MathDB

II Soros Olympiad 1995 - 96 (Russia)

Part of Soros Olympiad in Mathematics

Subcontests

(30)
11.9
3

BC - AM = CA - BM = AB - CM, A-excircle related

In triangle ABCABC, the side BC=aBC = a and the radius rr of the circle tangent to the side BC and the extensions of ABAB and ACAC (AA-excircle) are known. It is also known that inside the triangle there is a point MM such that BC−AM=CA−BM=AB−CMBC - AM = CA - BM = AB - CM Find the radius of the circle inscribed in the triangle BMCBMC.

base 2 numerical system

Let us denote by b(n)b(n) the number of ways to represent nn in the form n=a0+a1⋅2+a2⋅22+...+ak⋅2k,n = a_0+a_1 \cdot 2 +a_2 \cdot 2^2+...+ a_k \cdot 2^k, where the coefficients at, r=1r = 1,22,......, kk can be equal to 00, 11 or 22. Find b(1996)b(1996).

x(2^{1-2x}-1)=2^{x-2x^2}-1 (II Soros Olympiad 1997-98 R3 11.8)

Solve the equation x(21−2x−1)=2x−2x2−1x(2^{1-2x}-1)=2^{x-2x^2}-1
11.8
1
11.10
2

population of the country consisted of inhabitants and satraps

One eastern country was ruled by an old Shah. The population of the country consisted of inhabitants and satraps. Each resident had his own place of residence (place of registration). Satraps moved around the country and carried out the decrees of the Shah. One day the Shah issued a decree containing the following points:
1) Some residents are bandits. 2) Every bandit must be destroyed. 3) Together with the bandit, all those residents who are located closer to the bandit than the Shah (in other words, than the location of the Shah’s palace) must be destroyed.
Finding out which of the residents was a bandit was entrusted to the Shah's adviser, known for his connections with one hostile state. Prove that:
a) if the country in question is on a plane, then the adviser has the opportunity to declare no more than six inhabitants bandits in such a way that all inhabitants of the country must be destroyed in accordance with the decrees;
b) if the country is located on a sphere, then you can get by with five bandits.

similar triangle by segments of triangle sides

All sides of triangle ABCABC are different. On rays BAB A and CAC A the segments BKB K and CMCM are laid out, equal to side BCBC. Let us denote by xx the length of the segment KMKM. In the same way, by plotting the side ACAC on the rays ABAB and CBCB from AA and CC, we obtain a segment of length yy, and by plotting the side AB on the rays ACAC and BCBC, we obtain a segment of length zz. a) Prove that a triangle can be formed from the segments xx, yy and zz, and this triangle is similar to triangle ABCABC. b) Find the radius of the circumcircle of a triangle with sides xx, yy and zz, if the radii of the circumscribed and inscribed circles of triangle ABCABC are equal to RR and rr respectively.
11.7
2
11.3
3

[2 sinx] =2cos (3x+\pi/4) (II Soros Olympiad 1995-96 R1 11.3)

Solve the equation [2sin⁡x]=2cos⁡(3x+π4)[2 \sin x] =2\cos \left(3x+\frac{\pi}{4} \right) ([x][x] is the integer part of xx, [x][x] is equal to the largest integer not exceeding xx. For example, [3,33]=3[3,33] = 3, [2]=2[2] = 2, [−3.01]=−4[- 3.01] = -4).

cards with numbers, 300 problems (II Soros Olympiad 1995-96 R2 11.3)

The math problem book contains 300300 problems. The teacher has cards with numbers. She pins these cards to a special stand and indicates the numbers of four problems that need to be solved during the lesson. What is the smallest number of cards that a teacher can use in order to be able to indicate the numbers of any four problems from the problem book?

