Solutions to the problems of the introductory test in the ZMSH 1970

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Problem solutions
introductory work control
in 3MSh 1970

A. L. Toom

In the first issue of our magazine (page 56), as well as in the Teacher’s Newspaper (February 26, 1970), problems were published that need to be solved in order to enter the Correspondence Mathematics School. ZMSH received about 6,000 letters from eight grade students with solutions to problems, about 3,000 of them were accepted at the school.

We think that the problems are being addressed not only by those who sent letters to the ZMS, but also by many other readers «Kvant», including future candidates for the ZMS, and therefore today we are publishing the solutions to the problems. Almost all problems are analyzed in great detail, “thoroughly”, in order to clarify the solution even for not very prepared readers. For the convenience of readers, we repeat the conditions of the problems here.

It’s not enough to solve a problem and get the correct answer: you also need to be able to write it down, which is not at all easy. Therefore, we advise even those who have solved the problems to read these solutions and compare them with their own. Of course, the solutions given here are not the only possible ones, but by reading them you can understand how to write solutions to problems accurately and logically.

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Entrance test
in 3MSh 1970 (problem conditions)

1. The mathematician walked home along the bank of the stream upstream at a speed one and a half times greater than the speed of the stream, and held a stick and a hat in his hands. At some points, he threw his hat into the stream, confusing it with a stick, and continued to walk up the stream at the same speed. After a while he noticed the mistake, threw the stick into the stream and ran back at twice the speed of his hat. Having caught up with the dusty hat, he instantly fished it out of the water, turned and went upstream *) at first speed. After 10 minutes, he came across a stick dusting along the stream. How much earlier would he have come home if he had not confused the stick with the hat?

2. For every triangle ABC, is there a point P such that all three points, symmetrical points P with respect to the lines AB, BC and AC, lie on the circumcircle of triangle ABC?

3. For what values ​​of a is the difference between the roots of the equation ax²+x-2=0 equal to 3?

4. The student took 31 exams over five years of study. In each subsequent year he took more exams than in the previous one. In the fifth year there are more second exams than in the first. How many exams are there in the fourth year?

5. The bases of an equal-volume trapezoid are 4 cm and 8 cm, its area is 21 cm². Which side does the bissertis of the angle intersect with a larger base: the smaller base or the lateral side of the trapezoid?

6. All the digits of a certain four-digit number that is a perfect square can be reduced by the same number so that the result is a four-digit number that is also a perfect square. Find all such numbers.

7. Can the sum of the distances from a point inside a convex quadrilateral to all its vertices be greater than its perimeter?

8**). One of three gangsters known

*) This word “up” was mistakenly left out in No. 1 “Kwaンタ”, which made it difficult for some readers to understand the terms of the problem.

**) In other publications this problem was published in the following formulation:

Word reached King Pea that someone had finally killed the Mountain Serpent. Tsar Gorokh knew that this could have been done either by Ilya Muromets, or Dobrynya Nikitich, or Alyosha Popovich. And here they are, covered in dust, they came to the court. The king began to ask them. Every time the rich man held the ribs. And they said this:

And Lya Muromets: 1) It was not I who killed the Serpent of the Mountain. 2) I went to overseas countries. 3) The Serpent Gornitsa was killed by Alyosha Popovich.

Before brynya Nikitich: 1) Snake Gornytsa in the city of M under the clicks of Archie, Boss and Wesley, stole a briefcase (with money). During interrogation, each of them made three statements:

And rch: 1) I didn’t take the briefcase.
2) On the day of the theft, I left the city of M.
3) The briefcase hit Wesley.
4) Voss: 1) The briefcase stung Wesley.
2) Even if I had taken it, I would not have confessed.
3) I already have a lot of money.
Voss: 1) I didn’t take the briefcase.
2) I’ve been looking for a good briefcase for a long time.
3) Archie is right when he says he got the hang of M.

During the investigation, it was found that out of three statements of each guarantee, two were true and one was incorrect. Who stole the briefcase?

9. Find numbers x and y such that x>y>0 and x³+7y=y³+7x.

10. A square area measuring 100 m×100 m is paved with 1 m×1 m square slabs of four colors: white, red, black and gray—so that no two slabs of the same color collide with each other (i.e., do not have a common side or top). How many red tiles can there be?

11. Solve the equation
    (x²+6x-4)(x²+6x-3)=12.

12. Given triangle ABC. Find points K and H lying respectively on sides AB and BC, also that BK=KH=HC.

13. Find all numbers p such that p²+13 is also prime.

SOLUTIONS TO PROBLEMS

1. First of all, we note that the time lost by the mathematician, that is, the time that needs to be found, consists of two components: a) the time that he ran back and which we denote by t (minutes); b) the time he spent walking the distance he had to travel; since he ran twice as fast as he walked, this time is equal to 2 t, and all the lost time is equal to 3 t.

