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 $$r_1^2+r_2^2=|S_1{S}_2{|}^2$$ 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 $△IBD$, 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, $A^{\prime }=P$, $B^{\prime }=Q$ and $C^{\prime }=R$. Therefore the radius of the circumcircle of $△A^{\prime }{B}^{\prime }{C}^{\prime }$ is half the radius of $k$. ☐

Let $k(I,r)$ be the incircle of $△ABC$. 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 $△IQD\sim △IDB$ and $△IRD\sim △IDC$, 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, $A^{\prime }=P,B^{\prime }=Q$ and $C^{\prime }=R$. Therefore the radius of the circumcircle of $△A^{\prime }{B}^{\prime }{C}^{\prime }$ is half the radius of $k$. $\square$