Given the orthocenter $H(c,d)$ and the circumcenter $O(a,b)$. Let $A(\lambda,\mu), B(x_{2},y_{2}),C(x_{3},y_{3})$. The centroid $Z$ satisfies $HZ=2 \cdot ZO$, hence $\left\{\begin{array}{ll} x_{3}=2a+c-\lambda-x_{2} \\ y_{3}=2b+d-\mu-y_{2} \end{array}\right.$ $N(\frac{a+c}{2},\frac{b+d}{2})$ is the midpoint of the nine-points circle. The radius of this nine-points circle, $R_{1}$, equals $\frac{1}{2}R$ (the radius of the circumcircle). The midpoint $A'(\frac{\lambda+x_{2}}{2},\frac{\mu+y_{2}}{2})$ of $AB$ lies on the nine-points circle, so $A'N=\frac{1}{2} \cdot AO$, so we have a first relation $2(\lambda-a)(x_{2}-c)+(x_{2}-c)^{2}+2(\mu-b)(y_{2}-d)+(y_{2}-d)^{2}=0$. The midpoint $A''(\frac{2a+c-x_{2}}{2},\frac{2b+d-y_{2}}{2})$ of $AC$ lies on the nine-points circle, so $A''N=\frac{1}{2} \cdot AO$, so we have a second relation $(x_{2}-a)^{2}+(y_{2}-b)^{2}=(\lambda-a)+(\mu-b)^{2}$. Subtracting, we have $2x_{2}(\lambda-c)+2y_{2}(\mu-d)-2(\lambda-a)c+c^{2}-2(\mu-b)d+d^{2}=-\lambda^{2}\mu^{2}+2a\lambda+2b\mu$, so we can calculate $x_{2}$ as a function of $y_{2}$, the given $a,b,c,d$ and the two parameters $\lambda,\mu$. The value of $x_{2}$ in the second relation gives two values of $y_{2}$: $y_{2}=\frac{\sqrt{4a^{2}-8a\lambda+4b^{2}-8b\mu-c^{2}+2c\lambda-d^{2}+2d\mu+3\lambda^{2}+3\mu^{2}} \cdot \left|c-\lambda\right|+(2b+d-\mu)\sqrt{(\lambda-c)^{2}+(\mu-d)^{2}}}{2\sqrt{(\lambda-c)^{2}+(\mu-d)^{2}}}$. And then $x_{2}=\frac{(d-\mu)\sqrt{4a^{2}-8a\lambda+4b^{2}-8b\mu-c^{2}+2c\lambda-d^{2}+2d\mu+3\lambda^{2}+3\mu^{2}} \cdot \left|c-\lambda\right|-(c-\lambda)(2a+c-\lambda)\sqrt{(\lambda-c)^{2}+(\mu-d)^{2}}}{2(\lambda-c)\sqrt{(\lambda-c)^{2}+(\mu-d)^{2}}}$. We find $x_{2}$ and $y_{2}$ as a function of given $a,b,c,d$ and the two parameters $\lambda,\mu$. But, we see a square root in both expressions: $4a^{2}-8a\lambda+4b^{2}-8b\mu-c^{2}+2c\lambda-d^{2}+2d\mu+3\lambda^{2}+3\mu^{2} \geq 0$ or $(\lambda+\frac{c-4a}{3})^{2}+(\mu+\frac{d-4b}{3})^{2}-\frac{4}{9} \cdot [(a-c)^{2}+(d-b)^{2}] \geq 0$. It's not allowed to choose $(\lambda,\mu)$ inside the circle with midpoint $T(\frac{4a-c}{3},\frac{4b-d}{3})$ (see drawing) and radius $\frac{2}{3} \cdot \sqrt{(a-c)^{2}+(d-b)^{2}}$. I have use $a=7,b=1,c=2,d=5,\lambda=1,\mu=-2$, then $Z(\frac{16}{3},\frac{7}{3})$ and $S(12,-3)$. Attachment of a bad translation, from Russian to English, of the solution of this question (2007 Sharygin)

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