MathDB

Let

ABC

be a triangle with

\angle A,\angle B,\angle C <120^{\circ}

T

is its Fermat-Torricelli point. Consider a point

P

lying on the same plane with

\triangle ABC

. Prove that: a)

\dfrac{\overrightarrow {TA}}{TA}+\dfrac{\overrightarrow {TB}}{TB}+\dfrac{\overrightarrow {TC}}{TC}=\overrightarrow {0}.

PA + PB + PC \ge \frac{{\overrightarrow {PA} \overrightarrow {.TA} }}{{TA}} + \frac{{\overrightarrow {PB} .\overrightarrow {TB} }}{{TB}} + \frac{{\overrightarrow {PC} \overrightarrow {.TC} }}{{TC}}.

PA + PB + PC \ge TA + TB + TC

and the equality occurs iff

P\equiv T

Problem 3