The First Moscow Olympiad

The first Moscow Math Olympiad was conducted in 1935. Today, eighty years later, I decided to check it out. Most of the problems look standard, but some of the stereometry problems look too complicated. I found four problems that I really like: all of them are geometry problems.

Problem 1. The lengths of the sides of a triangle form an arithmetic progression. Prove that the radius of the inscribed circle is one third of one of the heights of the triangle.

Problem 2. A median, bisector, and height all originate from the same vertex of a triangle. You are given the three points that are the intersection points of the aforementioned median, bisector, and height with the circumscribed circle. Construct the triangle.

Problem 3. Find the set of points P on the surface of a cube such that the main diagonal subtends the smallest possible angle if viewed from P. Prove that the main diagonal subtends larger angles if viewed from other points on the surface. [Clarification: the two corners the main diagonal passes through don’t count.]

Problem 4. Given three parallel straight lines, construct a square such that three of its vertices belong to these lines.

Each of these problems has a powerful idea that solves it. You can try and solve these problems, but if you want help, the ideas are presented below as hints in a scrambled order.

  • Hint. Rotate by 90 degrees.
  • Hint. Consider a circumscribed sphere.
  • Hint. The line connecting the intersection point of the bisector with the circle and the circle’s center is parallel to the height.
  • Hint. Use Heron’s formula.
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3 Comments

  1. paolo:

    Call sides in arit. progress. (a-b), a, (a+b). now 2S = ((a-b)+a+(a+b))R = 3aR. but 2S = a H_a,
    so that R = H_a/3

  2. paolo:

    prob.4 consider the common perpendicular p to the three parallel lines q, r, s. p meets q, r, s in Q, R, S
    (r is between q, s)
    Now
    take on q a segment QA = RS and
    take on S a segment SB = QR
    on the same side of p.
    angle ARB measures 90 by construction. the fourth vertex C is found in the same way.
    the required square is the one inscribed in a (bigger) square with side QS

  3. paolo:

    sorry. there are two more squares.
    suppose we have already drawed the perpendicular to the three parallel lines.
    now point the drawing compass in Q with aperture QR and rotate until the compass cuts line q point, say, A.
    the segment AS is a side of a square. report AS with the compass and rotate until cuts line r in point, say, B. AS in perpendicular to aB.

    repeat the same procedure on the other end of the perpendicular, i.e. in S taking SR with the compass and
    cutting line s in a point, and going on with the same operations as for the previus case-

    Prob. 3 You must imgine the cube inscribed in a sphere. only the 8 vetices touch the sphere.
    now think about the diagonal as an a rotation axis and a plne passing through it. let rotae the spere
    together with the cube. you see the shape of a section of our cube rotaing, but the angle is the smallest when a vertex touches the sphere (90 degree). so only from 6 points see the diagonal under the smallest angle.

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