Given a triangle ABC, how can you find a point P on AC and a point Q on BC, such that AP=PQ=QB?
N.B. A construction method is sought, and only compass and straightedge are allowed.
(In reply to Two solutions for one triangle!
It is true in general that there are two solutions for any given triangle. The positioning is algebraically found via solutions to a quadratic equation: (1 - 2 cos C) x^2 + (a+b)(2 cos C - 2) x + c^2 = 0.
In the program to find the measured-off distance from A or B to measure toward C that I presented in my comment, two solutions are found for a quadratic. In some instances, such as the case you provide, none of the solutions have points that both lie on the sides of the triangle, and that program was written to reject such solutions. In other instances, one of the sets of points has both P and Q on their respective sides of the triangle (between A and C and between B and C), while the other has one of the points on an extended side.
For example, one of the cases I gave was
which is an isosceles triangle with AC=BC=1 and AB=.5. The solution the program gives is 1/3 unit from A toward C on AC and 1/3 unit from B toward C on BC, making PQ also have length 1/3. However, there's another solution that the program rejected for failing to have both points within the bounds of the sides of the triangle: -1. That is, measure 1 unit from A away from C, on CA extended, and one unit from B away from C on CB extended; PQ is then also 1 unit.
Another example I gave was
where AB is 1, BC is 1.8 and AB is 2. If .89945... is measured from A toward C for P, and from B toward C for Q, then PQ will also be .89945..., and both P and Q will actually be on the sides of the triangle, between A and C and between B and C respectively. But 5.131314615009747 was also a solution, meaning measure 5.131314615009747 from A toward C, thus winding up 4.131314615009747 beyond C on AC, and 5.131314615009747 from B toward C, winding up 3.331314615009747 beyond C on BC, and PQ will also be 5.131314615009747.
It looks as if among the four points, two are on the actual sides of the triangle and two on the sides extended. In the case you found, one of each PQ pair was on a side and the other was on a side extended, while in the cases I show, one PQ pair had both points on a respective side, and the other had them both on extensions of the sides. That was a selection on my part originally, to show only those where both points were on non-extended sides.
Posted by Charlie
on 2005-05-07 20:31:19