Problem 1642

### Part (a)

1. Define the Skew-Center:
For a right triangle $\triangle ABC$ with a right angle at $C$, the skew-center is defined as the second intersection of the circles with diameters $AC$ and $BD$, where $D$ is the foot of the altitude from $C$.

2. Lemma:
The skew-center of a right triangle lies inside the triangle. This can be shown by angle chasing, but we will assume it as given.

3. Base Case:
Let $P$ be the skew-center of $\triangle A_1A_2A_3$. We need to show that $P$ is the skew-center of $\triangle A_{n-2}A_{n-1}A_n$ for all $n \geq 3$.

4. Inductive Step:
Assume $P$ is the skew-center of $\triangle A_{k-2}A_{k-1}A_k$. By definition, $P$ lies on the circles with diameters $A_{k-2}A_k$ and $A_{k-1}A_{k+1}$.

5. Angle Calculation:
Since $A_{k+2}$ lies on the circle with diameter $A_{k-1}A_{k+1}$, we have:
$$\angle A_kPA_{k+2} = 2\pi - \angle A_{k+1}PA_{k+2} - \angle A_kPA_{k+1}$$
Using the fact that $\angle A_{k+1}PA_{k+2}$ and $\angle A_kPA_{k+1}$ are supplementary, we get:
$$\angle A_kPA_{k+2} = (\pi - \angle A_{k+1}PA_{k+2}) + (\pi - \angle A_kPA_{k+1}) = \angle A_{k+1}A_k + \angle A_{k+1}A_{k-1}A_{k-2} = \frac{\pi}{2}$$
Thus, $P$ lies on the circle with diameter $A_kA_{k+2}$.

6. Conclusion:
Since $P$ lies on both circles, it is the skew-center of $\triangle A_{k-1}A_kA_{k+1}$. By induction, $P$ is the skew-center of all triangles $\triangle A_{n-2}A_{n-1}A_n$.

7. Uniqueness:
As the side lengths of $\triangle A_{n-2}A_{n-1}A_n$ tend to zero as $n \to \infty$, any other point $Q$ would eventually be outside the triangle, proving $P$ is unique.

$\blacksquare$

### Part (b)

1. **Locus of $P$:**
By definition, the locus of $P$ is a subset of the circle with diameter $A_1A_3$, excluding $A_1$ and $A_3$.

2. Brocard Angle:
By similarity of triangles $\triangle A_1A_2A_3$ and $\triangle A_3A_4A_5$, and the fact that $P$ is the skew-center, we have:
$$\angle PA_1A_3 = \angle PA_3A_5 = \angle PA_3A_2$$
Similarly, from $\triangle A_3A_4A_5$ and $\triangle A_5A_6A_7$, we get:
$$\angle PA_3A_5 = \angle PA_5A_7$$
Since $PA_5A_2A_4$ is cyclic, we have:
$$\angle PA_5A_7 = \angle PA_2A_1$$
Thus, $\angle PA_1A_3 = \angle PA_3A_2 = \angle PA_2A_1 = \omega$, the Brocard angle of $\triangle A_1A_2A_3$.

3. Brocard Angle Calculation:
$$\cot \omega = \cot A_3 + \cot A_1 + \cot A_2 = 0 + \cot A_1 + \frac{1}{\cot A_1} \geq 2 \implies \tan \omega \leq \frac{1}{2} \implies \omega \leq \arctan \frac{1}{2}$$
Equality can be achieved, and $A_2$ can lie on either side of $A_1A_3$.

4. Final Locus:
The desired locus is an arc of the circle with diameter $A_1A_3$ subtending an angle of $2\arctan \frac{1}{2}$ at $A_1$, excluding $A_3$.

$$\boxed{\text{Locus of } P \text{ is an arc of the circle with diameter } A_1A_3 \text{ subtending an angle of } 2\arctan \frac{1}{2} \text{ at } A_1, \text{ excluding } A_3.}$$