# Pythagoras Point :: Challenge

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

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
3 years, 4 months ago

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

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

$$\displaystyle{E={ \left| { Z }_{ 1 }-{ Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 2 }-{ Z }_{ 3 } \right| }^{ 2 }+{ \left| { Z }_{ 3 }-{ Z }_{ 1 } \right| }^{ 2 }\\ E=({ Z }_{ 1 }-{ Z }_{ 2 })(\overset { \_ }{ { Z }_{ 1 } } -\overset { \_ }{ { Z }_{ 2 } } )+({ Z }_{ 2 }-{ Z }_{ 3 })(\overset { \_ }{ { Z }_{ 2 } } -\overset { \_ }{ { Z }_{ 3 } } )+({ Z }_{ 3 }-{ Z }_{ 1 })(\overset { \_ }{ { Z }_{ 3 } } -\overset { \_ }{ { Z }_{ 1 } } )\\ E=2({ \left| { Z }_{ 1 } \right| }^{ 2 }+{ \left| { Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 3 } \right| }^{ 2 })-2Re({ Z }_{ 1 }\overset { \_ }{ { Z }_{ 2 } } +{ Z }_{ 2 }\overset { \_ }{ { Z }_{ 3 } } +{ Z }_{ 3 }\overset { \_ }{ { Z }_{ 1 } } )\quad \quad .\quad .\quad .\quad (1)}$$

Now Consider $$\displaystyle{{ \left| { Z }_{ 1 }{ +Z }_{ 2 }+{ Z }_{ 3 } \right| }^{ 2 }\quad \ge 0\\ ({ Z }_{ 1 }{ +Z }_{ 2 }+{ Z }_{ 3 })(\overset { \_ }{ { Z }_{ 1 } } +\overset { \_ }{ { Z }_{ 2 } } +\overset { \_ }{ { Z }_{ 3 } } )\ge 0\\ { \left| { Z }_{ 1 } \right| }^{ 2 }+{ \left| { Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 3 } \right| }^{ 2 }+2Re({ Z }_{ 1 }\overset { \_ }{ { Z }_{ 2 } } +{ Z }_{ 2 }\overset { \_ }{ { Z }_{ 3 } } +{ Z }_{ 3 }\overset { \_ }{ { Z }_{ 1 } } )\ge 0\\ { (2Re({ Z }_{ 1 }\overset { \_ }{ { Z }_{ 2 } } +{ Z }_{ 2 }\overset { \_ }{ { Z }_{ 3 } } +{ Z }_{ 3 }\overset { \_ }{ { Z }_{ 1 } } )) }_{ min }=-({ \left| { Z }_{ 1 } \right| }^{ 2 }+{ \left| { Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 3 } \right| }^{ 2 })\quad .\quad .\quad .\quad (2)\\ \\ { E }_{ max }=3({ \left| { Z }_{ 1 } \right| }^{ 2 }+{ \left| { Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 3 } \right| }^{ 2 })\\ { E }_{ max }=3({ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 })\\ \boxed { { E }_{ max }=150 } }$$

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

- 3 years, 4 months ago

nice

- 3 years, 2 months ago

This is awesome !

- 3 years, 4 months ago

How did you wrote in a white box?

- 3 years, 4 months ago

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

- 3 years, 4 months ago

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

- 3 years, 4 months ago

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

- 3 years, 4 months ago

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!

- 3 years, 4 months ago

Your solution is far more elegant than mine. My approach was to first let the "central" angles at $$P$$ opposite sides $$AB, BC,CA$$ be $$\alpha, \beta, \gamma$$, respectively. I then used the Cosine Law to establish the equation

$$AB^{2} + BC^{2} + CA^{2} = 100 - 24\cos(\alpha) - 40\cos(\beta) - 30\cos(\gamma) = f(\alpha, \beta, \gamma)$$.

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

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

- 3 years, 4 months ago

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

- 3 years, 4 months ago

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.

- 3 years, 4 months ago

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

- 3 years, 4 months ago

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.

- 3 years, 4 months ago

I think this would give a maximum value of $$9^{2} + 8^{2} + 1^{2} = 81 + 64 + 1 = 146$$, just shy of the maximum of $$150$$ found previously.

- 3 years, 4 months ago

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

- 3 years, 4 months ago

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.

- 3 years, 4 months ago

Thank you for the method.

- 3 years, 4 months ago