Fermat distances

by math_explorer, Feb 25, 2011, 2:21 AM

(the bashy part from SMT 2011 Geo.10)

The problem: we have $\triangle ABC$, whose angles all don't exceed $120^\circ$, and $P$, the Fermat point, and want a neat, symmetric formula for $PA + PB + PC$.

The well-known symmetric angles about $P$ are a good place to start.
Suppose the area of $\triangle ABC$ is $S$.

Apply the Law of Sines area formula to $PAB$, $PBC$, $PCA$, noting that $\sin 120^\circ = \sqrt{3}/2$:
$S = \sqrt{3}/4 (\sum PB \times PC)$ (sum cyclic across $A, B, C$)
$4S/\sqrt{3} = \sum PB \times PC$

Next, Law of Cosines in $PBC$:
$a^2 = PB^2 + PC^2 + PB \times PC$
Summing across $A, B, C$ and dividing by 2:
$\frac{\sum a^2}{2} = \sum PA^2 + \sum PB \times PC / 2$

To adjust this so that the right side becomes $(\sum PA)^2$, we'll add $2\sqrt{3}S$.

$\frac{\sum a^2}{2} + 2\sqrt{3}S = \sum PA^2 + 2\sum PB \times PC$

Now just take the square root
$\sqrt{\frac{\sum a^2}{2} + 2\sqrt{3}S} = PA + PB + PC$
This post has been edited 1 time. Last edited by math_explorer, May 25, 2011, 2:13 PM

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