Variant for Year IX.

Problem 1.

Solve the system of equations \[\begin{cases} x^3 + y^3 = 9, \\ x^2 y + y^2 x = 6. \end{cases}\]

Solution.

from sympy import *
x, y = symbols('x y', real=True)

solve((x**3 + y**3 - 9, x**2 * y + y**2 * x - 6), x, y)

## [(1, 2), (2, 1)]

Problem 2.

In Triangle ABC Points P, Q and R are the foots of the altitudes, dropped from Vortices A, B and C, respectively. Prove that \(\angle ABQ = \angle APR\).

Solution

Drawing

Let the \(F\) is an intersection of altitudes - orthocenter and consider the quadrilateral triangle \(BRFP\). Since \(\angle BRC = \angle APB = \pi/2 \implies \angle BRC + \angle APB = \pi\).

Since the sum of angles equals \(\pi\) then \(BRFP\) can be inscribed into the circumference. But it is known that if two angles are inscribed on the same chord and on the same side of the chord, then they are equal. Hence \(\angle RBQ= \angle RPF\) and \(\angle ABQ = \angle APR\), c.e.d.

Problem 3.

What year do people born, who in 1969 became of the same age, as the sum of digits of their birth year.

Solution

for (i in 1:99) {
  born <- 1969 - i
  digit_sum <- sum(as.integer(unlist(strsplit(as.character(born), ""))))
  if (i == digit_sum) {
    print(i)
    break
  }   
}

## [1] 17

Problem 4

Why the fog which consists of the transparent water droplets is not transparent?

Solution

Light traveling through the fog gets randomly scattered, mainly by bouncing of the droplets.

Problem 5

The glass of water is located in on the spring scales. The cork with volume \(V\) and density \(\rho\) floats in the water. The cork was pushed underwater completely so it was fully submerged. What would the scales show?

Solution

The difference of weights will be equal to the change of Archimedes’ force. So

Before submersion:

\[F_A^{before}=g\rho_{water}\frac{\rho}{\rho_{water}}V=g\rho V\]

After submersion: \[F_A^{after}=g\rho_{water}V\] Difference:

\[P = F_A^{after} - F_A^{before}=g\rho_{water} - g\rho V= gV(\rho_{water}-\rho)\]

Variant for Year X.

Problem 1.

Solve the system of equations \[\begin{cases} x (y + z) = 3, \\ y (x + z) = 4, \\ z (x + y) = 5. \end{cases}\]

from sympy import *
x, y, z = symbols('x y z', real=True)

solve((x * (y + z) - 3, y * (x + z) - 4, z * (x + y) - 5), x, y, z)

## [(-sqrt(6)/3, -sqrt(6)/2, -sqrt(6)), (sqrt(6)/3, sqrt(6)/2, sqrt(6))]

Problem 2.

In trapezium \(ABCD\) with bases \(AD=a,BC=b (a > b)\) they draw \(MN\), parallel to the bases and dividing the area of trapezium in half. Find the length of this interval.

Solution

Drawing

Lets drop altitudes \(BH=x, HI=y, JK= h\) then solve the system of quadratic equations with definitions of a trapezium area (for 3 trepezia) and condition that \(x + y = h\).

from sympy import *
x, y, S, h, a, b, c, z = symbols('x, y, S, h, a, b, c, z', real=True)

sol = solve((
  (a + b) / 2 * h - S, 
  x + y - h, 
  (a + c) / 2 * x - S / 2,
  (c + b) / 2 * y - S / 2),
  x, y, h, c)

sol[0][3]

## sqrt(2)*sqrt(a**2 + b**2)/2

Problem 3

At the bottom of one of the сommunicating vessels there is a piston with the weight \(P\). Its friction force across the walls is \(F\). How much water should be poured into another vessel, so the piston shift to the height \(h\).

