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)r

^{2}, and, consequently, the area of ΔPQT = (32/25)r

^{2}.

Comments:

(1) Students should not find this overly challenging using standard methods for solving circle problems (and the fact that it is closely related to the previous question).

(2) Of course, what really intrigued me is how, once again, the 3-4-5 triangle recurs! Ask your students to find a triangle in the diagram similar to 3-4-5. They need to draw something but this should occur naturally from the standard solution to the problem.

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