## 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\(\displaystyle angle\)*=90^{o}, 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 \(\displaystyle alpha\)+\(\displaystyle beta\)+\(\displaystyle gamma\)+\(\displaystyle delta\)90^{o}. By symmetry, it can be assumed that +45^{o}.

As can be read from the Figure, *r*(cos+cos)=*r*(cos+cos)=1. The concavity of the cosine function yields

and hence .

Thus, the radius of the hole can be at most ; its diameter can be at most metre.

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

(1) |

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

*S*_{33}=*a*^{3}*b*^{3}+*a*^{3}*c*^{3}+*a*^{3}*d*^{3}+*b*^{3}*c*^{3}+*b*^{3}*d*^{3}+*c*^{3}*d*^{3};

*S*_{321}=*a*^{3}*b*^{2}*c*+*a*^{3}*b*^{2}*d*+...+*bc*^{2}*d*^{3};

*S*_{3111}=*a*^{3}*bcd*+*ab*^{3}*cd*+*abc*^{3}*d*+*abcd*^{3};

*S*_{222}=*a*^{2}*b*^{2}*c*^{2}+*a*^{2}*b*^{2}*d*^{2}+*a*^{2}*c*^{2}*d*^{2}+*b*^{2}*c*^{2}*d*^{2};

*S*_{2211}=*a*^{2}*b*^{2}*cd*+*a*^{2}*bc*^{2}*d*+*a*^{2}*bcd*^{2}+*ab*^{2}*c*^{2}*d*+*ab*^{2}*cd*^{2}+*abc*^{2}*d*^{2}.

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

(2) | 2S_{33}+6S_{321}+12S_{3111}15S_{222}+24S_{2211}. |

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

2*S*_{33}=(*a*^{3}*b*^{3}+*a*^{3}*c*^{3}+*b*^{3}*c*^{3})+(*a*^{3}*b*^{3}+*a*^{3}*d*^{3}+*b*^{3}*d*^{3})+

+(*a*^{3}*c*^{3}+*a*^{3}*d*^{3}+*c*^{3}*d*^{3})+(*b*^{3}*c*^{3}+*b*^{3}*d*^{3}+*c*^{3}*d*^{3})

3*a*^{2}*b*^{2}*c*^{2}+3*a*^{2}*b*^{2}*d*^{2}+3*a*^{2}*c*^{2}*d*^{2}+3*b*^{2}*c*^{2}*d*^{2}=3*S*_{222}.

We similarly obtain that *S*_{321}6*S*_{222}, *S*_{321}4*S*_{2211} and 3*S*_{3111}2*S*_{2211}. Thus

2*S*_{33}+6*S*_{321}+12*S*_{3111}=2*S*_{33}+2^{.}*S*_{321}+4^{.}*S*_{321}+4^{.}3*S*_{3111}

3*S*_{222}+2^{.}6*S*_{222}+4^{.}4*S*_{2211}+4^{.}2*S*_{2211}=15*S*_{222}+24*S*_{2211}.

**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*)=4*x*^{3}-3(*a*+*b*+*c*+*d*)*x*^{2}+

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

=4(*x*^{3}-(*u*+*v*+*w*)*x*^{2}+(*uv*+*uw*+*vw*)*x*-*uvw*).

Thus

and

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, p*5*, the number *

(1) |

*is divisible by p*^{2}.

*Vojtech Jarnik Mathematics Competition*, Ostrava, 2001

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

(2) |

is divisible by *p*.

For an arbitrary integer 1*x*<*p*, let *x*^{-1} be the multiplicative inverse of *x* modulo *p*.

For all 0<*k*<*p*,

(-1)^{.}(-2)^{.}...^{.}(-*k*+1)^{.}1^{-1}^{.}2^{-1}^{.}...^{.}*k*^{-1}(-1)^{k}^{.}*k*^{-1} (mod *p*),

and