Question

در مثلث ABC رابطه زیر برقرار است: $$ a^{2} + b^{2} =2023 c^{2} $$ مطلوب است محاسبه: $$ \frac{cotC}{cotA+cotB} $$

Answer 1

$$cotC= \frac{cosC}{sinC} = \frac{2abcosC}{2absinC} , a^{2} + b^{2} -2abcosC= c^{2} ,2abcosC = a^{2} + b^{2} - c^{2} ,2absinC =4S \Longrightarrow cotC= \frac{ a^{2} + b^{2} - c^{2} }{4S} \Longrightarrow cotB= \frac{ a^{2} + c^{2} - b^{2} }{4S} ,cotA= \frac{ b^{2} + c^{2} - a^{2} }{4S} \Longrightarrow \frac{cotC}{cotA+cotB} = \frac{ a^{2} + b^{2} - c^{2} }{ 2c^{2} } = \frac{2022 c^{2} }{2 c^{2} } =1011$$

Answer 2

در هر مثلث داریم:

$ \frac{a}{SinA} = \frac{b}{SinB} = \frac{c}{SinC} =2R , c^2=a^2+b^2-2abCosC$

$CotA= \frac{CosA}{SinA} = \frac{ \frac{b^2+c^2-a^2}{2bc} }{ \frac{a}{2R} } = \frac{R(b^2+c^2-a^2)}{abc} $

$ \Rightarrow \frac{CotC}{CotA+CotB}= \frac{ \frac{R(b^2+a^2-c^2)}{abc}}{ \frac{R(b^2+c^2-a^2)}{abc}+ \frac{R(a^2+c^2-b^2)}{abc}} = \frac{b^2+a^2-c^2}{b^2+c^2-a^2+a^2+c^2-b^2} = \frac{b^2+a^2-c^2}{2c^2}= \frac{2023c^2-c^2}{2c^2}= \frac{2022c^2}{2c^2} $

$=1011$

$ \Box $