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 O₁. 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 S₁. Since FS:FA=FS₁:FI=FO₁:FA'=1:3, the points S, S₁ and O₁ lie on the same line. In other words, the Euler line of triangle BCI, which is O₁S₁, passes through points S, the centroid of ABC.

Solution 2. Let O₁, O₂ and O₃ be the circumcenters of triangles BCI, CAI and ABI, respectively, and let M₁, M₂, M₃ be the orthocenters. The lines O₁O₂, O₂O₃, O₃O₁ 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 O₁O₂O₃ 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 O₁O₂O₃. Due to the symmetry, it is sufficient to deal with one of the triangles, say BCI. Since BO₁=IO₁=CO₁, the lines O₁O₂ and O₁O₃ are angle bisectors in BO₁I$\displaystyle \angle$ and IO₁C$\displaystyle \angle$, and thus BO₁C$\displaystyle \angle$=2O₃O₁O₂$\displaystyle \angle$=2^.60^o=120^o.

Figure 3

Let point U be the intersection of lines BI and CM₁, and let point V be the intersection of CI and BM₁. From the quadrilateral M₁VIU we obtain CM₁B$\displaystyle \angle$=60^o. The quadrilateral M₁BO₁C is cyclic, since CM₁B$\displaystyle \angle$+BO₁C$\displaystyle \angle$=60^o+120^o=180^o. Then, from BO₁=O₁C we obtain CM₁O₁$\displaystyle \angle$=O₁M₁B$\displaystyle \angle$=30^o.

Finally M₁O₁O₂$\displaystyle \angle$= O₁M₁B$\displaystyle \angle$ and O₃O₁M₁$\displaystyle \angle$=CM₁O₁$\displaystyle \angle$, thus M₁O₁O₂$\displaystyle \angle$=O₃O₁M₁$\displaystyle \angle$=30^o. Hence, the Euler line of triangle BCI, which is O₁M₁, is the angle bisector of O₃O₁O₂=$\displaystyle \angle$ and passes through the center of triangle O₁O₂O₃.

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$\displaystyle \ne$ $\emptyset$ and n $\ge$ 4 (If n $\le$ 2 and N= $\emptyset$ , then the statement fails since $\root3\of{6n}>n$ .)

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 $\root3\of{6n}$ elements. Then any maximal good subset proves the statement.

Let M be an arbitrary maximal good subset of A, and |M|=k. Clearly k $\ge$ 3, because all 3-element subsets are good. If $x\in A\setminus M$ is an arbitrary element, then the set M $\cup$ {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 $x,y\in A\setminus M$ are two different elements, then the sets H₁=h(x) $\cup$ {x} and H₂=h(y) $\cup$ {y} are elements of N. These two elements are different, because, for example, x $\in$ H₁ but $x\not\in H_2$ . Then H₁ and H₂ have at most 2 common elements. This implies h(x) $\ne$ h(y).

Consider now all triplets h(x). We have n-k such triplets, they are pairwise different. The set M has ${k\choose3}$ 3-element subsets, thus $n-k\le{k\choose3}$ . From this inequality we have

$n-k\le{k\choose3}$

6n $\le$ k³-3k²+8k=k³-k(3k-8)<k³

$k>\root3\of{6n}.$

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