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Question Number 94144 by Tony Lin last updated on 17/May/20

There is a moving point P in a triangle   ABC of which sides are a,b,c and a>b>c  find the minimum and maximum  of AP+BP+CP

$${There}\:{is}\:{a}\:{moving}\:{point}\:{P}\:{in}\:{a}\:{triangle} \\ $$ $$\:{ABC}\:{of}\:{which}\:{sides}\:{are}\:{a},{b},{c}\:{and}\:{a}>{b}>{c} \\ $$ $${find}\:{the}\:{minimum}\:{and}\:{maximum} \\ $$ $${of}\:{AP}+{BP}+{CP} \\ $$

Commented byTony Lin last updated on 17/May/20

thanks sir I almost forgot R=((abc)/(4Δ))

$${thanks}\:{sir}\:{I}\:{almost}\:{forgot}\:{R}=\frac{{abc}}{\mathrm{4}\Delta} \\ $$

Commented bymr W last updated on 17/May/20

minimum=3R=((3abc)/(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c))))  when P is the circumcenter.    maximum=a+b  when P is the vertex C.

$${minimum}=\mathrm{3}{R}=\frac{\mathrm{3}{abc}}{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}} \\ $$ $${when}\:{P}\:{is}\:{the}\:{circumcenter}. \\ $$ $$ \\ $$ $${maximum}={a}+{b} \\ $$ $${when}\:{P}\:{is}\:{the}\:{vertex}\:{C}. \\ $$

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