So, it is enough to calculate t. Let us denote the flow velocity by v. Then the walking speed is $\frac{3}{2}v$ and the running speed is 3v. Let’s consider the situation from the moment when the mathematician, throwing a stick into the water, marked it, until the moment when he met a puffy stick towards him. He fled now and killed Alyosha Popovich. 2) If I had killed him, I would not have told him. 3) There is still a lot of evil spirit left.

A Lesha A Popovich: 1) It was not I who killed the Serpent Gornyts. 2) I have been looking for a long time for some feat to accomplish in overseas countries.

Then King Pea found out that twice the rich man told each one the truth, and once he lied. Who killed the Mountain Serpent?

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time $t$ at a speed $3v$ and, therefore, ran a distance $t \cdot 3v$. Then he walked back at a speed of $\frac{3}{2}v$ and covered a distance of $10 \cdot \frac{3}{2}v$ in $10$ minutes. The stick danced all this time, that is, $(t+10)$ minutes, at a speed of $v$ and barked $(t+10) \cdot v$. Obviously, the first distance calculated here is equal to the sum of the other two:

$$t \cdot 3v = 10 \cdot \frac{3}{2}v + (t+10) \cdot v.$$

Reducing by $v$, we find that $t=12,5$, hence $3t=37,5$.

Answer: $37,5$ min.

This problem can also be solved using graphs (see Fig. 1; here $S$ is the distance from the river upstream, $T$ is time; the red line is the mathematician’s path; the blue segments are the floating hat and stick). By condition $\frac{AC}{CD} = 2$ and $BC=DE$, from where we can find the ratio of the segments $AB:BC:CD:DE=6:4:5:4$ and $\frac{AD}{DE} = \frac{AD}{10\text{ мин}} = \frac{15}{4}$, from where $AD = 37,5$ min.

2. Point $P$ must satisfy the third option:

1. the point $P_1$, symmetrical to $P$ with respect to $AB$, lies on the circumcircle;
2. the point $P_2$, symmetrical to $P$ with respect to $BC$, lies on the circumcircle;
3. the point $P_3$, symmetrical to $P$ with respect to $CA$, lies on the circumcircle.

Let us first find all points $P$ that satisfy the third option 1). All such points $P$ form a circle symmetrically circumscribed by $AB$ (drawn on figure 2 in white).

Thus, requirement 1) imposed on the point $P$ can be reformulated as follows:

1’) point $P$ lies on the circle $O_1$, symmetrical to the circumcircle with respect to $AB$.

In a completely similar way, we are convinced that requirements 2), 3) can be reformulated in the requirement that the point $P$ lie on the circle $O_2, O_3$, symmetrical to the circumcircle with respect to $BC$ and $CA$.

We need to find out whether for every triangle there is a point $P$ lying on all three circles $O_1, O_2, O_3$ at once, in other words, for every triangle these three circles $O_1, O_2, O_3$ pass through one point. Let us prove that this is true for any triangle $ABC$.

It is quite difficult to come up with such a miracle that would be suitable for all triangles $ABC$ at once (the fact is that we do not know in advance where the point $P$ of the intersection of circles $O_2$ and $O_3$ lies - inside the triangle or outside it, on which side of each of the straight lines $AB, BC$ and $CA$, etc., and if the miracle depends on location of the point $P$, then it is necessary to analyze several special cases). But to save space, we will still present just such a solution, which was suitable for all cases at once, albeit somewhat artificial.

Note that the circle $O_1$ can be obtained from the circle $O$ not only of the symmetric relative side $AB$, but also of the symmetric relative point $K$ - the middle of the side $AB$, i.e. by repeating the relative point $K$ on $180^\circ$. What happens if, together with the circle $O$, we rotate the triangle $ABC$? It is clear that it is transferred to $\triangle ABC_1$, which complements $\triangle ABC$ to a parallelogram $ABC_1C$, and $\triangle ABC_1$ is inscribed in the circle $O_1$ (Fig. 3). Having completed the same constructions of the final sides $BC$ and $CA$, we get four equal triangles, together making up one

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a 2l < 3l.

What is written here, only in another, unnecessary void, can be considered a solution.

The exact solution to this problem “for a written exam” may look, for example, like this: construct a quadrilateral ABCD and a point O inside it, as shown in figure 6, where BD ⊥ AC, AM=BM=DM=1, OM=100, CM=101. Then

$$OA + OB + OC + OD = 101 + 2 \cdot BO + 1,$$

$$AB + BC + CD + DA = 2\sqrt{2} + 2 \cdot BC.$$

Since BC < OB + OC = OB + 1, then 2\cdot BC < 2\cdot OB + 2 and therefore AB + BC + CD + DA < 2\sqrt{2} + 2\cdot BO + 2 < 100 + 2\cdot BO + 2 = OA + OB + OC + OD.

8*). There are three possible assumptions about who stole the briefcase. Let’s look at them one by one in order to find out which of them contradict the conditions.

