$ABC$ be a triangle $R_a$, $R_b$, $R_c$ are the radii of Lucas Circles of $ABC$. Prove that: $$R_a+R_b+R_c\geq \dfrac {8.\triangle}{(1+\sqrt {3})^2.R}$$ where $\triangle$ and $R$ are area and circumradius of $ABC$ respectively.

I couldn't get anything to even start. Please help me.

Edit: I read about Lucas Circles from https://artofproblemsolving.com/community/c2899h1233399_lucas_circles I didn't understand the explanation from this point further: Then draw perpendiculars to the side from each of $...$

  • $\begingroup$ what are the Lucas circles in a triangle? $\endgroup$ – Dr. Sonnhard Graubner Apr 17 '17 at 13:13
  • $\begingroup$ @Dr.SonnhardGraubner, No idea about that. $\endgroup$ – pi-π Apr 17 '17 at 13:21
  • $\begingroup$ see here scribd.com/document/318116922/Lucas-s-Inner-Circles $\endgroup$ – Dr. Sonnhard Graubner Apr 17 '17 at 13:24
  • $\begingroup$ What is the purpose of solving this problem if you do not even know anything about its subject? $\endgroup$ – Jack D'Aurizio Apr 17 '17 at 13:27
  • $\begingroup$ @JackD'Aurizio, I am preparing for exams where such ques. might be asked! $\endgroup$ – pi-π Apr 17 '17 at 13:28

Lucas circles arise in the problem of inscribing a square inside a triangle:

$\hspace{3cm}$enter image description here

The radius of the $A$-Lucas circle is given by $R_A=\frac{R}{1+\frac{2aR}{bc}}$ and these circles are pairwise tangent. Additionally, they are tangent to the circumcircle of $ABC$.

Hint: what happens by applying Descartes' circle theorem to the configuration given by the Lucas circles and the circumcircle?


Let $a^2=x$, $b^2=y$ and $c^2=z$.

Hence, we need to prove that: $$\sum_{cyc}\frac{R}{1+\frac{2a\cdot\frac{abc}{4\Delta}}{bc}}\geq\frac{8\Delta}{(1+\sqrt3)^2R}$$ or $$\sum_{cyc}\frac{1}{2\Delta+a^2}\geq\frac{2}{(2+\sqrt3)R^2}.$$ Now, by C-S $$\sum_{cyc}\frac{1}{2\Delta+a^2}\geq\frac{9}{6\Delta+a^2+b^2+c^2}.$$ Thus, it remains to prove that $$9(2+\sqrt3)R^2\geq2(a^2+b^2+c^2+6\Delta)$$ or $$\frac{9(2+\sqrt3)a^2b^2c^2}{16\Delta^2}\geq2(a^2+b^2+c^2)+12\Delta$$ or $$\frac{9(2+\sqrt3)xyz}{\sum\limits_{cyc}(2xy-x^2)}\geq2(x+y+z)+3\sqrt{\sum\limits_{cyc}(2xy-x^2)},$$ which is true because $$9xyz\geq(x+y+z)\sum\limits_{cyc}(2xy-x^2)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur and $$3\sqrt3xyz\geq\sqrt{\left(\sum\limits_{cyc}(2xy-x^2)\right)^3}$$ or $$27x^2y^2z^2\geq\left(\sum\limits_{cyc}(2xy-x^2)\right)^3$$ or $$x^2+y^2+z^2+3\sqrt[3]{x^2y^2z^2}\geq2(xy+xz+yz)$$ is Schur again: $$\sum_{cyc}\left((x^2+\sqrt[3]{x^2y^2z^2}\right)\geq\sum_{cyc}\left(\sqrt[3]{x^4y^2}+\sqrt[3]{x^2y^4}\right)\geq2(xy+xz+yz).$$ Done!

  • $\begingroup$ How did you use $\sum_{cyc}$? $\endgroup$ – pi-π Apr 17 '17 at 14:56
  • $\begingroup$ @Ramanujan For example $\sum\limits_{cyc}x=x+y+z$, $\sum\limits_{cyc}(x^2y+x^2z)=x^2y+x^2z+y^2z+y^2x+z^2x+z^2y$ and similar $\endgroup$ – Michael Rozenberg Apr 17 '17 at 15:10
  • $\begingroup$ How did you get the LHS of first statement? $\endgroup$ – pi-π Apr 17 '17 at 15:40
  • $\begingroup$ @Ramanujan It's just which we need to prove: $R_a+R_b+R_c\geq...$ $\endgroup$ – Michael Rozenberg Apr 17 '17 at 17:44

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