Find $\frac{a}{b}$ in the Circle of Balls

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Discussion Overview

The discussion revolves around finding the ratio $\frac{a}{b}$ in the context of a geometric problem involving circles, specifically a larger circle and smaller circles that are tangential to it. The scope includes mathematical reasoning and problem-solving related to geometry.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • Some participants propose a method to express $\frac{a}{b}$ in terms of other variables, leading to the formulation $\frac{a}{b}=\frac{c}{c+r}$, where $c$ and $r$ are defined in relation to $a$ and $b$.
  • One participant suggests using the Pythagorean theorem in specific triangles to derive expressions for $XY^2$ that relate $a$, $b$, and $r$.
  • Another participant questions whether $\frac{a}{b}=\frac{\sqrt{2}}{2}$ is the correct answer, which is later affirmed by another participant.

Areas of Agreement / Disagreement

There is a lack of consensus on the correctness of the derived expressions and the final ratio, although one participant confirms the ratio $\frac{a}{b}=\frac{\sqrt{2}}{2}$ as correct.

Contextual Notes

The discussion includes various assumptions about the relationships between the variables $a$, $b$, $c$, and $r$, and relies on specific geometric configurations that may not be fully detailed.

maxkor
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maxkor said:
How find $\frac{a}{b}$

Hi maxkor! (Smile)

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?
 
View attachment 4533
Let 1/2b radius of the big circle, let r radius of the smaller circle
Let $c=1/2a=1/2b−r,
b=2c+2r,
a=2c.$
So $\frac{a}{b}=\frac{2c}{2c+2r}=\frac{c}{c+r}$
Small circles respectively tangential to the large circles so
$z=c+2r,t=a−r=2c−r$

Is this right?
 

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Use Pythagoras in the triangles $CXY$, $DXY$ (where $Y$ is the centre of one of the footballs) to find two expressions for $XY^2$ in terms of $a$, $b$ and $r$. Putting those expressions equal to each other will give you an equation connecting $a$, $b$ and $r$.

You already know that $r = \frac12(b-a)$ (from your equation $c = \frac12a = \frac12b-r$). Substitute that value of $r$ into your equation, and it will give you the connection between $a$ and $b$.
 

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Last edited:
Is $\frac{a}{b}=\frac{\sqrt{2}}{2}$ correct answer?
 
maxkor said:
Is $\frac{a}{b}=\frac{\sqrt{2}}{2}$ correct answer?
Yes! (Happy)
 

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