, L, M$ are the points of tangency of the sides of a given triangle $ABC$ with the inscribed circle (see figure; if $OK=OL=OM$ are the radii of the inscribed circle

*) See the article “Around the Pick Formula”, «Kvant», 1974, No. 12.

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druga touching each other in pairs (at points K, L, M); b) if through the middle of each arc KL, LM, MK lying inside the triangle, draw a tangent to it, then four triangles are formed, the area of ​​one of which (the central one) is equal to the sum of the areas of the other three.

The caital line drawn through the midpoint $C'$ for KL (with center C) cuts off from $\triangle ABC$ an equilateral triangle equal in area to the quadrilateral $CKOL$: the height $CC'$ cuts this equilateral triangle into two right triangles on the left, equal to $\Delta KCO = \Delta LCO$. Here it should also be noted that this tangent (at figure is red) always intersects sides CA and CB (angle COB is obtuse, since the half-sum of two angles B and C of triangle ABC is less than $90^\circ$, therefore the hypotenuse CO of triangle COK is less than CB; similar is $\angle COA > 90^\circ$, and therefore $CO < CA$). A similar rearrangement for the other two red tangents turns the equilateral triangles with altitudes $AA'$ and $BB'$ into quadrilaterals $ALOM$ and $BMOK$. Thus, the sum of the areas of three equilateral triangles with red bases is equal to the area of ​​triangle $ABC$. This implies the last statement of the problem: the area not covered by triangles is equal to the area covered by them twice.

A. A. Gorbachev

M1118. a) Prove that the equation $(x+y+z)^2=x^2+y^2+z^2$ (*) has infinite multisolutions in four numbers $(x, y, z)$ that do not have a common divisor greater than 1.
b) How many solutions are there for which $z=1988$?

a) Let us rewrite equation (*) in the form
$$xy+yz+zx=0.$$
Note that if one of the unknowns is 0, then some others are also 0; In what follows we will not consider such trivial solutions. Let $d$ be the most common divisor of $x$ and $y$, $x=da$, $y=ab$. then $z=-xy/(x+y)=-dab/(a+b)$, where $a, b$ and $a+b$ are pairwise coprime, so $d$ is divisible by $a+b:d=k(a+b)$. From here we obtain a general formula for nontrivial solutions:
$$x=ka(a+b), y=kb(a+b), z=-kab,$$
where $a, b, k$ are integers, and $a$ and $b$ are coprime. For $k=1$ we obtain all possible different solutions from numbers that do not have a common divisor greater than 1.
b) Answer: 16. The question boils down to how many ways can one represent 1988 as the product $-ab$, where $a$ and $b$ are coprime integers. Since 1988 = $4 \cdot 7 \cdot 71$, in the role of $a$ we can take the product of some numbers - 1, 4, 7, 71 (then $b$ will be the product of the rest; the product of the “empty set” can naturally be considered 1.)
S. G. Mamikonyan, N. V. Vasiliev

M1119. Let us call a $k$-star a figure on a plane consisting of $k$ rays with a common beginning of partition of the plane into $k$ equal

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ny angles (360°/k).
For what k>2 is the following statement true: for any k points of a generic plane (no three of which lie on the same line) there exists a k-star, each of the k corners of which contains exactly one of these k points?

Fig. 1.

Fig. 2.

Zadatnik, Kanma

For k=4, consider two cases. If the points lie at the vertices of a convex quadrilateral, then the center of the 4-star can be taken as the point of intersection of its diagonals, and the rays can be directed along the bisectors of the angles between them (Fig. 2). If one of the points D lies inside the triangle ABC with vertices in the other three and ∠ADB>90°, then as O we can choose the point of intersection of the segment DC and a certain semicircle, the diameter of which A₁B₁ lies entirely on the segment AB, and the rays can be directed along straight lines OA₁ and OB₁ (Fig. 3).

To prove that for k=5 (and k>5) the statement is false, it is enough to give an example; 5 points lying on one arc of value α<72° are suitable as such. In fact, if the arc α intersects four points of the rays of a 5-star with center O, then α>72°. (since the half-sum of arcs AB′ and A′B on figure 4 is equal to ∠AOB′=36°.

