Solutions to the problems of the introductory test in ZMSH

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Solutions to the problems of the introductory test in the ZMS

A. L. Toom

In the first issue «Kvant» the problems of the introductory test for the Correspondence Mathematical Scale were published. ZMSH received about 6,000 letters from eighth-graders. About 2,500 of those who sent decisions were accepted into ZMSH and its branches.

In this issue we publish solutions to the most difficult problems.

Problem conditions

5. Four statements have been made about triangle ABC:

1. triangle ABC is right-angled;
2. ∠A = 30°; 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+b+c<0 and that the equation ax²+bx+c=0 has no real roots. Determine what sign the number c has.

7. The following is known about points A, B, 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 point A lies on segment BC.

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

9. Is it possible to write down nine numbers 1,2,3,…, 9 in a circle in such an order that the sum of no two adjacent numbers is divisible by 3, 5, or 7?

10. Residents of the city of Pravdin always tell the truth, and residents of the city of Liars always tell lies.

In one of these cities, a football match took place between the teams of these cities. After the match, near the stadium where it was played, the following happened:

A (to B and C): “I wasn’t at the match. Tell me, who won?
B: “G told me that their team lost.”
B: “Our team won.”
Γ (getting into a bus going to another city, and addressing A): “Let’s go together, A, because we are from the same city.”
A: “You are mistaken, G. I live here. This is B from your city."
Determine who is from each city, where the match takes place and which team won.

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

Problem solutions

5. We are told that out of four statements some two are true, but we are not told which ones are true and which are not. Therefore, let’s consider several cases. In each such case we will assume that some two certain statements are true and the other two are true. In order not to miss any case, we first find out

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Case 1.2. Since $\Rightarrow A=30^\circ$, the right angle is either $B$ or $C$. If $\Rightarrow B$ is direct, then $AC=2BC$, which should not be the case, since statement 3 is false in this case. If $\Rightarrow C$ is straight, then $AB=2BC$, which also contradicts the conditions. So case 1.2 is impossible.

Case 1.3. Angle $A$ cannot be right, since the opposite side does not lie opposite it. If $\Rightarrow C=90^\circ$, then $\Rightarrow A=30^\circ$, which cannot be in this case. So, $\Rightarrow B=90^\circ$. then statement 2 and 4 are false. This means that such a case is possible. Then, by the Pythagorean theorem, $AC=\sqrt{5}$, and the perimeter is $3+\sqrt{5}$.

Case 1.4. It is analyzed similarly to case 1.3. Possible if $\Rightarrow C=90^\circ$. The perimeter is $3+\sqrt{5}$.

Case 2.3. If $\Rightarrow A=30^\circ$ and $AB=2BC$, then $\Rightarrow C=90^\circ$, which contradicts the conditions.

Case 2.4. Similar to case 2.3 is impossible.

Case 3.4. Statement 1.2 is incorrect. So this case is possible. The perimeter is 5.

Answer. a) $3+\sqrt{5}$; b) 5.

6. Since the square triangular $ax^2+bx+c$ has no real roots, then $b^2-4ac<0$, hence $ac>\frac{b^2}{4}$. Let us prove that $c(a+b+c)>0$.

Really,

$$c(a+b+c)=ac+bc+c^2>\frac{b^2}{4}+bc+c^2=\left(\frac{b}{2}+c\right)^2\geq 0.$$

Since in a positive product one factor $a+b+c$ is negative, then the other factor $c$ is negative. So, $c<0$.

Solution option. It is known that if a square triangular $ax^2+bx+c$ is not equal to zero for any $x$, then its sign is the same for all $x$. Note that the number $a+b+c$ is the value of our triangle at $x=1$, and the number $c$ is its value at $x=0$. This means that these two numbers have the same sign. Since $a+b+c<0$, so is $c<0$.

7. Let us prove the required contradiction. Let us assume that the point $A$ does not lie on the segment $BC$, and we will come to a contradiction with the conditions. That is, we will prove that then it is not true that “for any point $M$ on the plane, the segment $AM$ is less than at least one of the segments $BM$ and $CM$.” In other words, let’s find a point $M$ on the plane such that the segment $AM$ is not less than one of the segments $BM$ and $CM$. Let’s consider two cases.

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the columns are numbered from 1 to 9. If two numbers can stand next to each other, then at the intersection of the row and column with these numbers there is a square, and if they cannot, then minutes. You can do otherwise. Let’s draw nine circles and place numbers from 1 to 9 in them. Let’s connect two linear circles, the numbers in which can be next to each other. And if two numbers cannot be adjacent, then we will not connect them. We will redraw the resulting diagram in order to “unravel” it as much as possible. The result is depicted in figure 4. Once such figure is constructed, the question can be sorted as follows: is it possible, and for only along lines, to go around all the circles of this figure once at a time and return them to their original place? We will solve the problem by looking at this drawing, although we could use a table in the same way.

