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There are two touching circles inscribed in a $60^\circ$ angle. The distance between the vertex of the angle and the center of the smaller circle is $5j$. What is the ratio of the surfaces of the two circles?

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    $\begingroup$ What do you mean by 5j? $\endgroup$ – Sawarnik Apr 18 '14 at 10:11
  • $\begingroup$ that is the distance from vertex to the center of the smaller circle $\endgroup$ – slimDeviant Apr 18 '14 at 11:08
  • $\begingroup$ No, 5 j? Is it an unit? Or just a variable? $\endgroup$ – Sawarnik Apr 18 '14 at 11:12
  • $\begingroup$ I don't know... I wrote the problem exactly as it goes. I suppose as a unit because it has no meaning in the context of this problem. What do you think? $\endgroup$ – slimDeviant Apr 18 '14 at 11:15
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    $\begingroup$ Where did you get the problem from? $\endgroup$ – Sawarnik Apr 18 '14 at 11:18
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The radius of the large circle is three times that of the small circle. This is easy to see if you draw things on a triangular grid:

Construction

Since area scales with the square of the radius, you get an area ratio of $9$. The scale of things, in particular that distance $AD$, is irrelevant for this computation.

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  • $\begingroup$ Your answer about the ratio is correct, the ratio is indeed 9, but something else is worying me... Is there a slight chance that this relationship could be derived from the data that I've given, more precisely from the distance of vertex of the angle to the center of the smaller circle? Since your answer is correct and since it matches with the answer given in this book, do you care to explain your reasoning for this soluton, to show me how you got there, preferably without the use of the grid? $\endgroup$ – slimDeviant Apr 18 '14 at 15:09
  • $\begingroup$ @slimDeviant: I got there by doing the construction in Cinderella. But you could get there e.g. by computing coordinates. If you choose them well enough, the solution will be as simple as the one with the grid. If not, you'll have square roots all over the place. One key element in my construction is that I constructed $FG$ parallel to $EB$, the rest should be clear from the drawing. $\endgroup$ – MvG Apr 18 '14 at 15:14
  • $\begingroup$ @slimDeviant: You could also start with the fact that $ABD$ is half of an equilateral triangle, so $AD=2\,BD$. With $DB=DE=DF$ you get $AF=3\,AE$. The rest follows by similarity. $\endgroup$ – MvG Apr 18 '14 at 15:18
  • $\begingroup$ It's maybe rude, but could you please show me the similarity because I'm not able to see? $\endgroup$ – slimDeviant Apr 18 '14 at 15:37
  • $\begingroup$ @slimDeviant: The similarity transformation characterized by $$A\mapsto A, B\mapsto G, D\mapsto H, C\mapsto K, E\mapsto F$$ is a homothety with center $A$ and scale factor $\frac{AF}{AE}=3$. It maps the small circle onto the large one. $\endgroup$ – MvG Apr 18 '14 at 15:47

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