## Solutions for advanced problems "A" in September, 2003 |

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. 323.** *I* is the isogonic point of a triangle *ABC* (the point in the interior of the triangle for which \(\displaystyle \angle\)*AIB*=\(\displaystyle \angle\)*BIC*=\(\displaystyle \angle\)*CIA*=120^{o}). Prove that the Euler lines of the triangles *ABI*, *BCI* and *CAI* are concurrent.

**Solution 1 (Rácz Béla András, Budapest).** We shall prove that the three Euler lines pass through the centroid of triangle *ABC*. By the symmetry, it is sufficient to prove this for the Euler line of triangle *BCI*.

Figure 1

Draw a regular triangle onto side *BC* outside; denote its third vertex by *A*', its center by *O*_{1}. The quadrilateral *IBA*'*C* is cyclic since *BA*'*C*\(\displaystyle \angle\)+*CIB*\(\displaystyle \angle\)=60^{o}+120^{o}=180^{o}. Due to *A*'*B*=*A*'*C*, line *A*'*I* is the angle bisector of \(\displaystyle \angle\)*CIB* and this is the same line as *AI* (see Figure 1).

Denote the midpoint of *BC* by *F*, the centroid of *ABC* by *S* and the centroid of *BCI* by *S*_{1}. Since *FS*:*FA*=*FS*_{1}:*FI*=*FO*_{1}:*FA*'=1:3, the points *S*, *S*_{1} and *O*_{1} lie on the same line. In other words, the Euler line of triangle *BCI*, which is *O*_{1}*S*_{1}, passes through points *S*, the centroid of *ABC*.

**Solution 2.** Let *O*_{1}, *O*_{2} and *O*_{3} be the circumcenters of triangles *BCI*, *CAI* and *ABI*, respectively, and let *M*_{1}, *M*_{2}, *M*_{3} be the orthocenters. The lines *O*_{1}*O*_{2}, *O*_{2}*O*_{3}, *O*_{3}*O*_{1} are the perpendicular bisectors of *CI*, *AI* and *BI*. Summing up the angles of the three shaded quadrilaterals in Figure 2, it can be obtained that all angles of triangle *O*_{1}*O*_{2}*O*_{3} are 60^{o}; this triangle is equilateral.

Figure 2

We shall show that the Euler lines of triangles *ABI*, *BCI* and *CAI* pass through the center of triangle *O*_{1}*O*_{2}*O*_{3}. Due to the symmetry, it is sufficient to deal with one of the triangles, say *BCI*. Since *BO*_{1}=*IO*_{1}=*CO*_{1}, the lines *O*_{1}*O*_{2} and *O*_{1}*O*_{3} are angle bisectors in *BO*_{1}*I*\(\displaystyle \angle\) and *IO*_{1}*C*\(\displaystyle \angle\), and thus *BO*_{1}*C*\(\displaystyle \angle\)=2*O*_{3}*O*_{1}*O*_{2}\(\displaystyle \angle\)=2^{.}60^{o}=120^{o}.

Figure 3

Let point *U* be the intersection of lines *BI* and *CM*_{1}, and let point *V* be the intersection of *CI* and *BM*_{1}. From the quadrilateral *M*_{1}*VIU* we obtain *CM*_{1}*B*\(\displaystyle \angle\)=60^{o}. The quadrilateral *M*_{1}*BO*_{1}*C* is cyclic, since *CM*_{1}*B*\(\displaystyle \angle\)+*BO*_{1}*C*\(\displaystyle \angle\)=60^{o}+120^{o}=180^{o}. Then, from *BO*_{1}=*O*_{1}*C* we obtain *CM*_{1}*O*_{1}\(\displaystyle \angle\)=*O*_{1}*M*_{1}*B*\(\displaystyle \angle\)=30^{o}.

Finally *M*_{1}*O*_{1}*O*_{2}\(\displaystyle \angle\)= *O*_{1}*M*_{1}*B*\(\displaystyle \angle\) and *O*_{3}*O*_{1}*M*_{1}\(\displaystyle \angle\)=*CM*_{1}*O*_{1}\(\displaystyle \angle\), thus *M*_{1}*O*_{1}*O*_{2}\(\displaystyle \angle\)=*O*_{3}*O*_{1}*M*_{1}\(\displaystyle \angle\)=30^{o}. Hence, the Euler line of triangle *BCI*, which is *O*_{1}*M*_{1}, is the angle bisector of *O*_{3}*O*_{1}*O*_{2}=\(\displaystyle \angle\) and passes through the center of triangle *O*_{1}*O*_{2}*O*_{3}.

**A. 324.** Prove that if *a*,*b*,*c* are positive real numbers then

\(\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc}. \)

**Solution 1.** The inequality can be transformed to this one:

*ab*(*b*+1)(*ca*-1)^{2}+*bc*(*c*+1)(*ab*-1)^{2}+*ca*(*a*+1)(*bc*-1)^{2}\(\displaystyle \ge\)0.

