Problem 1815

Let $A$, $B$, $C$ and $P$ have position vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ and
$$\mathbf{p} = \frac{\alpha}{\alpha + \beta + \gamma} \mathbf{a} + \frac{\beta}{\alpha + \beta + \gamma} \mathbf{b} + \frac{\gamma}{\alpha + \beta + \gamma} \mathbf{c},$$
where $\alpha$, $\beta$, $\gamma$ are the barycentric coordinates of $P$. Then $D$ has position vector
$$\frac{\beta}{\beta + \gamma} \mathbf{b} + \frac{\gamma}{\beta + \gamma} \mathbf{c},$$
so the circles on diameters $BC$ and $AD$ have equations
$$(\mathbf{r} - \mathbf{b}) \cdot (\mathbf{r} - \mathbf{c}) = 0 \quad \text{and} \quad (\mathbf{r} - \mathbf{a}) \cdot \left( \mathbf{r} - \frac{\beta}{\beta + \gamma} \mathbf{b} - \frac{\gamma}{\beta + \gamma} \mathbf{c} \right) = 0,$$
respectively; alternatively, but equivalently, the latter reads
$$\beta(\mathbf{r} - \mathbf{a}) \cdot (\mathbf{r} - \mathbf{b}) + \gamma(\mathbf{r} - \mathbf{a}) \cdot (\mathbf{r} - \mathbf{c}) = 0.$$
Any linear combination of the two is the equation of a circle or straight line through $a$ and $a'$. In particular, the linear combination formed by multiplying the first equation by $\beta\gamma$ and the second by $\alpha$, and adding, is
$$\alpha\beta(\mathbf{r} - \mathbf{a}) \cdot (\mathbf{r} - \mathbf{b}) + \beta\gamma(\mathbf{r} - \mathbf{b}) \cdot (\mathbf{r} - \mathbf{c}) + \gamma\alpha(\mathbf{r} - \mathbf{c}) \cdot (\mathbf{r} - \mathbf{a}) = 0.$$
The symmetry of this equation shows that this circle (or, possibly, straight line) also passes through $b$, $b'$ and $c$, $c'$. Finally, notice that $\alpha\beta + \beta\gamma + \gamma\alpha$ is positive when $P$ lies inside the triangle $ABC$, so the locus is indeed a circle centered at the point whose barycentric coordinates are $\alpha(\beta + \gamma)$, $\beta(\gamma + \alpha)$, $\gamma(\alpha + \beta)$, respectively.