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Question Number 205062 by BaliramKumar last updated on 07/Mar/24

Answered by cortano12 last updated on 07/Mar/24

P_(max) = 240

$$\:\:\mathrm{P}_{\mathrm{max}} \:=\:\mathrm{240}\: \\ $$

Commented by BaliramKumar last updated on 07/Mar/24

solution

$$\mathrm{solution} \\ $$

Answered by mr W last updated on 07/Mar/24

y^2 −x^2 =15^2 (y−x)(y+x)=15^2 =1×225 (y−x)_(min) =1 (y+x)_(max) =225 (x=112, y=113) ⇒P_(max) =(y+x)_(max) +15=225+15=240

$${y}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} \: \\ $$$$\left({y}−{x}\right)\left({y}+{x}\right)=\mathrm{15}^{\mathrm{2}} =\mathrm{1}×\mathrm{225} \\ $$$$\left({y}−{x}\right)_{{min}} =\mathrm{1} \\ $$$$\left({y}+{x}\right)_{{max}} =\mathrm{225} \\ $$$$\left({x}=\mathrm{112},\:{y}=\mathrm{113}\right) \\ $$$$\Rightarrow{P}_{{max}} =\left({y}+{x}\right)_{{max}} +\mathrm{15}=\mathrm{225}+\mathrm{15}=\mathrm{240} \\ $$

Commented by BaliramKumar last updated on 07/Mar/24

thanks

$$\mathrm{thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 07/Mar/24

If a,b∈N with a>b then a^2 −b^2 , 2ab , a^2 +b^2 are sides of right triangle i-e pathagorean triple. 2ab is certainly an even number and a^2 +b^2 is hypotenuse. Hence a^2 −b^2 =15 (a−b)(a+b)=1×3×5 { ((a−b=1 ∧ a+b=15⇒(a,b)=(8,7))),((a−b=3 ∧ a+b=5⇒(a,b)=(4,1))) :} determinant ((a,b,(a^2 −b^2 ),(2ab),(a^2 +b^2 ),(perimeter)),(8,7,(15),(112),(113),(240)),(4,1,(15),8,(17),(40))) The above formula works only for primitive triples.To find out non- primitive triples we can use this same formula to find out primitive triples, first, with one side 3 or 5: a^2 −b^2 =3⇒(a−b)(a+b)=1×3 a−b=1∧ a+b=3⇒(a,b)=(2,1) a^2 −b^2 =5⇒(a−b)(a+b)=1×5 a−b=1∧ a+b=5⇒(a,b)=(3,2) determinant ((a,b,(a^2 −b^2 ),(2ab),(a^2 +b^2 )),(2,1,3,4,5),(3,2,5,(12),(13))) (3,4,5)×5=(15,20,25)⇒p=60 (5,12,13)×3=(15,36,39)⇒p=80 Yet the max p=240

$${If}\:{a},{b}\in\mathbb{N}\:{with}\:{a}>{b}\:{then} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} ,\:\mathrm{2}{ab}\:,\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{are}\:{sides}\:{of} \\ $$$$\boldsymbol{{right}}\:\boldsymbol{{triangle}}\:{i}-{e}\:{pathagorean} \\ $$$${triple}. \\ $$$$\mathrm{2}{ab}\:{is}\:{certainly}\:{an}\:{even}\:{number} \\ $$$${and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{is}\:{hypotenuse}.\:{Hence} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{15} \\ $$$$\left({a}−{b}\right)\left({a}+{b}\right)=\mathrm{1}×\mathrm{3}×\mathrm{5} \\ $$$$\begin{cases}{{a}−{b}=\mathrm{1}\:\wedge\:{a}+{b}=\mathrm{15}\Rightarrow\left({a},{b}\right)=\left(\mathrm{8},\mathrm{7}\right)}\\{{a}−{b}=\mathrm{3}\:\wedge\:{a}+{b}=\mathrm{5}\Rightarrow\left({a},{b}\right)=\left(\mathrm{4},\mathrm{1}\right)}\end{cases} \\ $$$$\begin{array}{|c|c|c|}{{a}}&\hline{{b}}&\hline{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }&\hline{\mathrm{2}{ab}}&\hline{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }&\hline{{perimeter}}\\{\mathrm{8}}&\hline{\mathrm{7}}&\hline{\mathrm{15}}&\hline{\mathrm{112}}&\hline{\mathrm{113}}&\hline{\mathrm{240}}\\{\mathrm{4}}&\hline{\mathrm{1}}&\hline{\mathrm{15}}&\hline{\mathrm{8}}&\hline{\mathrm{17}}&\hline{\mathrm{40}}\\\hline\end{array}\: \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{above}}\:\boldsymbol{\mathrm{formula}}\:\boldsymbol{\mathrm{works}}\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{for}} \\ $$$$\boldsymbol{\mathrm{primitive}}\:\boldsymbol{\mathrm{triples}}.\boldsymbol{\mathrm{To}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{out}}\:\boldsymbol{\mathrm{non}}- \\ $$$$\boldsymbol{\mathrm{primitive}}\:\boldsymbol{\mathrm{triples}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{use}}\:\boldsymbol{\mathrm{this}} \\ $$$$\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{formula}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{out}}\:\boldsymbol{\mathrm{primitive}} \\ $$$$\boldsymbol{\mathrm{triples}},\:\boldsymbol{\mathrm{first}},\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{side}}\:\mathrm{3}\:\boldsymbol{\mathrm{or}}\:\mathrm{5}: \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{3}\Rightarrow\left({a}−{b}\right)\left({a}+{b}\right)=\mathrm{1}×\mathrm{3} \\ $$$$\:\:\:{a}−{b}=\mathrm{1}\wedge\:{a}+{b}=\mathrm{3}\Rightarrow\left({a},{b}\right)=\left(\mathrm{2},\mathrm{1}\right) \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{5}\Rightarrow\left({a}−{b}\right)\left({a}+{b}\right)=\mathrm{1}×\mathrm{5} \\ $$$$\:\:\:{a}−{b}=\mathrm{1}\wedge\:{a}+{b}=\mathrm{5}\Rightarrow\left({a},{b}\right)=\left(\mathrm{3},\mathrm{2}\right) \\ $$$$\begin{array}{|c|c|c|}{{a}}&\hline{{b}}&\hline{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }&\hline{\mathrm{2}{ab}}&\hline{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\\{\mathrm{2}}&\hline{\mathrm{1}}&\hline{\mathrm{3}}&\hline{\mathrm{4}}&\hline{\mathrm{5}}\\{\mathrm{3}}&\hline{\mathrm{2}}&\hline{\mathrm{5}}&\hline{\mathrm{12}}&\hline{\mathrm{13}}\\\hline\end{array} \\ $$$$\left(\mathrm{3},\mathrm{4},\mathrm{5}\right)×\mathrm{5}=\left(\mathrm{15},\mathrm{20},\mathrm{25}\right)\Rightarrow{p}=\mathrm{60} \\ $$$$\left(\mathrm{5},\mathrm{12},\mathrm{13}\right)×\mathrm{3}=\left(\mathrm{15},\mathrm{36},\mathrm{39}\right)\Rightarrow{p}=\mathrm{80}\: \\ $$$${Yet}\:{the}\:{max}\:{p}=\mathrm{240} \\ $$

Commented by BaliramKumar last updated on 07/Mar/24

thanks i think 4 integral side triangles possible

$$\mathrm{thanks} \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{4}\:\mathrm{integral}\:\mathrm{side}\:\mathrm{triangles}\:\mathrm{possible} \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 07/Mar/24

Yes sir, I′ve extended my answer to cover all possible triples now.

$${Yes}\:{sir},\:{I}'{ve}\:{extended}\:{my}\:{answer}\:{to} \\ $$$${cover}\:{all}\:{possible}\:{triples}\:{now}. \\ $$

Commented by BaliramKumar last updated on 07/Mar/24

Yes sir also y^2 − x^2 = 15^(2 ) = 225 (y−x)(y+x) = 1×225 (y−x)(y+x) = 3×75 (y−x)(y+x) = 5×45 (y−x)(y+x) = 9×25 (y−x)(y+x) = 15×15

$$\mathrm{Yes}\:\mathrm{sir} \\ $$$$\mathrm{also}\:\:\:\:\:{y}^{\mathrm{2}} \:−\:{x}^{\mathrm{2}} \:=\:\mathrm{15}^{\mathrm{2}\:} \:=\:\mathrm{225} \\ $$$$\left({y}−{x}\right)\left({y}+{x}\right)\:=\:\mathrm{1}×\mathrm{225} \\ $$$$\left({y}−{x}\right)\left({y}+{x}\right)\:=\:\mathrm{3}×\mathrm{75} \\ $$$$\left({y}−{x}\right)\left({y}+{x}\right)\:=\:\mathrm{5}×\mathrm{45} \\ $$$$\left({y}−{x}\right)\left({y}+{x}\right)\:=\:\mathrm{9}×\mathrm{25} \\ $$$$\left({y}−{x}\right)\left({y}+{x}\right)\:=\:\cancel{\mathrm{15}×\mathrm{15}} \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 08/Mar/24

An easy way!

$${An}\:{easy}\:{way}! \\ $$

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