**Solution 2 (Tamás Birkner, Budapest).** Multiply the inequality by (1+*abc*) and add 3 to both sides.

\(\displaystyle \left({1+abc\over a(1+b)}+1\right)+ \left({1+abc\over b(1+c)}+1\right)+ \left({1+abc\over c(1+a)}+1\right) \ge6\)

\(\displaystyle \left({1+a\over a(1+b)}+{b(1+c)\over1+b}\right)+ \left({1+b\over b(1+c)}+{c(1+a)\over1+c}\right)+ \left({1+c\over c(1+a)}+{a(1+b)\over1+a}\right)\ge6\)

\(\displaystyle \left({1+a\over a(1+b)}+{a(1+b)\over1+a}\right)+ \left({1+b\over b(1+c)}+{b(1+c)\over1+b}\right)+ \left({1+c\over c(1+a)}+{c(1+a)\over1+c}\right)\ge6.\)

In the last row, each parentheses have the form \(\displaystyle x+{1\over x}\) and is at least 2.

**Solution 3 (Dobrovolska Galyna, Kiev).** Multiplying by (1+*abc*),

\(\displaystyle {1\over a(1+b)}+{abc\over a(1+b)}+ {1\over b(1+c)}+{abc\over b(1+c)}+ {1\over c(1+a)}+{abc\over c(1+a)}\ge3\)

\(\displaystyle {1\over a(1+b)}+{ab\over1+a}+ {1\over b(1+c)}+{bc\over1+b}+ {1\over c(1+a)}+{ca\over1+c}\ge3\)

\(\displaystyle {1\over a}\cdot{1\over1+b}+b\cdot{1\over{1+{1\over a}}}+ {1\over b}\cdot{1\over1+c}+c\cdot{1\over{1+{1\over b}}}+ {1\over c}\cdot{1\over1+a}+a\cdot{1\over{1+{1\over c}}}\ge6. \eqno(1)\)

Since the pairs \(\displaystyle {1\over a}\), *b* and \(\displaystyle {1\over1+{1\over a}}\), \(\displaystyle {1\over1+b}\) are ordered oppositely,

\(\displaystyle {1\over a}\cdot{1\over1+b}+b\cdot{1\over{1+{1\over a}}} \ge{1\over a}\cdot{1\over{1+{1\over a}}}+b\cdot{1\over1+b} ={1\over1+a}+{b\over1+b}. \eqno(2)\)

Similarly,

\(\displaystyle {1\over b}\cdot{1\over1+c}+c\cdot{1\over{1+{1\over b}}} \ge{1\over1+b}+{c\over1+c},\eqno(3)\)

and

\(\displaystyle {1\over c}\cdot{1\over1+a}+a\cdot{1\over{1+{1\over c}}} \ge{1\over1+c}+{a\over1+a}.\eqno(4)\)

Summing up (2), (3) and (4), we obtain (1).

**A. 325.** We have selected a few 4-element subsets of an *n*-element set *A*, such that any two sets of four elements selected have at most two elements in common. Prove that there exists a subset of *A* that has at least \(\displaystyle \root3\of{6n}\) elements and does not contain any of the selected 4-tuples as a subset.

**Solution.** Let *N* be the set of all selected 4-element subsets. Assume *N*\(\displaystyle \ne\) and *n*4 (If *n*2 and *N*=, then the statement fails since .)

Call a subset of *A* ``good'' if no element of *N* is its subset. Trivially, the subsets of 3 or less elements, including the empty set, are all good. A good subset is called maximal if it has no good superset. There exists at least one maximal good subset, because starting from a good subset, new elements can be added one by one while it is possible.

We shall prove that each maximal good subset has more than elements. Then any maximal good subset proves the statement.

Let *M* be an arbitrary maximal good subset of *A*, and |*M*|=*k*. Clearly *k*3, because all 3-element subsets are good. If is an arbitrary element, then the set *M*{*x*} is not good. This means that *x* and some 3 elements of *M* form an element of *N*. Denote one of such triplets by *h*(*x*).

If are two different elements, then the sets *H*_{1}=*h*(*x*){*x*} and *H*_{2}=*h*(*y*){*y*} are elements of *N*. These two elements are different, because, for example, *x**H*_{1} but . Then *H*_{1} and *H*_{2} have at most 2 common elements. This implies *h*(*x*)*h*(*y*).

Consider now all triplets *h*(*x*). We have *n*-*k* such triplets, they are pairwise different. The set *M* has 3-element subsets, thus . From this inequality we have

6*n**k*^{3}-3*k*^{2}+8*k*=*k*^{3}-*k*(3*k*-8)<*k*^{3}