Pythagoras Point :: Challenge

Consider an Triangle ABC and "P" Point inside this Triangle such that : PA=3PA=3 and PB=4PB=4 and PC=5PC=5 . Then Find Maximum Possible Value of : (AB2+BC2+CA2)max\displaystyle{{ ({ AB }^{ 2 }+{ BC }^{ 2 }+{ CA }^{ 2 }) }_{ max }}


I got This Interesting situation from my friend ,But I couldn't able to solve it . Can You ? Take it as Challenge

Thanks!

Note by Karan Shekhawat
4 years, 9 months ago

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Okay , I done it in This way : Let Consider our Triangle frame in Argand Plane By assuming P(0),A(Z1),B(Z2)&C(Z3)P(0)\quad ,\quad A({ Z }_{ 1 })\quad ,\quad B({ Z }_{ 2 })\quad \& \quad C({ Z }_{ 3 }) So our Task is to find Emax{ E }_{ max }

Note : Here I will repeatedly Use : The Standard Properties : Z2=(Z)(Z_)Z+Z_=2Re(Z)\bullet { \left| { Z } \right| }^{ 2 }={ (Z) }(\overset { \_ }{ { Z } } )\\ \bullet Z+\overset { \_ }{ { Z } } =2Re(Z)

E=Z1Z22+Z2Z32+Z3Z12E=(Z1Z2)(Z1_Z2_)+(Z2Z3)(Z2_Z3_)+(Z3Z1)(Z3_Z1_)E=2(Z12+Z22+Z32)2Re(Z1Z2_+Z2Z3_+Z3Z1_)...(1)\displaystyle{E={ \left| { Z }_{ 1 }-{ Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 2 }-{ Z }_{ 3 } \right| }^{ 2 }+{ \left| { Z }_{ 3 }-{ Z }_{ 1 } \right| }^{ 2 }\\ E=({ Z }_{ 1 }-{ Z }_{ 2 })(\overset { \_ }{ { Z }_{ 1 } } -\overset { \_ }{ { Z }_{ 2 } } )+({ Z }_{ 2 }-{ Z }_{ 3 })(\overset { \_ }{ { Z }_{ 2 } } -\overset { \_ }{ { Z }_{ 3 } } )+({ Z }_{ 3 }-{ Z }_{ 1 })(\overset { \_ }{ { Z }_{ 3 } } -\overset { \_ }{ { Z }_{ 1 } } )\\ E=2({ \left| { Z }_{ 1 } \right| }^{ 2 }+{ \left| { Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 3 } \right| }^{ 2 })-2Re({ Z }_{ 1 }\overset { \_ }{ { Z }_{ 2 } } +{ Z }_{ 2 }\overset { \_ }{ { Z }_{ 3 } } +{ Z }_{ 3 }\overset { \_ }{ { Z }_{ 1 } } )\quad \quad .\quad .\quad .\quad (1)}

Now Consider Z1+Z2+Z320(Z1+Z2+Z3)(Z1_+Z2_+Z3_)0Z12+Z22+Z32+2Re(Z1Z2_+Z2Z3_+Z3Z1_)0(2Re(Z1Z2_+Z2Z3_+Z3Z1_))min=(Z12+Z22+Z32)...(2)Emax=3(Z12+Z22+Z32)Emax=3(32+42+52)Emax=150\displaystyle{{ \left| { Z }_{ 1 }{ +Z }_{ 2 }+{ Z }_{ 3 } \right| }^{ 2 }\quad \ge 0\\ ({ Z }_{ 1 }{ +Z }_{ 2 }+{ Z }_{ 3 })(\overset { \_ }{ { Z }_{ 1 } } +\overset { \_ }{ { Z }_{ 2 } } +\overset { \_ }{ { Z }_{ 3 } } )\ge 0\\ { \left| { Z }_{ 1 } \right| }^{ 2 }+{ \left| { Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 3 } \right| }^{ 2 }+2Re({ Z }_{ 1 }\overset { \_ }{ { Z }_{ 2 } } +{ Z }_{ 2 }\overset { \_ }{ { Z }_{ 3 } } +{ Z }_{ 3 }\overset { \_ }{ { Z }_{ 1 } } )\ge 0\\ { (2Re({ Z }_{ 1 }\overset { \_ }{ { Z }_{ 2 } } +{ Z }_{ 2 }\overset { \_ }{ { Z }_{ 3 } } +{ Z }_{ 3 }\overset { \_ }{ { Z }_{ 1 } } )) }_{ min }=-({ \left| { Z }_{ 1 } \right| }^{ 2 }+{ \left| { Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 3 } \right| }^{ 2 })\quad .\quad .\quad .\quad (2)\\ \\ { E }_{ max }=3({ \left| { Z }_{ 1 } \right| }^{ 2 }+{ \left| { Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 3 } \right| }^{ 2 })\\ { E }_{ max }=3({ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 })\\ \boxed { { E }_{ max }=150 } }

But I Didn't able to find AB , BC , AC as individual at this condition !

