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Question Number 214629 by ajfour last updated on 14/Dec/24

Commented by ajfour last updated on 14/Dec/24

say for a=3, b=2.

$${say}\:{for}\:{a}=\mathrm{3},\:{b}=\mathrm{2}. \\ $$

Commented by Ghisom last updated on 24/Dec/24

I get r=((ab)/(2(a+b)))

$$\mathrm{I}\:\mathrm{get} \\ $$$${r}=\frac{{ab}}{\mathrm{2}\left({a}+{b}\right)} \\ $$

Commented by ajfour last updated on 24/Dec/24

let me try!

$${let}\:{me}\:{try}! \\ $$

Commented by mr W last updated on 24/Dec/24

that′s correct!

$${that}'{s}\:{correct}! \\ $$

Answered by ajfour last updated on 24/Dec/24

Commented by ajfour last updated on 25/Dec/24

a=(r/(tan α))+(r/(tan ((π/4)−θ))) say tan α=t, tan θ=m ⇒ (1/t)=(a/r)−(((1+m)/(1−m))) or ((m+1)/(m−1))=(1/t)−(a/r) (1/m)=((((1/t)−(a/r))−1)/(((1/t)−(a/r))+1)) ....(i) ((2t)/(1−t^2 ))=(b/a)=(1/k) t^2 +2kt−1=0 t=(√(k^2 +1))−k (r/(tan θ))−(r/(tan ((π/4)+α)))=b (1/m)=(b/r)+((1−t)/(1+t)) ⇒ using ..(i) we get ((((1/t)−(a/r))−1)/(((1/t)−(a/r))+1))=(b/r)+((1−t)/(1+t)) eg. a=4, b=3 ⇒ k=(4/3), t=(1/3) ((2−(4/r))/(4−(4/r)))=(3/r)+(1/2) 2−(4/r)=((12)/r)+2−((12)/r^2 )−(2/r) ⇒ 14r=12 r=(6/7)

$${a}=\frac{{r}}{\mathrm{tan}\:\alpha}+\frac{{r}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)} \\ $$$${say}\:\:\mathrm{tan}\:\alpha={t},\:\:\mathrm{tan}\:\theta={m} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{{t}}=\frac{{a}}{{r}}−\left(\frac{\mathrm{1}+{m}}{\mathrm{1}−{m}}\right)\:\:\:{or} \\ $$$$\frac{{m}+\mathrm{1}}{{m}−\mathrm{1}}=\frac{\mathrm{1}}{{t}}−\frac{{a}}{{r}} \\ $$$$\frac{\mathrm{1}}{{m}}=\frac{\left(\frac{\mathrm{1}}{{t}}−\frac{{a}}{{r}}\right)−\mathrm{1}}{\left(\frac{\mathrm{1}}{{t}}−\frac{{a}}{{r}}\right)+\mathrm{1}}\:\:\:\:....\left({i}\right) \\ $$$$\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{{b}}{{a}}=\frac{\mathrm{1}}{{k}} \\ $$$${t}^{\mathrm{2}} +\mathrm{2}{kt}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}−{k} \\ $$$$\frac{{r}}{\mathrm{tan}\:\theta}−\frac{{r}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\alpha\right)}={b} \\ $$$$\frac{\mathrm{1}}{{m}}=\frac{{b}}{{r}}+\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\:\:\Rightarrow\:\:\:{using}\:..\left({i}\right)\:{we}\:{get} \\ $$$$\frac{\left(\frac{\mathrm{1}}{{t}}−\frac{{a}}{{r}}\right)−\mathrm{1}}{\left(\frac{\mathrm{1}}{{t}}−\frac{{a}}{{r}}\right)+\mathrm{1}}=\frac{{b}}{{r}}+\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}} \\ $$$${eg}.\:\:{a}=\mathrm{4},\:{b}=\mathrm{3}\:\:\Rightarrow\:\:\:{k}=\frac{\mathrm{4}}{\mathrm{3}},\:{t}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\mathrm{2}−\frac{\mathrm{4}}{{r}}}{\mathrm{4}−\frac{\mathrm{4}}{{r}}}=\frac{\mathrm{3}}{{r}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\cancel{\mathrm{2}}−\frac{\mathrm{4}}{{r}}=\frac{\mathrm{12}}{{r}}+\cancel{\mathrm{2}}−\frac{\mathrm{12}}{{r}^{\mathrm{2}} }−\frac{\mathrm{2}}{{r}} \\ $$$$\Rightarrow\:\:\mathrm{14}{r}=\mathrm{12} \\ $$$$\:\:\:\:{r}=\frac{\mathrm{6}}{\mathrm{7}} \\ $$

