Two rays from the right-angled vertex of an isosceles right triangle split the hypotenuse into three sgments.

Prove that the three segments can be used to form a right triangle (the middle segment as the hypotenuse) if and only if the angle between the rays is 45°.

Call the second right angled triangle ABC, labelling AB the hypotenuse, h, opposite C. A circle with h as diameter will intersect with the circle, radius a, on B, and the circle, radius b, on A, iff ACB is a right angle.

There is no loss of generality for this right-angled case, if we extend AB to points R and L on the edges of the circles radius a and b, which gives a 3-part line of the sort referred to in the original problem.

Assume however that ACB is not a right angle. The crucial difference here is that there will then be 3 separate intersection points between the 3 circles we have just drawn; this means that there will not, and cannot, be any single one point K of the sort we are about to construct. So the problem 'works' iff ACB is a right angle.

Construct a circle, X, with diameter RL. The angle bisector of ACB crosses X at two points, (1) J, on the SAME side of RL as ABC, where it creates a larger copy of ABC, and (2) K, on the OPPOSITE side of ABC, where it creates an isoceles right angled triangle:

Construct the perpendiculars P1 and P2 to RL from points A and B. These cross KR and KL at A'and B'; since these points also fall on the circles radius a and radius b, the two non-right angles of ARA' and BLB' respectively must each be equal and must be 45 degrees.

Now the angle bisector of LKR will pass through the origin of X, and must be parallel to the perpendiculars P1 and P2. It follows immediately that the rays KA and KB must be 45 degrees apart.

Posted by broll on 2011-04-04 07:06:30