Attention eighth graders! Correspondence mathematics school announces admission of students

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ATTENTION
EIGHTH-GRADER STUDENTS!
CORRESPONDENCE MATH SCHOOL
ANNOUNCES ADMISSION OF STUDENTS

Only students of the highest grades are admitted to ZMSH. Schoolchildren living in Moscow, Leningrad and their suburbs are not accepted into ZMSH. Classes at the school will begin on September 1.

ZMSH has two courses. School education is free.

This issue «Kvant» offers problems that serve as an introductory test for correspondence mathematics schools at Moscow State University and Leningrad State University. Those who want to enter the ZMSH must send the solution to these problems no later than March 10, 1971. After a performance review (around July 1971), you will be notified whether you have been accepted into the ZMSH.

Although some of the introductory problems differ in appearance from regular school ones, no additional knowledge of mathematics is required to solve them.

In order to be accepted at school, it is not necessary to solve every problem without exception. When evaluating work, it will take into account not only the number of problems solved, but also the quality of the solution. The solution to each problem must be justified. The answer without any further explanation may not be read. If there are several possible divisions in a problem, then all of them must be indicated.

Work must be completed in Russian in a student’s notepad in Klatek. Introductory works will not be sent out.

We ask you not to fold the terradi when sending it. In the envelope along with the terradi you need to put a piece of paper measuring 14x6 cm with your postal address written on it (we will paste it on the envelope when we send the answer).

Paste a sheet of checkered paper onto the cover of the terradi, draw it and fill it out according to the following sample (your work will not be checked!):

Region:
Last name, first name, year of birth:
School (full name):
Class:
Last name, first name, patronymic of mathematics teacher:
Parent position:
Omets - driver,
mother is a housewife
e. Tomma, st. Lenin,
3, apt. 8

CHECK RESULTS
(to be completed by the inspector)

Schoolchildren living in the Arkhangelsk, Kaliningrad, Leningrad, Murmansk, Novgorod, Pskov regions, Komi and Karalian USSR, Belarus, Latvian, Lithuanian and Estonian USSR must send work to the address: Leningrad, P-228, st. Savushkina, 61, Specialized School at Leningrad State University. Correspondence mathematics school. To the competition.

Schoolchildren living in other regions and republics of the USSR must send their work to the address: Moscow, V-234, Moscow State University, mechanics, ZMSh. To the competition.

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1. A semicircle of diameter 2 contains two circles that touch each other. Each one touches a diameter and a different circle. Find the diameter of one of them if the diameter of the other is 1.

2. Find the most common divisor of the numbers 987654321 and 123456789.

3. The bases of the trapezoid $ABCD$ are equal: $AB=a$, $CD=b$. O is the point of intersection of the diagonals. Determine the ratio of the area of ​​the triangle $OAB$ to the area of ​​the trapezoid.

4. It is known that $x+\frac{1}{x}$ is an integer.

Prove that then the number $x^8+\frac{1}{x^8}$ is also an integer.

5. The following 4 statements have been made about the triangle $ABC$:

1. Triangle $ABC$ is rectangular;
2. $\Rightarrow A=30^\circ$;
3. $AB=2BC$;
4. $AC=2BC$.

We know that two of these statements are true and the other two are false. Find the perimeter of triangle $ABC$ if $BC=1$.

6. It is known that $a+\theta+c<0$ and that the equation $ax^2+\theta x+c=0$ has no real roots. Determine what sign the number $c$ has?

7. The following is known about the points $A, B$ and $C$:

for any point $M$ on the plane, the segment $AM$ is less than at least one of the segments $BM$ and $CM$. Prove that the point $A$ lies on the segment $BC$.

8. What is the divisor of the number 1971$^{1971}$? (That is, how many different positive integers are there that this number is divisible by?)

9. Is it possible to wrap a cube with edge 1 in a square piece of paper with edge 3?

10. Is it possible to write down nine numbers 1, 2, 3 …, 9 in a circle in such an order that the sum of any two adjacent numbers is not divisible by 3, 5, or 7?

11. Residents of the city of Pravdin always tell the truth, and residents of the city of Liars always lie. In one of these cities, a football match is arranged between the teams of these cities. After the match, next to the stadium where he was lost, the following conversation took place:

$A$ (addressing $B$ and $B$): I wasn’t at the match. Tell me who won?

