Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

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## #1 Re: Help Me ! » Geometric Progression » 2006-03-29 06:10:03

You setup the equation right, but your algebra is incorrect.

162 / n = n / 18
162 = n^2 / 18      (this is where you went wrong)
162 * 18 = n^2
2916 = n^2
54 = n

You should be able to work out the rest for yourself.

## #5 Re: Help Me ! » Tangent to a Circle » 2006-03-14 05:56:43

Your equation for the gradient of the radius is correct.  Note that it doesn't matter if you take p - c or c - p as long as you are consistent (sign switches in both numerator and denominator and thus cancels).

## #6 Re: Puzzles and Games » Think different! » 2006-03-13 11:12:34

My solution was to connect D to the midpoint of IJ, for the exact same reason.

## #7 Re: Puzzles and Games » Testing Puzzle » 2006-03-13 11:00:58

I don't think I would have ever

it without the hint.

## #8 Re: Puzzles and Games » a cool maths problem I saw in the latest issue of Hugi » 2006-03-13 10:23:27

The probability is exactly 2/3.  Actually, this is a very simple puzzle if you attack it right (I got the answer in less than one minute.).

Proof:

Inscribe a regular hexagon inside the circle such that one of the vertices corresponds to one of the randomly chosen points.  Each of the edges of the inscribed hexagon is a cord with the exact same length as the circle's radius (This is obvious because a regular hexagon can be subdivided into six equilateral triangles and, in this case, two of those equilateral edges corresponds to the circles radius and the cord is the third.).

Therefore, the vertices of the hexagon have subdivided the circumference of the circle into six equal segments AND we know that all points that fall within the two circle segments that are adjacent to the vertex on that first random point form cords with it that are less than the circle's radius and all points that are beyond those two adjacent circle segments form cords that are longer than the radius of the circle.  Since four of the six equal length segments aren't adjacent to the original random point.  The probability that the second randomly selected point falls in one of them is 2/3.