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
Δ*AGD* 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
Δ*AGD* 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* = 2*r* and *BG* = 20, so *EG* = 2*r* – 20. The square side *BD*
is a bisected chord, so the circle intersecting chord property gives

20 • (2*r* – 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
Δ*AGD* 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
Δ*AGD* 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.

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