M. Khovanov

M1120. a) The sequence a₀, a₁, a₂, … is given by the Fibonacci correspondence a₀=0, a₁=1, a₂=1, a₃=2, …, given the conditions: aₙ+1=aₙ+aₙ-1, n=1,2,…

b) Prove a similar statement for the Fibonacci sequence a₀=0, a₁=1, a₂=1, a₃=2, …, given the conditions: aₙ+1=aₙ+aₙ-1, n=1,2,…

Let us prove the general statement: if the sequence a₀=0, a₁, a₂, … integers is such that

GCD(aₘ, aₖ)=GCD(aₘ-k, aₖ) (1)

for any m>k≥1, then

GCD(aₘ, aₖ)=aₖ, (2)

where d=GCD(m, k).

Indeed, for any pair of initial values ​​m, k, repeating the operation

(m, k)→(m-k, m)

(from the larger number of the pair, the smaller one is subtracted) leads to pale (d, 0), where d=GCD(m, k); This is a version of the Euclid algorithm for finding the most general divisor of the numbers m, n, presented in the textbook with information for grade 9 (usually the Euclid algorithm is called its “reduced” form, consisting of a sequence of divisions with a remainder.

Thus, from equalities (1) it follows that

GCD(aₘ, aₖ)=GCD(aₘ, a₀)=GCD(aₘ, 0)=aₖ

• this is (2).

a) Let P(P(P(…(P(x))…))=Pₙ(x); this is a polynomial with integer coefficients, and aₘ=Pₘ(a₀) and aₘ=Pₘ-k(aₖ) for m>k. Note that Pₙ(x)=aₙ+xQₙ(x), where Qₙ(x) is also a polynomial with by integer coefficients. Let’s check (1): for m>k≥1

GCD(aₘ, aₖ)=GCD(Pₘ-k(aₖ), aₖ)=GCD(aₘ-k+

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F1128. A uniform thin hoop of mass M and radius R rests on a horizontal surface. The horizontal diameter of the hoop is a light, smooth tube in which a ball of mass m is placed, attached to the hoop by two springs of stiffness k (Fig. 1). Keeping the hoop motionless, the ball was deflected to the left by amounts x, after which the system was left to its own devices. I will clear the acceleration of the center of the hoop at the initial moment of time. There is no slipping of the hoop.

figure 2 shows the forces acting on the ball and the hoop.

Let the acceleration of the center of the hoop be equal to a. Then the equation expressing Newton’s second law for the construction motion of the hoop (in projections to the horizontal direction) will look like:

$$Ma = 2kx - F_{тр}$$

Here 2kx is the total strength force of both springs, $F_{тр}$ is the friction force from the horizontal surface.

If the acceleration of the center of the hoop is equal to a, then the vertical acceleration of the ball will be equal to $ax/R$, and the equation of motion of the ball (in projections to the vertical direction) is written as:

$$ma \frac{x}{R} = mg - N.$$

Here N is the reaction force on the side of the smooth tube. Finally, let’s write down the equation that describes the rotational motion of the hoop around the center:

$$MRa = F_{тр}R + Nx.$$

Solving the system of three equations, we get

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F1129. A coil shaped like a narrow pole (Fig. 1) is put on a cylindrical permanent magnet in the region of its pole. If the path of the coil is such that it oscillates harmoniously along the axis OO’ with an amplitude of $A=1$ mm (which is much smaller than the dimensions of the magnet and coil) and a frequency of $f=1000$ Hz, then an induced emf with an amplitude of $\mathcal{E}_0=5$ V is induced in it. What force will act on a stationary coil if a current of $I=200$ mA is passed through it?

When the coil moves along the axis of the magnet, the flux of magnetic induction through the coil changes. As a result, an induced emf $\mathcal{E}=N\Delta\Phi/\Delta t$ appears in it, where $N$ is the number of turns of the coil, $\Phi$ is the flux of magnetic induction through one turn.

