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Question Number 19586 by ajfour last updated on 13/Aug/17

$$\mathrm{Given}\mathrm{in}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}\mathrm{a}$$$$\mathrm{lateral}\mathrm{side}\mathrm{b}\mathrm{and}\mathrm{the}\mathrm{base}\mathrm{angle}\alpha .$$$$\mathrm{Compute}\mathrm{the}\mathrm{distance}\mathrm{from}\mathrm{the}$$$$\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{inscribed}\mathrm{circle}\mathrm{to}\mathrm{the}$$$$\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{circumscribed}\mathrm{circle}.$$

Commented by Tinkutara last updated on 13/Aug/17

$$R=\frac{b}{2\sin \alpha },r=\frac{b\cos \alpha }{\cos \frac{\alpha }{2}}$$

Commented by ajfour last updated on 13/Aug/17

Commented by Tinkutara last updated on 13/Aug/17

$$\mathrm{I}\mathrm{found}\mathrm{this}\mathrm{on}\mathrm{Wikipedia}\mathrm{by}{\mathrm{Euler}}^{\prime }\mathrm{s}$$$$\mathrm{Theorem}\mathrm{in}\mathrm{Geometry}.$$$$\mid IO\mid =\sqrt{R(R-2r)}.\mathrm{Proof}\mathrm{is}\mathrm{given}\mathrm{on}$$$$\mathrm{the}\mathrm{same}\mathrm{page}.$$

Commented by ajfour last updated on 13/Aug/17

$$\mathrm{required}\mathrm{to}\mathrm{find}\mathrm{answer}\mathrm{in}\mathrm{terms}$$$$\mathrm{of}\mathrm{b},\mathrm{and}\alpha .$$

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