- #1
Albert1
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find the radius of the small circle O_2:
View attachment 1660
View attachment 1660
Find the radius of the small circle O_2
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Discussion Overview
The discussion revolves around finding the radius of a small circle labeled O_2, involving geometric relationships and the application of the cosine rule. The context includes mathematical reasoning and exploration of the problem through a diagram linked in the thread.
Discussion Character
- Mathematical reasoning, Technical explanation
Main Points Raised
- One participant proposes that the radius of the green circle, denoted as $r$, can be determined by constructing a triangle with vertices at the ends of a baseline and the center of the green circle, leading to the equation derived from the cosine rule.
- The cosine of an angle $\theta$ is expressed as $\cos\theta = \dfrac r{36-r}$, which is used in the derivation of the equation.
- The participant concludes that the radius $r$ simplifies to 6 based on their calculations.
- Another participant humorously acknowledges the first responder's promptness in providing a solution, indicating a light-hearted tone in the discussion.
- A further reply introduces a playful comment about the participant's avatar and its imaginary qualities, contributing to the informal atmosphere of the thread.
Areas of Agreement / Disagreement
The discussion does not present a consensus on the radius, as it primarily reflects one participant's calculations and others' humorous responses without further elaboration or challenge to the proposed solution.
Contextual Notes
The discussion lacks detailed exploration of the assumptions underlying the geometric relationships and the cosine rule application. There is no verification of the calculations presented.
- #2
I like Serena
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Re: find the radius of the small circle O_2
We can draw a triangle from the left bottom corner, to the center of the small circle, to the right bottom corner.
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):
$$(36-x)^2 - x^2 = (36+x)^2 - (36-x)^2$$
$$(36+x)^2 - 2(36-x)^2 + x^2 = 0$$
$$(36^2+2\cdot 36 x +x^2) - 2(36^2-2\cdot 36 x + x^2) + x^2 = 0$$
$$6\cdot 36 x = 36^2$$
$$x = 6$$
$\blacksquare$
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):
$$(36-x)^2 - x^2 = (36+x)^2 - (36-x)^2$$
$$(36+x)^2 - 2(36-x)^2 + x^2 = 0$$
$$(36^2+2\cdot 36 x +x^2) - 2(36^2-2\cdot 36 x + x^2) + x^2 = 0$$
$$6\cdot 36 x = 36^2$$
$$x = 6$$
$\blacksquare$
- #3
Opalg
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Re: find the radius of the small circle O_2
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 $r=6.$[/sp]
Edit. Yet again, ILS got there first (and I never even noticed).
[sp]Albert said:
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 $r=6.$[/sp]
Edit. Yet again, ILS got there first (and I never even noticed).
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- #4
I like Serena
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Re: find the radius of the small circle O_2
Must be my new avatar.
Sometimes it makes me feel imaginary.
Opalg said:
Must be my new avatar.
Sometimes it makes me feel imaginary.
- #5
alyafey22
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Re: find the radius of the small circle O_2
The invisible : not recognizable neither by name nor by avatar living in his imaginary complex paradigm , just kidding (Punch) .
I like Serena said:
The invisible : not recognizable neither by name nor by avatar living in his imaginary complex paradigm , just kidding (Punch) .
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