$B$: $\Gamma$ told me that their team lost.

$B$: Our team won.

$\Gamma$ (getting into a bus, going to another city and turning to $A$): Let’s serve together, $A$, because we are from the same city.

$A$: You’re wrong, $\Gamma$. I live here. This is $B$ from your city.

Determine who from each city, where the match took place and which team won.

12. Trucks travel along the highway in one direction at a constant speed at regular intervals without stopping. However, the man walked 4 km along the highway, and during this time he was overtaken by 6 trucks. Another time he walked 7 km, and during this time he was overtaken by 8 auto carriers. The third time he walked 17 km. How many cars did he overtake? (All three times the person walked at the same speed).

13. On the base $AB$ of the equilateral triangle $ABC$, point $E$ is taken, and circles are inscribed in the triangle $ACE$ and $ECB$, which touch the segment $CE$ at points $K$ and $H$. Find the length of the segment $KH$ if $AE=a, EB=b$.

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To the article “Rotational motion of bodies.”

1. In kilograms multiplied by a square meter ($\kappa \cdot m^2$).
2. The rolling times of the cylinders are the same, since their moments of inertia are equal. (Prove it!)
3. The work done by the friction force $A = M \cdot \varphi$ (where $\varphi = 2\pi \cdot 5 = 10\pi$ radan) must be equal to the energy of the wheel before braking

$$M\varphi = \frac{I\omega_0^2}{2}.$$

Hence $M = \frac{I\omega_0^2}{2\varphi} \approx 0,05 \text{ nm}.$

4. Let us first find what angular velocity the Earth will have after 100 lem. Since the time required for one revolution is related to the angular velocity by the formula $\tau = \frac{2\pi}{\omega}$, which is $\Delta\tau = \frac{2\pi}{\omega_2} - \frac{2\pi}{\omega_1}$, where $\omega_1$ is the angular velocity of the Earth now ($\omega_1 = \frac{2\pi}{24\cdot 3600} \text{ рад/c} = 7,3 \cdot 10^{-5} \text{ рад/c}$), $\omega_2$ is its angular velocity after 100 years and $\Delta\tau = 1$ s.
   Let us now write down the equation of motion of the Earth. $M_{\text{Tp}} = I \alpha$. I is the moment of inertia of the Earth. It is equal

$$\frac{2}{5} \cdot 5,9 \cdot 10^{24} \kappa \cdot 4 \cdot 10^{13} m^2 = 9,4 \cdot 10^{37} \kappa \cdot m^2.$$

Substituting $\alpha$ into the equation of the Earth’s motion together with its expression through $\omega_1$ and $\omega_2$, we obtain $M_{\text{Tp}} = I \frac{\omega_2 - \omega_1}{t} = 2,6 \cdot 10^{19} \text{ nm}.$

5. The work that needs to be done in the time $t = 10$ hour = 3.6·10⁴ s is equal to the change in the kinetic energy of the stabilizer

$$\Delta K = \frac{I\omega^2}{2}.$$

Pozny

$$N = \frac{I\omega^2}{2t} = \frac{Mr^2\omega^2}{2t} = 4,5 \cdot 10^4 \text{ em}.$$

6. The friction force $F_{\text{Tp}} = kmg$ acting on the ball replaces the translational motion of its center of mass and increases the speed of its rotation. Pozny

$$F_{\text{Tp}} = ma \text{ и } F_{\text{Tp}} \cdot r = I\alpha,$$

where $a$ is the acceleration of the translational motion of the ball, $\alpha$ is the acceleration of the rotating ball, $n$ is the module (n-1) - 1.

Movements and $I = \frac{2}{5}mr^2$ – moment of inertia of the ball.

