Find the Radius of 4th Circle When All are Tangent: Hint d/2

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Discussion Overview

The discussion revolves around determining the radius of a fourth circle that is tangent to three other circles, with the centers of the three circles aligned on a line and the center of the fourth circle positioned at a distance d from that line. Participants explore the geometric relationships and mathematical principles involved in this configuration.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests that the radius of the fourth circle is $$d/2$$, but seeks clarification on the meaning of d in the context of the problem.
  • Another participant expresses confusion regarding the definition of d and requests a solution, indicating a lack of clarity in the provided information.
  • A different participant notes that the radius $$d/2$$ appears to hold for all proportions of the circles, based on their visual interpretation, but admits uncertainty about how to begin solving the problem.
  • One participant elaborates on using Descartes' theorem to relate the radii of the circles, suggesting that the relationship between the radii can be derived from the curvatures of the circles and the geometry of the configuration.
  • This participant also mentions using the area of triangle formed by the centers of the circles to connect the radius of the fourth circle with the distance d, proposing a method to derive the radius based on geometric principles.

Areas of Agreement / Disagreement

Participants express uncertainty about the definition of d and the initial conditions of the problem. There is no consensus on the radius of the fourth circle, as some participants propose different approaches and interpretations without reaching a definitive conclusion.

Contextual Notes

The discussion highlights the need for clearer definitions and visual representations of the problem, as well as the potential complexity involved in applying geometric theorems to derive the radius of the fourth circle.

Andrei1
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The centers of three circles are situated on a line. The center of the fourth circle is situated at given distance d from that line. What is the radius of the fourth circle if we know that each circle is tangent to other three. Please give me a hint, if you can. Answer: $$d/2.$$
 

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Re: four circles

It is not clear what is d from the picture .
 
Re: four circles

ZaidAlyafey said:
It is not clear what is d from the picture .

Andrei said:
The center of the fourth circle is situated at given distance d from that line.

I am also interested in the solution.

Andrei said:
The center of the fourth circle is situated at given distance d from that line.

I am also interested in the solution.

Edit:

I also don't know where to start with this one.

But I really wonder why this works for all proportions in which the centre of the biggest circle is divided into.

It looks like d/2 is the radius of the 4th circle for all proportions (by my visualisation).

But to start working on it,I think I have to get to paper and pencil.

At first I would be looking for the case in which the centre of the big circle is divided into two equal parts as I assume it is easier to understand.
 
Last edited:
Andrei said:
The centers of three circles are situated on a line. The center of the fourth circle is situated at given distance d from that line. What is the radius of the fourth circle if we know that each circle is tangent to other three. Please give me a hint, if you can. Answer: $$d/2.$$
fourcircles.png


Write $r_1,\ r_2,\ r_3,\ r_4$ for the radii of the circles centred at $O_1,\ O_2,\ O_3,\ O_4$ respectively. Notice that $r_1 = r_2+r_3$. To calculate $r_4$ in terms of $r_2$ and $r_3$, use Descartes' theorem. For that, you need to use the curvatures $k_i = \pm1/r_i$ $(i=1,2,3,4)$. The first of these, $k_1$, will be negative (because the large circle touches the other three internally). Descartes' theorem says that $k_4 = k_1+k_2+k_3 \pm\sqrt{k_1k_2 + k_2k_3 + k_3k_1}$. When you substitute $k_1 = -1/(r_2+r_3)$, $k_i = 1/r_i$ for $i=2,3,4$, you find that the part inside the square root sign is zero (this corresponds to the fact that the centres $O_1,\ O_2,\ O_3$ are collinear). That gives you a formula for $r_4$ in terms of $r_2$ and $r_3$.

Now you have to bring in the distance $d$. I think the neatest way to do that is to use the fact that the area of triangle $O_2O_3O_4$ is $\frac12d(r_2+r_3)$. It is also given by Heron's formula in terms of $r_2,\ r_3,\ r_4$. Compare the two results and you will find that $r_4 = \frac12d$.
 
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