XII International Mathematical Olympiad for Schoolchildren

54

XII INTERNATIONAL MATHEMATICAL OLYMPIAD FOR SCHOOLCHILDREN

From July 8 to July 22, the XII International Mathematical Olympiad was held in Budapest and Keszthely (Hungary). Teams from Austria, England, Bulgaria, Hungary, Holland, East Germany, Mongolia, Poland, Romania, Soviet Union, France, Czechoslovakia, Sweden and Yugoslavia took part in it. Each team consisted of eight students.

On July 9, our team arrived in Budapest. On the evening of July 9 and the day of July 10, we got acquainted with the city, and on July 11, all the participants in the Olympiad moved to Kesztel, where on July 13 and 14 they took part in tasks-solving tasks.

As usual, on each perfect day the participants had to solve 3 problems in 4 hours. You will find the texts of these tasks at the end of the note. After each problem, it is indicated which country proposed the problem and how many points its solution was worth.

On July 18, members of the Jorny summed up the results of the northern competitions and discussed questions about the place and procedure for holding the next Olympiad. Czechoslovakian leaders announced the decision to organize the Olympics in Czechoslovakia. Proposals were made for individual changes to the regulations. In particular, it is proposed to give some tasks in two versions, from which participants can choose the one they like best. A proposal was also made to give problems from modern mathematics: group theory, algebraic structures, analytical geometry, vector algebra.

On July 19, all teams returned to Budapest, where on July 21, diplomas were awarded to the winners and the closing of the Olympiad took place.

Members of the Soviet team at the XII International Mathematical Olympiad (from left to right): Alexander Linetsky, Andrey KhodULEV, Alexander Korlyukov, Pavel Kopylov, I. S. Petrakov (deputy team leader), Sergey Semenkov, M. I. Serov (team leader), Alexey Alexandrov, Vello Altleis, Arkady Klimov.

Based on the results of the Olympiad, 7 first-degree diplomas, 11 second-degree and 40 third-degree diplomas were awarded.

First degree diplomas were awarded to participants who scored from 37 to 40 points. 40 points were scored by Andrey KhodULEV (USSR, integrative school at Moscow State University; from Kalinin), Wolfeane Burmeister (GDR), Imre Ruzhe (Hungary). Silverman Bernard (England), Arkady Klimov (USSR, integrative school at Moscow State University; from Arzamas), Erwin Baymots (Hungary) scored 39 points each. Hungarian schoolboy Iimtvan Gonizy scored 37 points.

II degree diplomas were awarded to participants who scored from 30 to 35 points. Of our schoolchildren

55

received Alexey Aleksadrov, who scored 35 points (9th grade at a boarding school at Leningrad State University).

Diplomas of the 11th degree were awarded to participants who scored from 19 to 28 points. In our team, they were received by Serdey Semenkov, who scored 27 points (10th grade at a boarding school at Leningrad State University), Vello Alpleis, who scored 24 points (11th grade at the News Secondary School of the Tartu District of the Estonian SSR) and Aleksadr Korlyukov, who scored 24 points (10th grade at a boarding school at Moscow State University; from Grodno).

Pavel Kolylov (Voronezh, school No. 58) and Aleksadr Linetsky (Kharkov, school No. 27) were unable to demonstrate all their capabilities and scored 15 and 17 points.

In general, Hungary’s team scored 233 points, the USSR and the GDR - 221 points each, Yugoslavia - 209 points, Romania - 208, England - 180, Bulgaria and Czechoslovakia - 145 each, France - 141, Sweden - 110, Poland - 105, Austria - 104, Holland - 87, MPR - 78.

Now all the members of the Soviet team who graduated from school participated in Moscow or Leningrad universities, and Aleksadrov is finishing 10th grade and is thinking about going to the XIII International Mathematical Olympiad in Czechoslovakia. I. S. Petrakov

Tasks proposed on the first day

1. Given a triangle ABC, M is the interior point of side AB. Let $r_1, r_2, r$ be the radii of circles inscribed in triangles AMC, BMC, ABC, respectively; $\rho_1, \rho_2, \rho$ are the radii of circles that lie inside the angle ACB, and is external for triangles AMC, BMC, ABC, respectively.

Prove that $\frac{r_1 r_2}{\rho_1 \rho_2} = \frac{r}{\rho}$ (Poland, 5 points).

2. Let $a, b$ and $n$ be natural numbers greater than ones. The numbers $a$ and $b$ are the main two number systems. The numbers $A_n, B_n$ have the same representation $\frac{x_n x_{n-1}}{x_1 x_0}$ in number systems with the basic $a$ and $b$ being $x_n \neq 0, x_{n-1} \neq 0$. The numbers obtained after crossing out the first digit $x_n$ will be called $A_{n-1}, B_{n-1}$.

Prove that $a > b$ if and only if $\frac{A_{n-1}}{A_n} < \frac{B_{n-1}}{B_n}$ (Romania, 7 points).

3. The sequence of real numbers $a_0, a_1, \dots, a_n, \dots$ satisfies the condition

$$1 = a_0 \leqslant a_1 \leqslant \dots \leqslant a_n \leqslant \dots$$

(1)

The sequence $b_1, b_2, \dots, b_n, \dots$ is defined as follows:

$$b_n = \sum_{k=1}^{n} \left(1 - \frac{a_k-1}{a_k}\right) \frac{1}{\sqrt{a_k}}.$$

a) Prove that $0 \leqslant b_n < 2$ for all $n$.

b) Prove that for a given $c$ such that $0 \leqslant c < 2$, there exists a sequence $a_0, a_1, \dots, a_n, \dots$ satisfying (1) also that $b_n > c$ for an infinite set of indices $n$ (Sweden, 8 points).

Problems proposed on the second day

4. Find all positive numbers $n$ that have the following properties: divide the set into two sets so that the product of all elements of one of them is equal to the product of all elements of the other (Czechoslovaks, 6 points).

5. In the terrahedron ABCD DB ⊥ DC and the main perpendicular dropped from D onto the plane of triangle ABC coincides with the orthocenter of this triangle.

Prove that $(AB + BC + CA)^2 \leqslant 6(AD^2 + BD^2 + CD^2)$. For which terrahedra is there a tie (Bulgaria 6 points)?

6. There are 100 points given on a plane, no three of which lie on the same line. All possible triangles with vertices at these points are considered.

Prove that among them there will be no more than 70% acute triangles (Soviet Union, 8 points).