About the written exam at the Faculty of Mechanics and Mathematics of Moscow State University in 1969

50

N. S. BAKHVALOV, N. N. KUZNETSOV

The written exam in mathematics is the main exam for applicants to the Faculty of Mechanics and Mathematics of Moscow State University. Often there comes a case about its excessive difficulty, that only graduates of Moscow mathematical schools can get even a “C” on this exam.

These cases are completely unfounded *). The tasks (the so-called “variants”, consisting of 4 problems) offered in the written exam are composed so that in each variant there is one problem

*) In 1969, out of 922 Moskovey applicants, 469 (about 50%) received a positive mark (three or higher) on the written exam, and out of 2,356 nonresidents - 912 people (about 40%).

are quite accessible to those with average mathematical training. Solving such problems already guarantees a three. Therefore, to successfully pass the written exam, you only need a solid knowledge of the school curriculum in mathematics, and only those applicants who have not shown preparation enough to successfully study at the faculty receive an unsatisfactory grade.

Of course, I will break with easy problems; the version also includes tasks whose solutions require sufficient mathematical culture (but not additional knowledge that goes beyond the scope of the program). Therefore, it is impossible to get a high mark on this exam.

Option 1

1. Points A and B are on roads intersecting at an angle ACB=60°. You can get from point A to B by bus - first along one road to intersection C, then along the other - spending 11 minutes. If you go from A to B in the right direction, it will take 1 hour. 10 minutes, and if you first take the shortest route to the road on which point B is located, and then take the bus, then even more time, even if there is a bus right away.

What is the distance from point A to the intersection if the pedestrian speed is 3 km/h and the bus speed is 30 km/h? (Roads are considered straight.)

2. In triangle ABC, bissertrices AD and BE are drawn, intersecting at point O. It is known that segment OE has length 1, and vertex C lies on the circle passing through points E, D, O. Find the sides and angles of triangle EDO.

3. Find all values ​​of x that simultaneously satisfy the following conditions:

$$\cos 13x = \cos x,$$
$$\cos 2x + \sin 5x = 1,$$
$$|x| < 3.$$

4. The primordial processes of a plane quadrilateral into two mutually perpendicular planes are squares with sides of 2. Find the perimeter of the quadrilateral, knowing that one of its sides is equal to $\sqrt{5}$.

51

This requires special preparation and the use of appropriate literature *).

Here we will consider, as an example, one typical set of problems in the version presented in the written exam for MEXMATH in 1969 (varnant 1 on p. 50). He was given 4 hours to decide. To receive a grade of “3” it was enough to solve at least one problem; a score of “5” was given if there was a complete solution to any three problems, and four were in the main solution.

We will analyze these problems in detail and note the characteristic mistakes made by those who arrived when solving them, and in conclusion we will present the readers of Kvanat with another option for independent solution.

TASK ANALYSIS

Task 1

Let’s make a drawing (Fig. 1). $D$ is the point of line $BC$ close to $A$. Note that the location of point $B$ relative to $D$ is unknown in advance (the drawing shows two location options). Let $x=AC$, $y=CB$ (these distances are measured in kilometers). Writing down the conditions of the problem, we obtain a system of two equations:

$$\begin{cases} x+y=5,5 \\ x^2+y^2-xy=12,25 \end{cases}$$

and inequality:

$$10x\sqrt{3}+|2y-x|>70$$

*) As suitable problem books, we point out, for example, the books by E. B. Vakhovsky and A. A. Ryvkin “Tasks in elementary mathematics of increased difficulty” (about it, see the article “How to solve a problem” on page 60 of this issue of “Kvanat”), N. Kh. Rozov, M. K. Potapov, G. V. Dorofeev, “A manual on mathematics for university applicants”, V. B. Lidsky, P. V. Ovsyannikova, A. N. Tulaikova and M. I. Shabunna “Tasks in elementary mathematics.”

System of equation (1) has two solutions: a) $x=4$, $y=1,5$; 6) $x=1,5$, $y=4$, and we must choose from them the one that satisfies the additional condition (2). The second solution to this condition does not satisfy, or $15\sqrt{3}+6,5<30+6,5<70$. The first decision satisfies him. To verify this, you need to check that $40\sqrt{3}+1>70$, or $40\sqrt{3}>69$. By squaring the last inequality, we obtain $4800>4781$ and thereby verify that it is true. So, the solution to the problem is $x=4$ km.

Let us note the typical mistakes that applicants made when solving this problem.

Most of them managed to compile and solve the system of equation (1); however, choosing the desired solution using the additional condition (2) turned out to be an insurmountable obstacle for many. Please, the most convenient way to understand the order is the incorrect visual representation of the relative position of the point $B$ and $D$, as a result of which, instead of inequality (2), the incorrect inequality was checked

$$10x\sqrt{3}+2y-x>70.$$

At the same time, quite often the examiner is limited to checking only one of the solutions found. For example, having established that the second solution of system (1) ($x=1,5$; $y=4$) does not satisfy inequality (3), they automatically did on this basis

52

the conclusion is that the first solution is the one we are looking for (although it is easy to see that it also does not satisfy inequality (3). Let us note one more mistake that a considerable part of those constructed made: when checking condition (2), the value $\sqrt{3}$ was replaced by an approximate rational value, for example, 1.7, and the equality was checked incorrectly

$$17x + |2y - x| > 70$$

(it is easy to check that both solutions of the system of equations (1) also do not satisfy it).

