Misc 19

Transcript

Misc 19 The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that a : b = (m + √(𝑚^2−𝑛^2 )) : (m – √(𝑚^2−𝑛^2 ) ) Introduction Componendo dividendo If 𝑥/𝑦 = 𝑎/𝑏 Applying componendo dividendo (𝑥 + 𝑦)/(𝑥 − 𝑦) = (𝑎 + 𝑏)/(𝑎 − 𝑏) Eg: Taking 1/2 = 4/8 (1+ 2)/(1 − 2) = (4 + 8)/(4 − 8) 3/(−1) = 12/(−4) -3 = -3 Misc 19 The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that a : b = (m + √(𝑚^2−𝑛^2 )) : (m – √(𝑚^2−𝑛^2 ) ) Here, the two numbers be a and b. Arithmetic Mean =AM= (a+b)/2 & Geometric Mean=GM= √ab According to the question, AM/( GM" " ) = 𝑚/𝑛 (𝑎 + 𝑏)/(2 √𝑎𝑏 " " ) = 𝑚/𝑛 Applying componendo dividendo (𝑎+𝑏+2√𝑎𝑏)/(𝑎+𝑏 −2√𝑎𝑏) = (𝑚 + 𝑛)/(𝑚 − 𝑛 ) ((√𝑎)2+(√𝑏)2+2(√𝑎×√𝑏))/((√𝑎)2+(√𝑏)2−2(√𝑎×√𝑏) ) =(𝑚 + 𝑛)/(𝑚 − 𝑛 ) Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy (√𝑎 + √𝑏)2/(√𝑎 − √𝑏)2 = (𝑚 + 𝑛)/(𝑚 − 𝑛 ) ((√𝑎 + √𝑏)/(√𝑎 − √𝑏))^2 = (𝑚 + 𝑛)/(𝑚 − 𝑛 ) (√𝑎 + √𝑏)/(√𝑎 − √𝑏) = √((𝑚 + 𝑛)/(𝑚 − 𝑛 )) (√𝑎 + √𝑏)/(√𝑎 − √𝑏) = √(𝑚 + 𝑛)/(√(𝑚 − 𝑛) ) Applying componendo dividendo ((√𝑎 + √𝑏) + (√𝑎 − √𝑏))/((√𝑎 + √𝑏) − (√𝑎 − √𝑏) ) = (√(𝑚 + 𝑛) + √(𝑚 − 𝑛))/(√(𝑚 + 𝑛) −√(𝑚 − 𝑛)) (2√𝑎)/(2√𝑏) = (√(𝑚 + 𝑛) + √(𝑚 − 𝑛))/(√(𝑚 + 𝑛) −√(𝑚 − 𝑛)) √𝑎/√𝑏 = (√(𝑚 + 𝑛) + √(𝑚 − 𝑛))/(√(𝑚 + 𝑛) −√(𝑚 − 𝑛)) Squaring both sides (√𝑎/√𝑏)^2 = ((√(𝑚 + 𝑛) + √(𝑚 − 𝑛))/(√(𝑚 + 𝑛) −√(𝑚 − 𝑛)))^2 (√𝑎)^2/(√𝑏)^2 = (√(𝑚 + 𝑛) + √(𝑚 − 𝑛))^2/(√(𝑚 + 𝑛) −√(𝑚 − 𝑛))^2 Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy 𝑎/𝑏 = ((√(𝑚 + 𝑛) )^2+(√(𝑚 − 𝑛) )^2+ 2(√(𝑚+𝑛))(√(𝑚−𝑛)))/((√(𝑚 + 𝑛) )^2+(√(𝑚 − 𝑛) )^2− 2(√(𝑚+𝑛))(√(𝑚−𝑛)) ) 𝑎/𝑏 = (𝑚 + 𝑛 + 𝑚 − 𝑛 + 2√((𝑚+𝑛)(𝑚−𝑛) ))/(𝑚 + 𝑛 + 𝑚 − 𝑛 − 2√((𝑚+𝑛)(𝑚−𝑛) )) 𝑎/𝑏 = (𝑚 +𝑚 + 𝑛 − 𝑛 + 2√((𝑚^2− 𝑛^2 ) ))/(𝑚 +𝑚 + 𝑛 − 𝑛 − 2√((𝑚^2− 𝑛^2 ) )) 𝑎/𝑏 = (2𝑚 + 2√((𝑚^2− 𝑛^2 ) ))/(2𝑚 − 2√((𝑚^2− 𝑛^2 ) )) 𝑎/𝑏 = 2(𝑚 + √((𝑚^2− 𝑛^2 ) ))/2(𝑚 − √((𝑚^2− 𝑛^2 ) )) 𝑎/𝑏 = (𝑚 + √((𝑚^2− 𝑛^2 ) ))/(𝑚 − √((𝑚^2− 𝑛^2 ) )) Thus, a : b = (m + √(𝑚^2−𝑛^2 )) : (m – √(𝑚^2−𝑛^2 ) ) Hence proved