Kvant

Rough automatic draft, low quality: expect mistakes and awkward wording. It will be replaced by a better translation. Original →

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ity" to the unusual form of the problem, although recently quite a lot of such problems have been considered in the literature for schoolchildren.

Here is the solution to this problem: if xx satisfies the first inequality, then 5<x<1635 < x < \frac{16}{3}, hence 23<6−x<1\frac{2}{3} < 6 - x < 1 and, therefore, 3<36−x<923 < \frac{3}{6 - x} < \frac{9}{2}. Since π2<3<92<3π2\frac{\pi}{2} < 3 < \frac{9}{2} < \frac{3\pi}{2}, the angle 36−x\frac{3}{6 - x} lies in either the second or third quadrant, and therefore its cosine is negative.

So, we have presented a solution to our problems, some difficult, some easy. You can evaluate their difficulty, and in turn, your strength, by solving problems from the two options below.

Options

  1. Solve the equation
    log⁡x+3(3−1−2x+x2)=12.\log_{x+3}(3 - \sqrt{1 - 2x + x^2}) = \frac{1}{2}.

  2. Solve inequality
    6sin⁡2x2ctg2x⩽2cos⁡2x+3ctg2x.6\sin^2\frac{x}{2}\mathrm{ctg}^2x \leqslant 2\cos^2x + 3\mathrm{ctg}^2x.

  3. At the base of the base SABCSABC lies ΔABC\Delta ABC with an angle C=90∘C=90^\circ and an acute angle 30∘30^\circ. The lateral edges of the prim are inclined to the plane of the base at an angle of 45∘45^\circ, the height of the prim is HH. Find the cross-sectional area of ​​the prim by a plane passing through the point CC parallel to ABAB and dividing the faces ASBASB into parts of equal area.

  4. Prove that if a line connecting opposite vertices of an inscribed quadrilateral passes through the point of intersection of tangents drawn at two other vertices, then the product of opposite sides is equal.

  5. Prove that if x2−5x+6x2−11x+30<0\frac{x^2 - 5x + 6}{x^2 - 11x + 30} < 0, then sin xx and cos xx have different symbols.
    (MGPI named after V.N. Lenin)

  6. In a regular triangular triangle, take the side of the base equal to aa, the angles between the edges at its vertex equal to α\alpha (α⩽90∘\alpha \leqslant 90^\circ). Determine the area of ​​the section drawn through the sides of the base perpendicular to the opposite side edge.

  7. Solve the equation
    lg⁡2x+lg⁡2(x+10)=2lg⁡11.\lg^2 x + \lg^2 (x + 10) = 2\lg 11.

  8. Solve the equation
    sin⁡3x⋅sin⁡x+1=0.\sin 3x \cdot \sin x + 1 = 0.

(Ivanovo Pedagogical)

UNITY IS STRENGTH

“While algebra and geometry each developed in its own way, its progress was slow and its applications limited.

But when these sciences united, they energetically supported each other and quickly marched towards perfection.”

Lagrange
NOT ENEMIES, BUT FRIENDS

“It is a mistake to think that rigor in proof is the enemy of simplicity. For example, you can apply the fact that the strict method is at the same time simpler, easier and more accessible. Every effort in the rigor of rigor is directed towards finding the simplest methods of proof.”

Gilbert
4, 5 and 6

4 identical balls can be placed so that each one touches the other three.

5 identical balls can be placed so that each one touches the other four.

Is it possible to arrange 6 identical unsharpened pencils so that each one touches the other five?