Deepanshu Gupta - 4 years, 9 months ago

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This is awesome !

Karan Shekhawat - 4 years, 9 months ago

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How did you wrote in a white box?

U Z - 4 years, 8 months ago

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@U Z I Don't Know , Beacuse I didn't Do that

Karan Shekhawat - 4 years, 8 months ago

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nice

Omar El Mokhtar - 4 years, 7 months ago

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Interesting situation ! I'am getting 150 is it correct ? I used Complex numbers @KARAN SHEKHAWAT

Deepanshu Gupta - 4 years, 9 months ago

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I get the same result, with AB=5,BC=73AB = 5, BC = \sqrt{73} and AC=52AC = \sqrt{52}.

Brian Charlesworth - 4 years, 9 months ago

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How did You get Individual Values of all sides ? Can You Please show this ? Because I'am able To find Maximum Value , But I'am unable To find What are Individual Values of Sides . Thanks!

Deepanshu Gupta - 4 years, 9 months ago

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@Deepanshu Gupta Your solution is far more elegant than mine. My approach was to first let the "central" angles at PP opposite sides AB,BC,CAAB, BC,CA be α,β,γ\alpha, \beta, \gamma, respectively. I then used the Cosine Law to establish the equation

AB2+BC2+CA2=10024cos(α)40cos(β)30cos(γ)=f(α,β,γ)AB^{2} + BC^{2} + CA^{2} = 100 - 24\cos(\alpha) - 40\cos(\beta) - 30\cos(\gamma) = f(\alpha, \beta, \gamma).

I then applied Lagrange multipliers under the condition g(α,β,γ)=α+β+γ=2πg(\alpha, \beta, \gamma) = \alpha + \beta + \gamma = 2\pi. Noting the negative signs in the expression for f(α,β,γ)f(\alpha, \beta, \gamma) I further applied the condition that each of the three angles were π2\ge \frac{\pi}{2}, (more of a working condition than a formal one).

This method gave me that ff was maximized when cos(α)=0,cos(β)=0.8\cos(\alpha) = 0, \cos(\beta) = -0.8 and cos(γ)=0.6\cos(\gamma) = -0.6, for which fmax=150f_{\max} = 150. Plugging these values back into the individual Cosine Law equations gave me the side values noted above.

Brian Charlesworth - 4 years, 8 months ago

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@Brian Charlesworth Thanks Sir ! , But Sir Is There any Other Pure Geometric Approach for it ? Thanks Sir

Karan Shekhawat - 4 years, 8 months ago

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@Karan Shekhawat I'm pretty sure there is one, but I haven't found it yet. If and when I do find it I will post it here.

Brian Charlesworth - 4 years, 8 months ago

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Yes My Freind Also Say's that answer is 150 . But Still How did you Solve it by complex number ? :O

Karan Shekhawat - 4 years, 9 months ago

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146 when the triangle collapses into a straight line in the limit . That is BC=5+4=9. CA=5 + 3 +8, AB=4-3 =1. So the sum of squares = 81+64+1=146. With say, vertical line CP=5, draw to locii of B and A as circles center P and radii 4 and 3. P should be in the triangle. Collapsed with P would be CPAB.

Niranjan Khanderia - 4 years, 9 months ago

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I think this would give a maximum value of 92+82+12=81+64+1=1469^{2} + 8^{2} + 1^{2} = 81 + 64 + 1 = 146, just shy of the maximum of 150150 found previously.

Brian Charlesworth - 4 years, 9 months ago

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Can you show your method? I have corrected my calculation.

Niranjan Khanderia - 4 years, 9 months ago

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@Niranjan Khanderia I've outlined my method above in my response to Deepanshu. His solution, however, is far more elegant than mine. Since the cosine values I found above were so "nice", I suspect that there is a geometric solution as well.

Brian Charlesworth - 4 years, 8 months ago

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@Brian Charlesworth Thank you for the method.

Niranjan Khanderia - 4 years, 8 months ago

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