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Question Number 224629 by Abdulazim last updated on 22/Sep/25
Answered by Abdulazim last updated on 22/Sep/25
Help
$${Help} \\ $$
Answered by fantastic last updated on 23/Sep/25
Commented by fantastic last updated on 23/Sep/25
(r/x)=tan ((α/2)) x=(r/(tan ((α/2)))) B^′ C=b−x=b−(r/(tan ((α/2)))) (r/(B^′ C))=tan ((β/2)) B^′ C=(r/(tan ((β/2)))) ∴b−(r/(tan ((α/2))))=(r/(tan ((β/2)))) b=r((1/(tan ((β/2))))+(1/(tan ((α/2))))) ∴r=((b/(((1/(tan ((α/2))))+(1/(tan ((β/2))))))))=((btan ((a/2))tan ((β/2)))/(tan ((α/2))+tan ((β/2))))
$$\frac{{r}}{{x}}=\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$${x}=\frac{{r}}{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)} \\ $$$${B}^{'} {C}={b}−{x}={b}−\frac{{r}}{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)} \\ $$$$\frac{{r}}{{B}^{'} {C}}=\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right) \\ $$$${B}^{'} {C}=\frac{{r}}{\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)} \\ $$$$\therefore{b}−\frac{{r}}{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)}=\frac{{r}}{\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)} \\ $$$${b}={r}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)}+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)}\right) \\ $$$$\therefore{r}=\left(\frac{{b}}{\left(\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)}+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)}\right)}\right)=\frac{{b}\mathrm{tan}\:\left(\frac{{a}}{\mathrm{2}}\right)\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)}{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)+\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)} \\ $$
Commented by mr W last updated on 24/Sep/25
r=(Δ/s) with Δ=Δ_(ABC) =(√(s(s−a)(s−b)(s−c))), s=((a+b+c)/2)
$${r}=\frac{\Delta}{{s}}\: \\ $$$${with}\:\Delta=\Delta_{{ABC}} =\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)},\: \\ $$$${s}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\ $$
Commented by fantastic last updated on 24/Sep/25
didnt know that
$${didnt}\:{know}\:{that} \\ $$
Commented by mr W last updated on 24/Sep/25
Δ=((ar)/2)+((br)/2)+((cr)/2)=(((a+b+c)r)/2)=sr
$$\Delta=\frac{{ar}}{\mathrm{2}}+\frac{{br}}{\mathrm{2}}+\frac{{cr}}{\mathrm{2}}=\frac{\left({a}+{b}+{c}\right){r}}{\mathrm{2}}={sr} \\ $$
Commented by fantastic last updated on 24/Sep/25
thanks sir. now i understand it
$${thanks}\:{sir}. \\ $$$${now}\:{i}\:{understand}\:{it} \\ $$
Answered by fantastic last updated on 23/Sep/25
S_(A^′ B^′ C^′ ) =(1/2)r^2 (sin θ+sin φ+sin ψ) where r=(((btan ((β/2))tan ((α/2)))/(tan ((α/2))+tan ((β/2))))) θ=180^0 −γ φ=180^0 −β ψ=180^0 −α γ=cos^(−1) (((a^2 +c^2 −b^2 )/(2ac))) β=cos^(−1) (((a^2 +b^2 −c^2 )/(2ab))) α=cos^(−1) (((b^2 +c^2 −a^2 )/(2bc)))
$${S}_{{A}^{'} {B}^{'} {C}^{'} } =\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \left(\mathrm{sin}\:\theta+\mathrm{sin}\:\phi+\mathrm{sin}\:\psi\right) \\ $$$${where}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{r}=\left(\frac{{b}\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)+\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)}\right) \\ $$$$\theta=\mathrm{180}^{\mathrm{0}} −\gamma\: \\ $$$$\phi=\mathrm{180}^{\mathrm{0}} −\beta \\ $$$$\psi=\mathrm{180}^{\mathrm{0}} −\alpha \\ $$$$\gamma=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}\right) \\ $$$$\beta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}\right) \\ $$$$\alpha=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right) \\ $$
Commented by mr W last updated on 24/Sep/25
sin ψ=sin A=((2Δ)/(bc))
$$\mathrm{sin}\:\psi=\mathrm{sin}\:{A}=\frac{\mathrm{2}\Delta}{{bc}} \\ $$
Commented by fantastic last updated on 24/Sep/25