Solution

Let \(H\) is the initial height which was filled in, \(S\) the piston area.

\[E_1 = \rho SHg/2\]
\[E_2 = Ph +\rho gSh^2/2-Fh+\rho gSh^2/2 + \rho g S(H-h)^2/2\]

\(E_1 = E_2\)

from sympy import *
P, h, rho, g, S, F, H= symbols('P, h, rho, g, S, F, H', real=True)

solve(P * h + rho * g * S * h**2 / 2 - F * h +  rho * g * S * h**2 / 2 + 
  rho * g * (H - h)**2 * S / 2 - rho * S * H**2 * g / 2, H)

## [(-F + P + 3*S*g*h*rho/2)/(S*g*rho)]

Problem 4

The box is put on the conveyour belt with zero initial speed. The belt is moving with the speed \(v\). The friction coefficient is \(k\). What distance will be covered by the box relatively to the belt until his speed becomes equal to \(v\).

Solution

All kinetic energy transferred to the work of friction force on the path \(l\) \(W_{fr}=F_{fr}l = mgkl\).

from sympy import *
m, k, v, g, l= symbols('m, k, v, g, l', real=True)

solve(m * v**2 / 2 - m * g * k * l, l)

## [v**2/(2*g*k)]

Problems on oral exams

Problem 1

Find the sum of squared roots of equation, without solving it: \[x^2 + px +q = 0\]

Solution

Let \(x_1, x_2\) be roots of the equation then: \[x_1^2+px_1+q=0, x_2^2+px_2+q=0\] then let’s add two equations \[(x_1^2 + x_2^2) + p(x_1+x_2) + 2q = 0 \implies x_1^2 + x_2^2 = p^2 - 2q\]

Problem 2

Which of two numbers is greater: \(2^{300}\) or \(3^{200}\).

Solution

\[2^{300} = (2^3)^{100} < (3^2)^{100} = 3^{200}\] ## Problem 3

The plot of quadratic function \(y = ax^2+bx+c\). Which signs of \(a\), \(b\) and \(c\) numbers?

curve(-x^2 + 2 * x + 2, xlim = c(-1, 3), ylim = c(-3, 6))
abline(h = 0)
abline(v = 0)

Solution

Since the parabola opens downward - \(a <0\), as instersection with \(Oy\) is positive then \(c >0\). Since \(x_2>x_1>0 \land a<0 \implies b > 0\).

So \(a<0, b>0, c>0\).

Problem 4

Prove that the triangle median sum is less than its perimeter and bigger than \(3/4\) of its perimeter.

Solution

test

sum of triangle’s medians is less than its perimeters.

Let us construct a parallelogram on \(AB\) and \(AC\) lines. It is known that parallelogram’s diagonals are intersecting in the middle. Then \(AC =2AF < AB+BE = AB+AC\) then carry out the same procedure for \(2CG < BC + AC\) and \(2BD<AB=BC\), then sum all 3 inequality \(2(AC + CG +BD) < 2(AB + AC + BC) \implies AC + CG +BD < AB + AC + BC\), i.e. sum of medians is less than its perimeter.

sum of triangle’s medians is greater than \(3/4\) of its perimeters.

Let \(O\) is a point of median intersection. \(AO + OC > AC, AO+OB>AB, OB+OC>BC\). Sum three equations together. \(2AO +2BO +2CO > AB+AC+BC\) since \(O\) divides medians in the ratio \(2:1\) then \(2\cdot\frac{2}{3}(AC + CG +BD) > AB+AC+BC \implies AC + CG +BD > \frac{3}{4}(AB+AC+BC)\).

Sum of medians is \(\Sigma m\), then \(\frac{3}{4}p<\Sigma m<p\), c.e.d.

Problem 5

The bases of the trapezium are \(a\) and \(b\) (\(a>b\)). Find the segment length, which parallel to the trapezium bases, which ass through the diagonal intersections and located between its legs.

Solution

Trapezium \[ AD = b, BC = a\]

Let Point \(O\) is an diagonals intersection, \(TL\) is a segment parallel to the trapezium bases, it passes through Point \(O\), the end of which is located on the legs. To prove this statement let us consider four pairs of similar triangles.

\(\triangle TBO\) and \(\triangle ABD\) are similar, then \(TO/b = BO/BD\);

\(\triangle OCL\) and \(\triangle ACD\) are similar hence \(OL/b= OC/AC\);

\(\triangle DOL\) and \(\triangle BDC\) are similar hence \(OL/a= OD/BD\) ;

\(\triangle TAO\) and \(\triangle BAC\) are similar hence \(TO/a= AO/AC\).

