Entrance exams in mathematics at the Faculty of Mathematics of the Moscow State Pedagogical Institute. V. I. Lenina

Kvant 1970, No. 3 p. 48-51 Tag TCFK

Rough automatic draft, low quality: expect mistakes and awkward wording. It will be replaced by a better translation. Original →

ENTRANCE EXAMINATIONS
IN MATHEMATICS
AT THE FACULTY OF MATHEMATICS
MOSCOW STATE
PEDAGOGICAL INSTITUTE
THEM. V. I. LENIN
Mr. V. DOROFEEV

A student of a pedagogical institute must go through many floors of a high-rise building of modern mathematics in order to become a good teacher who knows his job. But it is supposed to study the steep stairs of this building, which is completely devoid of elevators, maybe only those who are already firmly entrenched in its foundation at school.

There are such applicants, and there are quite a lot of them. This is a large part who have already understood the attractiveness of mathematics and appreciated its importance and significance for modern life. But, the proud properties of mathematical conditions (and, unfortunately, too many) decide that their only purpose in life is to become professional mathematicians or physicists.

Meanwhile, specialists who are excellent in mathematics are now needed practically in all areas of life. And to prepare them, we need only new, experienced teachers. In turn, the qualifications of teachers themselves depend on the mathematical literacy of those who enter pedagogical institutes, on their interest in their future profession, and their love for it.

Last year, at the Faculty of Mathematics of Moscow State Pedagogical Institute. V.I. Lenin received 475 applications for 175 places - about 2.7 applications per place. For comparison, we note that a significant majority were the number of people who wanted to become non-mathematics teachers and non-teacher mathematicians: for example, at the Faculty of Russian Language and Literature of our institute there were 8.3 applications for a place, and at the Faculty of Mechanics and Mathematics of Moscow State University - 6.

In the written exam in 1969, applicants were given several options, each of which contained 5 problems. All work was given 4 hours. “Test”, one-

However, it was carried out not on all problems, but on four that were best solved.

The task was chosen in such a way that a geometry lover who did not want to bother with any formal calculations could get an A, but in order to prove his “right” to neglect technical transformations - this is also one of the components of mathematical art - he had to overcome special geometric difficulties. In the same way, an “algebraist” or “analyst” who dislikes geometry could spare himself the trouble of understanding a relatively complex geometric configuration, but “in return” was obliged to solve a non-standard problem of a formal nature.

To get a grade of “3” it was enough to solve two problems. Despite this not very strong criterion, the number two turned out to be internal - 224*). Some unsuccessful works were written, unfortunately, by medal winners and graduates of mathematical schools, including Moscow ones.

But what is especially unpleasant is that most bad works contain errors that cannot be called anything other than ugly. The imagination of the authors of such works is too great to be able to list all these “pearls.” However, we will still mention one that surprised even seasoned examiners: in one of the works on the drawing for the problem where I decided on the common chord of two circles, these circles (different!) also had a common arc, and outside this arc they could diverge in different directions.

All this means that not all applicants to colleges know mathematics even at an elementary level; many of them have significant problems in mathematical processing. And it’s not so much about the breadth, but about the depth of knowledge, in the mind to note that both Muscovite and nonresident applicants did about the same about the mathematics exam (written and oral).

Option

Solve the equation
$\sqrt{1 - \log_2 x^2} \cdot \log_2 x = -1.$
Solve inequality
$2 \cos x (\cos x - \sqrt{8} \mathrm{tg} x) < 5.$
The rectangle $ABCD$ from the sides $a$ and $2a$ lies at the base of the rectangle, all of whose lateral edges are equal to $\sqrt{3}$ . Find the cross-sectional area of the straight line passing through the diagonal of the base $BD$ parallel to the edge $AS$ .
A straight line is drawn through the end $A$ of the common chord $AB$ of two circles, intersecting the first circle at point $C$ , and the second circle at point $D$ . Prove that the intersection point of the tangents drawn through points $C$ to the first circle and through points $D$ to the second circle lies on the same circle with points $B$ , $C$ and $D$ .
Prove that if
$3x^2 - 31x + 80 < 0,$
That
$\cos \frac{3}{6-x} < 0.$

Solutions

Task 1. Since $\log_2 x^2 = 2 + \log_2 x$ and $\log_2 x^4 = \frac{1}{\log_2 x^2} = -\frac{1}{4 + \log_2 x}$ are given, the equation is reduced to the form
$\sqrt{3 - \log_2 x} = -4 - \log_2 x.$
Denoting $3 - \log_2 x$ by $y$ , we obtain the equation
$\sqrt{y} = y - 1.$

Surprisingly, the solution to even such a simple equation was not

many applicants can do it. Everyone agrees to square both sides and obtain a quadratic equation with roots $y_{1,2} = \frac{3 \pm \sqrt{5}}{2}$ , but then different “points of view” are revealed; Some people don’t even think about the fact that with such a solution, roots may appear; others know that this can happen, but do not know why, and, as a rule, are confident that those roots that are not included in the permissible value range will be constructed; Still others try to check the roots by directly substituting them into the equation - this method is correct in principle, but is convenient only for checking “good” roots, and $y_1$ and $y_2$ are not very “good”.

However, it is absolutely inconceivable that since the left side of equation () is non-negative, then of the numbers $y_1$ and $y_2$ the root will be only that for which the right side is also non-negative. Therefore, the only root of equation () is $y = \frac{3 + \sqrt{5}}{2}$ , and to find $x$ we get the equation $\log_x 2 = \frac{-(9 + \sqrt{5})}{2}$ , which implies $x = 2^{-\frac{2}{9 + \sqrt{5}}}$ .

We emphasize: when solving, we completely dispensed with finding the ODZ of the original equation. Meanwhile, recently there has been a widespread prejudice among applicants that the solution to any equation or equality must be indicated starting with the calculation of the ODZ. This opinion, however, has no theoretical basis, and in practice it does a disservice: those who followed this path had to solve not only this equation, but also the logarithmic equation $-1 - \log_x 2x^2 \geq 0$ , which he himself invented - a problem no less difficult. In general, there is no strictly defined rule whether it is necessary or not to find the ODZ at the beginning of the decision, and already here, when choosing a path, the applicant can show how much he can think concretely and independently.

Problem 2. We will not solve this inequality in detail. Opening the brackets, we easily arrive at the quadratic inequality for sin $x$ :

$2\sin^2x + 4\sqrt{2}\sin x + 3 > 0,$

solving which we obtain the simplest trigonometric inequality $\sin x > -\frac{\sqrt{2}}{2}$ . But… this is precisely what most applicants failed to cope with. We will not analyze the data and answers, which are often completely meaningless, and simply draw the attention of future applicants to such tasks. The solution method is sufficiently described in the textbook and in many manuals, and you just need to be aware of it. Let us also note that even those who correctly solved the simplest inequality did not always remember that from the resulting solutions it was necessary to discard the values $x$ for which tg $x$ does not make sense.

Problem 3. It turned out to be too “treacherous” for those applicants who are accustomed to solving problems only on regular geometric bodies. They were unable to reject patterns and erroneously imagined a given geometric configuration. Thus, many people believed “out of habit” that the height of $\triangle BKD$ is the segment $KO$ (Fig. 1).

But everyone knows that geometric equation is only an auxiliary means of solving the problem, and the facts “seen” in the drawing still need to be strictly confirmed.

show. But, unfortunately, many people know this purely abstractly, but in practice they neglect the proofs, sometimes limiting themselves to phrases like “this is true by the theorem of three perpendiculars” even in the case when it is completely unclear which perpendiculars we are talking about.

Meanwhile, a small attempt to prove the perpendicularity of $KO$ and $BD$ immediately leads to opposition. Indeed, if this were so, then the diagonal $BD$ would be perpendicular to $SO$ and $KO$ , that is, to the plane $ASC$ , and therefore to the diagonal $AC$ . But the diagonals of the base, as is absolutely clear from the condition, are not perpendicular.

And to correctly construct the height at $\triangle BKD$ , it is necessary from point $K$ to carry out a perpendicular to the plane of the base of the paramid, prove that its base $O_1$ lies on $AC$ (having previously proven that the height of the paramid passes through the point $O$ of intersection of the diagonals), and from point $O_1$ to carry out a perpendicular to $BD$ . The segment $KL$ , according to the theorem of three perpendiculars, will have a height of $\triangle BKD$ . To complete the task, one more statement will come up - the parallelism of $OK$ and $AS$ , and this also needs to be proven. After this, the computational part of the problem proceeds without difficulty.

Problem 4. Among the proposed problems in planology, this turned out to be one of the most difficult for applicants: out of 93 who solved the corresponding option, only 3 people completely solved it. Meanwhile, it has a very simple and, most importantly, perfect natural solution in two replacement books by D. Polya “How to solve a problem.”

Indeed, what does it mean that the point $K$ (Fig. 2) lies on the indicated circle? This means that a circle can be circumscribed around the quadrilateral $BCKD$ . When can a circle be drawn around a quadrilateral? When the sum of its perpendicular angles is equal to $180^\circ$ . Therefore, from the conditions of the problem it is necessary to extract some statement about the equality of angles.

What theorems about the equality of angles can come to mind in connection with this problem? Only theorems on angles between tangents and chords and inscribed angles. Based on these theorems, the angles marked identically on figure 2 are equal. And now one cannot help but report that the sum of the opposite angles of the quadrilateral $BCKD$ at the vertices $B$ and $K$ is equal to the sum of the angles $\Delta CKD$ , that is, $180^\circ$ . Of course, it already follows from this that the sum of the other two opposite angles is also equal to $180^\circ$ , and the required statement is proven.

The failure of the majority of those who solved this problem is apparently simply explained by the fact that schoolchildren in the 9th and 10th grades largely forget the simplest planimetric theorems and do not bother to fill this gap in the process of direct preparation for exams. It is quite likely that 8th grade graduates would cope with this task with greater success than applicants who are wise in general knowledge.

Problem 5. Tasks of this type turned out to be of medium difficulty, and most of those who took on their solution completed it more or less correctly. It must be said, however, that many people did not take on this task, although it is simpler than others. Here, apparently, the “psychological unpreparedness” still manifests itself.