This is a bit of fun geometry that doesn’t have much to do with what’s going on in class, but does reflect on mathematical thinking.
An article promoting the use of technology in the classroom began: “Draw a perfect circle. Now bisect that with a 45-degree angle, the perfect slice of geometric pizza. Now, using your drawing, find the area of the rest of the circle. ”
What they meant was that the 45-degree angle divides the circle (or, more properly speaking, the region contained by the circle*). “Bisect” has a rigorous definition in mathematics: It means to divide an object into two objects of the same exact size. So if we bisect a circular region with anything, then the area of the two pieces will have half the area of the circular region, by definition.
Rather than accepting this and moving on, though, I asked myself: Can you bisect a circular region with a 45-degree angle? I shifted this to: Given a circle and an angle whose vertex is on or in the circle, what is the smallest angle that will cut out half of its region?
My conjecture is that the shaded region, where angles BEA and CEA are congruent, represents the largest portion of the circular region that can be covered by angle BEC. Let \(\alpha = m\angle BEC\).
So what is the area of the shaded region? First, we have sector BAC. The area of a sector is \(\beta r^2/2\), where \(\beta\) is the measure of the central angle (in radians). Since it’s a unit circle, \(r^2 = 1\). Since angle BEC is the inscribed angle that corresponds to the central angle BAC, it has half the measure, and the area of the sector is \(\beta/2 = \alpha\).
Since AE bisects angle BEC, the two triangles are congruent. Since AC, AE, and AB are all radii, the two triangles are isosceles. We can determine the area of triangle BAE by first dropping a line perpendicular to BE. This divides the triangle into two congruent right triangles, AFE and AFB.
Call \(\gamma = m\angle AEF\), so \(\gamma = \alpha/2\). To find the area of \(\Delta AFE\), we need a height and a width. Since AE is a radius, the height AF is \(\sin\gamma\) and the width FE is \(\cos\gamma\). This gives an area of \(\sin\gamma\cos\gamma/2\).
I’ll write about the double- and half-angle trigonometric formulas in a separate post, but one of these is: \[\sin 2\theta = 2\sin\theta\cos\theta\]
Applying this gives us an area of triangle AEF of \((\sin 2\gamma)/4 = (\sin\alpha)/4\). Since there are four such congruent triangles, the total area of the two larger shaded triangles is \(\sin\alpha\), and the area of the entire shaded region is \(\sin\alpha + \alpha\).
If the shaded region is half the area bounded by the circle, and since the area bounded by a unit circle is \(\pi\), \(\sin\alpha + \alpha = \pi/2\). I’m not sure how or if that can be solved analytically, but we can use a calculator to graph \(\sin\alpha + \alpha – \pi/2\) and find its solution.
The solution function of the calculator gives a zero at 0.8317112.
Throughout this post, I have worked in radians. For the Algebra 2 students, we haven’t gotten there yet. Radians represent a different way to measure angles, but it’s a straightforward conversion: \(2\pi = 360^o\), and \(1 = 180^o/\pi\). So we have \(0.8317112 \times 180^o/\pi \approx 47.65^o\).
So: An inscribed angle slightly larger than 47.65 degrees bisected by a diameter will create a rounded wedge that bounds half as many points as a circle.
Going back to the original inspiration for this item, this means that the 45 degree angle with a vertex that is inside a circle cannot bisect the circle’s region.
I started with a conjecture. This conjecture can be proven, but the formal proof requires more trigonometry, so I’ll leave it for the reader, or for another time.
* Since this article is about rigorous definitions, I’m using “circle” to refer to the set of points equidistant from a point (i.e., \((x-h)^2 + (y-k)^2 = r^2\)) and “circular region” to refer to all points satisfying \((x-h)^2 + (y-k)^2 \le r^2\). In high school, we often conflate these two with the word “circle”.