angle chasing and sine law...

Trigonometry finishes off this question very quickly. Let the center of $O$ be $X$ and let the radii of $O$ and $O_1$ be $R$ and $R_1$ respectively. Let $XP$ intersect $O$ at $D$. We see that $O_1$ is tangential to $O$ iff $2R = 2R_1 + PD = 2AP \tan \frac{\alpha}{2} + AP \tan \frac{\gamma}{2} = R(2 \sin \gamma + \tan \frac{\alpha}{2} + \sin \gamma \tan \frac{\gamma}{2})$. Using the substitution $\tan \frac{\alpha}{2} = t, \tan \frac{\gamma}{2} = u$, this reduces to $tu = \frac{1}{2}$. Now apply the result to $O_2$.

Since the circle tangent to $AB$ (at its midpoint) is also tangent to $AC$( uppose at point $Y$) and to $O$, from the Casey's theorem we get: $$CY=\frac{a-b}{2}\Longleftrightarrow a+c=3b (1)$$ (where $a,b,c$ are the sides of $\triangle{ABC}$) Now if $O_2$ is tangent to $AC$ at $T$ and to $BC$ at $Q$ then, using (1) and the fact that $CT=CQ=\frac{a}{2}$, we obtain: $$CT\cdot{c}=b\cdot{BQ}+a\cdot{AT} (2)$$ However, the only circle which is tangent to $CA,BC$ and the circumcircle of $\triangle{ABC}$ should also satisfy $(2)$ (from the Casey's theorem), therefore $O_2$ is tangent to $O$.

This is another intresting property involving a triangle with $a+c=3b$

Yeah,that triangle has many many good properties.

what a boring solution with an inversion on point a with the AB*AC it is obviously

ooiler wrote: what a boring solution with an inversion on point a with the AB*AC it is obviously Congratulations for a magnificent solution!

ooiler wrote: what a boring solution with an inversion on point a with the AB*AC it is obviously How? I don't understand?

ooiler wrote: what a boring solution with an inversion on point a with the AB*AC it is obviously

Sailor wrote: Since the circle tangent to $AB$ (at its midpoint) is also tangent to $AC$( uppose at point $Y$) and to $O$, from the Casey's theorem we get: $$CY=\frac{a-b}{2}\Longleftrightarrow a+c=3b (1)$$ (where $a,b,c$ are the sides of $\triangle{ABC}$) Now if $O_2$ is tangent to $AC$ at $T$ and to $BC$ at $Q$ then, using (1) and the fact that $CT=CQ=\frac{a}{2}$, we obtain: $$CT\cdot{c}=b\cdot{BQ}+a\cdot{AT} (2)$$ However, the only circle which is tangent to $CA,BC$ and the circumcircle of $\triangle{ABC}$ should also satisfy $(2)$ (from the Casey's theorem), therefore $O_2$ is tangent to $O$. Can you just state "by Casey's Theorem..." Is it well-known enough to use without proof? For example, in APMO #3, Wolstenholme's Theorem need proof! Can someone tell me which theorems you need proof for?

Wait you actually don't need anything complicated (it's all straightforward bashing): *My actual proof's a bit long so I'll just write my thought processes: 1) First you rephrase the problem statement: Triangle ABC is drawn such that the A mixtilinear circle touches AB at the midpoint, P. Prove that the C mixtilinear circle touches BC at the midpoint Q. 2) Then you find a series of equivalent statements. The path I went along used the fact that the tangency pts of mixtilinear circles on the sides of the triangle are bisected by the incenter. Anyways, it suffices to prove that the incenter I bisects C-mixtilinear tangency points $R_2$ and Q, which reduces to $R_2, I, Q$ need to be collinear. 3) After some simplifications you arrive at a+c=3b. This statement is independent of everything I just did and is essentially a much easier problem than the one we started with. 4) Specifically, the new problem states that "the A mixtilinear circle is tangent to AB at the midpoint P. Prove that a+c = 3b". Here, you note that AIP is a right triangle and you length bash. (using [ABC]=rs, and other incircle stuff) This works out in the end and you're done.

It's very easy using bary first note that the problem equivalet to "$\triangle ABC$ is drawn such that the A-mixtilinear circle touches $AB$ at the midpoint, Prove that the C-mixtilinear circle touches $BC$ at the midpoint Let the A-mixtilinear touches $(ABC)$ at $M$ and $E$ the ex-touch point on $BC$ then easitly get $M(-a:\frac{b^2}{s-b}:\frac{c^2}{s-c})$ (using that $AM$ and $AE$ are isogonal but $MP$ is the perpindacular bisetor of $AB$ thus $M$ is the midpoint of arc $ACB$ then $M(-a:b:\frac{c^2}{a-b})$ $b=s-b \implies a+c=3b$ which is symmetric in $A,C$ so this is equivalent to "the C-mixtilinear circle touches $BC$ at the midpoint "

It is obvious that $O_1$ is $A$-mixtilinear circle. The problem is equivalent to proving that if the $A$-mixtilinear circle touches $AB$ at its midpoint, then $C$-mixtilinear circle touches BC at its midpoint. Let $F$, $G$ be touchpoints of $A$-mixtilinear and $C$-mixtilinear with $AC$ and $Q$ touchpoint of $C$-mixtilinear and $BC$. Since $GQ$ and $FP$ intersect each other at middle (the incenter-well-known), we get that FGPQ is parallelogram, thus PQ|| AC. $P$ is midpoint => $Q$ is midpoint.