A quick remark: it's very similar to the 2'nd problem from the first Romanian TST in 2003: it was the same except for the values, which were PA=1, PB=2, PC=3.

I think it's from Rumania, There is someone ask me to solve it, I got the answer is 36+22 \sqrt 3 is right?

Ehm, were the numerical data in the problem (post #1) changed? Else, I don't understand Grobber's grobber wrote: it was the same except for the values, which were PA=1, PB=2, PC=3. Darij

Yes, I think they were changed. (In fact, my guess is that sigma copied the wrong values at first, and then, thanks to grobber's post, they were corrected )

I think the properties$PA=1,PB=2,PC=3$ can be replace by$PA=x,PB=y,PC=z$for $x,y,z$are given positive number.My solution only use CauChy inequality

maybe rotating the triangle about point $A$ 60 degrees would help...

With my solution the property$\angle BAC=60$can be rewrited $\angle BAC=\alpha$,but the maximum of the area of triangle must satisfy a 4-degree equation

mumath wrote: With my solution the property$\angle BAC=60$can be rewrited $\angle BAC=\alpha$,but the maximum of the area of triangle must satisfy a 4-degree equation What is your solution?

I think that the Erdos- Mordel inequality will help: $\ OA+OB+OC \geq 2(OP+OQ+OR)$

I hope it's correct. let's take $P$ and $A$ fix. so the locus of $B$ and $C$ will be two circle with center $P$ and radius $2\ (\omega_2)$ and $3\ (\omega_3)$. we have $S_{ABC}=\frac 12.b.c\sin 60^{\circ}$ so area will be maximum when $b.c$ be maximum. now rotate $\omega_3$ wrt $A\ \ 120^{\circ}$ name it $\omega_3'$. with this rotaion $C\rightarrow C'$ and $B,A,C'$ are collinear.(SEE FIGURE 1) now probelm become: We have two intersecting circle$(\omega_2\ ,\omega_3')$ and a point $A$ in these circles.line $\ell$ pass through $A$ and intersect these cicles at $P$ and $Q$. find line $\ell$ such that $AP.AQ$ be Maximum.(SEE FIGURE 2) proof.consider a circle $(\mathcal C)$ which is tangent to these circles and the tangency points are collinear with $A$. name this tangency points $M,N$. now it's clear for every points $P,Q$ that $A,P,Q$ are collinear we have $AM.AN\ge AP.AQ$. for drawing this circle it's sufficient that $MN$ pass throuh Ex-homothecy center of these circles. hence we find triangle $ABC$ . Click on the picture to see larger

Attachments:

Quote: ...now we have that: PA^2+PD^2-PB^2-PC^2=2PO^2+1/2*AD^2-2PQ^2-1/2BC^2=... dear lagrangia I cannot find points O,Q.May you draw a figure and, may you explain the following equalities: 1) Quote: PA^2+PD^2-PB^2-PC^2=2PO^2+1/2*AD^2-2PQ^2-1/2BC^2=2*AB*AC*cos(60) and 2) Quote: PD^2=12+AB*AC thanks.

Can someone explain how we find the area of the triangle after doing what Amir did? I don't know how you would find the circle tangent at points collinear with A, and how you would do the rest.

Any solution?