To calculate $\Phi$ we proceed as follows. Let’s stretch some surface onto a coil and count the number of magnetic field induction lines $\bar{B}$ intersecting this surface. In this case, it is convenient to use a surface in the form of a cylinder with a lid (Fig. 2). In this case, we will keep in mind that when mixing the coil, the cover remains in place, but the distance between it and the coil changes. Let the coil move down by $\Delta x$. Then a “belt” with an area of ​​$2\pi R\Delta x$ is added to the side surface of the cylinder, where $R$ is the radius of the coil. Corresponding to $\Phi$ is added $\Delta\Phi$ - magnetic flux through this “belt”:

$$\Delta\Phi=2\pi R\Delta x\cdot B_r,$$

where $B_r$ is the radial (i.e., perpendicular to the side surface of the cylinder) projection of the vector $\bar{B}$. Thus,

$$\mathcal{E}=N\cdot 2\pi R\frac{\Delta x}{\Delta t}B_r=N\cdot 2\pi RvB_r,$$

where $v$ is the coil speed. With harmonic oscillations, the velocity amplitude

$$v_0=\omega x_0=2\pi fA,$$

And

$$\mathcal{E}_0=N\cdot 2\pi R\cdot 2\pi fAB_r.$$

On the other hand, the required force is equal to

$$F=\sum\Delta F=\sum N\cdot I\Delta l\cdot B_r=N\cdot 2\pi RIB_r.$$

From the last two equalities we finally obtain

$$F=\frac{\mathcal{E}_0I}{2\pi fA}\approx 0,16\text{ H.}$$

M. M. Tsyplin

F1130. The oscillatory circuit consists of a vacuum capacitor with a capacity $C$, the distance between the plates of which is $d$, and an inductor. The natural part of the circuit oscillations is equal to $\omega_0$. What will be your own

Let us write down the energy of the system at an arbitrary moment in time:

$$W=\frac{LI^2}{2}+\frac{CU^2}{2}+\frac{mv^2}{2}-\frac{qUx}{d},$$

where $L$ is the inductance of the coil, $I$ is the current in the coil, $U$ is the voltage on the capacitor, $v$ is the speed of the particle, $x$ is its coordinate. The equation of particle motion in the electric field of a capacitor has the form:

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Frequency if a free precise frequency of mass $m$ and charge $q$ is placed between the plates of a capacitor? There is also no strength. The marginal effects and strength of the “electrostatic device” are claimed.

$$ma = \frac{qU}{d}.$$

If $U$ changes according to the harmonic law $U = U_0 \cos \omega t$, then
$$v = \left( qU_0 / (md\omega) \right) \sin \omega t,$$
$$x = -\left( qU_0 / (md\omega^2) \right) \cos \omega t,$$
$$I = -CU_0 \omega \sin \omega t.$$

According to the law of conservation of energy
$$W = U_0^2 \left( \frac{L}{2} (C\omega)^2 + \frac{m}{2} \left( \frac{q}{m\omega d} \right)^2 \right) \sin^2 \omega t +$$
$$+ U_0^2 \left( \frac{C}{2} + \frac{q^2}{m(\omega d)^2} \right) \cos^2 \omega t = \text{const.}$$

This is only possible if the coefficients of $\cos^2 \omega t$ and $\sin^2 \omega t$ are equal. From here we get
$$\omega^2 = \omega_0^2 / 2 + \left( \left( \omega_0^2 / 2 \right) \right)^2 + \left( q\omega_0 / d \right)^2 / (mC)^{1/2}.$$

Let us evaluate under what conditions it is possible to apply with one “electrostatic choice”* (as was done above). This force, equal in order of magnitude to $q^2 x / (\varepsilon_0 d^3)$, should be much less than the “ordinary” electrostatic force $qU/d$, whence
$$\frac{q^2}{\varepsilon_0 d^3 m\omega^2} \ll 1.$$

Since the area of ​​the capacitor plates is $S \gg d^2$ and $C = \varepsilon_0 S / d$, formula (*) is valid if the introduction of a particle into the capacitor does not greatly change the frequency of the oscillations in the circuit.