3x3 system sin \pi}/ 2 xy =z (II Soros Olympiad 1997-98 R3 11.3)

Solve the system of equations {sin⁡π2xy=zsin⁡π2yz=xsin⁡π2zx=y   ?\begin{cases} \sin \frac{\pi}{2}xy =z \\ \sin \frac{\pi}{2}yz =x \\ \sin \frac{\pi}{2}zx =y \end{cases} \,\,\, ?
11.6
2
11.5
3

1995th 7-digit number (II Soros Olympiad 1995-96 R1 11.5)

Let's consider all possible natural seven-digit numbers, in the decimal notation of which the numbers 11, 22, 33, 44, 55, 66, 77 are used once each. Let's number these numbers in ascending order. What number will be the 1995th1995th ?

polyhedron has exactly 6 right dihedral angles.

66 points are taken on the surface of the sphere, forming three pairs of diametrically opposite points on the sphere. Consider a convex polyhedron with vertices at these points. Prove that if this polyhedron has one right dihedral angle, then it has exactly 66 right dihedral angles.

plane intersevting uinit cubes, filling space

The space is filled in the usual way with unit cubes. (Each cube is adjacent to 66 others that have a common face with it.) On three edges of one of the cubes emerging from one vertex, points are marked at a distance of 1/191/19, 1/91/9 and 1/71/7 from it, respectively. A plane is drawn through these points. Let's consider the many different polygons formed when this plane intersects with the cubes filling the space. How many different (unequal) polygons are there in this set?
11.4
3

points M(a, b), with x^4+ax+b=0, 0<=x<=1 (II Soros Olympiad 1995-96 R1 11.4)

Draw on the coordinate plane a set of points M(a,b)M(a, b) such that the equation x4+ax+b=0x^4+ax+b=0 has a unique root satisfying the condition 0≤x≤10 \le x \le 1.

x^6 - 100x+1 = 0 (II Soros Olympiad 1995-96 R2 11.4)

Prove that the equation x6−100x+1=0x^6 - 100x+1 = 0 has two roots, and both of these roots are positive. a) Find the first non-zero digit in the decimal notation of the lesser root of this equation. b) Find the first two non-zero digits in the decimal notation of the lesser root of this equation.

>=6 tangents for a point to y = (1 -x^2)^3 (II Soros Olympiad 1997-98 R3 11.4)

Consider the graph of the function y=(1−x2)3y = (1 -x^2)^3. Find the set of points M(x,y)M(x,y) through which you can draw at least 66 lines touching this graph.
11.2
3
11.1
3
10.10
2

each deputy quarreled with exactly 3 other deputies

Each deputy of the Academic Duma quarreled with exactly three other deputies. The President ordered the Speaker to divide the deputies into n factions so that agreement reigned within one faction. For what smallest nn is this always possible? (This means that there is such nn that deputies could always be divided into nn factions, but not always into (n−1)(n- 1) factions.)

min area, semicircle and triangle related

The Order "For Faithful Service" of the 77th degree in shape is a combination of a semicircle with a diameter AB=2AB = 2 and a triangle AMBAM B. The sidesAM AM and BMBM intersect the semicircle (the border of the semicircle). The part of the circle outside the triangle and the part of the triangle outside the circle are made of pure copper. What should the side of the triangle be equal to in order for the area of the copper part to be the smallest?
10.9
2

comp. geo with cyclic quad, related to Newton−Gauss line

The opposite sides of a quadrilateral inscribed in a circle intersect at points KK and LL. Let FF be the midpoint of KLKL, EE and GG be the midpoints of the diagonals of the given quadrilateral. It is known that FE=aFE = a, FG=bFG = b. Calculate KLKL in terms of aa and b.b.
(It is known that the points FF, EE and GG lie on the same straight line. This is true for any quadrilateral, not necessarily inscribed. The indicated straight line is sometimes called the Newton−Gauss line. This fact can be used without proof in proving the problem, as it is known).

