## Tuesday, May 20, 2008

### Geometric "Connections" - How one problem leads to another...

Answers to Problems from Previous Post:

(a) Substitute 0 for x, -1 for y in both equations.

(b) a > 1/2

(c) Two points above x-axis: a > 1; Below x-axis: 1 > a > 1/2; On x-axis: a = 1
Note: If 1/2 > a > 0, then the only point of intersection would be (0,-1).

(d) x = ±[√(2a-1)]/a; y = (a-1)/a

(e) Points: (±4/5 , 3/5); a = 5/2
Note: Pls check for accuracy!

Now for the connection...

In the previous post, we were given that the radius of the circle was 1 and the area of the triangle was 32/25. From this it can be shown that PQ = 8/5 and, in fact, the quadrilateral PQRS shown in the figure at the left is a square whose area is 64/25. The fact that this was a square intrigued me. I hypothesized that, up to similarity, these numbers were unique. This led me to the diagram at the left and the following converse of the previous problem.
Note: This problem is now unrelated to the parabola.

In the diagram above, points P and Q are on the circle, PQRS is a square and segment SR is tangent to the circle at T.
If the radius of the circle is r, show that the area of the square is (64/25)
r2, and, consequently, the area of ΔPQT = (32/25)r2.