By the moment $t$ the ball will have a translational speed $v = v_0 - at = v_0 - \frac{F_{\text{Tp}}}{m}t = v_0 - kgt$ and an angular velocity of rotation

$$\omega = \alpha t = \frac{F_{\text{Tp}}r}{I} t = \frac{kmgr}{2\frac{5}{mr^2}} t = \frac{5}{2} \frac{kgt}{r}.$$

If at this moment the slippage stops, then $\omega = \frac{v}{r}$, or

$$\frac{5}{2} \frac{kgt}{r} = \frac{v_0 - kgt}{r}.$$

Hence $t = \frac{2}{7} \frac{v_0}{kg}.$

At moment $t$ the speed of translational motion of the ball is equal to $v_0 - kg \frac{2v_0}{7kg} = \frac{5}{7} v_0.$

Let the distance traveled by the ball in time $t$ be equal to $v_0 t - \frac{at^2}{2} = \frac{12}{49} \frac{v_0^2}{kg}.$

7. Let’s write down the law of conservation of angular momentum

$$I\omega = mR^2\omega_1,$$

where $\omega$ is the angular speed of the ship, and $\omega_1$ is the angular speed of the astronaut and $I = \frac{2}{3} MR^2$ is the moment of inertia of the ship.

This is equal to the familiarized moment $(MR^2 + 2mR^2)\omega = [MR^2 + 2m(R + l)^2]\omega.$

Since $\omega_1 = \frac{1}{n} \omega$, then after simple transformations of this equation we find that

$$l = R \left( \sqrt{n + \frac{M}{2m}} (n - 1) - 1 \right).$$

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By the way, “Combinatorics”

1. The number $p_k$ can enter into the divisor $q$ with the exponent $0, 1, \dots, \alpha_k$ in only $\alpha_k+1$ ways. According to the product rules, the number of divisors is equal to
   $(\alpha_1+1)(\alpha_2+1)\dots(\alpha_k+1)$.
2. In permutations where 4 letters $e$ are in a row, they can be combined and counted as one letter. The number of such permutations is 5!, leaving $P(4, 1, 1, 1, 1) - 5! = 1560$.
3. Let’s write out the vowels in this order. Then for the letter $k$ we have 5 places; after it is entered, for the letter $m - 6$ places, for $n - 7$ places. There are $5\cdot 6\cdot 7 = 210$ ways in total.
4. Similar to problem $2P(3, 1, 1, 1) - 4! = 96$.
5. Each digit in each digit appears $4^2=16$ times. Therefore, the sum of the digits of the first digit is 16 $(1+2+3+4)=160$, the second - 1600 and the third - 16,000. In total we get 17,760.
6. Each digit appears $P_4/4=6$ times in each digit. As in problem 5, we get 66,660.
7. Here the total number of permutations is 12, with the numbers 1 and 5 appearing in each digit 3 times, and the number 2-6 times. The sum of the digits in each digit is $3\cdot 1+3\cdot 5+6\cdot 2=30$. The total amount is 33,300.
8. a) This is the number of all subsets of a set consisting of $n$ elements. Each element is either included in the subset or not (2 ways). Using the product rules, we obtain the required equality.

6. Here we write the number of ways to select two subsets from a set consisting of $n$ elements: one from $k$ elements, the other from $m - k$ elements. You can first select a $k$-subset, then from the remaining elements - a $(m - k)$-subset, or first a $m$-subset, and then a $k$-subset within it.

By the way, “Movement of planets”
Optical problem
If the source is at focus $F_1$ (Fig. 1), then for any point $A$ on the ellipson where the ray coming from $F_1$ hits, the distance $F_2F_1'$ between focus $F_2$ and the image $F_1'$ of the source is $2a$ ($F_1' - \text{отражение } F_1$ relative to the tangent $t$). The fact that $F_1'AF_2$ is straight was proven in the statistic. Therefore, the image of the source $F_1$ will be a sphere of radius $2a$ centered at the focus $F_2$.

Geometric problem
Let the large circle at figure 1 be a “guide”, and the small one pass through the focus $F_1$ and touch the larger one $A_1$—

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3. Draw a circle through points A, P, M, Q and rewrite the condition $\angle APM = \angle AQM$ for each (the fact that they are equal to 90° is not even used, as long as points A, P, M, Q lie on the same circle).

4. Draw a circle through points A, M, C. The triangles are similar at two angles; based on arcs AM and AN.

5. Draw a circle through the points C, M, N, H. Use the fact that CH is a diameter and prove that $\angle BAC = \angle CNM$.

