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Question Number 190340 by ajfour last updated on 01/Apr/23
Commented by a.lgnaoui last updated on 01/Apr/23
Commented by a.lgnaoui last updated on 01/Apr/23
Commented by mr W last updated on 01/Apr/23
Commented by mr W last updated on 01/Apr/23
there are infinite equilateral triangles. one is the smallest (red one) and one is the largest (green one).
$${there}\:{are}\:{infinite}\:{equilateral}\:{triangles}. \\ $$$${one}\:{is}\:{the}\:{smallest}\:\left({red}\:{one}\right)\:{and}\:{one} \\ $$$${is}\:{the}\:{largest}\:\left({green}\:{one}\right). \\ $$
Answered by a.lgnaoui last updated on 01/Apr/23
soit MN Droites tengent aux cercles MN∣∣ EA et EA∣∣BC C_1 (O_1 ,b) et cercleC_2 (O_2 ,a) triangle EAO_2 (Equilaterale) OA=R point A tel que O_2 A=EA=R R=a+b tan (π/6)=(R/(2h))=((√3)/3)⇒h=(((a+b)(√3))/2) △ABC AEO_2 semblables (h/(S+a))=((√3)/2)⇒S=((2h)/( (√3)))−a S=(a+b)−a S=b
$${soit}\:{MN}\:{Droites}\:{tengent}\:{aux}\:{cercles} \\ $$$${MN}\mid\mid\:{EA}\:\:{et}\:\:{EA}\mid\mid{BC} \\ $$$${C}_{\mathrm{1}} \left({O}_{\mathrm{1}} ,{b}\right)\:\:{et}\:{cercleC}_{\mathrm{2}} \left({O}_{\mathrm{2}} ,{a}\right) \\ $$$${triangle}\:{EAO}_{\mathrm{2}} \left({Equilaterale}\right) \\ $$$${OA}={R} \\ $$$${point}\:{A}\:{tel}\:{que}\:{O}_{\mathrm{2}} {A}={EA}={R} \\ $$$${R}={a}+{b} \\ $$$$\mathrm{tan}\:\frac{\pi}{\mathrm{6}}=\frac{{R}}{\mathrm{2}{h}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\Rightarrow{h}=\frac{\left({a}+{b}\right)\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\bigtriangleup{ABC}\:\:\:{AEO}_{\mathrm{2}} \:\:{semblables} \\ $$$$\:\frac{{h}}{{S}+{a}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow{S}=\frac{\mathrm{2}{h}}{\:\sqrt{\mathrm{3}}}−{a} \\ $$$${S}=\left({a}+{b}\right)−{a} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{S}}=\boldsymbol{{b}} \\ $$
Answered by ajfour last updated on 01/Apr/23
Commented by ajfour last updated on 07/Apr/23
B(0,b) A(0,−a) Q(bsin φ, b−bcos φ) P(asin θ,−a+acos θ) let M(h,k) bsin φ=h+(s/2)cos δ b−bcos φ=k+(s/2)sin δ asin θ=h−(s/2)cos δ −a+acos θ=k−(s/2)sin δ ⇒ 2k=b−a+acos θ−bcos φ ssin δ=b+a−acos θ−bcos φ 2h=bsin φ+asin θ scos δ=bsin φ−asin θ T(p,q) p=h+((s(√3))/2)sin δ q=k−((s(√3))/2)cos δ (h+((s(√3))/2)sin δ)^2 +(k−((s(√3))/2)cos δ−b+a)^2 =(a+b)^2 θ and φ are then related herein {bsin φ+asin θ+(√3)(b+a−acos θ−bcos φ)}^2 +{(b−a+acos θ−bcos φ) −(√3)(bsin φ−asin θ)−2(b−a)}^2 =4(a+b)^2 while s^2 =(b+a−acos θ−bcos φ)^2 +(bsin φ−asin θ)^2 .....
