Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Solutions for problems "A" in September, 2001

In this page only the sketch of the solutions are published; in some cases only the final results. To achieve the maximum score in the competition more detailed solutions needed.


A. 269. A round hole is to be completely covered with two square boards. The sides of the squares are 1 metre. In what interval may the diameter of the hole vary?

Solution. It is easy to see that a hole of radius \(\displaystyle 2-\sqrt2\) (and all smaller holes) can be covered by two unit squares, see Figure 1.

Figure 1

We will prove that if a unit square covers at least half of the perimeter of a circle of radius r, then \(\displaystyle r\le2-\sqrt2\). This implies that the diameter of the hole cannot exceed \(\displaystyle 4-2\sqrt2\approx1.17\) metre.

Let K be a circle of radius r, and N be a unit square, which covers at least the half of K. If r\(\displaystyle ge\)1/2, then N can be translated such that each side of N or its extension has at least one common point with K and the covered part of the curve is increasing, see Figure 2.

Figure 2

So, it can be assumed that r>1/2 and (the extension of) each side of N has a common point with K.

First, consider the case when a vertex of N is inside K. Let the vertices of N be A, B, C, D and suppose that A is inside K. Denote the intersecion of K and the half-lines AB and AD by P and Q, respectively. (Figure 3.)

Figure 3

Because PAQ\(\displaystyle angle\)=90o, the centre of K and vertex A are on the same side of line PQ. This implies that the uncovered arc PQ is longer than the half of K. This is a contradiction, thus this case is impossible.

We are left to deal the case when there are no vertices of N inside K and each side or its extension has a common point with K. It is easy to check that these common points cannot be on the extension of the sides of N; they must lie on the perimeter of N.

Denote by \(\displaystyle alpha\), \(\displaystyle beta\), \(\displaystyle gamma\) and \(\displaystyle delta\) the angles drawn in Figure 4.

Figure 4

The total of the uncovered arcs is at most the half of K, so \(\displaystyle alpha\)+\(\displaystyle beta\)+\(\displaystyle gamma\)+\(\displaystyle delta\)le90o. By symmetry, it can be assumed that alpha+gammale45o.

As can be read from the Figure, r(cosalpha+cosgamma)=r(cosbeta+cosdelta)=1. The concavity of the cosine function yields

1=r(\cos\alpha+\cos\gamma)\ge r\big(1+\cos(\alpha+\gamma)\big)\ge r(1+\cos45^\circ)=r\left(1+{\sqrt2\over2}\right),

and hence r\le{1\over1+\sqrt2/2}=2-\sqrt2.

Thus, the radius of the hole can be at most 2-\sqrt2; its diameter can be at most 4-2\sqrt2\approx1.17 metre.


A. 270. Prove that if a, b, c, d are positive numbers then

(1)\root3\of{abc+abd+acd+bcd\over4}\le\sqrt{ab+ac+ad+bc+bd+cd\over6}

Solution 1. Define the folowing symmetric expressions of a, b, c, d:

S33=a3b3+a3c3+a3d3+b3c3+b3d3+c3d3;

S321=a3b2c+a3b2d+...+bc2d3;

S3111=a3bcd+ab3cd+abc3d+abcd3;

S222=a2b2c2+a2b2d2+a2c2d2+b2c2d2;

S2211=a2b2cd+a2bc2d+a2bcd2+ab2c2d+ab2cd2+abc2d2.

Taking the 6th power of (1), we obtain

(2)2S33+6S321+12S3111ge15S222+24S2211.

By the A.M.-G.M. inequality

2S33=(a3b3+a3c3+b3c3)+(a3b3+a3d3+b3d3)+

+(a3c3+a3d3+c3d3)+(b3c3+b3d3+c3d3)ge

ge3a2b2c2+3a2b2d2+3a2c2d2+3b2c2d2=3S222.

We similarly obtain that S321ge6S222, S321ge4S2211 and 3S3111ge2S2211. Thus

2S33+6S321+12S3111=2S33+2.S321+4.S321+4.3S3111ge

ge3S222+2.6S222+4.4S2211+4.2S2211=15S222+24S2211.

Solution 2. The polynomial f(x)=(x-a)(x-b)(x-c)(x-d) has four positive (not necessariliy different) roots. Thus, its derivative has three positive roots; denote them by u, v and w.

By Viete's formulae,

f'(x)=4x3-3(a+b+c+d)x2+

+2(ab+ac+ad+bc+bd+cd)x-(abc+abd+acd+bcd)=

=4(x3-(u+v+w)x2+(uv+uw+vw)x-uvw).

Thus

{abc+abd+acd+bcd\over4}=uvw

and

{ab+ac+ad+bc+bd+cd\over6}={uv+uw+vw\over3}.

Substituting these in (1), it becomes the inequality between the arithmetic and geometric means of uv, uw and vw.


A. 271. Prove that for any prime, pge5, the number

(1)\sum_{0<k<{2p\over3}}{p\choose k}

is divisible by p2.

Vojtech Jarnik Mathematics Competition, Ostrava, 2001

Solution. Each term of (1) is divisible by p, so our goal is to prove that

(2)\sum_{0<k<{2p\over3}}{1\over p}{p\choose k}

is divisible by p.

For an arbitrary integer 1lex<p, let x-1 be the multiplicative inverse of x modulo p.

For all 0<k<p,

{1\over p}{p\choose k}={(p-1)(p-2)\dots(p-k+1)\over1\cdot2\cdot\dots\cdot k}\equiv

equiv(-1).(-2).....(-k+1).1-1.2-1.....k-1equiv(-1)k.k-1 (mod p),

and

\sum_{0<k<{2p\over3}}{1\over p}{p\choose k}\equiv\sum_{0<k<{2p\over3}}(-1)^k\cdot k^{-1}=

=\sum_{0<k<{2p\over3}}k^{-1}-2\sum_{0<k<{p\over3}}(2k)^{-1}\equiv

\equiv\sum_{0<k<{2p\over3}}k^{-1}-\sum_{0<k<{p\over3}}{k}^{-1}=\sum_{{p\over3}<k<{2p\over3}}k^{-1}=

=\sum_{{p\over3}<k<{p\over2}}(k^{-1}+(p-k)^{-1})\equiv\sum_{{p\over3}<k<{p\over2}}(k+(p-k))\cdot k^{-1}\cdot(p-k)^{-1}=

=\sum_{{p\over3}<k<{p\over2}}p\cdot k^{-1}\cdot(p-k)^{-1})\equiv0\pmod p.