Problem 2

Let’s connect the points $C$ and $O$ (see Fig. 2). Line $CO$ is the bisector of angle $C$, or it passes through the intersection point of bisectors $AD$ and $BE$. That’s why

$$\angle ECO = \angle DCO$$

(1)

Angles inscribed in a circle $OECD$

Fig. 2.

$DEO$ and $DCO$ are equal, as if they rest on the same arc. The same is true for angles $EDO$ and $ECO$. Based on (1) we conclude that

$$\angle DEO = \angle EDO,$$

i.e., the triangle $EDO$ is equilateral.

The inscribed angles $ECD$ and $EOD$ are supported by additional arcs; therefore their sum is $180^\circ$. On the other hand, we have an integer equality:

$$\angle EOD = \angle AOB =$$

$$= 180^\circ - (\angle OAB + \angle OBA) =$$

$$= 180^\circ - \frac{1}{2} (\angle A + \angle B) =$$

$$= 180^\circ - \frac{1}{2} (180^\circ - \angle C) =$$

$$= 90^\circ + \frac{1}{2} (\angle C).$$

Because

$$180^\circ - (\angle C) = 90^\circ + \frac{1}{2} (\angle C),$$

where

$$\angle C = 60^\circ.$$

From the reasoning given above it follows that

$$\angle DEO = \angle EDO = 30^\circ,$$

$$\angle EOD = 120^\circ,$$

so the sides of triangle $EDO$ are 1, 1 and $\sqrt{3}$.

Nibaulier’s typical mistake in this problem was to “prove” that $\triangle ABC$ is equilateral. Some of the examinees, based on a geometric illusion, assumed that the center of the circle $OECD$ lies on the straight line $CO$. It follows that the angles $OEC$ and $ODC$ are right angles, so the bisectors $AD$ and $BE$ are simultaneously altitudes, which is only possible in an equilateral triangle.

Another part of those who prove the equilaterality of the triangle $ABC$, based on the following reasoning: let us drop from the point $O$ the perpendiculars $OE'$ and $OD'$ to the sides $AC$ and $BC$, respectively. The distance is two options for the location of the point $D'$ and $E'$ relative to the point $D$ and $E$ (see Fig. 3). In case a) the angles $CDO$ and $CEO$ are obtuse, and in case b) they are both acute. Therefore, the vector of points $C$, $D$, $O$, $E$ can describe a circle only when the point $E'$ coincides with $E$, and $D'$ with $D$.

This leads to the conclusion that $AD \perp BC$ and $BE \perp AC$.

53

We want to leave it as an exercise for readers to find the error in the above reasoning.

It should be noted that the number of examinees who correctly solved this problem (and similar problems of other options) was relatively low: approximately 1 out of 9 people solved it.

Task 3

Solution of the first equation of this system:

$$\cos x - \cos 13x = 0$$

or

$$2\sin 7x\sin 6x = 0$$

gives two series of values

a) $x = \frac{1}{7}n\pi$ ($n = 0, \pm 1, \pm 2, \ldots$),

6. $x = \frac{1}{6}m\pi$ ($m = 0, \pm 1, \pm 2, \ldots$).

Selecting from these values ​​those that satisfy the inequality $|x| < 3$, we find:

a) $x = \frac{1}{7}n\pi$, $|n| \leq 6$,

6. $x = \frac{1}{6}m\pi$, $|m| \leq 5$.

Now we must choose those values ​​of $x$ that satisfy the second equation of the system. This equation can be represented as:

$$1 = \sin 5x + \cos 2x = \sin 5x + \sin \left(\frac{\pi}{2} + 2x\right) = 2\sin \left(\frac{7x}{2} + \frac{\pi}{4}\right)\cos \left(\frac{3x}{2} - \frac{\pi}{4}\right).$$

Let us first substitute series (a) into it. Since $7x = n\pi$ ($|n| \leq 6$), the equality must be satisfied:

$$\sin \left(\frac{n\pi}{2} + \frac{\pi}{4}\right)\cos \left(\frac{3n\pi}{14} - \frac{\pi}{4}\right) = \frac{1}{2}$$

Since $\sin \left(\frac{n\pi}{2} + \frac{\pi}{4}\right) = \pm \frac{V^2}{2}$, then $\cos \left(\frac{3n\pi}{14} - \frac{\pi}{4}\right) = \pm \frac{V^2}{2}$ and, therefore, $\frac{3n\pi}{14} - \frac{\pi}{4}$ must differ from $\frac{\pi}{4}$ by an initial multiple of $\frac{\pi}{2}$. However, $\frac{3n\pi}{14} = \frac{3n}{7}\cdot \frac{\pi}{2}$, so $\frac{3n}{7}$ must be an integer. Under the constraint $|n| \leq 6$ this occurs only at $n = 0$. Substituting the value $n = 0$ into equation (1), we make sure that it satisfies it. So from series (a) only $x = 0$ is a solution.

