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Given an isosceles triangle with equal sides of length $b$ and base angle $A$ $(<π/4)$. $O$ and $I$ are the circumcenter and incenter of the triangle respectively. Then true or false $$OI= \left|\frac{b \cos(3A/2)}{2\sin(A) \cos (A/2)}\right|.$$

I have found out the values of the inradius and circumradius to be $b \sin(2 A)/2(1+\cos A)$ and $b/2 \sin(A)$ respectively and also that the isosceles triangle is obtuse angled. I also know that in an isosceles triangle the circumcenter and incenter lie on the same straight line. But I have no idea how to find $OI$. Please help.

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I think you are right.

Let in $\Delta ABC$ we have: $AC=CB=b$, $\measuredangle A=\alpha$,

$D$ be midpoint of $AB$ and $E$ be a midpoint of $BC$.

Thus, $CD=b\sin\alpha$, $CO=\frac{b}{2\sin\alpha}$, $ID=b\cos\alpha\tan\frac{\alpha}{2}$ and $$OI=|CD-CO-ID|=\left|b\sin\alpha-\frac{b}{2\sin\alpha}-b\cos\alpha\tan\frac{\alpha}{2}\right|=$$ $$=\frac{b}{2\sin\alpha\cos\frac{\alpha}{2}}\left|2\sin^2\alpha\cos\frac{\alpha}{2}-\cos\frac{\alpha}{2}-2\sin\alpha\cos\alpha\sin\frac{\alpha}{2}\right|=$$ $$=\frac{b}{2\sin\alpha\cos\frac{\alpha}{2}}\left|\cos2\alpha\cos\frac{\alpha}{2}+\sin2\alpha\sin\frac{\alpha}{2}\right|=$$ $$=\frac{b\left|\cos\frac{3\alpha}{2}\right|}{2\sin\alpha\cos\frac{\alpha}{2}},$$ which you got.

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  • $\begingroup$ By ID you mean inradius right? $\endgroup$ – Lokesh Sangewar Sep 12 '17 at 17:20
  • $\begingroup$ @Lokesh Sangewar Yes, of course. $\endgroup$ – Michael Rozenberg Sep 12 '17 at 17:22
  • $\begingroup$ hahah....ok thanks a lot for your help. $\endgroup$ – Lokesh Sangewar Sep 12 '17 at 17:23
  • $\begingroup$ @Lokesh Sangewar You are welcome! $\endgroup$ – Michael Rozenberg Sep 12 '17 at 17:23
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this distance is given by $$d=R^2-2Rr$$ this is the formula of Euler

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  • $\begingroup$ Any short and sweet proof for this? $\endgroup$ – Lokesh Sangewar Sep 12 '17 at 16:38
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    $\begingroup$ see here youtube.com/watch?v=AxJy9V0gPZ0 i hope this will help you $\endgroup$ – Dr. Sonnhard Graubner Sep 12 '17 at 16:40
  • $\begingroup$ I'm having problem simplifying the expression. Can you please help me by writing down the simplification? $\endgroup$ – Lokesh Sangewar Sep 12 '17 at 16:46
  • $\begingroup$ which expression do you mean? $\endgroup$ – Dr. Sonnhard Graubner Sep 12 '17 at 16:49
  • $\begingroup$ d=R^2-2Rr. I have reached the denominator part....having some trouble $\endgroup$ – Lokesh Sangewar Sep 12 '17 at 16:50

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