Let $S$ be circumcenter of triangle $ABC$ and let angle bisector at $A$ cut circumcircle of triangle $ABC$ at $A'$. Since $A,M,N$ and $S$ are concyclic with $\angle MSA = 90^{\circ}$ it is enought to prove that $I$ bisect $AA'$. But this could be easly proved by Ptolomy: $AA'\cdot BC = AB\cdot A'C+AC\cdot A'B$ Since $A'I = A'B =A'C$ this follows.

From $BC= \frac{AB+AC}{2}$ we have that $BC = BM+CN$. Let $E$ be the point in $BC$ such that $BM= CE$ and $CN = CE$. Since $BI$ and $CI$ are angle bisectors of angles $\angle ABC$ and $\angle ACB$, respectively, and triangles $BME$ and $CNE$ are isosceles we have that $BI$ and $CI$ are the perpendicular bisectors of $ME$ and $NE$, respectively. Hence $I$ is the circumcenter of triangle $MEN$. Therefore, since $I$ lies on the perpendicular bisector of $MN$ and on the angle bisector of $\angle MAN$ we have that $AMIN$ is cyclic.

Let AI meets BC at D and the circumcircle of ABC at K. We have AB/BD = c/(ac/(b+c)) = 2. Since ABD is similar to AKC, then AK=2KC. This also implies AK=2KI. The homothety centered at A of ratio 1/2 brings K,B,C to I,M,N, and also the circumcircle of ABC to the circle passing through A,M,I,N, so they are concyclic, as desired.

Btw, I had this problem on my geometry test..

master_Hjom wrote: Btw, I had this problem on my geometry test.. We know that $AIXZ$ is cyclic, where $X,Z$ are the points of perpendiculars from $I$ to $AB$ and $AC$ respectively. So we need to prove is $\angle XIM=\angle ZIN$. From the condition $2BC=AB+AC$ we get that $XM=ZN$. Thus $\triangle INZ= \triangle IMX$. So we get that $\angle XIM=\angle ZIN$, from which follows that $AMIN$ is cyclic.

The problem is equivalent to prove that $IM=IN$. i.e.$IM^2=IN^2$ Use barycentric coordinate and if we compare it remains to prove that $S_B(s-2b)=S_C(s-2c)$ where $S_B=\frac{1}{2}(a^2+c^2-b^2)$ and like wise $S_C$. If we calculate $\frac{S_B}{S_C}=\frac{5c-3b}{5b-3c}$ which is nothing but $\frac{s-2c}{s-2b}$. [While calculating we use the fact $a=\frac{b+c}{2}$] Hence proved.

Let line AI intersect (ABC) at A and P. It follows that BP = CP = IP = r by the incenter excenter lemma. Applying a homothety about A with scale factor 2 maps M to B and N to C. If AMIN were cyclic, the homothety would also map I to P. So it is sufficient to prove that AI = IP, or AP = 2r. Let AB = c, AC = b, and BC = a. The problem's condition tells us that 2a = b+c. Applying Ptolemy's on ABPC gives: $$AB \cdot PC + BP \cdot AC = AP \cdot BC \rightarrow cr+br=a \cdot AN \rightarrow AN = 2r$$as desired.

Dear Mathlinkers, I being the midpoint of AA'...we are done by homothety... Sincerely Jean-Louis

MAN IS MANI We use barycentric coordinates, with $ABC$ as the reference triangle. We have $M = \left(\frac{1}{2}, \frac{1}{2}, 0\right), N = \left(\frac{1}{2}, 0, \frac{1}{2}\right), I = (a:b:c)$. Consider the circumcircle of $AMN$. We must show that $I$ also lies on this circle. Plugging in each of the points $A, M, N$ into the circle formula, we get $u=0, w=\frac{b^2}{2}, v=\frac{c^2}{2}$. The equation of $(AMN)$ is $$-a^2yz-b^2zx-c^2xy+\left(\frac{c^2y+b^2z}{2}\right)(x+y+z)=0.$$Now, plugging in $(a:b:c)$, we have \begin{align*} -a^2bc-ab^2c-abc^2+\left(\frac{bc^2+b^2c}{2}\right)(x+y+z)=0 \\ abc(a+b+c)=\frac{(bc^2+b^2c)(a+b+c)}{2} \\ \frac{bc^2+b^2c}{2}=abc \\ a=\frac{1}{2}(b+c), \\ \end{align*}which is clear from the problem statement, so we are done. $\blacksquare$

Literally same as @above, but posting for storage. Clearly $M,N$, the midpoints of $AB,AC,$ in addition to the circumcenter $O$ and $A$ are concyclic with diameter $OA.$ Thus, it suffices to show that $I$ lies on $(AMN)$. The general equation of a circle is $$-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0$$for some constants $u,v,w$. To find the constants $u,v,w$ which determine $\Omega$, we can plug in each of $A,M,N$ in the general form and solve for $u,v,w$. Note that the equation of a circle is homogenous so we can use homogenized coordinates here. Plugging in $A=(1,0,0)$ gives $u=0$. Plugging in $M=(1:0:1)$ gives $u+w=\frac{b^2}2$. Plugging in $N=(1:1:0)$ gives $u+v=\frac{c^2}2$. Plugging in these constants in the general equation, we find that $\Omega$ has equation $$-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{c^2}2\cdot y+\frac{b^2}2\cdot z\right)=0.$$Hence, using Barycentric Power of a Point, we can compute $$\text{Pow}_{\Omega}(I)=-(a+b+c)(abc)+(a+b+c)\left(\frac{c^2}2\cdot b+\frac{b^2}2\cdot c\right)$$since $I=(a:b:c)$. But this is $0$ since $a=\frac{b+c}2$. Done.