Dear Alexey!!!

Who is the proposer of this wonderful problem?

and what is his proof?

My proof is the following:

The point that minimizes the sum PA^p + PB^p + PC^p

is the point with homogeneous barycentric coordinates

P = (x : y : z) = ( PA^(p-2) : PB^(p-2) : PC^(p-2) )

The triangle A'B'C' that minimizes the sum

(B'C')^q + (C'A')^q + (A'B')^q is the pedal triangle of the point

Q = ( a^q.QA^(q-2) : b^q.QB^(q-2) : c^q.QC^(q-2) )

If Q is the isogonal conjugate of P then

QA / PA : QB / PB : QC / PC = x / a : y / b : z / c or

QA : QB : QC = PA^(p-1) / a : PB^(p-1) / b : PC^(p-1) / c or

Q = ( aa.PA^(p-1)(q-2) : bb.PB^(p-1)(q-2) : cc.PC^(p-1)(q-2) )

Since P, Q are isogonal conjugates we must have

(p - 1)(q - 2) = 2 - p or

1 / p + 1 / q = 1

Best regards

Nikos Dergiades

> Dear colleagues!

> Let a triangle ABC be given. Consider two extremal

> problems.

> (1) Find a point X_p minimizing the sum

> X_pA^p+X_pB^p+X_pC^p

> (2) Find a points A', B', C' on BC, CA, AB respectively

> minimizing the sum A'B'^q+B'C'^q+C'A'^q.

> It is evident then the sought triangle A'B'C' in problem (2)

> is a pedal triangle of some point Y_q. Now the main result.

> If 1/p+1/q=1 then X_p and Y_q are isogonally conjugated.

>

> Sincerely

>

> Alexey