Let’s now check series (6). Presenting the second equation of this system in the form:

$$\sin 5x = 1 - \cos 2x = 2\sin^2x,$$

we note that $\sin 5x$ must be non-negative, that is, the angle $5x = \frac{5}{6}m\pi$ ($|m| \leq 5$) is contained in the first two quadruples. Substituting possible integer values ​​$m$ from $-5$ to $+5$ into the expression $\frac{5}{6}m\pi$, we establish that only six are subject to verification

54

of which: $m=0, 1, 3, 5; -2, -4$. Calculating the left and right sides of equation (2) with these values, we obtain the table:

$$\begin{array}{c c c c c c c} m & 0 & 1 & 3 & 5 & -2 & -4 \\ \sin \frac{5m\pi}{6} & 0 & \frac{1}{2} & 1 & \frac{1}{2} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\ 2\sin^2 \frac{m\pi}{6} & 0 & \frac{1}{2} & 2 & \frac{1}{2} & \frac{3}{2} & \frac{3}{2} \end{array}$$

from which we obtain the required solutions: $m=0, 1, 5, t. e.$

$$x=0, \frac{\pi}{6}, \frac{5\pi}{6}.$$

Three values ​​of (3) give a complete solution to the problem, since the solution of series (a) gave the value $x=0$, already included in (3).

Note that in the same way, by brute force, we can check series (a).

When solving this problem, a typical mistake was the following: many forgot that the solution must satisfy all the conditions of the problem at the same time, and often simply did not understand this. Some limited themselves to solving one of the equations and selecting a solution based on inequality, others combined equations and ultimately solved only one equation that was not equivalent to the proposed system.

Task 4

Let $ABCD$ be the given quadrilateral, $AB=\sqrt{5}$. The primangular processes of any polygon onto parallel planes are equal to each other; Therefore, we will draw planes $P_1$ and $P_2$ parallel to the original planes through the vertex $A$ and in the future we will talk about the processes of the quadrilateral precisely on these planes.

For each point $M$ of space we will denote by $M_1,M_2$ and

55

Since $AB_1=2$, then from the right triangle $ABB_1$ there is $BB_1=\sqrt{5}-4=1$. Since in a right-angled triangle $AB_2B_3$ the hypotenuse $AB_2$ is equal to 2, and the leg $B_2B_3=BB_1=1$, then $\angle B_2AB_3=30^\circ$; therefore, the acute angle $AD_2D_3$, the sides of which are perpendicular to the sides of the angle $B_2AB_3$ ($AB_2\perp AD_2$, $D_2D_3\perp AB_3$), is also equal to $30^\circ$; therefore from the right triangle $AD_2D_3$ it is found that

$$D_2D_3=D_2A\cdot\frac{\sqrt{3}}{2}=\sqrt{3},$$

where

$$DD_1=D_2D_3=\sqrt{3}$$

And

$$AD=\sqrt{4}+3=\sqrt{7}.$$

Receptions $AB$ and $DC_2$ are the lines of intersection of parallel planes $ABB_1$ and $DD_1C_1C$ with the plane of a given quadrilateral $ABCD$; therefore $AB||DC$. We similarly conclude that $BC||AD$. Therefore, $ABCD$ is a parallelogram, and therefore its perimeter is equal to

$$2(\sqrt{5}+\sqrt{7}).$$

Please note that we wrote down the solution in such a way that we never had to refer to the drawing to justify the explanations (although in order to understand the solution, it is useful to make figure for yourself like 4 a, 6, c). Often, when solving such problems, they rely heavily on the drawing, forgetting to analyze all possible cases of one of them.

This problem has been solved very little; Therefore, it is impossible to talk about typical mistakes in connection with this task.

In conclusion, we offer the reader another version of the written task for the exercise.

Option 2

1. At points $A$ and $B$ two channels lead away from the river, crossing each other at right angles. The boat travels from $A$ to $B$ along the river for 1 hour 15 minutes, and along the canals for 1 hour 10 minutes. From point $C$ on the river, located between $A$ and $B$, one and a half times closer to $A$ than to $B$, you can get to the point of intersection of the canals using an automatic route in less than 13 minutes.

How long does it take the boat to travel from $A$ to the point of intersection of the channels, if the speed of the boat along the river is one third less, and the speed of the automatic vehicle is twice as high as the speed of the boat along the canal? (The river and canals are considered straight.)

2. In the isosceles triangle $KLM(KL=KM)$ the bissertrices $KN$, $LP$, $MQ$ are drawn. It is known that the vertex $K$ lies on the circle drawn through the points $N$, $P$, $Q$. Find $KM$ if $KP=2m$.

3. Find all values ​​of $x$ satisfying the simultaneous conditions:

$$2\cos^2 9x=1+\cos 10x,$$

$$\cos 9x+\sin 5x=1,$$

$$|x|<5.$$

4. The primugal processes of triangle $ABC$ onto two mutually perpendicular planes are regular triangles with sides equal to 1. The median $AD$ of triangle $ABC$ is equal to $\sqrt{\frac{9}{8}}$. Find $BC$.