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Forum URL: http://www.cut-the-knot.org/cgi-bin/dcforum/forumctk.cgi
Forum Name: High school
Topic ID: 248
Message ID: 14
#14, RE: Cut Triangle in Half
Posted by Bractals on Jul-15-03 at 06:56 PM
In response to message #13
Hi Golland,

In calculating the x = w + sqrt(...) value one must use the u2,v2 values and for the x = w - sqrt(...) value one must use the u1,v1 values.

With respect to the deltoid, consider the following figure ( unlabeled - but consider the triangles labeled as in my previous figure ).

http://geocities.com/bractals/deltoid.gif

In the upper left corner the yellow area represents all the points P satisfying the inequality c/2 < x < c, where the x is the w + sqrt(...) value. In the upper right corner the blue area represents all the points P satisfying the same inequality c/2 < x < c, where the x is the w - sqrt(...) value.

When you overlay these figures you get the middle one with the small dark triangle representing points P which have two different lines cutting sides AB/AC as solutions.

About this dark triangle. The "side", which is not along a median of triangle ABC, is not a straight line. It is a hyperbolic arc concaved in towards the centroid. The equation for this arc is u = bc/(8v); which comes from setting the radicand equal to zero.

When you add in the points for the lines that cut sides AB/BC and AC/BC you add two more small triangles making up the deltoid.

After going through this further analysis - I want to modify my conjecture on the deltoid. All points in the interior have three solutions, all points in the exterior have one solution, and all points on the boundary have two solutions ( except for the deltoid vertices which have only one solution ). This can be seen by noting that a line cutting sides AB/BC can intersect the small dark triangle in my figure.

I hope this puts the problem to rest.

Bractals