In the days preceding a recent lunar eclipse, my daughter saw an illustration that seemed to show that the Moon's diameter was just about equal to the width of the penumbra. She conjectured that if that were true, it may be because of the fact that the apparent diameter of the Moon (as seen from Earth) is approximately equal to the apparent diameter of the Sun. I made a GeoGebra applet to help me think about this question, and found that she was right.

In this post, I will explain the thinking that went into making this construction. Stop reading if you want to figure it out on your own!

**How to construct the tangent to a circle through a given point:**Of course, this is impossible if the point is inside the circle, and it is easy if the point is on the circle. But what if the point is outside the circle?

GeoGebra has a "Tangent" tool (under the "Perpendicular Line" tool,) but in my construction unit, I ask geometry students to do this in the "Basic Geometry" environment, which does not have that tool.

The key, as in many construction challenges, is to start by imagining the problem is already solved, and to study the resulting figure. We are given point P outside a circle with center O:

All we need to construct the tangent is to find T, the point of tangency. But the radius is perpendicular to the tangent at the point of tangency. It follows that the point of tangency is on the circle with diameter OP:

**How to construct the common tangent to two circles:**This is trickier, and moreover, GeoGebra does not have a built-in tool to do it. Again, imagine the problem is solved:

If we can find point P, we can use what we learned earlier to construct the tangent. The key is to study the similar triangles in the figure, and see that the ratio of similarity is the ratio of the radii. But this ratio would be the same in this figure:

Fortunately, this figure is easy to construct. Draw any radius in one circle, and a parallel radius in the other. Join the points where they meet the circles, and extend the line connecting the centers. P is at the intersection of those two lines.

A similar method yields the common tangents in between the circles. Both are needed in the Penumbra figure.

--Henri

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