6-th Czech-Slovak Match

Let $I$ and $r$ denote the center and the radius of circle $k$. Let $D, E$, and $F$ denote the points where $k$ touches $BC, AC$, and $AB$, respectively. Let $P, Q$, and $R$ denote the midpoints of $EF, DF$, and $DE$ respectively. We will use the well known lemma:
LEMMA. The circles to $k_1(S_1, r_1)$ and to $k_2(S_2, r_2)$ are orthogonal if and only if
$$r_1^2 + r_2^2 = |S_1S_2|^2$$
First of all we prove that point $Q$ and $R$ lie on circle $k_a$. Obviously $BDIF$ is a deltoid, so $Q$ is the foot of a perpendicular from point $D$ to $BI$. Thus, applying the first Euclidean's theorem to triangle $\triangle IBD$, we have $|IQ| \cdot |IB| = |ID|^2 = r^2$. Similarly $|IR| \cdot |IC| = r^2$. Thus $|IQ| \cdot |IB| = |IR| \cdot |IC|$, so the points $B, C, R, Q$ lie on a circle which we denote $\gamma_a$.
The points $Q \in IB$ and $R \in IC$, so that point $I$ lies outside the circle $\gamma_a$. By the above relations, we obtain that the power of the point $I$ to the circle $\gamma_a$ is $r^2$, which means that the circles $k$ and $\gamma_a$ are orthogonal. From the uniqueness of $k_a$ it follows that $k_a = \gamma_a$. Thus $k_a$ contains $Q$ and $R$. Similarly $k_b$ contains $P$ and $R$ and $k_c$ contains $P$ and $Q$. Hence, $A' = P$, $B' = Q$ and $C' = R$. Therefore the radius of the circumcircle of $\triangle A'B'C'$ is half the radius of $k$. ☐

Let $k(I, r)$ be the incircle of $\triangle ABC$. Let $D, E$, and $F$ denote the points where $k$ touches $BC, AC$, and $AB$, respectively. Let $P, Q$, and $R$ denote the midpoints of $EF, DF$, and $DE$ respectively. We prove that $k_a$ passes through $Q$ and $R$.
Since $\triangle IQD \sim \triangle IDB$ and $\triangle IRD \sim \triangle IDC$, we obtain $IQ \cdot IB = IR \cdot IC = r^2$. We conclude that $B, C, Q$, and $R$ lie on a single circle $\gamma_a$. Moreover, since the power of $I$ with respect to $\gamma_a$ is $r^2$, it follows for a tangent $IX$ from $I$ to $\gamma_a$ that $X$ lies on $k$ and hence $k$ is perpendicular to $\gamma_a$. From the uniqueness of $k_a$ it follows that $k_a = \gamma_a$. Thus $k_a$ contains $Q$ and $R$. Similarly $k_b$ contains $P$ and $R$ and $k_c$ contains $P$ and $Q$. Hence, $A' = P, B' = Q$ and $C' = R$. Therefore the radius of the circumcircle of $\triangle A'B'C'$ is half the radius of $k$. $\square$