Turkey — Team Selection Test

We will show that $L_A, L_B, L_C$ are the midpoints of the sides, and $K_A, K_B, K_C$ are the tangent points of the incircle to the sides in the triangle $ABC$. Note that the desired equality holds in this case. Indeed, if this is the case, we obtain

$$|L_A K_A| = |BL_A| - |BK_A| = \frac{|AC| - |AB|}{2}$$
($|AB| < |BC| < |CA|$). Similarly, we also get
$$|L_B K_B| = \frac{|BC| - |AB|}{2} \quad \text{and} \quad |L_C K_C| = \frac{|AC| - |BC|}{2},$$

which clearly implies $|K_A L_A| = |K_B L_B| + |K_C L_C|$. Let $L'_A, K'_A \in BC$ be points such that $|BL'_A| = |CL'_A|$ and $IK'_A \perp BC$. Define $K'_B, L'_B, K'_C, L'_C$, similarly. Now we can complete the proof in two different ways:

First Way:

We claim that $H_B K'_A \perp I L'_A$. This completes the proof since it also means $H_C K'_A \perp I L'_A$ by symmetry, which implies $I L'_A \perp H_B H_C$ and $K'_A \in H_B H_C$, so we can conclude that $L_A = L'_A$ and $K_A = K'_A$. Let $P$ be the point of intersection of $H_B K'_A$ and $I L'_A$, and let $S$ be the point on $IH_B$ such that $S L'_A \perp IH_B$. Since $ISK'_A L'_A$ is a cyclic quadrilateral, we find $\overline{P L'_A S} = \overline{I L'_A S} = \overline{I K'_A S}$. Therefore, if we prove that $\overline{I K'_A S} = \overline{K'_A H_B I}$, we get $\overline{P H_B S} = \overline{P H_B I} = \overline{P L'_A S}$, which means $L'_A P S H_B$ is cyclic and $\overline{H_B P L'_A} = \overline{H_S L'_A} = 90^\circ$. As a result, it suffices to show that $\overline{I K'_A S} = \overline{K'_A H_B I}$, which is equivalent to $|IS||IH_B| = |I K'_A|^2 = r^2$ where $r$ is the inradius of the triangle $ABC$. Since $\overline{AIC} = 90^\circ + \frac{\hat{B}}{2}$, the Sine theorem gives $\frac{|IH_B|}{|AC|} = \tan \frac{\hat{B}}{2} = \frac{r}{u - |AC|}$. On the other hand, we have $|IS| = \frac{h_B}{2} - r = \frac{ur}{|AC|} - r$ because $L'_A S \parallel AC$. Therefore, we get $|IS||IH_B| = \frac{|AC|r}{u - |AC|} \frac{r(u - |AC|)}{|AC|} = r^2$, so the result follows.

Second Way:

We claim $H_B H_C$ is the polar line of $L'_A$ with respect to the incircle, which completes the proof easily. Let $N$ be the point on $IC$ such that $BN \perp IC$. Since $|BL'_A| = |CL'_A|$, we have $\overline{NL'_A B} = 2 \cdot \overline{NCB} = \hat{C}$. This implies $NL'_A \parallel AC$, in other words we get $N \in L'_A L'_C$. On the other hand, we obtain $180^\circ - \overline{NK'_C I} = \overline{NBI} = 90^\circ - \overline{IBC} - \overline{ICB} = IK'_C K'_B$ since $N, K'_C, I, B$ are concyclic, so we obtain $N \in K'_B K'_C$.

Now, $N \in K'_B K'_C$ means that $N$ lies on the polar line of $A$, so we can say that $A$ lies on the polar line of $N$. Since $AH_B \perp IN$, we get $AH_B$ is the polar line of $N$, which implies $N$ lies on the polar line of $H_B$. On the other hand, we have $L'_A L'_C \perp IH_B$ and $N \in L'_A L'_C$, so we can conclude that $L'_A L'_C$ is the polar line of $H_B$. Similarly, we obtain $L'_A L'_B$ is the polar line of $H_C$, which implies that $H_B H_C$ is the polar line of $L'_A$, so the result follows.