Find all positive integers $x,y$ satisfying the equation \[ 9(x^2+y^2+1) + 2(3xy+2) = 2005 . \]

Click for solution The equation is equivalent to :$9[(x+y)^2-2xy+1]+2(3xy+2)=2005$ Now we put $x+y=k$ $xy=l$ and this is equivalent $9k^2-18l+9+6l+4=2005$ <=> $3k^2=4(l+166)$ => $k=even$. The above is equivalent to $2k^2+k^2-4l-664=0$ (1) . The number $k^2-4l$ is the discriminant of $a^2-ka+l=0$ and to have roots it should be $\geq 0$. So $k^2-4l\geq 0$. So from (1) we have that $2k^2\leq 664$ <=> $k\leq 18,..$ so $k\leq 18$ because $k$ is an integer. Now if $k=18$ then from (1) we have that $l=77$ and $x=7$ or $11$ or $y=7$ or $11$. For $k=16$ we have $l=1044$ and discriminant$<0$ ,no roots If $k\leq 14$ , $3k^2<664$ and so $l<0$ absurb. So the only roots are $x=7$ or $11$,$y=7$ or $11$

Let $ABC$ be an acute-angled triangle inscribed in a circle $k$. It is given that the tangent from $A$ to the circle meets the line $BC$ at point $P$. Let $M$ be the midpoint of the line segment $AP$ and $R$ be the second intersection point of the circle $k$ with the line $BM$. The line $PR$ meets again the circle $k$ at point $S$ different from $R$. Prove that the lines $AP$ and $CS$ are parallel.

Click for solution Consider an inversion of pole $P$ which invariates circle $k$. Consequently, $SC$ will transform into the circle circumscribed to $\triangle{BRP}$. Since $MA^2=MR\cdot{MB}=MP^2$ we get that the line $AP$ is tangent to the aforementioned circle and we are done!

Prove that there exist (a) 5 points in the plane so that among all the triangles with vertices among these points there are 8 right-angled ones; (b) 64 points in the plane so that among all the triangles with vertices among these points there are at least 2005 right-angled ones.

Click for solution a) Choose the four vertices of a square and its center b) Just consider a $8 \times 8$ grid. For each point, say $M$ of the grid, there are $7$ other points in the same column and $7$ in the same row. Thus, there are at least $49$ right-angled triangles with right-angle in $M$. Thus, there are at least $49 \times 64 =3136$ right-angled triangles. And much more, if we consider the triangles whose sides re not parallel to the sides of the grid... Pierre.

Find all 3-digit positive integers $\overline{abc}$ such that \[ \overline{abc} = abc(a+b+c) , \] where $\overline{abc}$ is the decimal representation of the number.

Click for solution The condition is equivalent to: $100a+10b+c=abc(a+b+c)$ <=> $9(11a+b)=(a+b+c)(abc-1)$ Now the cases: i) if $a+b+c=9k$ with $k\leq 3$ .If $a+b+c=9$ the solutions are $a=1,b=3,c=5$ and another solution $a=1,b=c=4$. If $a+b+c=18$ there are no solutions. If $a+b+c=27$ then again there exist no solutions. ii) if $abc-1=9m$, then again there are no solutions iii) if $a+b+c=3n$, $n\leq 8$ and the same time $abc-1=3l$. After all the tests and tries we have one more solution for $a=1,b=c=4$. So there exist two numbers. The numbers $135$ and $144$. This is a problem very boring why you always try and test cases