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Question-54068




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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