Problem 1740

1. Define Points and Setup:
Let the feet of the altitudes from points $A$ and $C$ to $BC$ and $AB$ be $H_a$ and $H_c$ respectively. Let $H$ be the orthocenter of $\triangle ABC$ and $M$ be the midpoint of $AB$. Define $O_1$ as the intersection of $AH_a$ and the perpendicular bisector of $AB$ (denoted as $OM$).

2. Intersection Point:
We need to prove that $O_1$ lies on $KT$. Using Menelaus' theorem for $\triangle AH_cH$ with points $T \in HH_c$, $K \in AH_c$, and $O_1 \in AH$, we have:
$$O_1 \in KT \iff \left(\frac{HT}{TH_c}\right)\left(\frac{H_cK}{KA}\right)\left(\frac{AO_1}{O_1H}\right) = 1$$

3. **Calculate $\frac{HT}{TH_c}$:**
$$\frac{TH_c}{BH_c} = \tan{\gamma} = \frac{\sin{\gamma}}{\cos{\gamma}}$$
$$\frac{HH_c}{BH_c} = \tan(90^\circ - \alpha) = \frac{\cos\alpha}{\sin\alpha}$$
$$\frac{HT}{TH_c} = 1 - \frac{HH_c}{TH_c} = 1 - \frac{\left(\frac{HH_c}{BH_c}\right)}{\left(\frac{TH_c}{BH_c}\right)} = 1 - \frac{(\cos\alpha)(\cos\gamma)}{(\sin\alpha)(\sin\gamma)} = \frac{(\sin\alpha)(\sin\gamma) - (\cos\alpha)(\cos\gamma)}{(\sin\alpha)(\sin\gamma)}$$
$$\boxed{\frac{HT}{TH_c} = \frac{(\sin\alpha)(\sin\gamma) - (\cos\alpha)(\cos\gamma)}{(\sin\alpha)(\sin\gamma)}}$$

4. **Calculate $\frac{KH_c}{AK}$:**
$$\frac{KH_c}{CH_c} = \tan(\beta - \alpha) = \frac{\sin(\beta - \alpha)}{\cos(\beta - \alpha)} = \frac{\sin\beta\cos\alpha - \sin\alpha\cos\beta}{\cos\alpha\cos\beta + \sin\alpha\sin\beta}$$
$$\frac{AH_c}{CH_c} = \tan(90^\circ - \alpha) = \frac{\cos\alpha}{\sin\alpha}$$
$$\frac{KH_c}{AK} = \frac{KH_c}{AH_c - KH_c} = \frac{\left(\frac{\sin\beta\cos\alpha - \sin\alpha\cos\beta}{\cos\alpha\cos\beta + \sin\alpha\sin\beta}\right)}{\left(\frac{\cos\alpha}{\sin\alpha}\right) - \left(\frac{\sin\beta\cos\alpha - \sin\alpha\cos\beta}{\cos\alpha\cos\beta + \sin\alpha\sin\beta}\right)}$$
$$= \frac{(\sin\alpha)(\sin\beta\cos\alpha - \sin\alpha\cos\beta)}{(\cos\alpha)(\cos\alpha\cos\beta + \sin\alpha\sin\beta) - (\sin\alpha)(\sin\beta\cos\alpha - \sin\alpha\cos\beta)}$$
$$= \frac{\sin\alpha(\sin\beta\cos\alpha - \sin\alpha\cos\beta)}{\cos\beta(\sin^2\alpha + \cos^2\alpha)} = \frac{\sin\alpha(\sin\beta\cos\alpha - \sin\alpha\cos\beta)}{\cos\beta}$$
$$\boxed{\frac{KH_c}{AK} = \frac{\sin\alpha(\sin\beta\cos\alpha - \sin\alpha\cos\beta)}{\cos\beta}}$$

5. **Calculate $\frac{AO_1}{O_1H}$:**
$$\frac{AO_1}{O_1H} = \frac{AM}{MH_c}$$
Since $O_1M \perp AB \perp HH_c$, we have:
$$BH_c = a\sin(90^\circ - \beta) = a\cos\beta$$
$$\frac{AO_1}{O_1H} = \frac{AM}{MH_c} = \frac{\left(\frac{AB}{2}\right)}{\left(\frac{AB}{2}\right) - BH_c} = \frac{\left(\frac{c}{2}\right)}{\left(\frac{c}{2}\right) - a\cos\beta}$$
$$= \frac{2R\sin\gamma}{2R\sin\gamma - 4R\sin\alpha\cos\beta} = \frac{\sin{\gamma}}{\sin\gamma - 2\sin\alpha\cos\beta}$$
$$\boxed{\frac{AO_1}{O_1H} = \frac{\sin{\gamma}}{\sin\gamma - 2\sin\alpha\cos\beta}}$$

6. Combine Results:
$$\left(\frac{HT}{TH_c}\right)\left(\frac{H_cK}{KA}\right)\left(\frac{AO_1}{O_1H}\right) = \left(\frac{(\sin\alpha)(\sin\gamma) - (\cos\alpha)(\cos\gamma)}{(\sin\alpha)(\sin\gamma)}\right)\left(\frac{\sin\alpha(\sin\beta\cos\alpha - \sin\alpha\cos\beta)}{\cos\beta}\right)\left(\frac{\sin{\gamma}}{\sin\gamma - 2\sin\alpha\cos\beta}\right)$$
$$= \left(\frac{(\sin\alpha)(\sin\gamma) - (\cos\alpha)(\cos\gamma)}{\cos\beta}\right)\left(\frac{\sin\beta\cos\alpha - \sin\alpha\cos\beta}{\sin\gamma - 2\sin\alpha\cos\beta}\right)$$

7. Simplify:
Using the identity $\cos\gamma = \cos(180^\circ - \alpha - \beta) = \cos(180^\circ - \alpha)\cos(\beta) - \sin(180^\circ - \alpha)\sin\beta = \sin\alpha\sin\beta - \cos\alpha\cos\beta$, we get:
$$\frac{(\sin\alpha)(\sin\gamma) - (\cos\alpha)(\cos\gamma)}{\cos\beta} = 1$$
Also, using $\sin\gamma = \sin\beta\cos\alpha + \sin\alpha\cos\beta$, we have:
$$\sin\gamma - 2\sin\alpha\cos\beta = (\sin\beta\cos\alpha + \sin\alpha\cos\beta) - 2\sin\alpha\cos\beta = \sin\beta\cos\alpha - \sin\alpha\cos\beta$$
Thus, the second fraction is also $1$.

8. Conclusion:
$$\left(\frac{HT}{TH_c}\right)\left(\frac{H_cK}{KA}\right)\left(\frac{AO_1}{O_1H}\right) = 1$$
Therefore, $O_1$ lies on $KT$, proving that the perpendicular bisector of $AB$, the altitude through $A$, and the segment $KT$ are concurrent.

$\blacksquare$