Since 4 is connected only to 7 and 9, it means that 7 and 9 are its neighbors in location on the circle. Similarly, neighboring 2 are the numbers 6 and 9. This means that 2 and 4 are neighboring 9, and 7 and 8 are obviously not 相邻 9. This means that the connections of 9 with 7 and 8 can be crossed out (which is what was done on figure). Similarly, 1 has neighbors 3 and 7. This means that the neighbors of 7 are numbers 1 and 4, and the connection between 7 and 6 can be crossed out. Now 8 is connected only to 3 and 5, which means that the connection between 3 and 5 can be crossed out. The remaining lines will process a closed path, which constitutes the only (up to the direction of traversal) solution.

11. Let’s make a table:

| And from Pravdin? | G from Pravdin? |
| :— | :— |
| B from Pravdin? | Did the match take place in Pravdin? |
| B from Pravdin? | Did Pravdin’s team win? |

Solving a problem means filling in the empty columns of this table with the words “yes” or “no” so as not to conflict with the conditions. More precisely, we need to find all such filling methods, since there may be more than one of them. At worst, you can rework all possible filling methods (there are 64 in total) and check for each one whether it contradicts the conditions or not. But this, of course, is an ugly way to solve it. We will decide otherwise.

1. From statement B it follows that Pravdin’s team won. Let’s prove it. If B is from Pravdin, then what he said is true, and Pravdin’s team won. If B is from Liar, then what he said was not true, and again Pravdin’s team won.

2. From B’s statement it follows that B is a Liar, since his statement is completed incorrectly. Let’s prove it. We already know that Pravdin’s team won.

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Dina. If $\Gamma$ is from Pravdin, then he could not say that their team lost, since that would be a lie. If $\Gamma$ is from Liar, then he also could not say this, since it would be the truth.

3. From the statement $\Gamma$ it follows that $A$ is from Pravdin. Let’s prove it. If $\Gamma$ is from Pravdin, then $A$ is valid from the same city, that is, from Pravdin. If $\Gamma$ is from Lgunov, then $A$ is not from the same city, that is, $A$ is again from Pravdin.

4. Now it’s easy to answer all the questions. From the last statement $A$ it follows that the case is happening in Pravdin, $B$ and $\Gamma$ - from Lgunov.

Answer. $A$ - from Pravdin, $B$, $B$, $\Gamma$ - from Lgunov, the match was hosted in Pravdin, Pravdin’s team won.

12. It is clear that auto heights overtake a person at regular intervals. (Any such interval is equal to $\frac{v_1 t}{v_1 - v_2}$, where $t$ is the time interval between autovises, $v_1$ is the speed of autovises, $v_2$ is the speed of a person. However, we will not formulate this.) Therefore, between those moments when the autovises overtake him, the person has time to travel the same distance each time. Let’s denote this distance by $x$.

On figure 5, the segment $AB$, marked with a curly bracket, represents the 4 km that the person walked for the first time. Let’s mark on it those 6 points $P_1$, $P_2$, $P_3$, $P_4$, $P_5$, $P_6$, where it was cut by the autovis. There are 5 spaces between them, and the length of each is $x$. There remain two pieces of the path at the ends $AP_1$ and $P_6B$, which the person passed before the first autovis overtook him, and after he was overtaken by the last autovis (drawn with a red line at figure). The length of each of these segments is less than $x$. Let’s prove it. Let’s say, for example, that $P_6B \geq x$. Then on the segment $P_6B$ there will be another point $P_7$ such that $P_6P_7 = x$. At this point, of course, a person will also be overtaken by an autovis, and in total over these 4 km a person will be overtaken by more than six autovises, which is contrary to the conditions. The fact that $AP_1 < x$ proves similarly. Taking all this into account, we can write:

$$5x \leq 4 < 7x. \tag{1}$$

Similarly, since on a segment of 7 km there will be 8 points spaced $x$ from each other, and the end segments will again be less than $x$:

$$7x \leq 7 < 9x. \tag{2}$$

From (1) we obtain that $x \leq 4/5$, and from (2) we obtain that $x > 7/9$.

Now we must determine how many points, spaced $x$ km apart, can be located on a segment of 17 km.

Since $\frac{17}{4/5} = 21 \frac{1}{4}$ and $\frac{17}{7/9} = 21 \frac{6}{7}$, then $21 \frac{1}{4} \leq \frac{17}{x} < 21 \frac{6}{7}$ or

$$21 \frac{1}{4}x \leq 17 < 21 \frac{6}{7}x. \tag{3}$$

Let us prove that on a segment of 17 km a person will be overtaken by 21 or 22 cars. Indeed, if a person is passed by 20 autovises (or less), then there will be 19 intervals of length $x$ between them (or less), hence $17 < 21x$, which contradicts (3). If a person is overtaken by 23 autovises (or more), then $17 > 22x$, which also contradicts (3).

Answer. 21 or 22 autovis.