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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

- Thread starter Albert
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- Jan 25, 2013

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- Mar 5, 2012

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And then split it into 2 rectangular triangles.

Let's call the radius of the small circle x.

Then the left rectangular triangle has hypotenuse (36-x) and horizontal side at the x-axis (x).

And the right rectangular triangle has hypotenuse (36+x) and horizontal side at the x-axis (36-x).

Since they share their third side, the following equation must hold (Pythagoras):

$$(36-x)^2 - x^2 = (36+x)^2 - (36-x)^2$$

$$(36+x)^2 - 2(36-x)^2 + x^2 = 0$$

$$(36^2+2\cdot 36 x +x^2) - 2(36^2-2\cdot 36 x + x^2) + x^2 = 0$$

$$6\cdot 36 x = 36^2$$

$$x = 6$$

$\blacksquare$

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- #3

- Feb 7, 2012

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find the radius of the small circle O_2:

View attachment 1660

Let $r$ be the radius of the green circle. Draw the triangle whose vertices are the two ends of the red baseline and the centre of the green circle (the points labelled O, O_2 and O-1 in the diagram). The lengths of its sides are $36-r$, $36+r$ and $36$. The angle labelled $\theta$ has $\cos\theta = \dfrac r{36-r}$. The cosine rule then gives the equation $(36+r)^2 = 36^2 + (36-r)^2 -2\cdot36(36-r)\dfrac r{36-r},$ which simplifies to $r=6.$

Last edited:

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- #4

- Mar 5, 2012

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Must be my new avatar.Edit.Yet again, ILS got there first (and I never even noticed).

Sometimes it makes me feel imaginary.

- Jan 17, 2013

- 1,667

The invisible : not recognizable neither by name nor by avatar living in his imaginary complex paradigm , just kidding .Must be my new avatar.

Sometimes it makes me feel imaginary.