Bariz - Problem

Source: Bulgaria 2013 NMO p5

Tags: geometry, Triangle, circumradius, symmetry, area of a triangle

parmenides51

28.05.2019 16:46

Consider acute $\triangle ABC$ with altitudes $AA_1, BB_1$ and $CC_1$ ($A_1 \in BC,B_1 \in AC,C_1 \in AB$). A point $C' $ on the extension of $B_1A_1$ beyond $A_1$ is such that $A_1C' = B_1C_1$. Analogously, a point $B'$ on the extension of A$_1C_1$ beyond $C_1$ is such that $C_1B' = A_1B_1$ and a point $A' $ on the extension of $C_1B_1$ beyond $B_1$ is such that $B_1A' = C_1A_1$. Denote by $A'', B'', C''$ the symmetric points of $A' , B' , C'$ with respect to $BC, CA$ and $AB$ respectively. Prove that if $R, R'$ and R'' are circumradiii of $\triangle ABC, \triangle A'B'C'$ and $\triangle A''B''C''$, then $R, R'$ and $R'' $ are sidelengths of a triangle with area equals one half of the area of $\triangle ABC$.

Here is the official solution of this problem, solved by barely 1 out of 40+ contestants. Let $O$ be the circumcentre of $ABC$ and $S$ be its reflection with respect to $AB$. It is well known that $SO = CH$ and $CO \perp A_1B_1$, so $SOHC$ is a parallelogram, $SH \parallel A_1B_1$ and $SH = R$. Next, let $\rho$ be the inradius of $A_1B_1C_1$ and $q$ be its semiperimeter. Then if $T$ is the feet from $S$ to $A_1B_1$, we have $HT = \rho$, $ST = R + \rho$, $TC' = q$ and hence $HC' = \sqrt{\rho^2 + q^2}$. Analogously we get the same expression for $HA'$ and $HB'$, so $H$ is the circumcenter of $A'B'C'$ and $R' = HC'$. On the other hand, $OC" = SC' = \sqrt{(R+\rho)^2 + q^2}$ and so $O$ is the circumcenter of $A"B"C"$ and $R" = SC"$. Overall, $R$, $R'$, $R"$ are equal to the sides of $HSC'$, which has area $\frac{Rq}{2} = \frac{1}{2}(S_{A_1OB_1C} + S_{B_1OC_1A} + S_{C_1OA_1B}) = \frac{S_{ABC}}{2}$, as desired.