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Question-82878




Question Number 82878 by mr W last updated on 25/Feb/20
Commented by behi83417@gmail.com last updated on 25/Feb/20
Beautiful quistion.waiting for answer...
$$\mathrm{Beautiful}\:\mathrm{quistion}.\mathrm{waiting}\:\mathrm{for}\:\mathrm{answer}… \\ $$
Commented by ajfour last updated on 25/Feb/20
so please answer Behi Sir..
$$\mathrm{so}\:\mathrm{please}\:\mathrm{answer}\:\mathrm{Behi}\:\mathrm{Sir}.. \\ $$
Commented by behi83417@gmail.com last updated on 25/Feb/20
dear Ajfour sir! i am waiting for answer  of my master: mrW.(also your answer  too,most welcome.)
$$\mathrm{dear}\:\mathrm{Ajfour}\:\mathrm{sir}!\:\mathrm{i}\:\mathrm{am}\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{answer} \\ $$$$\mathrm{of}\:\mathrm{my}\:\mathrm{master}:\:\mathrm{mrW}.\left(\mathrm{also}\:\mathrm{your}\:\mathrm{answer}\right. \\ $$$$\left.\mathrm{too},\mathrm{most}\:\mathrm{welcome}.\right) \\ $$
Answered by mr W last updated on 26/Feb/20
Commented by mr W last updated on 26/Feb/20
ΔAFG:  semi perimeter=u  incircle radius =n  excircle radius =a  u=((p+q+s)/2)  n=(Δ_(AFG) /u)  ⇒(1/n)=(u/Δ_(AFG) )  a=(Δ_(AFG) /(u−s))  ⇒(1/a)=((u−s)/Δ_(AFG) )    ΔAGE:  semi perimeter=v  incircle radius =b  excircle radius =m  v=((q+r+t)/2)  similarly  ⇒(1/b)=(v/Δ_(AGE) )  ⇒(1/m)=((v−t)/Δ_(AGE) )    (1/a)+(1/b)=((u−s)/Δ_(AFG) )+(v/Δ_(AGE) )=(1/Δ_(AFG) )(u−s+(Δ_(AFG) /Δ_(AGE) )v)  ⇒(1/a)+(1/b)=(1/Δ_(AFG) )(u−s+(s/t)v)  (1/n)+(1/m)=(u/Δ_(AFG) )+((v−t)/Δ_(AGE) )=(1/Δ_(AFG) )[u+(Δ_(AFG) /Δ_(AGE) )(v−t)]  (1/n)+(1/m)=(1/Δ_(AFG) )[u+(s/t)(v−t)]  ⇒(1/n)+(1/m)=(1/Δ_(AFG) )(u+(s/t)v−s)  ⇒(1/a)+(1/b)=(1/m)+(1/n)
$$\Delta{AFG}: \\ $$$${semi}\:{perimeter}={u} \\ $$$${incircle}\:{radius}\:={n} \\ $$$${excircle}\:{radius}\:={a} \\ $$$${u}=\frac{{p}+{q}+{s}}{\mathrm{2}} \\ $$$${n}=\frac{\Delta_{{AFG}} }{{u}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}=\frac{{u}}{\Delta_{{AFG}} } \\ $$$${a}=\frac{\Delta_{{AFG}} }{{u}−{s}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}=\frac{{u}−{s}}{\Delta_{{AFG}} } \\ $$$$ \\ $$$$\Delta{AGE}: \\ $$$${semi}\:{perimeter}={v} \\ $$$${incircle}\:{radius}\:={b} \\ $$$${excircle}\:{radius}\:={m} \\ $$$${v}=\frac{{q}+{r}+{t}}{\mathrm{2}} \\ $$$${similarly} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{b}}=\frac{{v}}{\Delta_{{AGE}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{{m}}=\frac{{v}−{t}}{\Delta_{{AGE}} } \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{{u}−{s}}{\Delta_{{AFG}} }+\frac{{v}}{\Delta_{{AGE}} }=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left({u}−{s}+\frac{\Delta_{{AFG}} }{\Delta_{{AGE}} }{v}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left({u}−{s}+\frac{{s}}{{t}}{v}\right) \\ $$$$\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{m}}=\frac{{u}}{\Delta_{{AFG}} }+\frac{{v}−{t}}{\Delta_{{AGE}} }=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left[{u}+\frac{\Delta_{{AFG}} }{\Delta_{{AGE}} }\left({v}−{t}\right)\right] \\ $$$$\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{m}}=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left[{u}+\frac{{s}}{{t}}\left({v}−{t}\right)\right] \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{m}}=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left({u}+\frac{{s}}{{t}}{v}−{s}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{m}}+\frac{\mathrm{1}}{{n}} \\ $$
Commented by ajfour last updated on 28/Feb/20
Thanks sir, what a presentation!
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{what}\:\mathrm{a}\:\mathrm{presentation}! \\ $$

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