# John Lindal's Blog

## Math/Physics

**Light bulb physics**

The floodlight is a stable equilibrium because when we tip it, the center of mass rises. When we let it go, it falls back to the local minimum of its potential energy surface. The standard light bulb is an unstable equilibrium because the center of mass falls as it rolls over. This suggests the answer to making a neutral equilibrium: The center of mass must remain at the same height as the bulb rotates. The way to do this is to sculpt the bulb so its radius of curvature is equal to the distance between the center of mass and the bulb — no mean feat since the center of mass obviously depends on the shape of the bulb! But it should be possible for at least some portion of the bulb. After all, it works just fine for basketballs.

**Circles & Triangles**

**Q**: How does one inscribe an equilateral triangle inside a circle?

**A**: Draw a chord through the center of the circle. Construct an equilateral triangle on each side of this chord. Pick one end of the chord and bisect both angles. Extend these bisection lines until they intersect the circle. Connect these two points to get the third edge of the inscribed, equilateral triangle.

**Q**: How does one circumscribe an equilateral triangle with a circle?

**A**: Bisect two of the angles inside the triangle. The intersection of these two lines is the center of the circle. The three corners of the triangle obviously lie on the circle itself.

**Q**: Can an arbitrary triangle be circumscribed by a circle?

**A**: It seems reasonable, given that both triangles and circles have three degrees of freedom. To prove it, pick any two points in the plane. Construct the smallest circle that contains these two points: The diameter of the circle is the line segment between the two points. Now imagine expanding the circle while still including the two original points. Expand it first in one direction, then the other direction. The circle will sweep across all points in the plane except the limit, which is the line containing the two points. This proves that given any two points, there exists a circle which passes through any third, non-colinear point.

**Q**: Great! How do I do it?

**A**: All three sides of the triangle must be chords of the circumscribing circle. If you bisect any chord of a circle, the resulting line passes through the center of the circle. So, bisect any two edges of the triangle. The intersection point of these bisection lines is the center of the circumscribing circle.

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