computational with inscribed trapezoid

Trapezoid ABCDABCD with bases ADAD and BCBC is inscribed in a circle, MM is the intersection of of its diagonals. A straight line passing through MM perpendicular to the bases intersects BCBC at pointK K, and the circle at point LL, where LL is the one of the two intersection points for which MM lies on the segment KLKL. It is known that MK=aMK = a, LM=bLM = b. Find the radius of the circle tangent to the segments AMAM, BMBM and the circle circumscribed around ABCDABCD.
10.8
2
10.7
2

inradius , 3 lines divide plane in 7 parts, distances of 3 points from lines

Three straight lines ℓ1\ell_1, ℓ2\ell_2 and ℓ3\ell_3, forming a triangle, divide the plane into 77 parts. Each of the points M1M_1, M2M_2 and M3M_3 lies in one of the angles, vertical to some angle of the triangle. The distance from M1M_1 to straight lines ℓ1\ell_1, ℓ2\ell_2 and ℓ3\ell_3 are equal to 7,37,3 and 11 respectively The distance from M2M_2 to the same lines are 44, 11 and 33 respectively. For M3M_3 these distances are 33, 55 and 22. What is the radius of the circle inscribed in the triangle?
[hide=second sentence in Russian]Каждая из точек М_1, М_2 и М_з лежит в одном из углов, вертикальном по отношению к какому-то углу треугольника.

2x2 system, <19x>=, <96y>=x (II Soros Olympiad 1997-98 R3 10.7)

Let us denote by <a><a> the distance from aa to the nearest integer. (For example, <1,2>=0.2<1,2> = 0.2, <3>=2−3<\sqrt3> = 2-\sqrt3) How many solutions does the system of equations have {<19x>=y<96y>=x   ?\begin{cases} <19x>=y \\ <96y>=x \end{cases} \,\,\, ?
10.6
2

plane flew along a diamond-shaped route (II Soros Olympiad 1995-96 R1 10.6)

The sports plane flew along a diamond-shaped route in windy weather. He flew through the first three sides of the rhombus in aa , bb and cc hours, respectively. How long did it take him to cover the fourth side of the diamond? (The speed of an aircraft is a vector equal to the sum of two vectors: the aircraft’s own speed and the wind speed. Wind speed is a constant vector. The aircraft’s own speed is a vector of constant length).

inradius wanted, 3 equal inradii related

On sides BCBC, CACA and ABAB of triangle ABCABC, points A1A_1, B1B_1, C1C_1 are taken, respectively, so that the radii of the circles inscribed in triangles A1BC1A_1BC_1, AB1C1AB_1C_1 and A1B1CA_1B_1C are equal to each other and equal to rr. The radius of the circle inscribed in triangle A1B1C1A_1B_1C_1 is equal to r1r_1. Find the radius of the circle inscribed in triangle ABCABC.
10.4
3

x^2+ [y]=10, y^2+[x]=13 (II Soros Olympiad 1995-96 R1 10.4)

Solve the system of equations {x2+[y]=10y2+[x]=13\begin{cases} x^2+ [y]=10 \\ y^2+[x]=13 \end{cases} ([x][x] is the integer part of xx, [x][x] is equal to the largest integer not exceeding xx. For example, [3,33]=3[3,33] = 3, [2]=2[2] = 2, [−3.01]=−4[- 3.01] = -4).

common tangent circle, parabola (II Soros Olympiad 1995-96 R2 10.4)

Find the equation of the line tangent to the parabola y=1/3(x2−2x+4)y = 1/3(x^2-2x+4) and a circle of unit radius centered at the origin. (List all solutions.)

2x2 system x+(x-y)/(x^2+y^2)=2 (II Soros Olympiad 1997-98 R3 10.4)

Solve system of equations {x+x+yx2+y2=1x+x−yx2+y2=2\begin{cases} x+\dfrac{x+y}{x^2+y^2}=1 \\ x+\dfrac{x-y}{x^2+y^2}=2 \end{cases}
10.5
3
10.3
3

(12x-1)(6x-1)(4x-1)(3x -1) = 5 (II Soros Olympiad 1995-96 R1 10.3)

Solve the equation (12x−1)(6x−1)(4x−1)(3x−1)=5.(12x-1)(6x-1)(4x-1)(3x -1) = 5.

