## Re: [EMHL] two extremal problems

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• 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 +
Message 1 of 3 , May 25, 2012
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

> 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
• Dear Nikos! Who is the proposer of this wonderful problem? and what is his proof? My proof is the same. Problems (1), (2) were formulated by Protasov and
Message 2 of 3 , May 27, 2012
Dear Nikos!

Who is the proposer of this wonderful problem?
and what is his proof?

My proof is the same.
Problems (1), (2) were formulated by Protasov and Tihomirov ("Kvant" N2,
2012). They indicate that the solutions of both problems are known only for
p, q equal to 1, 2 or infinity. Since fot these values the main result is
true I supposed that it may be correct for all p, q.

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

Sincerely Alexey
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