Could you, pls, check your problem? How possible for three concurrent lines to make a triangle?? Is it about the triangle made by the intersection of $PA$ and $BC$, a.s.o.? Best regards, sunken rock

I thought he meant that "$ ABC $ is a triangle in the plane. Find the locus of point $ P $ also on the same plane such that the lengths of the segments $ PA, PB $ and $ PC $ form a triangle whose area is equal to one third of the area of triangle $ ABC $."

@Vo Duc Dien: You are right, of course! Best regards, sunken rock

See if you can derive the solution from this problem http://www.artofproblemsolving.com/Forum/viewtopic.php?p=841206&sid=baba23efc478703ce1dda682c636cc43#p841206

Given $\triangle ABC, A(0,0),B(b,0),C(c,d)$, area $= \frac{1}{2}bd$. Point $P(x,y)$. Area triangle, formed with $PA,PB,PC : \sqrt{s(s-PA)(s-PB)(s-PC)}$. Condition: $(PA+PB+PC)(PB+PC-PA)(PA+PC-PB)(PA+PB-PC)=\frac{4}{9}b^{2}d^{2}$. \[27x^{4}-36x^{3}(b+c)+18x^{2}(3y^{2}-2dy-b^{2}+4bc-c^{2}+d^{2})\] \[-36x(y^{2}(b+c)+2dy(c-b)-b^{3}+b^{2}c+b(c^{2}+d^{2})-c(c^{2}+d^{2}))+27y^{4}-36dy^{3}\] \[+18y^{2}(b^{2}+c^{2}-d^{2})-36dy(b^{2}-c^{2}-d^{2})-9b^{4}+2b^{2}(9c^{2}+7d^{2})-9(c^{4}+2c^{2}d^{2}+d^{4})=0\] No program plots this equation with parameters $b,c,d$. Therefore, choose e.g. $b=10,c=4,d=7$, see picture. The equation: \[27x^{4}-504x^{3}+18x^{2}(3y^{2}-14y+93)-504x(y^{2}-6y-15)+27y^{4}-252y^{3}+1206y^{2}-8820y-30625=0\] Control with points satisfying this equation: \[P(6,\frac{7}{3}), P(5,2.582742863), P(7,8.112465714), P(3.744107969,0)\]

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Bright idea. It's a good way to use the computer to verify the solution. But I don't think it was intended to be solved that way with the computer. On the other hand, I believe you could've made some errors during the rigorous calculation because if you pick the westernmost point on the closed curve, all PA, PB and BC could be greater than AB, BC and AC and, therefore, the area of ABC cannot be three time that of the other. Please check.

Westpoint $P(-4.945668188,3.5)$, line segments $PA,PB,PC$, see picture. Construct $PF=PC$ and $BF=PA \rightarrow \triangle PBF$ with area one third !!

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Isn't that triangle supposed to be equilateral? The curve is generally too weird to show up in real competition...

Assuming the triangle is equilateral, let $l$ be it's sidelength,$S$ its area and $S'$ the area of the triangle with sidelenghts $AP,BP,CP$. It is known (by an argument of rotation of $60$ degrees around $A$) that $2S=3S'+\frac{\sqrt 3}{4}(AP^2+BP^2+CP^2)$. If we want $S'=\frac{S}{3}$ we must have $S=\frac{\sqrt 3}{4}l^2=\frac{\sqrt 3}{4}(AP^2+BP^2+CP^2)$. However, if $O$ is the circumcenter of $ABC$, $\sum AP^2=\sum (\vec{P}-\vec{A})\cdot (\vec{P}-\vec{A})=3\vec{OP}\cdot\vec{OP}-2\vec{OP}\cdot\sum{\vec{A}}+\sum \vec{A}\cdot\vec{A}=$ $3OP^2-0+3R^2\geq 3R^2=3(\frac{2}{3}\frac{\sqrt 3}{2}l)^2=l^2$ so the equality can only be achieved if $P\equiv O$.