OK, well maybe that title isn’t really what this problem is about, but many teachers might think this problem is too hard for a lot of geometry students. Initially, we were hesitant to include it in our discussion of circles, since we don’t discuss the all-to-familiar unit circle really at all in our geometry course discusison of trigonometry. But this problem is really more of a discussion of the special right triangle (45-45-90) or if students choose the prior method of the intersection of the line and circle (see previous blog entry). This is a really interesting problem with different solutions, so it’s fun to see how different students approach it.

The vertices of a square with sides parallel to the coordinate axes, lie on a circle centered at the origin with radius 5. Find the vertices of the square.

What have we discussed up until now? The students have formalized the idea of the equation of a circle centered at the origin as x^2+y^2=r^2 (from the Pythagorean Theorem) so they should be able to come up with the equation of the circle. They also know the terms of the radius, diameter, coordinate axes and show be able to set up the picture.

When I did this problem in class, many students started by drawing rectangles inscribed in the circle with dimensions of 6 x 8, realizing that the diagonal would be 10. This was a great start. They realized that somewhere in between the 6 x8 and 8 x 6 rectangle was the square they were looking for. They were then able to actually draw the diagram with that square in the circle. They could then approximate the coordinates which helped a lot in the class discussion. I felt very proud of students who were able to come to class with even that done. However, other students got even farther.

This is where I wish I were able to draw some diagrams on my blog. I need to do a little more research with Edublogs to see if this is possible. Most students then drew the diameter as the hypotenuse of an isosceles right triangle and either used the Pythagorean theorem to solve for the side, or remembered the ratio of the 45-45-90 triangle, and simply divided 10 by sqrt(2). Once they realized the side of the triangle would be 10/sqrt(2), they just needed to transfer that information to the coordinate plane. This is where a good diagram was important.

Since the origin was at the center of the circle, and the radius was half the diameter, most students could easily extrapolate the fact that you would simply divide the side of the isosceles triangle in half to obtain the coordinates. In other words, you’d just go over 5/sqrt(2) and up 5/sqrt(2) to get to the vertex in the first quadrant and do the same for the other quadrants (taking into account the negatives for left, right, up and down).

This was an amazing problem for discussion since another student did the problem by finding the intersection of the line y=x and the circle x^2+y^2=25. This student had been able to independently transfer the skill of substituting the line equation into the circle equation from a previous problem. This was the minority of the class however, but still someone was else to do it. I was impressed. If you’d like to try this problem it was #8 on p.50 at my curriculum