## Use Your Mind to Save Yourself Time

I’m looking forward to Games Night at the Math Prize for Girls coming up this weekend. I’ve been working with the author of “Mess or Math?” to create a really fun math activity and I hope many of you who are going will participate!

Most math contests are timed making time precious. You don’t want to waste it. Yet, often I see students waste time performing unnecessary computations or going down laborious detours that eat up that precious time. So here’s one small piece of advice to help you use your time more effectively: Before you plow through a laborious computation, stop and think about whether you really need to do it. Use your mind to save yourself time.

To illustrate, consider problem #24 on the 2000 AMC 12: If circular arcs $AC$ and $BC$ have centers at $B$ and $A$, respectively, then there exists a circle tangent to both arc $AC$ and arc $BC$, and to segment $AB$. If the length of arc $BC$ is 12, then the circumference of the circle is…[multiple choices follow]

Here’s how one student solved this problem. First, the student exploited the key fact that at a point on a circle, the tangent at that point is perpendicular to the radius to that point. Therefore, the line segment from $A$ to the point where arc $BC$ and the inscribed circle are tangent, passes through the center of the inscribed circle and we have the situation shown in the figure at right.

The student then applied the Pythagorean theorem to obtain the equation $r^2 + (R/2)^2 = (R - r)^2$. At this point, the student noted that triangle $ABC$ is equilateral, and that, therefore, arc $BC$ is one-sixth of the full circle that contains it. This means that the full circumference of that circle is 6 times 12 (the given length of arc $BC$), or 72 units. Using the formula $2\pi R$ for the circumference of a circle with radius $R$, the student deduced that $R = 36/\pi$. Substituting this value into the Pythagorean equation yields $r^2 + (18/\pi)^2 = (36/\pi)^2 - 2(36/\pi)r + r^2$.

Solving for $r$, the student found $r = \frac{(36/\pi)^2 - (18/\pi)^2)}{2(36/\pi)}$. The student then evaluated this expression, diligently computing the squares of 36 and 18, like so: $r = \frac{(36/\pi)^2 - (18/\pi)^2}{2(36/\pi)} = \frac{1}{\pi}\frac{36^2 - 18^2}{72} = \frac{1}{\pi}\frac{1296 - 324}{72} = \frac{1}{\pi}\frac{972}{72} = \frac{27}{2\pi}$.

Finally, the student used the formula $2\pi r$ to compute the circumference of the inscribed circle.

That’s certainly a valid solution and demonstrates a wealth of geometric knowledge. But during a contest, because of the time factor, you want to save as much time as you can. You might have noticed that the computation itself can be simplified a lot. For example, it wasn’t really necessary to compute the squares of 36 and 18. For instance, one could use the identity $a^2 - b^2 = (a+b)(a-b)$ to write $36^2 - 18^2 = (36 + 18)(36-18)$. But I’m trying to get at even greater simplification than this!

Stop and think: What am I being asked to find and what do I need to get the answer?

You have to find a circumference of a circle. You’re given one-sixth the circumference of a related circle. All circles are similar to each other, so if you can find the ratio of the radii of the two circles, you’re close to being done.

From the Pythagorean equation $r^2 + (R/2)^2 = (R - r)^2$, we simplify before substituting any values, and find $2rR = 3R^2/4$, or $r = 3R/8$. Therefore, the circumference of the inscribed circle is 3/8 that of the circle that contains the arc $BC$. Therefore, the answer is 3/8 of (6 times 12), which is 27. Done!

No computation of squares, no circumference formulas, no complex fractions…and plenty of time remaining to work out problem #25.

To all MPFG competitors: good luck, have fun, and I hope you make many new friends! 