5 points, 5 arcs, names (II Soros Olympiad 1995-96 R2 10.3)

Points AA, BB, CC, DD and EE are placed on the circle. In how many ways can the resulting five arcs be designated by the letters aa, bb, cc, dd and ee, if it is forbidden to designate an arc with the same letter as one of its ends? (For example, an arc with ends AA and BB cannot be designated by the letter aa or bb.)

triangle with sides a cosA, b cosB, c cosC

Each side of an acute triangle is multiplied by the cosine of the opposite angle. a) Prove that a triangle can be formed from the resulting segments. 6) Find the radius of the circle circumscribed around the resulting triangle if the radius of the circle circumscribed around the original triangle is equal to RR.
10.2
3
10.1
3
9.10
2

four proposals on how to select a president

At a meeting of students of the 9th "G" class, it was decided to declare the 9th "G" a presidential republic. Four blocs nominated their candidates for the presidency: “Our Street”, “Our Yard”, “Our House” and “Our Entrance”. When discussing how to select a president four proposals were made.
A. “What is there to think about! Have each student place a piece of paper with the name of the candidate they support in the box. Whoever gets the most votes is the president.”
B. “No, you can’t do that. If no one gets more than half the votes, a repeat vote must be held, in which the top two from the first vote must participate.”
C. “We must choose the one who is better than anyone else. How to do it? Let each person make a list: in the first place on his list he should put the best in his opinion, in the second place - the second, etc. If in most lists B is higher than A, then he is better than B. So, B is better than everyone if he is better than A, better than B and better than D.”
D. “Let everyone really make their own list, as V said. For the first place on the list, the candidate receives three points, for the second - 2, for the third - 1 point, and for the fourth - 0. Whoever scores the most points will he’s the president.”
As you can see, all four proposed methods are quite democratic. And yet, can it turn out that with method A one candidate wins, with method B another candidate wins, with C a third one, and in option D a fourth one? It is known that there are 29 people in the class, but the applicants (there are four of them) do not participate in the voting. Each student votes strictly according to his list (see speech B).
After a heated discussion, option B was adopted. Interestingly, if elections had been held immediately, then after two rounds a representative of the Our Yard bloc would have become president. However, elections were scheduled for a week later. Students belonging to the “Our Yard” bloc, not knowing the true state of affairs, based on the principle “you can’t spoil porridge with butter,” launched a vigorous campaign in support of their candidate. As a result of this agitation, many students did not change their opinion. True, in some lists the position of the representative of “Our Yard” has improved. (All the changes boiled down to the fact that only this candidate’s position improved.) But as a result, another was elected president. How could this happen?
[hide=might have typos, here is the original wording] На собрании учеников 9-го «Г» класса было принято решение — объявить 9-й «Г» президентской республикой. Своих кандидатов на пост президента выдвинули четыре блока: «Наша улица», «Наш двор», «Наш дом» и «Наш подъезд». При обсуждении способов выбора президента прозвучало четыре предложения. А. «Чего здесь думать! Пусть каждый ученик опустит в ящик бумажку с фамилией поддерживаемого им претендента. Кто наберет больше голосов, тот и президент.» Б. «Нет, так нельзя. Если никто не наберет больше половины голосов, надо устроить повторное голосование, в котором должны участвовать двое лучших по результатам первого голосования.» В. «Надо выбрать того, кто лучше любого другого. Как это сделать? Пусть каждый человек составит список: на первое место в своем списке он должен поставить самого лучшего по его мнению, на второе — второго и т.д. Если в большинстве списков В стоит выше А, то он лучше В. Значит, В лучше всех, если он лучше А, лучше Б и лучше Г.» Г. «Пусть и в самом деле каждый составит свой список, как сказал В. За первое место в списке кандидат получает три очка, за второе — 2, за третье — 1 очко, а за четвертое — 0. Кто наберет больше всех очков, тот и президент.» Как видим, все четыре предложенных способа вполне демократичны. И все же, может ли получиться так, что при способе А побеждает один кандидат, при способе Б — другой кандидат, при В — третий, ну, а в варианте Г — четвертый? Известно, что в классе 29 человек, но претенденты (их четверо) в голосовании не участвуют. Каждый ученик голосует строго в соответствии со своим списком (см. выступление В). После бурного обсуждения был принят вариант Б. Интересно, что если бы сразу же были проведены выборы, то после двух туров президентом стал бы представитель блока «Наш двор». Однако выборы были назначены на неделю позже. Ученики, входящие в блок «Наш двор», не зная ис тинного положения дел, исходя из принципа «кашу маслом не испортишь», развернули бурную агитацию в поддержку своего кандидата. В результате этой агитации многие ученики никак не изменили своего мнения. Правда, в некоторых списках улучшилось положение представителя «Наш двор». (Все изменения свелись к тому, что улучшилось положение только этого претендента.) Но в результате президентом был избран другой. Как такое могло случиться? [\hide]