1). Let Archie steal the briefcase. Then his first and third statements are incorrect, and the conditions say that only one of the statements is false. This means that this assumption contradicts the conditions; Archie couldn’t have stolen the briefcase.

2). Let the Boss steal the briefcase. Then his first statement is false, therefore the other two must be true. True, in his second statement it seems to be implied that he, the Boss, did not take the briefcase, but he does not say this directly, and therefore no contradiction will arise. Further, Archie’s third statement is then false, therefore the first two must be true. Since Archie’s second statement is true, then Wesley’s third statement is true, and since the first is true, then the second is false.

We see that there are no contradictions with

*) The problem about the Serpent Gornych is solved in a similar way. (ANSWER: Snake Gornych was killed by Dobrynya Nikitich.)

condition does not arise. So the Boss could have stolen the briefcase.

3). Let Wesley steal the briefcase. Then his first statement is false, and therefore the second and third are true. Since Wesley’s third statement is true, then Archie’s second statement is true. Archie’s first and third statements are also true. This means that all three of Archie’s statements are true, which contradicts the conditions. Wesley couldn’t have stolen the briefcase.

Answer: The briefcase was stolen by the Boss.

Note: Was it possible, after considering the assumption that the briefcase was stolen by the Boss, and making sure that it satisfies the conditions, to calm down and consider it proven? No.

After all, it could turn out that the data collected in the conditions is generally insufficient to determine who exactly stole the briefcase.

9. Let’s transform:

$$(x^3 - y^3) - 7(x - y) = 0,$$

$$(x - y)(x^2 + xy + y^2) - 7(x - y) = 0,$$

$$(x - y)(x^2 + xy + y^2 - 7) = 0.$$

Since $x > y$, then $x - y \neq 0$, which means the second bracket is equal to zero:

$$x^2 + xy + y^2 - 7 = 0,$$

$$x^2 + xy + y^2 = 7.$$

Which of the three numbers $x^2, xy, y^2$ is an integer, positive and less than seven. Then $x, y$ can only be equal to 1 or 2 (already at $x = 3, y = 9 > 7$). Since $x > y$, then $x = 2, y = 1$. By substitution we make sure that this is a solution to the problem.

Answer: $x = 2, y = 1$.

10. First of all, let’s make sure that the area can be laid out with slabs as required by the conditions. This can be done, for example, as shown in figure 7. The law of considering slabs over all areas is clear from the right-handed part. Bold lines are drawn to make this law easier to understand. With this consideration, there are 2500 pieces of slabs of all colors. This means that there can be 2500 red slabs. Let us prove that no other number of slabs is possible.

Let the area be somehow laid out with slabs in accordance with the conditions. Divide it into squares $2m \times 2m$, each containing 4 slabs, as was done in figure 7. There are 2500 such squares. Let’s consider one such square. The 4 slabs of which it consists are square to each other. Therefore, they should all be different colors, that is, one of each color. Since there is one red tile in each square, and there are 2500 squares, then there are 2500 red tiles.

11. Let us introduce the notation: $x^2 + 6x = y$. Then $(y - 4)(y - 3) = 12,$ $y^2 - 7y = 0,$ $y_1 = 0,$ $y_2 = 7.$

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Now we need to solve two equations:

1. $x^2 + 6x = 0$, $x_1 = 0$, $x_2 = -6$;
2. $x^2 + 6x = 7$, $x^2 + 6x - 7 = 0$, $x_1 = -7$, $x_2 = 1$.

Answer: $x_1 = -7$, $x_2 = -6$, $x_3 = 0$, $x_4 = 1$.

12. 1. Analysis. Since $KB = KH$, then $\angle KHB = \angle B$. Let’s do $KC$. Since $KH = HC$, then $\angle HKC = \angle HCK$. But their sum is equal to the external angle $\angle KHB$: $\angle HKC + \angle HCK = \angle KHB$ (see Fig. 8). This means that each of them is equal to half of it:

$$\angle CKH = \angle KCH = \frac{1}{2} \angle KHB = \frac{1}{2} \angle B.$$

2. Construction. Let $\triangle ABC$ be given. Divide $\angle B$ in half and construct an angle $\angle BCM = \frac{1}{2} \angle B$ so that the ray $CM$ lies on the same side of the line $BC$ as the point $A$. Let ray $CM$ intersect side $AB$ (not its extension) at point $K$. Let’s construct the angle $\angle CKN = \frac{1}{2} \angle B$ so that the ray $KN$ lies on the same side of the straight line $CK$ as the point $B$. Let ray $KN$ intersect side $BC$ (not its extension) at point $H$. The constructed points $K, H$ are exceptional.

3. Proof. By construction, $\angle HKC = \angle HCK$, therefore $KB = KH$.

Next,

$$\angle KHB = \angle HKC + \angle HCK = \frac{1}{2} \angle B + \frac{1}{2} \angle B = \angle B,$$

pose $KB = KH$.