understood
$${understood} \\ $$
Answered by mr W last updated on 24/Sep/25
Commented by fantastic last updated on 24/Sep/25
Sir is my solution right? please tell me
$${Sir}\:{is}\:{my}\:{solution}\:{right}? \\ $$$${please}\:{tell}\:{me} \\ $$
Commented by mr W last updated on 24/Sep/25
u+v=c v+w=a w+u=b ⇒u+v+w=((a+b+c)/2)=s ⇒u=s−a=((−a+b+c)/2) ⇒v=s−b=((a−b+c)/2) ⇒w=s−c=((a+b−c)/4) say area of ΔABC is Δ Δ_(AB′C′) =((p^2 sin A)/2)=(u^2 /(bc))×((bc sin A)/2)=((u^2 Δ)/(bc)) similarly Δ_(BC′A′) =((v^2 Δ)/(ca)) Δ_(CA′B′) =((w^2 Δ)/(ab)) Δ_(A′B′C′) =Δ−Δ_(AB′C′) −Δ_(AB′C′) −Δ_(CA′B′) =Δ(1−(u^2 /(bc))−(v^2 /(ca))−(w^2 /(ab))) =Δ[1−(((s−a)^2 )/(bc))−(((s−b)^2 )/(ca))−(((s−c)^2 )/(ab))] =Δ[1−(((−a+b+c)^2 )/(4bc))−(((a−b+c)^2 )/(4ca))−(((a+b−c)^2 )/(4ab))] with Δ=(√(s(s−a)(s−b)(s−c))) alternative method: Δ_(A′B′C′) =((r^2 (sin A+sin B+sin C))/2) =r^2 ((Δ/(bc))+(Δ/(ca))+(Δ/(ab))) =((4Δ^3 )/((a+b+c)^2 ))((1/(bc))+(1/(ca))+(1/(ab))) =((4Δ^3 )/(abc(a+b+c)))
$${u}+{v}={c} \\ $$$${v}+{w}={a} \\ $$$${w}+{u}={b} \\ $$$$\Rightarrow{u}+{v}+{w}=\frac{{a}+{b}+{c}}{\mathrm{2}}={s} \\ $$$$\Rightarrow{u}={s}−{a}=\frac{−{a}+{b}+{c}}{\mathrm{2}} \\ $$$$\Rightarrow{v}={s}−{b}=\frac{{a}−{b}+{c}}{\mathrm{2}} \\ $$$$\Rightarrow{w}={s}−{c}=\frac{{a}+{b}−{c}}{\mathrm{4}} \\ $$$${say}\:{area}\:{of}\:\Delta{ABC}\:{is}\:\Delta \\ $$$$\Delta_{{AB}'{C}'} =\frac{{p}^{\mathrm{2}} \:\mathrm{sin}\:{A}}{\mathrm{2}}=\frac{{u}^{\mathrm{2}} }{{bc}}×\frac{{bc}\:\mathrm{sin}\:{A}}{\mathrm{2}}=\frac{{u}^{\mathrm{2}} \Delta}{{bc}} \\ $$$${similarly} \\ $$$$\Delta_{{BC}'{A}'} =\frac{{v}^{\mathrm{2}} \Delta}{{ca}} \\ $$$$\Delta_{{CA}'{B}'} =\frac{{w}^{\mathrm{2}} \Delta}{{ab}} \\ $$$$\Delta_{{A}'{B}'{C}'} =\Delta−\Delta_{{AB}'{C}'} −\Delta_{{AB}'{C}'} −\Delta_{{CA}'{B}'} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\Delta\left(\mathrm{1}−\frac{{u}^{\mathrm{2}} }{{bc}}−\frac{{v}^{\mathrm{2}} }{{ca}}−\frac{{w}^{\mathrm{2}} }{{ab}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\Delta\left[\mathrm{1}−\frac{\left({s}−{a}\right)^{\mathrm{2}} }{{bc}}−\frac{\left({s}−{b}\right)^{\mathrm{2}} }{{ca}}−\frac{\left({s}−{c}\right)^{\mathrm{2}} }{{ab}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\Delta\left[\mathrm{1}−\frac{\left(−{a}+{b}+{c}\right)^{\mathrm{2}} }{\mathrm{4}{bc}}−\frac{\left({a}−{b}+{c}\right)^{\mathrm{2}} }{\mathrm{4}{ca}}−\frac{\left({a}+{b}−{c}\right)^{\mathrm{2}} }{\mathrm{4}{ab}}\right] \\ $$$${with}\:\Delta=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$$ \\ $$$${alternative}\:{method}: \\ $$$$\Delta_{{A}'{B}'{C}'} =\frac{{r}^{\mathrm{2}} \left(\mathrm{sin}\:{A}+\mathrm{sin}\:{B}+\mathrm{sin}\:{C}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:={r}^{\mathrm{2}} \left(\frac{\Delta}{{bc}}+\frac{\Delta}{{ca}}+\frac{\Delta}{{ab}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}\Delta^{\mathrm{3}} }{\left({a}+{b}+{c}\right)^{\mathrm{2}} }\left(\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ca}}+\frac{\mathrm{1}}{{ab}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}\Delta^{\mathrm{3}} }{{abc}\left({a}+{b}+{c}\right)} \\ $$
Commented by mr W last updated on 24/Sep/25
i don′t know.
$${i}\:{don}'{t}\:{know}. \\ $$
Commented by fantastic last updated on 24/Sep/25
why
$${why} \\ $$
Commented by TonyCWX last updated on 24/Sep/25
Problem is, no angle were given in the diagram. Hence, mrW′s answer will be a better option.
$${Problem}\:{is},\:{no}\:{angle}\:{were}\:{given}\:{in}\:{the}\:{diagram}. \\ $$$${Hence},\:{mrW}'{s}\:{answer}\:{will}\:{be}\:{a}\:{better}\:{option}. \\ $$
Commented by fantastic last updated on 24/Sep/25
you said right
$${you}\:{said}\:{right} \\ $$

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