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Question Number 54068 by ajfour last updated on 28/Jan/19

Commented by ajfour last updated on 28/Jan/19

locate P (x,y) such that △APB, △BPC, and △CPA each have the same perimeter. (in terms of a,b,c)

$${locate}\:{P}\:\left({x},{y}\right)\:{such}\:{that}\:\bigtriangleup{APB}, \\ $$$$\bigtriangleup{BPC},\:{and}\:\bigtriangleup{CPA}\:{each}\:{have}\:{the} \\ $$$${same}\:{perimeter}.\:\left({in}\:{terms}\:{of}\:{a},{b},{c}\right) \\ $$

Commented by mr W last updated on 28/Jan/19

this is (not always) possible.

$${this}\:{is}\:\left({not}\:{always}\right)\:{possible}. \\ $$

Answered by mr W last updated on 28/Jan/19

s=(√((x−x_A )^2 +(y−y_A )^2 ))+(√((x−x_B )^2 +(y−y_B )^2 ))+(√((x_A −x_B )^2 +(y_A −y_B )^2 )) s=(√((x−x_B )^2 +(y−y_B )^2 ))+(√((x−x_C )^2 +(y−y_C )^2 ))+(√((x_B −x_C )^2 +(y_B −y_C )^2 )) s=(√((x−x_C )^2 +(y−y_C )^2 ))+(√((x−x_A )^2 +(y−y_A )^2 ))+(√((x_C −x_A )^2 +(y_C −y_A )^2 )) 3 eqn. with 3 unknowns x,y,s. A(0,0) B(b,0) C(c,d) s=(√(x^2 +y^2 ))+(√((x−b)^2 +y^2 ))+b s=(√((x−b)^2 +y^2 ))+(√((x−c)^2 +(y−d)^2 ))+(√((b−c)^2 +d^2 )) s=(√((x−c)^2 +(y−d)^2 ))+(√(x^2 +y^2 ))+(√(c^2 +d^2 )) ⇒(√(x^2 +y^2 ))−(√((x−c)^2 +(y−d)^2 ))=(√((b−c)^2 +d^2 ))−b ⇒(√((x−b)^2 +y^2 ))−(√(x^2 +y^2 ))=(√(c^2 +d^2 ))−(√((b−c)^2 +d^2 )) example: A(0,0), B(5,0), C(2,4) ⇒P(2.7888,1.1056)

$${s}=\sqrt{\left({x}−{x}_{{A}} \right)^{\mathrm{2}} +\left({y}−{y}_{{A}} \right)^{\mathrm{2}} }+\sqrt{\left({x}−{x}_{{B}} \right)^{\mathrm{2}} +\left({y}−{y}_{{B}} \right)^{\mathrm{2}} }+\sqrt{\left({x}_{{A}} −{x}_{{B}} \right)^{\mathrm{2}} +\left({y}_{{A}} −{y}_{{B}} \right)^{\mathrm{2}} } \\ $$$${s}=\sqrt{\left({x}−{x}_{{B}} \right)^{\mathrm{2}} +\left({y}−{y}_{{B}} \right)^{\mathrm{2}} }+\sqrt{\left({x}−{x}_{{C}} \right)^{\mathrm{2}} +\left({y}−{y}_{{C}} \right)^{\mathrm{2}} }+\sqrt{\left({x}_{{B}} −{x}_{{C}} \right)^{\mathrm{2}} +\left({y}_{{B}} −{y}_{{C}} \right)^{\mathrm{2}} } \\ $$$${s}=\sqrt{\left({x}−{x}_{{C}} \right)^{\mathrm{2}} +\left({y}−{y}_{{C}} \right)^{\mathrm{2}} }+\sqrt{\left({x}−{x}_{{A}} \right)^{\mathrm{2}} +\left({y}−{y}_{{A}} \right)^{\mathrm{2}} }+\sqrt{\left({x}_{{C}} −{x}_{{A}} \right)^{\mathrm{2}} +\left({y}_{{C}} −{y}_{{A}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{3}\:{eqn}.\:{with}\:\mathrm{3}\:{unknowns}\:{x},{y},{s}. \\ $$$${A}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${B}\left({b},\mathrm{0}\right) \\ $$$${C}\left({c},{d}\right) \\ $$$${s}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }+\sqrt{\left({x}−{b}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }+{b} \\ $$$${s}=\sqrt{\left({x}−{b}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }+\sqrt{\left({x}−{c}\right)^{\mathrm{2}} +\left({y}−{d}\right)^{\mathrm{2}} }+\sqrt{\left({b}−{c}\right)^{\mathrm{2}} +{d}^{\mathrm{2}} } \\ $$$${s}=\sqrt{\left({x}−{c}\right)^{\mathrm{2}} +\left({y}−{d}\right)^{\mathrm{2}} }+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }+\sqrt{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} } \\ $$$$ \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }−\sqrt{\left({x}−{c}\right)^{\mathrm{2}} +\left({y}−{d}\right)^{\mathrm{2}} }=\sqrt{\left({b}−{c}\right)^{\mathrm{2}} +{d}^{\mathrm{2}} }−{b} \\ $$$$\Rightarrow\sqrt{\left({x}−{b}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\sqrt{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }−\sqrt{\left({b}−{c}\right)^{\mathrm{2}} +{d}^{\mathrm{2}} } \\ $$$$ \\ $$$${example}: \\ $$$${A}\left(\mathrm{0},\mathrm{0}\right),\:{B}\left(\mathrm{5},\mathrm{0}\right),\:{C}\left(\mathrm{2},\mathrm{4}\right) \\ $$$$\Rightarrow{P}\left(\mathrm{2}.\mathrm{7888},\mathrm{1}.\mathrm{1056}\right) \\ $$

Commented by mr W last updated on 28/Jan/19

that was an error.

$${that}\:{was}\:{an}\:{error}. \\ $$

Commented by ajfour last updated on 28/Jan/19

Thank you Sir .

$${Thank}\:{you}\:{Sir}\:. \\ $$

Commented by ajfour last updated on 28/Jan/19

you are seldom in error Sir.

$${you}\:{are}\:{seldom}\:{in}\:{error}\:{Sir}. \\ $$

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