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There is black semicircle, which radius is $ R $. The red circle is tangentially inward to the semicircle and to the diameter in its center. The yellow one is tangent externally to the red circle, internally to the semicircle and tangent to the diameter of the semicircle.

My question is :

What is the relationship between $R$ and the radius $r$ of yellow one?

My attempts :

I tried to use similarity of triangles, but always I had the third unknown number and two equations. I am sure that the radius of the red one is $ 0.5R$

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  • $\begingroup$ Draw a line passing through the centers of the black and of the yellow circle and a line connecting the centers of the yellow and red circles and then use Pythagorean Theorem. Does that work? $\endgroup$ – Matteo Mar 28 at 20:41
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Let the radius of the "black" semicircle be $[CK]=2R$. Thus, the radius of the "red" circumference will be $[DC]=R$. In virtue of the Pythagorean Theorem in the triangles $\triangle DGE$ and $\triangle ECF$ respectively

\begin{align*} GE^2=CF^2&=DE^2-DG^2=(R+r)^2-(R-r)^2=4Rr\\ &=CE^2-EF^2=(2R-r)^2-r^2=4R^2-4Rr\end{align*} Hence

$$4Rr=4R^2-4Rr\iff \color{blue}{2r=R}$$

The radius of the yellow circumference is, consequently, one-fourth of the radius of the black circle.

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  • $\begingroup$ :) already in my comment and in my hint below $\endgroup$ – Matteo Mar 28 at 21:09
  • $\begingroup$ Oh, see... We had the same idea... $\endgroup$ – Dr. Mathva Mar 28 at 21:10
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HINT

Consider the Figure below and note that

  1. $\overline{O_2H} = \frac{R}{2}-r$;
  2. $\overline{O_2O_3} = \frac{R}{2} + r$;
  3. $\overline{HO_1} = r$;
  4. $\overline{O_1O_3} = R-r$;

Use Pythagorean theoerm on $\triangle O_2HO_3$ and on $\triangle O_1HO_3$ to determine $\overline{HO_3}$ in two ways and thus get an equation, which, once solved, will give you the desider result of $r$ as a function of $R$.

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