Can you please help me solve this really difficult problem: Find R/r where R is the radius of the circumscribed circle of a trapezoid and r is the radius of the inscribed circle of this trapezoid.

Thank you very much for your trying to help, in advance !enter image description here

  • $\begingroup$ What are you given? I don' quite understand. If you know the radius's, then what do you have to find out? $\endgroup$ – Sawarnik Apr 7 '14 at 15:42
  • $\begingroup$ The only information given is: --a trapezoid --in this trapezoid we have an inscribed circle and also a circle circumscribed to it --- next, we have that the radius of inscribed circle is 'r' but of circumscribed circle is "R". So now, we have to find R/r, and I know that it should be an exact value. I hope I was clear if not please ask me , thank you that you looked at my problem. $\endgroup$ – Ivan Gandacov Apr 7 '14 at 15:49
  • $\begingroup$ I don't understand what you are given, the sides or angles of the trapezoid? if you have r and R then R/r follows immediately. $\endgroup$ – Sawarnik Apr 7 '14 at 15:51
  • $\begingroup$ I have only the radiuses of the circumscribed and inscribed (R and r )circles , and, in other words, I have to find the ratio between R and r/ $\endgroup$ – Ivan Gandacov Apr 7 '14 at 15:52
  • $\begingroup$ @Sawarnik We don't have the exact values of R and r their just, as variables $\endgroup$ – Ivan Gandacov Apr 7 '14 at 16:01

I believe your answer is given on these links: Isosceles tangential trapezoid and Bicentric quadrilateral.


This journal can answer your question, just click and it will automatically download the journal. I found this journal in the link that I gave to you. I hope this help.

Your question can be solved by using Fuss' theorem (you can see this theorem on the links that I gave to you). Fuss' theorem gives a relation between the inradius $r$, the circumradius $R$ and the distance $x$ between the center of the inner circle and center of the outer circle, for any bicentric quadrilateral (trapezoid is included). The relation is $$ \frac{1}{(R-x)^2}+\frac{1}{(R+x)^2}=\frac{1}{r^2}. $$ In this case, using your picture, we have $x=r$. Therefore $$ \frac{1}{(R-r)^2}+\frac{1}{(R+r)^2}=\frac{1}{r^2}.\tag1 $$ By using equation $(1)$, I think it is not difficult to obtain that: $$ \frac{R}{r}=\sqrt{2+\sqrt{5}}. $$


$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

  • $\begingroup$ No its not I need. In my problem I know only the radiuses and that's all $\endgroup$ – Ivan Gandacov Apr 7 '14 at 15:39
  • $\begingroup$ @JohnG. Are you sure? If so, then the answer is simply$$\frac{R}{r}\ge\sqrt{2}.$$Just please take a look this same link carefully. :) $\endgroup$ – Tunk-Fey Apr 7 '14 at 15:55
  • $\begingroup$ It's the direct answer but I would like to see the explanation. Thank you for the link ,I will try more! $\endgroup$ – Ivan Gandacov Apr 7 '14 at 15:59
  • $\begingroup$ @JohnG. Please read carefully my friend. The answer is directly in front of your eyes. Let me give you a hint: the equality holds only when the two circles are concentric (have the same center as each other); then the quadrilateral is a square. Now, using the hint, I believe you can derive that inequality. :) $\endgroup$ – Tunk-Fey Apr 7 '14 at 16:05
  • $\begingroup$ @JohnG. This journal can answer your question, just click and it will automatically download it. I found this journal in the link that I gave to you. $\endgroup$ – Tunk-Fey Apr 7 '14 at 16:18

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