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Question-130241

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Question Number 130241 by mr W last updated on 23/Jan/21
Commented by mr W last updated on 23/Jan/21
three circles with radii p, q, r have the same center. (p<q<r) what is the side length of the smallest and the largest square which has three of its vertexes on each of the three circles?
$${three}\:{circles}\:{with}\:{radii}\:{p},\:{q},\:{r}\:{have} \\ $$$${the}\:{same}\:{center}.\:\left({p}<{q}<{r}\right) \\ $$$${what}\:{is}\:{the}\:{side}\:{length}\:{of}\:{the}\:{smallest} \\ $$$${and}\:{the}\:{largest}\:{square}\:{which}\:{has} \\ $$$${three}\:{of}\:{its}\:{vertexes}\:{on}\:{each}\:{of}\:{the} \\ $$$${three}\:{circles}? \\ $$
Answered by ajfour last updated on 23/Jan/21
s=z_B +p ∣z_B ∣=∣s−p∣=q z_C =z_B +is ∣z_C ∣=∣z_B +iz_B +ip∣=r let s=t(cos θ−isin θ) z_B =(tcos θ−p)−itsin θ (tcos θ−p)^2 +t^2 sin^2 θ=q^2 ________________________ t^2 +p^2 −2ptcos θ=q^2 ...(i) ________________________ cos θ=((t^2 +p^2 −q^2 )/(2pt)) (tcos θ−p+tsin θ)^2 +(−tsin θ+tcos θ)^2 =r^2 ...(ii) ⇒ t^2 +p^2 +2t^2 sin θcos θ −2pt(cos θ+sin θ) +t^2 −2t^2 sin θcos θ=r^2 ⇒ 2t^2 +p^2 −2pt(cos θ+sin θ)=r^2 using (i) t^2 −2ptsin θ=r^2 −q^2 and ..(i) again t^2 −2ptcos θ=q^2 −p^2 ⇒ (t^2 +q^2 −r^2 )^2 +(t^2 +p^2 −q^2 )^2 =4p^2 t^2 ⇒ (z−a)^2 +(z−b)^2 =2cz 2z^2 −2(a+b+c)z+a^2 +b^2 =0 z=(1/2)(a+b+c)±(√((1/4)(a+b+c)^2 −(((a^2 +b^2 )/2)))) a=r^2 −q^2 ; b=q^2 −p^2 ; c=2p^2 sqare side=t=(√z) a+b+c=p^2 +r^2 a^2 +b^2 =r^4 +q^4 −2q^2 r^2 +q^4 +p^4 −2p^2 q^2 =p^4 +2q^4 +r^4 −2q^2 (p^2 +r^2 ) =(p^2 +r^2 −q^2 )^2 +q^4 −2p^2 r^2 2z=p^2 +r^2 ±(√((p^2 +r^2 )^2 −2(p^4 +2q^4 +r^4 −2q^2 p^2 −2q^2 r^2 ))) ________________________ 2z=p^2 +r^2 ±(√(4q^2 (p^2 +r^2 −q^2 )−(r^2 −p^2 )^2 )) ________________________ t=(√z) say for example: p=6, q=7, r=8 ∣s∣=(√(50±7(√(47))))
$${s}={z}_{{B}} +{p} \\ $$$$\mid{z}_{{B}} \mid=\mid{s}−{p}\mid={q} \\ $$$${z}_{{C}} ={z}_{{B}} +{is} \\ $$$$\mid{z}_{{C}} \mid=\mid{z}_{{B}} +{iz}_{{B}} +{ip}\mid={r} \\ $$$${let}\:\:{s}={t}\left(\mathrm{cos}\:\theta−{i}\mathrm{sin}\:\theta\right) \\ $$$${z}_{{B}} =\left({t}\mathrm{cos}\:\theta−{p}\right)−{it}\mathrm{sin}\:\theta \\ $$$$\left({t}\mathrm{cos}\:\theta−{p}\right)^{\mathrm{2}} +{t}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={q}^{\mathrm{2}} \:\: \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:\:\:\:{t}^{\mathrm{2}} +{p}^{\mathrm{2}} −\mathrm{2}{pt}\mathrm{cos}\:\theta={q}^{\mathrm{2}} \:\:…\left({i}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:\mathrm{cos}\:\theta=\frac{{t}^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{2}{pt}} \\ $$$$\left({t}\mathrm{cos}\:\theta−{p}+{t}\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$+\left(−{t}\mathrm{sin}\:\theta+{t}\mathrm{cos}\:\theta\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\Rightarrow \\ $$$${t}^{\mathrm{2}} +{p}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$$−\mathrm{2}{pt}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right) \\ $$$$+{t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta={r}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{t}^{\mathrm{2}} +{p}^{\mathrm{2}} −\mathrm{2}{pt}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)={r}^{\mathrm{2}} \\ $$$${using}\:\left({i}\right) \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{pt}\mathrm{sin}\:\theta={r}^{\mathrm{2}} −{q}^{\mathrm{2}} \\ $$$${and}\:\:..