2 equal circles inscribed in an angle, common tangent

Two disjoint circles are inscribed in an angle with vertex AA, whose measure is equal to aa. The distance between their centers is dd. A straight line tangent to both circles and not passing through AA intersects the sides of the angle at points BB and CC. Find the radius of the circle circumscribed about triangle ABCABC.
9.6
3

\sqrt[3]{5\sqrt{13}+18}-... (II Soros Olympiad 1995-96 R1 9.6)

Without using a calculator (especially a computer), find out what is more: 513+183−213+53   or   1\sqrt[3]{5\sqrt{13}+18}- \sqrt[3]{2\sqrt{13}+5} \,\,\, or \,\,\, 1

3 distances , equilateral (II Soros Olympiad 1995-96 R2 9.6)

There is a point inside a regular triangle located at distances 55, 66 and 77 from its vertices. Find the area of this regular triangle.

f(x)=x^2-6x+5, f(y)&lt;=f(x)&lt;= - f(y) (II Soros Olympiad 1995-96 R3 9.4)

Let f(x)=x2−6x+5f(x)=x^2-6x+5. On the plane (x,y)(x, y) draw a set of points M(x,y)M(x, y) whose coordinates satisfy the inequalities {f(x)+f(y)≤0f(x)−f(y)≥0\begin{cases} f(x)+f(y)\le 0 \\ f(x)-f(y)\ge 0 \end{cases}
9.9
2

locus of intersection of 2 circles

Two points AA and BB are given on the plane. An arbitrary circle passes through BB and intersects the straight line ABAB for second time at a point KK, different from AA. A circle passing through AA, KK and the center of the first circle intersects the first one for second time at point MM. Find the locus of points MM.

5 ingots of different weigths (II Soros Olympiad 1995-96 R3 9.9)

There are 55 ingots weighing 11, 22, 33, 44 and 55 kg with an unknown copper content that varies in different ingots. Each ingot must be divided into 55 parts and 55 new ingots of the same mass of 11, 22, 33, 44 and 55 kg must be made. This requires that the percentage of copper in all pieces be the same, regardless of what it was in the original pieces. What parts should each piece be divided into?
9.8
2

R=? h_a, m_a, d_a given (II Soros Olympiad 1995-96 R1 9.8)

The altitude, angle bisector and median coming from one vertex of the triangle are equal to 3\sqrt3, 22 and 6\sqrt6, respectively. Find the radius of the circle circumscribed round this triangle.

