From my last post, you know that I’m a big fan of problems that can be solved in multiple ways, especially for students of many ages. Here’s a surprisingly pretty geometry problem that I found on Twitter under #mathchat (https://twitter.com/hashtag/mathchat)—a phenomenal source of math conversations and professional support.

A square of side length 20 has two vertices on a circle and one side tangent to the circle. What is the circle’s diameter? (https://casmusings.files.wordpress.com/2014/12/circle1.jpg?w=500 )

What I particularly like about this problem is that it is accessible to middle schoolers with simple tools and also has value to older students with significant trigonometry skills. In addition, it has a couple of hidden gems.

Placing auxiliary lines is definitely an acquired skill, but they are key to this problem. I added diameter BE and radius AD.

Method 1

Segment BG is a symmetry line for the square, so BC = GD = 10 and BG is perpendicular to FD, making ΔAGD a right triangle. If radius AB = r, comparison of segments GB and DC gives AG = 20 – r. From here, use the Pythagorean theorem to determine r and double it to find the requested diameter.

Unexpected Surprise 1

Once you have a value for r, you can show that ΔAGD is a 3-4-5 right triangle. A lovely surprise.

Method 2

A more sophisticated geometric approach sees FD and BE as intersecting chords in a circle, and therefore the product of their intersected parts are equal. Diameter BE = 2r and BG = 20, so EG = 2r – 20. The square side BD is a bisected chord, so the circle intersecting chord property gives

20 • (2r – 20) = 10 • 10, a linear equation in r giving the same answer as method 1.

Unexpected Surprise 2

Without the Pythagorean theorem, triangle similarity establishes the method 2 chord relationship, and symmetry proves that ΔAGD is right. Generalizing the original square, we let GD = x, GA= y, and AD = r. Using method 2 again gives (r + y) • (r – y) = x • x. Expanding the left side and rearranging the terms give an unexpected proof of the Pythagorean theorem!

Method 3

Even more sophisticated geometry and algebra write the area of ΔAGD two different ways, one using Heron’s formula, to get

This becomes a perfect square quadratic when expanded and rearranged and like terms are collected.

Method 4

A definitely more complicated approach uses alternate interior angles and the law of cosines.
In right triangle BCD,

so applying the law of cosines to Δ ABD in isosceles triangle BAD gives

Substituting

ΔBCD again leads to the solution.

Method 5

This is the most complicated method, making use of the law of sines and some trigonometric identities on angles in Δ BAD:

Some of these methods are definitely more elegant, whereas others require complicated thought and exploration. In the end, I hope students (and teachers!) occasionally can use problems like this to develop deeper thinking, connections, sense making, and comparative reasoning along the lines of the inspirational and aspirational Common Core State Standards for Mathematics.

CHRIS HARROW, casmusings@gmail.com, a National Board Certified Teacher, is the mathematics chair at the Hawken School in Cleveland, Ohio. He blogs and presents nationally on the educational uses of technology and on Computer Algebra Systems (CAS) in precollegiate mathematics. He is also the recipient of a Presidential Award for Excellence in Mathematics and Science Teaching.