Let us add together these four equalities: \[TO/b+ OL/b+ TO/a+ OL/a= BO/BD+ OC/AC+ AO/AC+ OD/BD\]

Let us simplify:

\[TL/b+ TL/a= BD/BD+ AC/AC\] \[TL(1/b+ 1/a)=2 ;TL= 2ab/(a+b)\] The segment’s \(TL\) length is \(h\). Then \(h= 2ab/(a+b)\).

Problem 6.

Prove that circles constructed on the side of convex quadrilateral as on diameters, completely cover the quadrilateral.

Solution

test

Let us suppose, some point \(X\), located in the interior of the given quadrilateral and does not belong inside any given circle. Then form the point \(X\) any diameter of each circle is seen at the acute angle. Since quadrilateral is convex, then sum of these four angles should equal to \(2\pi\), and it is impossible since by initial supposition each of the angles is less than \(\pi/2\), c.e.d.

Problem 7

Two circumeferences with radii \(r_1\) and \(r_2\) touch the given line. Find the locus of points intersecting their common interior tangents with the condition that the circumferences can move along the given line arbitrarily.

Solution

Two circles and tangents

Two circles are moving on the line \(l\). Point \(M\) is located on the symmetry axes of these two circumferences - line \(O_1 O_2\), where \(O_1\) and \(O_2\) centers of circumferences. Hence locus is located on the intersection \(O_1 O_2\) and \(T_1T_2\). Draw the radii to the touch points \(O_1T_1\) and \(O_2T_2\). Point \(M\) split the segment \(O_1O_2\) in relation \(r_1/r_2\). Right triangles \(\triangle MO_1T_1\) and \(\triangle MO_2T_2\) are similar. It is clear that the set of centers \(O_1\) and set of \(O_2\) are lines parallel to line \(l\) The set of points \(M\), which divide segment \(O_1O_2\) with the ends at the ratio \(r_1/r_2\) which is a line parallel to \(l\). So the locus of tagents intersection is line parallel to \(l\) and located relatively \(l\) at the distance \(\frac{2r_1r_2}{r_1 + r_2}\).

Problem 8

An ice cube is floating in the glass of ware, there is a peace of lead in it. Ice melted. How will a water level change?

Solution

Since an ice cube with lead has a mass greater than a pure ice cube of the same volume, it is submerged deeper into the water than a pure ice cube, and displaces a larger volume of water than that which will be taken up by the water after ice melted. Therefore, when the ice melts, the water level will drop (a piece of lead will fall to the bottom, but its volume remains the same, and it does not directly change the water level).

Problem 9

A cannot shoot at the angle \(\alpha\). The initial speed of the projectile is \(v\). What distsnce \(L\) from the cannon there is a screen, from which the projectile is reflecting elastically. What distance from the cannot will the projectile fall?

Solution

from sympy import *
x, y, a, x0, y0, v0x, v0y, g, t, L, Lfall, v = symbols('x, y, a, x0, y0, v0x, v0y, g, t, L, Lfall, v')

x = v * cos(a) * t
y = v * sin(a) * t - g * t**2 / 2

vx = v * cos(a)
vy = v * sin(a) - g * t

t_screen = solve(x - L, t)[0]

xL = x.subs(t, t_screen) 
yL = y.subs(t, t_screen) 
vLx = -vx.subs(t, t_screen)
vLy = vy.subs(t, t_screen)

x2 = xL + vLx * t
y2 = yL + vLy * t - g * t**2 / 2

solve((x2 - Lfall, y2), t, Lfall)[1][1]

## 2*L - v**2*sin(2*a)/g

60_internat

Artem

2022-07-16

Variant for Year IX.

Problem 1.

Solution.

Problem 2.

Solution

Problem 3.

Solution

Problem 4

Solution

Problem 5

Solution

Variant for Year X.

Problem 1.

Problem 2.

Solution

Problem 3

Solution

Problem 4

Solution

Problems on oral exams

Problem 1

Solution

Problem 2

Solution

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6.

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution