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Misc 9 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is γ€–γ€–(π‘Žγ€—^(2/3) + 𝑏^(2/3)) γ€—^(3/2)Let P be the point on the hypotenuse AC Given point P is at distance a & b from sides of triangle Let’s construct PL βŠ₯ AB & PM βŠ₯ BC ∴ PL = a & PM = b Let ∠ ACB = ΞΈ Thus, ∠ APL = ΞΈ We need to find maximum length of the hypotenuse Let l be the length of the hypotenuse , ∴ l = AP + PC In βˆ† 𝑨𝑷𝑳 cos ΞΈ = (𝑆𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ πœƒ )/(π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ ) cos ΞΈ = 𝑃𝐿/𝐴𝑃 cos ΞΈ = π‘Ž/𝐴𝑃 AP = π‘Ž/cosβ‘πœƒ AP = 𝒂 π’”π’†π’„β‘πœ½ In βˆ† 𝑷𝑴π‘ͺ sin ΞΈ = (𝑆𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ πœƒ)/(π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ ) sin ΞΈ = 𝑃𝑀/𝑃𝐢 sin ΞΈ = (𝑏 )/𝑃𝐢 PC = 𝑏/sinβ‘πœƒ PC = 𝒃 𝒄𝒐𝒔𝒆𝒄 𝜽 Now, l = AP + PC l = a sec ΞΈ + b cosec ΞΈ Differentiating w.r.t ΞΈ 𝒅𝒍/π’…πœ½=𝑑(π‘Ž secβ‘γ€–πœƒ + 𝑏 π‘π‘œπ‘ π‘’π‘ πœƒγ€— )/π‘‘πœƒ 𝑑𝑙/π‘‘πœƒ= a sec ΞΈ tan ΞΈ – b cosec ΞΈ cot ΞΈ Putting 𝒅𝒍/π’…πœ½=𝟎 a sec ΞΈ tan ΞΈ – b cosec ΞΈ cot ΞΈ = 0 a 1/cosβ‘πœƒ . sinβ‘πœƒ/γ€– cosγ€—β‘πœƒ βˆ’π‘ . 1/sinβ‘γ€–πœƒ γ€— Γ— cosβ‘πœƒ/sinβ‘πœƒ =0 (π‘Ž sinβ‘πœƒ)/cos^2β‘πœƒ βˆ’(𝑏 cosβ‘πœƒ)/sin^2β‘πœƒ =0 (π‘Ž sinβ‘πœƒ)/cos^2β‘πœƒ = (𝑏 cosβ‘πœƒ)/sin^2β‘πœƒ a sin ΞΈ Γ— sin2 ΞΈ = b cos ΞΈ cos2 ΞΈ a sin3 ΞΈ = b cos3 ΞΈ sin^3β‘πœƒ/cos^3β‘πœƒ = 𝑏/π‘Ž tan3 ΞΈ = 𝑏/π‘Ž tan ΞΈ = (𝒃/𝒂)^(𝟏/πŸ‘) Finding (𝒅^𝟐 𝒍)/(𝒅^𝟐 𝜽) 𝑑𝑙/π‘‘πœƒ=π‘Ž sec⁑〖θ tanβ‘πœƒβˆ’π‘ π‘π‘œπ‘ π‘’π‘ πœƒ cotβ‘πœƒ γ€— Differentiating w.rt ΞΈ (𝑑^2 𝑙)/(𝑑^2 πœƒ)= 𝑑(π‘Ž secβ‘γ€–πœƒ tanβ‘γ€–πœƒ βˆ’ 𝑏 π‘π‘œπ‘ π‘’π‘πœƒ cotβ‘πœƒ γ€— γ€— )/π‘‘πœƒ = a 𝑑(secβ‘γ€–πœƒ tanβ‘πœƒ γ€— )/π‘‘πœƒβˆ’π‘ 𝑑(π‘π‘œπ‘ π‘’π‘ πœƒ cotβ‘πœƒ )/π‘‘πœƒ = a((secβ‘πœƒ )^β€² tanβ‘πœƒ+(tanβ‘πœƒ )^β€² secβ‘πœƒ )βˆ’π‘((π‘π‘œπ‘ π‘’π‘ πœƒ)^β€² cotβ‘πœƒ+(cotβ‘πœƒ )^β€² π‘π‘œπ‘ π‘’π‘ πœƒ) = a (secβ‘πœƒ.tanβ‘γ€–πœƒ.tanβ‘πœƒ+sec^2β‘γ€–πœƒ.secβ‘πœƒ γ€— γ€— )βˆ’π‘((βˆ’π‘π‘œπ‘ π‘’π‘ πœƒ.cotβ‘πœƒ ) cotβ‘πœƒ+(βˆ’π‘π‘œπ‘ π‘’π‘^2 πœƒ).π‘π‘œπ‘ π‘’π‘ πœƒ) = a(secβ‘γ€–πœƒ tan^2β‘γ€–πœƒ+sec^3β‘πœƒ γ€— γ€— )βˆ’π‘(βˆ’π‘π‘œπ‘ π‘’π‘ πœƒ cot^2β‘γ€–πœƒβˆ’π‘π‘œπ‘ π‘’π‘^3 πœƒγ€— ) = a sec ΞΈ (〖𝒕𝒂𝒏〗^πŸβ‘γ€–πœ½+〖𝒔𝒆𝒄〗^𝟐⁑𝜽 γ€— )+𝑏 π‘π‘œπ‘ π‘’π‘ πœƒ(〖〖𝒄𝒐𝒕〗^𝟐 πœ½γ€—β‘γ€–+𝒄𝒐𝒔𝒆𝒄^𝟐 πœ½γ€— ) Here, square terms – (〖𝒕𝒂𝒏〗^πŸβ‘γ€–πœ½+〖𝒔𝒆𝒄〗^𝟐⁑𝜽 γ€— ) (〖〖𝒄𝒐𝒕〗^𝟐 πœ½γ€—β‘γ€–+𝒄𝒐𝒔𝒆𝒄^𝟐 πœ½γ€— ) are always positive And Since ΞΈ is acute, i.e. 0 < ΞΈ < πœ‹/2 ∴ ΞΈ lies in 1st quadrant So, sec ΞΈ & cosec ΞΈ will be positive Thus, (𝒅^𝟐 𝒍)/(𝒅^𝟐 𝜽) > 0 at tan ΞΈ = (𝑏/π‘Ž)^(1/3) ∴ l is least when tan ΞΈ = (𝑏/π‘Ž)^(1/3) Now, tan ΞΈ = 𝒃^(𝟏/πŸ‘)/𝒂^(𝟏/πŸ‘) tan ΞΈ = π»π‘’π‘–π‘”β„Žπ‘‘/π΅π‘Žπ‘ π‘’ Height = 𝒃^(𝟏/πŸ‘) & base 𝒂^(𝟏/πŸ‘) Using Pythagoras theorem π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’^2=π»π‘’π‘–π‘”β„Žπ‘‘^2+π΅π‘Žπ‘ π‘’^2 π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’^2=(𝑏^(1/3) )^2+(π‘Ž^(1/3) )^2 π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’† = √(𝒂^(𝟐/πŸ‘) + 𝒃^(𝟐/πŸ‘) ) Least value of l Least value of l = a sec ΞΈ + b cosec ΞΈ = a Γ— (π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’† )/(𝑩𝒂𝒔𝒆 )+𝒃 "Γ— " (π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’† )/(π‘―π’†π’Šπ’ˆπ’‰π’• ) = a Γ— √(π‘Ž^(2/3) + 𝑏^(2/3) )/π‘Ž^(1/3) +b Γ—βˆš(π‘Ž^(2/3) + 𝑏^(2/3) )/𝑏^(1/3) l = √(π‘Ž^(2/3)+𝑏^(2/3) ) (π‘Ž^(1 βˆ’ 1/3)+𝑏^(1 βˆ’ 1/3) ) l = √(π‘Ž^(2/3)+𝑏^(2/3) ) (π‘Ž^(2/3)+𝑏^(2/3) ) l = (π‘Ž^(2/3)+𝑏^(2/3) )^(1/2 + 1) l = (𝒂^(𝟐/πŸ‘)+𝒃^(𝟐/πŸ‘) )^(πŸ‘/𝟐) Hence l = (𝒂^(𝟐/πŸ‘)+𝒃^(𝟐/πŸ‘) )^(πŸ‘/𝟐) Hence proved..

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.