## A Geometry Problem with Three Circles and a Point

I friend gave me a geometric problem which turns out to boil down to the figure above. We have three arbitrary circles C1, C2, C3. Now we construct the three green lines through the two intersection points of each pair of these circles. We get lines g11, g12, g13. These lines intersect in one point. Why?

As you know, the argument for the middle perpendicular lines on the sides of a triangle goes like this: Each middle perpendicular is the set of points which have the same distance to two of the corners. So if we intersect two of them in P then d(P,A)=d(P,B) and d(P,B)=d(P,C), which implies d(P,A)=d(P,C). As usual, d(P,A) denotes the distance from P to A. Thus P is also on the third middle perpendicular. Note that we need that P is on the middle perpendicular on AB if and only if d(P,A)=d(P,B).

A similar argument is possible for the angle bisectors. These rays are the set of points with equal distance to two sides. For the heights, such an argument is not available. The standard proof goes by constructing a bigger triangle where the heights are middle perpendiculars. By the way, this proof stops working in Non-Euclidean hyperbolic geometry, where the fact still holds.

Can we make up a proof similar to these proofs for our problem? It turns out that this is indeed possible. The correct value is the following:

$$f(P,C) = d(P,M_C)^2-r_C^2$$

where r(C) is the radius of the circle C, and M(C) is its center. To complete the proof, we only need to show that the line through the intersection of two circles C1 and C2 is the set of all points P such that f(P,C1)=f(P,C2). Then the proof is as easy as the proofs above.

There are several ways to see this.

We could use a result that I call chord-secant-tangent theorem which deals with products of distances of a point on a secant or chord to the circle. But it is possible to get a proof using the Pythagoras only. In the image above we have

$$d(P,Q)^2+d(M,Q)^2 = d(P,M)^2, \quad d(S,Q)^2 + d(M,Q)^2 = r^2$$

Thus

$$f(P,C) = d(P,M)^2 – r^2 = d(P,Q)^2-d(S,Q)^2$$

where C is the circle. Now, if we have two intersection circles, the right-hand side of the equation is the same for both circles, and thus also the left-hand side.

We have seen that f(P,C1)=f(P,C2) for all points on the green line.

But we have to prove the converse too. For this, we observe that D(P,C1)=D(P,C2) implies that P is on a circle around M(C1) and on another circle around M(C2). The two circles meet only in points on the green line.

There is also another way to see that f(P,C1)=f(P,C2) defines the green line. If you work out this equation analytically, you see that it is equivalent to an equation for a line. I leave that to you to check.

Note that there is a second situation where the result does hold too.

In this case, we need f(P,C) for P inside C. It will be negative, but f(P,C1)=f(P,C2) still holds for all points on the line through the intersection, and even if P is on the circles.

There is the following special situation.

It can be seen as a limit case of the previous situation. But it can also be proved by observing that all the tangents have the same length between the intersecting point and the tangent point.

Here is another situation.

The green lines are the sets of points such that f(P,C1)=f(P,C2) for two of the circles. It is quite interesting to construct these lines. I leave that to the reader.

25. Januar 2018 von mga010
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