\left({i}\right)\:{again} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{pt}\mathrm{cos}\:\theta={q}^{\mathrm{2}} −{p}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left({t}^{\mathrm{2}} +{q}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({t}^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{4}{p}^{\mathrm{2}} {t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left({z}−{a}\right)^{\mathrm{2}} +\left({z}−{b}\right)^{\mathrm{2}} =\mathrm{2}{cz} \\ $$$$\mathrm{2}{z}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}+{c}\right){z}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:{z}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}+{c}\right)\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}\left({a}+{b}+{c}\right)^{\mathrm{2}} −\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}\right)} \\ $$$${a}={r}^{\mathrm{2}} −{q}^{\mathrm{2}} \:\:;\:\:{b}={q}^{\mathrm{2}} −{p}^{\mathrm{2}} \:;\:{c}=\mathrm{2}{p}^{\mathrm{2}} \\ $$$${sqare}\:{side}={t}=\sqrt{{z}}\:\: \\ $$$${a}+{b}+{c}={p}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={r}^{\mathrm{4}} +{q}^{\mathrm{4}} −\mathrm{2}{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{q}^{\mathrm{4}} +{p}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={p}^{\mathrm{4}} +\mathrm{2}{q}^{\mathrm{4}} +{r}^{\mathrm{4}} −\mathrm{2}{q}^{\mathrm{2}} \left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)^{\mathrm{2}} +{q}^{\mathrm{4}} −\mathrm{2}{p}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$$\mathrm{2}{z}={p}^{\mathrm{2}} +{r}^{\mathrm{2}} \pm\sqrt{\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({p}^{\mathrm{4}} +\mathrm{2}{q}^{\mathrm{4}} +{r}^{\mathrm{4}} −\mathrm{2}{q}^{\mathrm{2}} {p}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} {r}^{\mathrm{2}} \right)} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\mathrm{2}{z}={p}^{\mathrm{2}} +{r}^{\mathrm{2}} \pm\sqrt{\mathrm{4}{q}^{\mathrm{2}} \left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)−\left({r}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${t}=\sqrt{{z}} \\ $$$${say}\:\:{for}\:{example}: \\ $$$${p}=\mathrm{6},\:{q}=\mathrm{7},\:{r}=\mathrm{8} \\ $$$$\mid{s}\mid=\sqrt{\mathrm{50}\pm\mathrm{7}\sqrt{\mathrm{47}}} \\ $$
Commented by mr W last updated on 23/Jan/21
all correct sir! thanks alot!
$${all}\:{correct}\:{sir}! \\ $$$${thanks}\:{alot}! \\ $$
Answered by mr W last updated on 23/Jan/21
Commented by mr W last updated on 23/Jan/21
cos α=((s^2 +q^2 −p^2 )/(2qs)) cos ((π/2)−α)=sin α=((s^2 +q^2 −r^2 )/(2qs)) (((s^2 +q^2 −p^2 )/(2qs)))^2 +(((s^2 +q^2 −r^2 )/(2qs)))^2 =1 let t=s^2 (t+q^2 −p^2 )^2 +(t+q^2 −r^2 )^2 =4q^2 t 2t^2 −2(p^2 +r^2 )t+p^4 +r^4 +2q^4 −2(p^2 +r^2 )q^2 =0 t=s^2 =((p^2 +r^2 ±(√(4(p^2 +r^2 −q^2 )q^2 −(r^2 −p^2 )^2 )))/2) example: p=6,q=7,r=8 s=(√(50±7(√(47))))=9.898, 1.417
$$\mathrm{cos}\:\alpha=\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}{qs}} \\ $$$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right)=\mathrm{sin}\:\alpha=\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{qs}} \\ $$$$\left(\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}{qs}}\right)^{\mathrm{2}} +\left(\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{qs}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${let}\:{t}={s}^{\mathrm{2}} \\ $$$$\left({t}+{q}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({t}+{q}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}{q}^{\mathrm{2}} {t} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){t}+{p}^{\mathrm{4}} +{r}^{\mathrm{4}} +\mathrm{2}{q}^{\mathrm{4}} −\mathrm{2}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){q}^{\mathrm{2}} =\mathrm{0} \\ $$$${t}={s}^{\mathrm{2}} =\frac{{p}^{\mathrm{2}} +{r}^{\mathrm{2}} \pm\sqrt{\mathrm{4}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −{q}^{\mathrm{2}} \right){q}^{\mathrm{2}} −\left({r}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${example}: \\ $$$${p}=\mathrm{6},{q}=\mathrm{7},{r}=\mathrm{8} \\ $$$${s}=\sqrt{\mathrm{50}\pm\mathrm{7}\sqrt{\mathrm{47}}}=\mathrm{9}.\mathrm{898},\:\mathrm{1}.\mathrm{417} \\ $$
Commented by mr W last updated on 23/Jan/21
we got the same formula.
$${we}\:{got}\:{the}\:{same}\:{formula}. \\ $$
Commented by ajfour last updated on 23/Jan/21
Thank you sir.
$${Thank}\:{you}\:{sir}. \\ $$

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