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I've been given the following school problem:

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ABC is an isosceles triangle (AB = AC). The radius of the incircle is R and of the other circle (which is tangent to the incircle and to the legs of the triangle) is r. We need to express the sides of the triangle using R and r.

I was able to show that the line joining the 2 circles' centers gets to A and is an altitude of the triangle, and that the line tangent to both circles is parallel to it (and to BC), but I don't know how to continue from there.

Thanx you!

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When you say edges,you mean sides of triangle? I'm going to write the important stuff and leave computation to you. Okay, denote $h$ as height of triangle , $a=BC,b=AB=AC$,D altitude of height h from vertex A, $S1$ center of bigger circle,$S2$ center of smaller circle, M feet of altitude from $S1$ to $AC$, $N$ feet of altitude from $S2$ to $AC$.

Triangles $\triangle ADC, \triangle AS1M, \triangle AS2N$ are similar (why?).It follows that $\frac{a}{2}:b= R:(h-R)=r:(h-2R-r) $

From $ R:(h-R)=r:(h-2R-r) \Rightarrow h=\frac{2R^2}{R-r} $

From $\frac{a}{2}:b= R:(h-R) \Rightarrow b=\frac{a(h-R)}{2R} = \frac{a(R+r)}{2(R-r)} $

Also from ADC we have $ \frac{a^2}{4}+h^2=b^2 $ , and plugging expression for h and b only leaves us in terms of a,R,r and we get

$ a=2R \sqrt{\frac{R}{r}}$ and $b=\frac{R\sqrt{R}(R+r)}{\sqrt{r}(R-r)} $

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  • $\begingroup$ Yes, I meant "sides", sorry.. I've fixed that. Thanx for the answer! $\endgroup$ – Yaron Cohen-Tal May 1 '16 at 17:00
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It's a good beginning. If you call the (yet unknown) height of the triangle $h$ and draw the perpendiculars from the two circle centres to $AC$, you get some similar right triangles that you can use to derive relations between $r$, $R$, and $h$.

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  • $\begingroup$ Your answer is pretty similar to @zezanjee.. thanx! $\endgroup$ – Yaron Cohen-Tal May 1 '16 at 17:02

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