Geometer's sketchpad- archimedean method
Sangeeta Gulati
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From: BSK <sa...@ict.sc.ru>
Newsgroups: geometry.puzzles
Subject: Re: Inscribed circle
Date: Sun, 19 Oct 2008 03:32:12 EDT
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> Find the equation of the inscribed circle of the
> triangle formed by the lines
> 4x-3y+9 = 0
> 12x+5y-43 = 0 and
> 7x+24y+19 = 0
(4x-3y+9)/5 = -(12x+5y-43)/13 = (7x+24y+19)/25 = R
Hence
x = 1
y = 1
R = 2
Scott Steketee
> I am trying to estimate value of pi using archimedean method with GSP. I have
> worked out the way to make inscribed polygons with iteration etc., but I can't
> get the circumscribed polygons. Please suggest the possible method.
You can construct the circumscribed polygon in a similar way, but with an additional step. The easiest thing to do is to make the construction of the inscribed polygon into a tool. I assume your inscribed polygon is based on two points (O, the center of the circle, and A, a point on the radius) and a parameter n, that you've rotated the point on the circle by 360/n, used the rotated point to construct the first side of the polygon, and iterated to depth n-1.
Make this construction into a tool by selecting the three given objects (point O, point A, and the parameter n) and the produced objects (the first side and the iterated images of that side that complete the polygon). You will use this tool in a moment to construct the circumscribed polygon. All you need to do is to locate one vertex of the circumscribed polygon and use the tool with that vertex.
Let the circumscribed polygon be tangent to the circle at A. To make the tangent, construct segment OA and the perpendicular at A. Then rotate A about O by 360/2n and construct the ray from O through A' (the new rotated point). Label the intersection with the tangent B, and use the tool on point O, point B, and parameter n.
Let me know if you have any questions about this construction.
[The following observations for Sangeeta may also be useful for other particularly intrepid readers.]
Since you worked out the construction of the inscribed polygon, it seems likely that the difficult aspect of the second construction was figuring out how to create that iteration based on a constructed point. As you know, iterations must be based on independent points and on parameters rather than constructed points or calculated values. In my directions above, I got around the independent-point requirement by embedding the iteration in a tool, so that I could apply it to the constructed intersection.
There's a second way to construct the circumscribed polygon that also would have worked. You could construct a second polygon based on center point O and a new independent point C. After constructing this polygon, you could merge independent point C with the constructed intersection B. By merging these two points, you turn the second polygon into the desired circumscribed polygon.
I sometimes use a similar technique when I want to do an iteration that uses a calculated value as a given of the construction. (I'm not talking about the depth, which can be a calculated value, but about an iteration in which a calculated value is to be one of the objects to be iterated.) In this case, I use a parameter instead to make the iteration. Once I've created the iteration, I edit the parameter to turn it into the desired calculation.
I hope this information is useful; let me know if you have any questions.
-Scott
Scott Steketee
Sketchpad Projects
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From: Avni Pllana <avn...@hotmail.com>
Newsgroups: geometry.puzzles
Subject: Re: Inscribed circle
Date: Mon, 20 Oct 2008 03:10:36 EDT
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> Can't we just let P(a,b) be the centre of the
> inscribed circle, then use the fact that the centre
> is the point which is the shortest distance from each
> line and this shortest distance is the radius, hence
> equal. Use the formula for the shortest distance to
> each line (thanks for the nice pythagorean triples),
> and solve these linear equations, which gives us the
> centre of the circle. The radius is easy to find, as
> we can just sub the values of the centre back into
> any one of the equations to determine the radius of
> the circle. That way, we never have to calculate the
> vertices of the the original triangle.
Hi Jack,
Your idea is quite good in the particular case of the incircle, as shown by BSK, but shy author used the very powerful technique of barycentric coordinates to solve the problem.
http://en.wikipedia.org/wiki/Barycentric_coordinates_(mathematics)
As shy author mentioned, the things get more interesting in 3D.
Best regards,
Avni