31.

Let x, y, z be positive real numbers for which x² + y² + z² = 1. Find the minimum value of

Solution 1. [S. Niu] Let a = yz/x, b = zx/y and c = xy/z. Then a, b, c are positive, and the problem becomes to minimize S = a + b + c subject to ab + bc + ca = 1. Since

Solution 2. We have that

32.

The segments BE and CF are altitudes of the acute triangle ABC, where E and F are points on the segments AC and AB, respectively. ABC is inscribed in the circle Q with centre O. Denote the orthocentre of ABC by H, and the midpoints of BC and AH be M and K, respectively. Let ÐCAB = 45^°.

(a) Prove, that the quadrilateral MEKF is a square.

(b) Prove that the midpoint of both diagonals of MEKF is also the midpoint of the segment OH.

(c) Find the length of EF, if the radius of Q has length 1 unit.

Consider a 90^° rotation with centre F that takes H ® B. Then FA ® FC, FH ® FB, so DFHA ® DFBC and K ® M. Hence FK = FM and ÐKFM = 90^°.

But FK = KE and FM = ME, so MEKF is an equilateral quadrilateral with one right angle, and hence is a square.

(b) Consider a 180^° rotation (half-turn) about the centre of the square. It takes K « M, F « E and H « H¢. By part (a), DFHA º DFBC and AH ^BC. Since KH || MH¢ (by the half-turn), MH¢^BC. Since AH = BC, BM = ¹/₂BC = ¹/₂AH = KH = MH¢, so that BMH¢ is a right isosceles triangle and ÐCH¢M = ÐBH¢M = 45^°. Thus, ÐBH¢C = 90^°. Since ÐBAC = 45^°, H¢ must be the centre of the circle through ABC. Hence H¢ = O. Since O is the image of H by a half-turn about the centre of the square, this centre is the midpoint of OH¢ as well as of the diagonals.

(c) |EF | = Ö2 |FM | = Ö2 |BM | = |OB | = 1.

Solution 2. [M. Holmes] (a) Consider a Cartesian plane with origin F (0, 0) and x-axis along the line AB. Let the vertices of the triangle be A (-1, 0), C (0, 1), B (b, 0). Since the triangle is acute, 0 < b < 1. The point E is at the intersection of the line AC (y = x + 1) and a line through B with slope -1, so that E = ( ¹/₂(b - 1),¹/₂(b + 1). H is the intersection point of the lines BE and CF, so H is at (0, b); K is the midpoint of AH, so K is at (-¹/₂, ^b/₂); M is the midpoint of BC, so M is at (^b/₂,¹/₂). It can be checked that the midpoints of EF and KM are both at (¹/₄(b-1), ¹/₄(b+1)). The slope of EF is (b+1)/(b-1) and that of KM is the negative reciprocal of this, so that EF ^KM. It is straightforward to check that the lengths of EF and KM are equal, and we deduce that EKFM is a square.

(b) O is the intersection point of the right bisectors of AB, AC and BC. The line x + y = 0 is the right bisector of AC and the abscissae of points on the right bisector of BC are all ¹/₂(b - 1). Hence O is at (¹/₂(b-1), ¹/₂(1 - b)). It can be checked that the midpoint of OH agress with the joint midpoint of EF and KM.

(c) This can be checked by using the coordinates of points already identified.

Comment. One of the most interesting theorems in triangle geometry states that for each triangle there exists a circle that passes through the following nine special points: the three midpoints of the sides; the three intersections of sides and altitudes (pedal points); and the three midpoints of the segments connecting the vertices to the orthocentre. This circle is called the nine-point circle. If H is the orthocentre and O is the circumcentre, then the centre of the nine-point circle is the midpoint of OH. Note that in this problem, the points M, E, F, K belong to the nine-point circle.

33.

Prove the inequality a² + b² + c² + 2abc < 2, if the numbers a, b, c are the lengths of the sides of a triangle with perimeter 2.

Solution 2. [L. Hong] The perimeter of the triangle is a + b + c = 2. We have that

34.

Each of the edges of a cube is 1 unit in length, and is divided by two points into three equal parts. Denote by K the solid with vertices at these points.

(a) Find the volume of K.

(b) Every pair of vertices of K is connected by a segment. Some of the segments are coloured. Prove that it is always possible to find two vertices which are endpoints of the same number of coloured segments.

(b) The polyhedron has 3 ×8 = 24 vertices. Each edge from a given vertex is joined to 23 vertices. The possible number of coloured segments emanating from a vertex is one of the twenty-four numbers, 0, 1, 2, ¼, 23. But it is not possible for one vertex to be joined to all 23 others and another vertex to be joined to no other vertex. So there are in effect only 23 options for the number of coloured segments emanating from each of the 24 vertices. By the Pigeonhole Principle, there must be two vertices with the same number of coloured segments emanating from it.

35.

There are n points on a circle whose radius is 1 unit. What is the greatest number of segments between two of them, whose length exceeds Ö3?

Recall Turan's theorem (see the solution of problem 23 in Olymon 1:4: Let G be a graph with n vertices. Denote by l(G) the number of its edges and t(G) the number of triangles contained in the graph. If t(G) = 0, then l(G) £ ën²/4 û. >From this theorem, it follows that the number of segments with chords exceeding Ö3 is at most ën²/4 û.

To show that this maximum number can be obtained, first construct points A, B, C, D on the circle, so that the disjoint arcs AB and CD subtend angles of 120^° at the centre. If n = 2k+1 is odd, place k points on the arc BC and k+1 points on the arc DA. Any segment containing a point in BC to a point in DA must subtend an angle exceeding 120^°, so its length exceeds Ö3. There are exactly k(k+1) = ë(2k+1)²/4 û such segments. If n = 2k is even, place k points in each of the arcs BC and CA, so that there are exactly k² = ë(2k)²/4 û such segments. In either case, the maximum number of segments whose length exceeds Ö3 is ën²/4û.

36.

Prove that there are not three rational numbers x, y, z such that

Solution. Suppose the x = u/m, y = v/m and z = w/m, where m is the least common multiple of the denominators of x, y and z. Then, multiplying the given equation by m² yields