$${B}\left(\mathrm{0},{b}\right)\:\:\:{A}\left(\mathrm{0},−{a}\right) \\ $$$${Q}\left({b}\mathrm{sin}\:\phi,\:{b}−{b}\mathrm{cos}\:\phi\right) \\ $$$${P}\left({a}\mathrm{sin}\:\theta,−{a}+{a}\mathrm{cos}\:\theta\right) \\ $$$${let}\:\:{M}\left({h},{k}\right) \\ $$$${b}\mathrm{sin}\:\phi={h}+\frac{{s}}{\mathrm{2}}\mathrm{cos}\:\delta \\ $$$${b}−{b}\mathrm{cos}\:\phi={k}+\frac{{s}}{\mathrm{2}}\mathrm{sin}\:\delta \\ $$$${a}\mathrm{sin}\:\theta={h}−\frac{{s}}{\mathrm{2}}\mathrm{cos}\:\delta \\ $$$$−{a}+{a}\mathrm{cos}\:\theta={k}−\frac{{s}}{\mathrm{2}}\mathrm{sin}\:\delta \\ $$$$\Rightarrow\:\:\mathrm{2}{k}={b}−{a}+{a}\mathrm{cos}\:\theta−{b}\mathrm{cos}\:\phi \\ $$$${s}\mathrm{sin}\:\delta={b}+{a}−{a}\mathrm{cos}\:\theta−{b}\mathrm{cos}\:\phi \\ $$$$\mathrm{2}{h}={b}\mathrm{sin}\:\phi+{a}\mathrm{sin}\:\theta \\ $$$${s}\mathrm{cos}\:\delta={b}\mathrm{sin}\:\phi−{a}\mathrm{sin}\:\theta \\ $$$${T}\left({p},{q}\right) \\ $$$${p}={h}+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\delta \\ $$$${q}={k}−\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\delta \\ $$$$\left({h}+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\delta\right)^{\mathrm{2}} +\left({k}−\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\delta−{b}+{a}\right)^{\mathrm{2}} \\ $$$$=\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\theta\:{and}\:\phi\:\:{are}\:{then}\:{related}\:{herein} \\ $$$$\left\{{b}\mathrm{sin}\:\phi+{a}\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\left({b}+{a}−{a}\mathrm{cos}\:\theta−{b}\mathrm{cos}\:\phi\right)\right\}^{\mathrm{2}} \\ $$$$+\left\{\left({b}−{a}+{a}\mathrm{cos}\:\theta−{b}\mathrm{cos}\:\phi\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:−\sqrt{\mathrm{3}}\left({b}\mathrm{sin}\:\phi−{a}\mathrm{sin}\:\theta\right)−\mathrm{2}\left({b}−{a}\right)\right\}^{\mathrm{2}} \\ $$$$=\mathrm{4}\left({a}+{b}\right)^{\mathrm{2}} \\ $$$${while} \\ $$$${s}^{\mathrm{2}} =\left({b}+{a}−{a}\mathrm{cos}\:\theta−{b}\mathrm{cos}\:\phi\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:+\left({b}\mathrm{sin}\:\phi−{a}\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$….. \\ $$
Answered by mr W last updated on 02/Apr/23
Commented by mr W last updated on 04/Apr/23
r=a+b C(r sin θ, r cos θ) (x−r sin θ)^2 +(y−r cos θ)^2 =s^2 point A: (x_A −r sin θ)^2 +(y_A −r cos θ)^2 =s^2 (x_A −b)^2 +y_A ^2 =a^2 (b−r sin θ)(2x_A −b−r sin θ)−r cos θ (2y_A −r cos θ)=s^2 −a^2 2(b−r sin θ)x_A −2r cos θ y_A =s^2 −a^2 +b^2 −r^2 ? ⇒y_A =((2(b−r sin θ)x_A −s^2 +a^2 −b^2 +r^2 )/(2r cos θ)) (x_A −b)^2 +[((2(b−r sin θ)x_A −s^2 +a^2 −b^2 +r^2 )/(2r cos θ))]^2 =a^2 x_A ^2 −2bx_A +b^2 +[(((b−r sin θ)x_A )/(r cos θ))−((s^2 −a^2 +b^2 −r^2 )/(2r cos θ))]^2 =a^2 [1+(((b−r sin θ)^2 )/(r^2 cos^2 θ))]x_A ^2 −2[b+(((b−r sin θ)(s^2 −a^2 +b^2 −r^2 ))/(2r^2 cos^2 θ))]x_A +(((s^2 −a^2 +b^2 −r^2 )^2 )/(4r^2 cos^2 θ))+b^2 −a^2 =0 ⇒x_A =(1/(1+(((b−r sin θ)^2 )/(r^2 cos^2 θ)))){b+(((b−r sin θ)(s^2 −a^2 +b^2 −r^2 ))/(2r^2 cos^2 θ))±(√([b+(((b−r sin θ)(s^2 −a^2 +b^2 −r^2 ))/(2r^2 cos^2 θ))]^2 −[1+(((b−r sin θ)^2 )/(r^2 cos^2 θ))][(((s^2 −a^2 +b^2 −r^2 )^2 )/(4r^2 cos^2 θ))+b^2 −a^2 ]))} point B: (x_B −r sin θ)^2 +(y_B −r cos θ)^2 =s^2 (x_B +a)^2 +y_B ^2 =b^2 −(a+r sin θ)(2x_B +a−r sin θ)−r cos θ (2y_B −r cos θ)=s^2 −b^2 2(a+r sin θ) x_B +2r cos θ y_B =−s^2 +b^2 −a^2 +r^2 ⇒y_B =((−2(a+r sin θ)x_B −s^2 +b^2 −a^2 +r^2 )/(2r cos θ)) (x_B +a)^2 +[((−2(a+r sin