Written exam in mathematics at the Faculty of Chemistry of Moscow State University in 1970

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Written exam
in mathematics at the Faculty of Chemistry of Moscow State University in 1970

N. N. Kolesnikov, V. M. Tikhomirov

In 1970, at the Faculty of Chemistry of Moscow State University, only a written exam was held in mathematics.

Each version of the written work contained five problems covering all the main sections of the school mathematics course - algebra, geometry and trigonometry.

Having analyzed the four problems of one of the options, we will see that solving them does not require any knowledge beyond the scope of the mathematics program for building a university.

In this regard, I would like to emphasize once again that applicants often have weak skills of sound logical reasoning and that they are given clearly insufficient attention to geometric problems and theorems.

Along with these two fundamental shortcomings, which I would like to draw the teachers’ attention to, great harm is caused to applicants by negligence and lack of composure when completing and preparing written work, and the inability to complete the solution of problems and examples. A problem in which an idea for a solution is outlined, but not brought to an answer with the necessary intermediate reasoning, cannot be considered completely solved. Carelessness also leads to various errors and clerical errors that distort the course of the decision and the results; sometimes the conditions of the task are misunderstood, but the task with changed conditions is not considered by the examination committee, even if it turns out to be more difficult.

Option

1. From point A to point B I left automatically and at the same time I left from point B to point A by magnitude. After the meeting they continued on their way. The car, having reached point B, immediately turned back and caught up with the left station two hours after the moment of the first meeting. How long after the first meeting did Velicidict travel to point A, if it is known that by the time of the second meeting he had traveled $2/5$ the entire distance from B to A? (Speeds are automatically and magnitude dict plotted.)

2. Solve the equation
   $$4\sqrt{3x^2-2x}+1+2=9.2\sqrt{3x^2-2x}$$

3. Find all values ​​$x$ lying in the interval $-1<x<4$ and satisfying

$$\log_{0,75}\sin x \geq \log_{9/10}0,75.$$

4. The chord $AB$ subtends another chord equal to $120^\circ$. Point $C$ lies on this arc, and point $D$ lies on chord $AB$. In this case, $AD=2$, $BD=1$, $CD=\sqrt{2}$. Find the area of ​​triangle $ABC$.

5. Given a straight triangular line $ABCA_1B_1C_1$ ($AA_1, BB_1, CC_1$ are lateral edges), which has $AC=6$ and $AA_1=8$. Through the vertex $A$ there is a plane intersecting the edges $BB_1$ and $CC_1$ corresponding at points $M$ and $N$. Find in what ratio this plane divides the volume, if it is known that $BM=MB_1$ and $AN$ are bissertis of the angle $CAC_1$.

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1. In this problem we indicate only the main points of the solution.

Let $S$ be the distance between points $A$ and $B$, $v_a$ the speed of the car, $v_b$ the speed of the bicycle, $t$ the time (in hours) from the start of the movement to the first meeting. For the four entered quantities ($S, v_a, v_b, t$), three equations can be created:

$$\left( v_a + v_b \right) t = S,$$
$$\frac{2}{5} S = v_b (t + 2),$$
$$\frac{7}{5} S = v_a (t + 2).$$

It is impossible to find the values ​​of all unknowns, but this is not required; you need to find the values ​​$\frac{S}{v_b} - t$, which are a combination of unknowns. For the third equation to the second, we get that $v_a = \frac{7}{2} v_b$.

Substituting this correspondence into the first equation and eliminating $\frac{S}{v_b}$ from it and the second equation, we find $t = \frac{5}{2}$ hours. From the second equation it turns out that $\frac{S}{v_b} = \frac{45}{4}$ hours, and the exclusive time $\frac{S}{v_b} - t$ is equal to $\frac{35}{4}$ hours, that is, 8 hours 45 minutes.

Note that this problem (like all the first problems in the versions given at the Faculty of Chemistry) is essentially arithmetic and allows solutions based on the simplest reasoning.

