GLAT #16
Seems like I am working in backwords order almost. Well, I asked Mr. Horne about number 16 today, he wasn’t quite sure about it, which was not very comforting, but After a few minutes of thought in the stimulating environment of a school bus, I saw the light. Before I go any further, I should probably Write the problem.
Given a triangle ABC, how would you use only a compass and a straight edge to find a point P such that triangles ABP, ACP, and BCP have equal permiters? (Assume that triangle ABC is constructed so that a solution does exist.)
Amazingly, I did this with no help from outside sources, and Even without a compass….All the bus had to offer was a Coke Bottle Cap. Here goes: Triangle ABC is a 45-45-90 triangle, Drawn Such that AC is the hypotenuse. Take your compass and place it on point C, such that the radius of the circle to be drawn is the length CB (equal to AB). Draw this circle, then place the compass on point A and to the same. You will notice that your Two circles Intersect at two points: B , and a new point (opposite B), which is Labeled P!. You have just created a perfect square! Of course, this is not nearly as trivial if the triangle is not a right isosceles triangle… in such case, one would need to find the isoperemetric point which exists at the center of the outer soddy circle.
June 14th, 2005 at 1557:12
But what if it’s NOT a right triangle like that? The solution still exists, but how can you be sure your method works?
(I think it does; that looks like the one I stumbled on. But have you considered it?)
June 15th, 2005 at 1200:55
In the case that it is not a right, you would use the isoperemetirc point of the triagle. The isoperemetric point is the center of both soddy cirlces. The outer soddy circle is tangent to three tangent circles drawn about each of the points.