θ)x_B −s^2 +b^2 −a^2 +r^2 )/(2r cos θ))]^2 =b^2 (x_B +a)^2 +[(((a+r sin θ)x_B )/(r cos θ))+((s^2 −b^2 +a^2 −r^2 )/(2r cos θ))]^2 =b^2 [1+(((a+r sin θ)^2 )/(r^2 cos^2 θ))]x_B ^2 +2[a+(((a+r sin θ)(s^2 −b^2 +a^2 −r^2 ))/(2r^2 cos^2 θ))]x_B +(((s^2 −b^2 +a^2 −r^2 )^2 )/(4r^2 cos^2 θ))+a^2 −b^2 =0 ⇒x_B =(1/(1+(((a+r sin θ)^2 )/(r^2 cos^2 θ)))){−a−(((a+r sin θ)(s^2 −b^2 +a^2 −r^2 ))/(2r^2 cos^2 θ))±(√([a+(((a+r sin θ)(s^2 −b^2 +a^2 −r^2 ))/(2r^2 cos^2 θ))]^2 −[1+(((a+r sin θ)^2 )/(r^2 cos^2 θ))][(((s^2 −b^2 +a^2 −r^2 )^2 )/(4r^2 cos^2 θ))+a^2 −b^2 ]))} AB=s: (x_A −x_B )^2 +(y_A −y_B )^2 =s^2 x_A ^2 +y_A ^2 +x_B ^2 +y_B ^2 −2x_A x_B −2y_A y_B =s^2 ⇒(a+x_A )(b−x_B )−y_A y_B =(s^2 /2)+ab ......
$${r}={a}+{b} \\ $$$${C}\left({r}\:\mathrm{sin}\:\theta,\:{r}\:\mathrm{cos}\:\theta\right) \\ $$$$\left({x}−{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\left({y}−{r}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$${point}\:{A}: \\ $$$$\left({x}_{{A}} −{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\left({y}_{{A}} −{r}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$$\left({x}_{{A}} −{b}\right)^{\mathrm{2}} +{y}_{{A}} ^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\left({b}−{r}\:\mathrm{sin}\:\theta\right)\left(\mathrm{2}{x}_{{A}} −{b}−{r}\:\mathrm{sin}\:\theta\right)−{r}\:\mathrm{cos}\:\theta\:\left(\mathrm{2}{y}_{{A}} −{r}\:\mathrm{cos}\:\theta\right)={s}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\mathrm{2}\left({b}−{r}\:\mathrm{sin}\:\theta\right){x}_{{A}} −\mathrm{2}{r}\:\mathrm{cos}\:\theta\:{y}_{{A}} ={s}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{r}^{\mathrm{2}} ? \\ $$$$\Rightarrow{y}_{{A}} =\frac{\mathrm{2}\left({b}−{r}\:\mathrm{sin}\:\theta\right){x}_{{A}} −{s}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}{r}\:\mathrm{cos}\:\theta} \\ $$$$\left({x}_{{A}} −{b}\right)^{\mathrm{2}} +\left[\frac{\mathrm{2}\left({b}−{r}\:\mathrm{sin}\:\theta\right){x}_{{A}} −{s}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}{r}\:\mathrm{cos}\:\theta}\right]^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${x}_{{A}} ^{\mathrm{2}} −\mathrm{2}{bx}_{{A}} +{b}^{\mathrm{2}} +\left[\frac{\left({b}−{r}\:\mathrm{sin}\:\theta\right){x}_{{A}} }{{r}\:\mathrm{cos}\:\theta}−\frac{{s}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{r}\:\mathrm{cos}\:\theta}\right]^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\left[\mathrm{1}+\frac{\left({b}−{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\right]{x}_{{A}} ^{\mathrm{2}} −\mathrm{2}\left[{b}+\frac{\left({b}−{r}\:\mathrm{sin}\:\theta\right)\left({s}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{\mathrm{2}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\right]{x}_{{A}} +\frac{\left({s}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}_{{A}} =\frac{\mathrm{1}}{\mathrm{1}+\frac{\left({b}−{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}}\left\{{b}+\frac{\left({b}−{r}\:\mathrm{sin}\:\theta\right)\left({s}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{\mathrm{2}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\pm\sqrt{\left[{b}+\frac{\left({b}−{r}\:\mathrm{sin}\:\theta\right)\left({s}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{\mathrm{2}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\right]^{\mathrm{2}} −\left[\mathrm{1}+\frac{\left({b}−{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\right]\left[\frac{\left({s}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right]}\right\} \\ $$$${point}\:{B}: \\ $$$$\left({x}_{{B}} −{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\left({y}_{{B}} −{r}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$$\left({x}_{{B}} +{a}\right)^{\mathrm{2}} +{y}_{{B}} ^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$−\left({a}+{r}\:\mathrm{sin}\:\theta\right)\left(\mathrm{2}{x}_{{B}} +{a}−{r}\:\mathrm{sin}\:\theta\right)−{r}\:\mathrm{cos}\:\theta\:\left(\mathrm{2}{y}_{{B}} −{r}\:\mathrm{cos}\:\theta\right)={s}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\mathrm{2}\left({a}+{r}\:\mathrm{sin}\:\theta\right)\:{x}_{{B}} +\mathrm{2}{r}\:\mathrm{cos}\:\theta\:{y}_{{B}} =−{s}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\Rightarrow{y}_{{B}} =\frac{−\mathrm{2}\left({a}+{r}\:\mathrm{sin}\:\theta\right){x}_{{B}} −{s}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}{r}\:\mathrm{cos}\:\theta} \\ $$$$\left({x}_{{B}} +{a}\right)^{\mathrm{2}} +\left[\frac{−\mathrm{2}\left({a}+{r}\:\mathrm{sin}\:\theta\right){x}_{{B}} −{s}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}{r}\:\mathrm{cos}\:\theta}\right]^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\left({x}_{{B}} +{a}\right)^{\mathrm{2}} +\left[\frac{\left({a}+{r}\:\mathrm{sin}\:\theta\right){x}_{{B}} }{{r}\:\mathrm{cos}\:\theta}+\frac{{s}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{r}\:\mathrm{cos}\:\theta}\right]^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\left[\mathrm{1}+\frac{\left({a}+{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}\right]{x}_{{B}} ^{\mathrm{2}} +\mathrm{2}\left[{a}+\frac{\left({a}+{r}\:\mathrm{sin}\:\theta\right)\left({s}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{\mathrm{2}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\right]{x}_{{B}} +\frac{\left({s}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}_{{B}} =\frac{\mathrm{1}}{\mathrm{1}+\frac{\left({a}+{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}}\left\{−{a}−\frac{\left({a}+{r}\:\mathrm{sin}\:\theta\right)\left({s}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{\mathrm{2}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\pm\sqrt{\left[{a}+\frac{\left({a}+{r}\:\mathrm{sin}\:\theta\right)\left({s}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{\mathrm{2}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\right]^{\mathrm{2}} −\left[\mathrm{1}+\frac{\left({a}+{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}\right]\left[\frac{\left({s}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right]}\right\} \\ $$$$ \\ $$$${AB}={s}: \\ $$$$\left({x}_{{A}} −{x}_{{B}} \right)^{\mathrm{2}} +\left({y}_{{A}} −{y}_{{B}} \right)^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$${x}_{{A}} ^{\mathrm{2}} +{y}_{{A}} ^{\mathrm{2}} +{x}_{{B}} ^{\mathrm{2}} +{y}_{{B}} ^{\mathrm{2}} −\mathrm{2}{x}_{{A}} {x}_{{B}} −\mathrm{2}{y}_{{A}} {y}_{{B}} ={s}^{\mathrm{2}} \\ $$$$\Rightarrow\left({a}+{x}_{{A}} \right)\left({b}−{x}_{{B}} \right)−{y}_{{A}} {y}_{{B}} =\frac{{s}^{\mathrm{2}} }{\mathrm{2}}+{ab} \\ $$$$…… \\ $$
Commented by ajfour last updated on 06/Apr/23
lengthy but you solved it, sir. Very commendable! thank you.
$${lengthy}\:{but}\:{you}\:{solved}\:{it},\:{sir}.\:{Very} \\ $$$${commendable}!\:{thank}\:{you}. \\ $$

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