# Olympiad geometry problem related to fermat point and another point which is formed when sides subtend 60 degrees added to the opposing angle

This is a question from BMOTC. I have spent a day trying to get into it but all I have found is a mess of variables. What I am looking for is an elegant solution, as this is an olympiad problem. Here it goes:

Let triangle $ABC$ have side lengths $a$, $b$, and $c$ as usual. Points $P$ and $Q$ have properties such that $\angle APB=\angle BPC=\angle CPA=120^{\circ}$ and $\angle AQB=60^{\circ}+ \angle C$, $\angle BQC=60^{\circ}+ \angle A$, $\angle CQA=60^{\circ}+ \angle B$. Prove that $$(|AP|+|BP|+|CP|)^3 \cdot|AQ| \cdot|BQ| \cdot|CQ|=(abc)^2$$

Take the unique points $A_1, B_1, C_1$ outside the triangle $ABC$ such that triangles $ABC_1, \, AB_1C, \, A_1BC$ are equilateral triangles. Then the point $P$ is the intersection point of $AA_1, \, BB_1, \, CC_1$ nd the angle between any pair of these lines is $60^{\circ}$ (or $120^{\circ}$ depending which angle you measure). It is also the intersection point of the circles $k_{AB}, \, k_{BC}, \, k_{CA}$ circumscribed around equilateral triangles $ABC_1, \, AB_1C, \, A_1BC$ respectively. Moreover, $$|AA_1| = |BB_1| = |CC_1| = l$$ The point $Q$ is the isogonal conjugate to $P$. What that last statement means is that if you reflect the line $AA_1$ with respect to the interior angle bisector of vertex $A$ of triangle $ABC$, the line $BB_1$ with respect to the angle bisector of vertex $B$ and the line $CC_1$ with respect to the angle bisector of vertex $C$, the three new lines also intersect at a common point, we denote by $Q'$. The point $Q'$ satisfies the angle conditions listed in the statement of the problem. However, these angle angle conditions can be satisfied by exactly one point, so $Q' \equiv Q$. This means that $\angle \, CAP = \angle \, BAQ$ and $\angle \, CAP = \angle \, BAQ$ and since $\angle \, AQB = 60 + \angle \, C = \angle \, ACA_1 = \angle \, BCB_1$, the triangle $\angle \, ABQ$ is similar to the two congruent triangles $\angle \, AA_1C \cong \angle \, BB_1C$. Therefore, $$\frac{|AQ|}{b} = \frac{|AQ|}{|AC|} = \frac{|AB|}{|AA_1|} = \frac{c}{l} \, .$$ Analogous arguments applied to triangles $BCQ$ (similar to triangles $CC_1A \cong BB_1A$) and $CAQ$ (similar to triangles $CC_1B \cong AA_1B$) yields $$\frac{|BQ|}{c} = \frac{|BQ|}{|AB|} = \frac{|BC|}{|BB_1|} = \frac{a}{l} \,$$ $$\frac{|CQ|}{a} = \frac{|BQ|}{|CB|} = \frac{|CA|}{|CC_1|} = \frac{b}{l} \, .$$ Finally, let us look at $l = |AA_1| = |AP| + |PA_1|$. Since $P$ lies on the circumcircle $k_{BC}$ of equilateral triangle $A_1BC$ it is a known fact that $|PA_1| = |PB| + |PC|$. Consequently, $l = |AA_1| = |AP| + |PA_1| = |PA| + |PB| + |PC|$.
Putting all of this together, we get $$l = |PA| + |PB| + |PC|$$ $$l = \frac{bc}{|AQ|}$$ $$l = \frac{ca}{|BQ|}$$ $$l = \frac{ab}{|CQ|}$$
raised to power three becomes $$\big(|PA| + |PB| + |PC|\big)^3 = l^3 = \frac{bc}{|AQ|} \, \frac{ca}{|BQ|} \, \frac{ab}{|CQ|} = \frac{(abc)^2}{|AQ| \, |BQ| \, |CQ|}$$ and there you have it
$$\big(|PA| + |PB| + |PC|\big)^3 \,\, |AQ| \, |BQ| \, |CQ| = {(abc)^2}$$
• @Shikhar_Mohan I added a short explanation about $Q$ being an isogonal conjugate point of $P$. Sep 26, 2016 at 17:52