{5{4{3{2{х}}}}}=1 nested fractional parts (II Soros Olympiad 1995-96 R3 9.8)

Let a{a} be the fractional part of the number aa, that is, {a}=a−[a]\{a\} = a - [a], where[a] [a] is the integer part of a a. (For example, {1.7}=1.7−1=0.7\{1.7\} = 1.7 -1 = 0.7,{−2}=−2−(−3)=3−2\{-\sqrt2 \}= -\sqrt2 -(-3) = 3-\sqrt2.) a) How many solutions does the equation have {5{4{3{2{x}}}}}=1  ? \{5\{4\{3\{2\{x\}\}\}\}\}=1\,\, ? b) Find its greatest solution.
9.7
2

300 people in a circle, last person standing wins

300300 people took part in the drawing for the main prize of the television lottery. They lined up in a circle, then, starting with someone who received number 11, they began to count them. Moreover, every third person dropped out every time. (So, in the first round, everyone with numbers divisible by 33 dropped out). The counting continued until there was only one person left. (It is clear that more than one circle was made). This person received the main prize. (It “accidentally” turned out to be the TV director’s mother-in-law). What number did this person have in the initial lineup?

min of largst areas by 2 straight lines through a point of triangle with area 1

Through a point located on a side of a triangle of area 11, two straight lines are drawn parallel to the two remaining sides. They divided the triangle into three parts. Let ss be the largest of the areas of these parts. Find the smallest possible value of ss.
9.3
3
9.5
3
9.4
3

x^2- 10[x] + 9 = 0 (II Soros Olympiad 1995-96 R1 9.4)

Solve the equation x2−10[x]+9=0x^2- 10[x] + 9 = 0.
([x][x] is the integer part of xx, [x][x] is equal to the largest integer not exceeding xx. For example, [3,33]=3[3,33] = 3, [2]=2[2] = 2, [−3.01]=−4[- 3.01] = -4).

100 schoolchildren, 4 tasks (II Soros Olympiad 1995-96 R2 9.4)

100100 schoolchildren took part in the Mathematical Olympiad. 44 tasks were proposed. The first problem was solved by 9090 people, the second by 8080, the third by 7070 and the fourth by 6060. However, no one solved all the problems. Students who solved both the third and fourth questions received an award. How many students were awarded?

colored lines x = k, y = m, 19x+96y= c

All possible vertical lines x=kx = k and horizontal lines y=my = m are drawn on the coordinate plane, where kk and mm are integers. Let's imagine that all these straight lines are black. A red straight line is also drawn, the equation of which is 19x+96y=c19x+96y= c. Let us denote by MM the number of segments of different lengths formed on the red line when intersecting with the black ones.(The ends of each segment are the intersection points of the red and black lines. There are no such intersection points inside the segment.) What values can MM take when cc changes?
9.2
3
9.1
3

exchange rates of Dollar + German mark (II Soros Olympiad 1995-96 R1 9.1)

The exchange rates of the Dollar and the German mark during the week changed as follows:
\begin{tabular}{|l|l|l|} \hline & Dollar & Mark \\ \hline Monday & 4000 rub. & 2500 rub. \\ \hline Tuesday & 4500 rub. & 2800 rub.\\ \hline Wednesday & 5000 rub. & 2500 rub.\\ \hline Thursday & 4500 rub. & 3000 rub.\\ \hline Friday & 4000 rub. & 2500 rub.\\ \hline Saturday & 4500 rub. & 3000 rub.\\ \hline \end{tabular}
What percentage was the maximum possible increase in capital this week by playing on changes in the exchange rates of these currencies? (The initial capital was in rubles. The final capital should also be in rubles. During the week, the available money can be distributed as desired into rubles, dollars and marks. The selling and purchasing rates are considered the same.)

(x+1)^2-5(x+1) \sqrt{x}+4x=0 (II Soros Olympiad 1995-96 R2 9.1)

Solve the equation (x+1)2−5(x+1)x+4x=0(x+1)^2-5(x+1) \sqrt{x}+4x=0

(x-1)(x^2-1) ...(x^{101}-1)&gt;=0 (II Soros Olympiad 1995-96 R3 9.1)

Solve the inequality (x−1)(x2−1)(x3−1)⋅...⋅(x100−1)(x101−1)≥0(x-1)(x^2-1)(x^3-1)\cdot ...\cdot (x^{100}-1)(x^{101}-1)\ge 0