At the same time (before the second meeting), the value is simple $2/5$, and the automatic value $7/5$ is always a path. This means that the ratio of their speeds is 2:7. Then, by the time of the first meeting, the value is simple $2/9$, and the automatic value $7/9$ is always on the way. But, after spending another two hours, the simple value $2/5$ is always the way. Therefore, in an hour he covers $\left( \frac{2}{5} - \frac{2}{9} \right): 2 = \frac{4}{45}$ hours of the entire journey. $7/9$ paths that still had to be covered prime value prime value prime value $x > 0$. Only a quarter of all solvers were able to correctly write down the solution to the inequality $\sin x \leq \frac{\sqrt{3}}{2}$ with the addition

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5. Let $CAC_1 = 2\alpha$ (Fig. 3). According to conditions $tg2\alpha = \frac{4}{3}$. Since $AN$ is the bisector of angle $CAC_1$, using known trigonometric formulas it is easy to find that $(\Rightarrow NAC = \alpha)$

$$\sin \alpha = \frac{1}{V_5}, \cos \alpha = \frac{2}{V_5}.$$

Let us denote $\Rightarrow ACB$ in triangle $CAB$ by $\beta$. Now it is easy to write the volume of this prism as follows:

$$V = \frac{1}{2}AC \cdot CB \sin \beta \cdot AA_1 = 24CB \sin \beta.$$

The plane passing through the points $A, M$ and $N$ cuts off the paramid $ABCNM$ from the prism, whose lateral face $ACB$ is perpendicular to the base, since the prism $ABCA_1B_1C_1$ is straight (by condition). The base $BCNM$ is a trapezoid, the plane sides $CN$ and $BM$ are parallel. From equalities

$$BM = \frac{BB_1}{2} = 4, CN = AC \cdot tg\alpha = 3$$

we find: $S_{BCNM} = \frac{7}{2}BC$ (BC is perpendicular to $CN$ and $BM$ and is the height of the trapezoid). The volume of the paramid is $ABCNM$, therefore equal to

$$V = \frac{1}{3}S_{BCNM} \cdot AC \sin \beta = 7CB \sin \beta.$$

The volume of the other part cut off by the plane from the prism is the difference between the volumes of the prism and paramid: $V_2 = V - V_1$. From here

$$V_1 : V_2 = V_1 : (V - V_1) = 7 : 17.$$

In this problem, it was not necessary to find the numerical values ​​of the volumes of the parts of the prism that interest us; the plane only needed their ratio. To get the answer, it was enough to express them through the same, although unknown to us, quantities $BC$ and $\beta$. However, this very circumstance was not understood by some of the applicants who are trying hard.

Fig. 3. $AC = 6, AA_1 = 8, BM = MB_1$.

We tried to find the numerical values ​​of the required volumes and, of course, were not successful (compare with the situation in the first problem, where it was also enough to find the relation of the unknowns to obtain the answer).

Try to solve the following option yourself.

1. Two tractors plow a field divided into two equal parts. Both tractors started at the same time and each is plowing its half. 5 hours after the moment when they jointly plowed half of the entire field, they found out that the first tractor had $1/10$ part of its plot left to plow, and the second - $4/10$ of its plot. How long will it take the second tractor to plow the entire field alone?
2. Solve the equation

$$3^2 \sqrt{x^2 + 3x} + 1 + 9 = 28 \cdot 3 \sqrt{x^2 + 3x}.$$

3. Find all values ​​of $x$ lying in the interval $-1 < x < 5$ and satisfying the inequality $\log_0.5 \sin x \geq \log_3(\frac{\pi}{3})$.

4. In triangle $ABC$, angle $C$ is $60^\circ$, and the radius of the circle circumscribed around this triangle is $2\sqrt{3}$. On the side $AB$ the point $D$ is taken so that $AD = 2DB$, and at the same time $CD = 2\sqrt{2}$. Find the area of ​​triangle $ABC$.

5. Given a triangular prism $ABCA_1B_1C_1$ ($AA_1, BB_1$ and $CC_1$ are the side edges). The plane intersects the edges $A_1B_1, B_1C_1$ and $BC$ of the corresponding one at points $M, N, P$. Find in what ratio this plane divides the volume of the prism, if it is known that $B_1M : A_1B_1 = 1:2, B_1N : B_1C_1 = 2:3$ and $BP : CB = 1:3$.