this result seems... really convenient.. Basically we are being asked to show that the midpoint of arc $BC$ is the circumcenter, taking a homothety from $(AI)$ to $(AB'C')$ we see we are just being asked to show that $\frac{AM}{\frac 12 AI} = \frac{ (b + c)}{s - a}$. Conveniently write $2AM = AI + AI_A$, so we want to show $\frac{AI_A}{AI} + 1 = \frac{2b +2c }{b + c - a}$. Writing $\frac{2b + 2c}{ b + c - a} = 1 + \frac{a+ b + c}{b + c -a} $ gives the desired equality as $\frac{AI_A}{AI} = \frac{a + b + c}{b + c - a} = \frac{s}{s -a}$, which is obviously true. The result $\frac{AM}{AI} = \frac{b + c }{b + c - a}$ seems way too good to be true.

In how many ways can you fill a 3 × 3 table with the numbers 1 through 9 (each used once) such that all pairs of adjacent numbers (sharing one side) are relatively prime? can anybody please provide its sol its ans is 2016 and can someone please explain how to post problems on aops

Let $D$ be the circumcenter of $ABC$, $E$ the circumcenter of $AB´C´$ and $\omega$ the circumscribed circle of $AB´C´$. We want to prove that $ABCE$ is a cyclic quadrilateral. See that $AB´=AC+CB´=AC+AB, AC´=AB+BC´ \Rightarrow AB´=AC´$. Let $G$ be such that $G=\omega \cap AE$. As a consequence of $AG$ being a diameter of $\omega$ then $A\widehat{B´}G=A\widehat{C´}G=90°$. In addition $cos(B´\widehat{A}G)=\frac{AC´}{AG}= cos(C´\widehat{A}G)= \alpha$. Because $EA$ and $EB´$ are radius of $\omega$ then $EA=EB´$. As a result $B´EA$ is isosceles and $E\widehat{B´}A=\alpha$. We have that $E\widehat{B´}A=\alpha=C´\widehat{A}G, B´C=AB$ and $EA=EB´$ so $B´EA$ and $CEB$ are rotomothetic triangles and $E\widehat{B}C=\alpha=B\widehat{A´}G$ and $ABCE$ is a cyclic quadrilateral.

Attachments:

Let $S$ be the $A$-superpoint of triangle $ABC$. We claim that this is the circumcenter of $AB'C'$. Let the reflection of $B$ in line $AS$ be $B_1$. Obviously, $B_1$ lies on $AC$. Now, as $SC=SB=SB_1$, $S$ lies on the perpendicular bisector of $B_1C$, but $AB_1=AB=CB'$ so the midpoint of $B_1C$ coincides with the midpoint of $AB'$. This implies that $S$ lies on the perpendicular bisector of $AB'$. Similarly, $S$ lies on the perpendicular bisector of $AC'$, as desired.