A. V. Andrianov, D. A. Kuptsov

F1131. It is difficult to determine the wind speed in the smerya using conventional meteorological instruments (since the smerya is not large in size and moves) and is not zonal. It is proposed to measure it from a distance using a portable radio (since there is a lot of dust and small objects inside the smert, it reflects radiovolts). The radar emits radiovolts at frequency $f_0 = 10^{10} \text{ Гц}$. The spectrum of reflection from the mertz is shown in figure ($\Delta f = f - f_0$). Find the maximum wind speed in Smereta.

The frequency of the reflected signal differs from the frequency of the emitter due to the Doppler effect. The maximum frequency shift will occur if the speed of the object reflecting radiovolts is directed exactly towards the radar (or away from it).

We will assume that the radar sends short radio pulses with a repetition frequency $f_0$, and the frequency of the pulses it receives is $f_1$. The time between arrivals at the radar antenna of the $n$th and $(n+1)$th pulses is equal to
$$\frac{1}{f_1} = \frac{1}{f_0} + T_{n+1} - T_n,$$
where $T_n$ is the propagation time of the $n$ pulse from the radar to the reflecting object and back, $T_{n+1}$ is the propagation time of the $(n+1)$ pulse.

Let us denote by $L_n$ the distance from the radar to the reflecting object at the moment of emission of the $n$ pulse, and by $L_{n+1}$ the same distance at the moment of emission of the $(n+1)$ pulse. It’s obvious that
$$T_n = \frac{2L_n}{v + c}, \quad T_{n+1} = \frac{2L_{n+1}}{v + c},$$

*) You can read about how this force can be determined for some specific cases in the note “Method of Electrostatic Isolations” (Kvant, 1987, No. 3, p. 39). (Editor’s note)

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where $v$ is the desired wind speed in the tornado, $c$ is the speed of propagation of radio waves (speed of light). The difference $L_n - L_{n+1}$ is equal to the displacement of the reflecting object in a time equal to $1/f_0$:

$$L_n - L_{n+1} = \frac{v}{f_0}.$$

From here we get

$$f_1 = f_0 \frac{(c + v)}{(c - v)} \approx f_0 \left(1 + \frac{2v}{c}\right),$$

And

$$v \approx \frac{c}{2} \frac{\Delta f}{f_0} \approx 100 \text{ m/c.}$$

D. A. Kuptsov

F1132. It has recently been established that the extinction of animal species on Earth is particularly intense during periodic periods of time lasting $T = 6,2$ million years. This pattern is explained by the hypothesis proposed by the existence of the star Nemesis, which is a satellite of the Sun. This faint and therefore invisible star moves in an orbit that is located within the so-called “Oort belt”, containing a supply of comets (Fig. 1). By disturbing the movement of comets, Nemesis causes a “comet house” on Earth, the duration of which practically coincides with the time of Nemesis’s stay inside the Oort belt.

Determine the greater half a of the orbit of Nemesis and the period of its revolution around the Sun, if it is assumed that the perihelion of the orbit is at a distance of 0.15a from the Sun. The area of ​​the ellipse is $S = \pi ab$, where $a$ and $b$ are the ellipse.

Over a time of $T = 6,2$ million years, the radius vector of Nemesis occupies the area protected by figure 2. This is the half of the area of ​​the ellipse minus the area of ​​the triangle with base $2b$ and height 0.85 $a$. The area of ​​this figure, as it is easy to calculate, is $(\pi ab/2 - 0,85ab)/\pi ab = 0,23$ the area of ​​the entire ellipse. Ta-

Fig. 1.

Fig. 2.

What part of the period is the time of Nemesis inside the Oort belt (this follows from Kepler’s second law). The total period of Nemesis’ orbit around the Sun is therefore $T_x = 27$ million years.

Comparing, with the help of Kepler’s third law, the motion of the Earth and Nemesis, we obtain a greater semi-presence of the invisible satellite of the Sun:

$$a = \left( T_x/T_0 \right)^{2/3} R_0 = \left( 27 \cdot 10^6 \right)^{2/3} a.e. = 9 \cdot 10^4 a.e. \approx 13,5 \cdot 10^{12} \text{km.}$$

Here $T_0$ is the Earth year, and $R_0$ is the radius of the Earth’s orbit, equal to one astronomical unit (1 AU), or 150 million km.

V. E. Belonuchkin