Answered by mr W last updated on 24/Dec/24

Commented by mr W last updated on 24/Dec/24

c=(√(a^2 +b^2 )) AE=AD=(r/(tan (A/2))) CF=CD=a−(r/(tan (A/2))) BK=BG=r tan (B/2) CH=CK=b+r tan (B/2) FH=b+r tan (B/2)−(a−(r/(tan (A/2)))) GE=c−r tan (B/2)−(r/(tan (A/2))) GE=FH c−r tan (B/2)−(r/(tan (A/2)))=b+r tan (B/2)−a+(r/(tan (A/2))) 2r(tan (B/2)+(1/(tan (A/2))))=c+a−b 2r(((1−cos B)/(sin B))+((1+cos A)/(sin A)))=c+a−b 2r(((c−b)/a)+((c+a)/b))=c+a−b 2r(((cb−b^2 +ac+a^2 )/(ab)))=c+a−b ((2r(a+b)(c+a−b))/(ab))=c+a−b ⇒r=((ab)/(2(a+b))) ✓

$${c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${AE}={AD}=\frac{{r}}{\mathrm{tan}\:\frac{{A}}{\mathrm{2}}} \\ $$$${CF}={CD}={a}−\frac{{r}}{\mathrm{tan}\:\frac{{A}}{\mathrm{2}}} \\ $$$${BK}={BG}={r}\:\mathrm{tan}\:\frac{{B}}{\mathrm{2}} \\ $$$${CH}={CK}={b}+{r}\:\mathrm{tan}\:\frac{{B}}{\mathrm{2}} \\ $$$${FH}={b}+{r}\:\mathrm{tan}\:\frac{{B}}{\mathrm{2}}−\left({a}−\frac{{r}}{\mathrm{tan}\:\frac{{A}}{\mathrm{2}}}\right) \\ $$$${GE}={c}−{r}\:\mathrm{tan}\:\frac{{B}}{\mathrm{2}}−\frac{{r}}{\mathrm{tan}\:\frac{{A}}{\mathrm{2}}} \\ $$$${GE}={FH} \\ $$$${c}−{r}\:\mathrm{tan}\:\frac{{B}}{\mathrm{2}}−\frac{{r}}{\mathrm{tan}\:\frac{{A}}{\mathrm{2}}}={b}+{r}\:\mathrm{tan}\:\frac{{B}}{\mathrm{2}}−{a}+\frac{{r}}{\mathrm{tan}\:\frac{{A}}{\mathrm{2}}} \\ $$$$\mathrm{2}{r}\left(\mathrm{tan}\:\frac{{B}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{{A}}{\mathrm{2}}}\right)={c}+{a}−{b} \\ $$$$\mathrm{2}{r}\left(\frac{\mathrm{1}−\mathrm{cos}\:{B}}{\mathrm{sin}\:{B}}+\frac{\mathrm{1}+\mathrm{cos}\:{A}}{\mathrm{sin}\:{A}}\right)={c}+{a}−{b} \\ $$$$\mathrm{2}{r}\left(\frac{{c}−{b}}{{a}}+\frac{{c}+{a}}{{b}}\right)={c}+{a}−{b} \\ $$$$\mathrm{2}{r}\left(\frac{{cb}−{b}^{\mathrm{2}} +{ac}+{a}^{\mathrm{2}} }{{ab}}\right)={c}+{a}−{b} \\ $$$$\frac{\mathrm{2}{r}\left({a}+{b}\right)\left({c}+{a}−{b}\right)}{{ab}}={c}+{a}−{b} \\ $$$$\Rightarrow{r}=\frac{{ab}}{\mathrm{2}\left({a}+{b}\right)}\:\checkmark \\ $$

Commented by Ghisom last updated on 25/Dec/24

yes.

$$\mathrm{yes}. \\ $$

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