6. a) $90^\circ; 60^\circ; 30^\circ$.

6. $90^\circ; 22^\circ; 5; 67^\circ, 5$.

7. The triangle is determined up to similarity. Take an arbitrary point on the bisector and consider it the intersection point of the bisector with the circumcircle. A perpendicular dropped from this point to the side of the triangle (it is parallel to the height) hits the base of the medina and passes through the center of the circle. The third point on the circle will be reflected in this perpendicular. Be sure to conduct a feasibility analysis and consider all possible cases.

8. The key to solving this problem is given by auxiliary circles circumscribed about equilateral triangles and intersecting at one point, through which all three straight lines $AA_1, BB_1$ and $CC_1$ pass.

9. Let a circle be constructed on side AB of triangle ABC as a diameter. Prove that if angle C is obtuse, then its vertex lies inside the circle.

Now think about how to construct a circle to solve problem 9.

10. The circle drawn through A, B, C must touch the side of the angle at point C. Prove that $\angle SCA = \angle SBC$. Triangles SAC and SBC are similar. To determine SC, you can draw any circle through points A and B and determine the length of the tangent segment drawn from point S, from S to the point of tangency. This will be SC.

11. When solving this problem, you can apply a formula that expresses the side of a triangle through the diameter of the circumscribed circle and the opposite angle.

To the article “Entrance exams in mathematics at Moscow State Pedagogical Institute named after. V.I. Lenin"

Option I

$$V = \frac{p^9 \cos^2 \alpha tg^2 \frac{\alpha}{2} tg \beta}{12 \cos^4 \frac{\alpha}{2}}.$$

2. Writing an equation

$$v_1^2 - \frac{2\pi R}{a} v_1 - \frac{(2\pi R)^2}{at} = 0;$$

his solution: $v_1 = \frac{\pi R}{a} \left( \sqrt{1 + \frac{4a}{t}} + 1 \right) -$

speed of the first car, $v_2 = \frac{\pi R}{a} \times \left( \sqrt{1 + \frac{4a}{t}} - 1 \right) -$ speed of the second car. The second root does not satisfy the conditions of the problem.

3. If $a = 0$, then $x = 0$; if $a > 1$, then $x = \frac{-1 + \sqrt{4a - 3}}{2}$; if $0 < a < 1$, then there are no roots.

4. $x = \frac{\pi}{4} + \frac{\pi k}{2}$; $x = 2\pi k$;

$$x = \frac{\pi}{2} + 2\pi k$$

$(k = 0, \pm 1, \pm 2, \dots)$

Option 2

$$V = \frac{h^3}{\sqrt{2} \sin 2\alpha \cos \frac{\alpha}{2} \cos \left( \frac{\pi}{4} - \frac{\alpha}{2} \right)}$$

2. Writing an equation

$$\frac{a(m - x) + bx}{m} = \frac{b(n - x) + ax}{n},$$

where $a, b$ is the amount of copper in 1 kg corresponding to the 1st and 2nd pieces.

After eliminating $a$ and $b$ from the equation we get $(m + n)(b - a)x = (b - a)mn, b > a,$

$$x = \frac{mn}{m + n}.$$

3. $1 < x \leq \sqrt{2}, \frac{1}{\sqrt{2}} \leq x < \frac{1}{5\sqrt{4}}.$

4. $x = \frac{\pi}{8} + \frac{\pi k}{2}$ $(k = 0, \pm 1, \pm 2, \dots)$.

To the article “Research exams in physics at NSU”

Option 3

1. Since the voltmeter has a high value, the point through it can be transformed and potentially the points $c$ and $d$ can be determined as if the voltmeter did not exist at all. Therefore, assuming that the potential of the point $a$ is equal to zero, we can write that

$$\varphi_c - \varphi_d = \Delta U = \frac{UR_1}{R_1 + R_2} - \frac{UR_3}{R_3 + R_4} = U \frac{R_1 R_4 - R_2 R_3}{(R_1 + R_2)(R_3 + R_4)}$$

2. Having found the position of the center of gravity of the body, we write down the equality of the moments of the forces of water pressure and the force of gravity acting on the body, taking into account the fact that water does not approach its lower boundary. We select the edge of the lower boundary as the axis