Great Problems Keep on Giving
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
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.
auxiliary lines is definitely an acquired skill, but they are key to this problem.
I added diameter BE and radius AD.
is a symmetry line for the square, so BC
= GD = 10 and BG is perpendicular to FD,
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
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.
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!
Even more sophisticated geometry and algebra
write the area of
different ways, one using Heron’s formula, to get
This becomes a perfect
square quadratic when expanded and rearranged and like terms are collected.
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
ΔBCD again leads to the
This is the most complicated method, making use of the law of sines and
some trigonometric identities on angles in
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.
email@example.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.