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Question Number 150177 by mr W last updated on 10/Aug/21

Commented by mr W last updated on 10/Aug/21

$$anotherwaytosolveQ149894$$

Commented by mr W last updated on 10/Aug/21

$$thelocusoftheupperrightvertex$$$$oftheequilateralofsidelengths$$$$isanellipsewithmajoraxisat45°$$$$witha=\frac{(\sqrt{3}+1)s}{2},b=\frac{(\sqrt{3}-1)s}{2}.$$$$thereforethequestionnowistofind$$$$thetwoellipseswhichtangentthe$$$$circle.$$

Commented by mr W last updated on 10/Aug/21

Commented by mr W last updated on 11/Aug/21

$$s=sidelengthofequilateral$$$$a=\frac{\sqrt{3}+1}{2}s=\alpha s$$$$b=\frac{\sqrt{3}-1}{2}s=\beta s$$$$\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=s^2$$$${(x-p)}^2+{(y-q)}^2=r^2$$$$sayP(\alpha s\cos \theta ,\beta s\sin \theta )$$$$\tan \phi =\frac{\beta }{\alpha \tan \theta }$$$$for{s}_{min}:$$$$x_C=\alpha s\cos \theta +r\sin \phi$$$$x_C=\alpha s\cos \theta +r\frac{\beta }{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }}=p$$$$\Rightarrow s_{min}=\frac{1}{\alpha \cos \theta }(p-r\frac{\beta }{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }})$$$$y_C=\beta s\sin \theta +r\cos \phi$$$$y_C=\beta s\sin \theta +r\frac{\alpha \tan \theta }{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }}=q$$$$\Rightarrow s_{min}=\frac{1}{\beta \sin \theta }(q-r\frac{\alpha \tan \theta }{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }})$$$$\frac{1}{\alpha \cos \theta }(p-r\frac{\beta }{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }})=\frac{1}{\beta \sin \theta }(q-r\frac{\alpha \tan \theta }{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }})$$$$\Rightarrow \frac{\alpha q}{\tan \theta }-\frac{({\alpha }^2-{\beta }^2)r}{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }}=\beta p$$$$$$$$similarlyfor{s}_{max}:$$$$s_{max}=\frac{1}{\alpha \cos \theta }(p+r\frac{\beta }{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }})$$$$s_{max}=\frac{1}{\beta \sin \theta }(q+r\frac{\alpha \tan \theta }{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }})$$$$\frac{\alpha q}{\tan \theta }+\frac{({\alpha }^2-{\beta }^2)r}{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }}=\beta p$$$$$$$$withp=\frac{k+h}{\sqrt{2}},q=\frac{k-h}{\sqrt{2}}$$$$$$$$example:h=6,k=8,r=2$$$$p=7\sqrt{2},q=\sqrt{2}$$$$s_{max}=11.9086✓$$$$s_{min}=6.0829✓$$

Commented by mr W last updated on 12/Aug/21

$$summary:$$$$\alpha =\frac{\sqrt{3}+1}{2},\beta =\frac{\sqrt{3}-1}{2}$$$$p=\frac{k+h}{\sqrt{2}},q=\frac{k-h}{\sqrt{2}}$$$$s_{max/min}=\frac{1}{\alpha }(\frac{p}{\cos \theta }\pm \frac{\beta r}{\sqrt{{\alpha }^2{\sin }^2\theta +{\beta }^2{\cos }^2\theta }})$$$$with\theta from$$$$\frac{\alpha q}{\tan \theta }\pm \frac{({\alpha }^2-{\beta }^2)r}{\sqrt{{\beta }^2+{\alpha }^2{\tan }^2\theta }}=\beta p$$$$$$$$(+for{s}_{max}and-for{s}_{min})$$

Answered by mr W last updated on 10/Aug/21

Commented by mr W last updated on 10/Aug/21

$$example:h=6,k=8,r=2$$$$s_{max}=11.9088$$$$s_{min}=6.0828$$

Commented by mr W last updated on 10/Aug/21

Commented by mr W last updated on 10/Aug/21

$$C(h,k)$$$$D(\frac{s}{2}\cos \varphi ,\frac{s}{2}\sin \varphi )$$$$x_P=\frac{s}{2}\cos \varphi +\frac{\sqrt{3}s}{2}\sin \varphi =s\sin (\varphi +\frac{\pi }{6})$$$$y_P=\frac{s}{2}\sin \varphi +\frac{\sqrt{3}s}{2}\cos \varphi =s\sin (\varphi +\frac{\pi }{3})$$$${[s\sin (\varphi +\frac{\pi }{6})-h]}^2+{[s\sin (\varphi +\frac{\pi }{3})-k]}^2=r^2$$$$[{\sin }^2(\varphi +\frac{\pi }{6})+{\sin }^2(\varphi +\frac{\pi }{3})]s^2-2[h\sin (\varphi +\frac{\pi }{6})+k\sin (\varphi +\frac{\pi }{3})]s+(h^2+k^2-r^2)=0$$$$u\left(\varphi \right)s^2-2v\left(\varphi \right)s+(h^2+k^2-r^2)=0$$$$s=\frac{v\left(\varphi \right)\pm \sqrt{v^2\left(\varphi \right)-(h^2+k^2-r^2)u\left(\varphi \right)}}{u\left(\varphi \right)}$$$$with$$$$u\left(\varphi \right)={\sin }^2(\varphi +\frac{\pi }{6})+{\sin }^2(\varphi +\frac{\pi }{3})$$$$v\left(\varphi \right)=h\sin (\varphi +\frac{\pi }{6})+k\sin (\varphi +\frac{\pi }{3})$$

Commented by mr W last updated on 10/Aug/21

Commented by mr W last updated on 10/Aug/21

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