This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
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Math
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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
The general formula is R=sin(nπ)r[1+sin(nπ)] where r,R and n are the radius of the inner circle, radius of the outer circle and number of circles inscribed within the larger circle respecticely.
Substituting we get the radius of the outer circle ≈2.1547 cm.
Try proving the formula of your own!
An other way using basic knowledge is like this:
Join centres of small circle to obtain an equilateral triangle. Then simply find length of centroid using basic trigonometry or Pythagoras theorem. When that length is found add the radius of small circle to it. As the centroid of triangle is the center of bigger circle.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
The general formula is R=sin(nπ)r[1+sin(nπ)] where r,R and n are the radius of the inner circle, radius of the outer circle and number of circles inscribed within the larger circle respecticely.
Substituting we get the radius of the outer circle ≈2.1547 cm. Try proving the formula of your own!
An other way using basic knowledge is like this: Join centres of small circle to obtain an equilateral triangle. Then simply find length of centroid using basic trigonometry or Pythagoras theorem. When that length is found add the radius of small circle to it. As the centroid of triangle is the center of bigger circle.