Let $ABC$ be an acute-angled triangle and $M$ a point inside it. We denote by $C_1$, $C_2$, $C_3$ the centers of the Nine-Point circles corresponding to the triangles $BMC$, $CMA$, and $AMB$, respectively. Show that $$(R+r)^2 \leq C_1C_2^2 + C_2C_3^2 + C_3C_1^2 \leq 2R^2 + r^2,$$

where $R$ is the radius of the circumcircle of $ABC$ and $r$ is the radius of the incircle $ABC$ and $M \neq H$ where H is the orthocenter.

I got stuck at this problem. Can it be solved using complex numbers? I would appreciate any help with finding the correct solution.

  • $\begingroup$ Of course you can solve it through putting any coordinates and bash the algebra hard enough. For example, you can compute the barycentric coordinate of the nine point centre of a triangle knowing all three angles or all three side lengths. But is it the nicest way? Probably not. $\endgroup$ Commented Oct 1, 2023 at 16:49
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    $\begingroup$ What are $R$ and $r$? $\endgroup$ Commented Oct 1, 2023 at 20:55
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    $\begingroup$ If you choose $M$ to be an orthocenter of $ABC$ then all the Euler circles of small triangles will be also Euler circle of the triangle $ABC$ itself (by definition, Euler circle of $ABC$ passes through all the midpoints of segments from triangle vertices to the orthocenter, so in our case it will pass through all the midpoints of the small triangles sides and so will be also their Euler circle). So you have in this case $C_1 = C_2 = C_3$ and $C_1C_2^2 + C_2C_3^2 + C_3C_1^2 = 0$, which is contradiction to the first inequality $\endgroup$ Commented Oct 4, 2023 at 10:26
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    $\begingroup$ @math.enthusiast9 okay maybe, but note that this also implies (due to continuity) that you can choose $M$ close enough to the orthocenter of $ABC$ to get arbitrarily small value for sum $C_1C_2^2 + \dots + C_3C_1^2$ $\endgroup$ Commented Oct 4, 2023 at 12:57
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    $\begingroup$ From the algebraic-geometry tag description: "This tag should not be used for elementary problems which involve both algebra and geometry." Please follow the usage guidance. $\endgroup$
    – KReiser
    Commented Nov 30, 2023 at 20:01

1 Answer 1


If $0,z_1,z_2\in\Bbb C$ are vertices of a triangle then its circumcenter is $$z_O=\frac{z_1z_2(\bar z_1-\bar z_2)}{\bar z_1z_2- z_1\bar z_2}.\tag1$$ For example if we let now $B=0, A=z_1=r_1e^{i\alpha_1}$ and $M=r_2e^{i\alpha_2}$ then $$C_1=Z_O(\triangle BMA)=\frac{r_1e^{-i\alpha_1}-r_2e^{-i\alpha_2}}{e^{-i2\alpha_1}-e^{-2i\alpha_2}}.\tag2$$ From $(2)$, we see that when $M$ approaches side $BA$ so that $r_1\neq r_2$, then $\alpha_2\to\alpha_1$ and $C_1\to\infty$ whearas $C_2$ and $C_3$ does not go to infinity. Hence, $C_1C_2^2+C_2C_3^2+C_3C_1^2\to\infty$ and your claim is false. Of course from law of sines it is more easy to this.

Also, the orthocenter of triangle with vertices $0,z_1,z_2$ is $$z_H=\frac{(\bar z_1z_2+ z_1\bar z_2)(z_2-z_1)}{\bar z_1z_2- z_1\bar z_2}.$$ Now our task is to show that when $M=H,$ $z_O(\triangle BMA)=z_O(\triangle BMC)$ so that $C_1C_2^2+C_2C_3^2+C_3C_1^2=0$ follows. Can you do that? It looks complicated. I gave up.


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