Circle tangent to circular arcs

By Brad Rodgers (P1930) on Tuesday, January 23, 2001 - 11:37 pm :

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 inner circle is what? It really is hard to make images in paint, isn't it?

Diagram

Any hints or answers are greatly appreciated.

By Tom Hardcastle (P2477) on Wednesday, January 24, 2001 - 07:13 pm :

A first step would be to find the radius of the arcs. This can be done by considering that

r θ = 12

(from the information in the question, where

θ

is the angle through which the arc is traced, in radians) and that

r \cos θ = 0.5 r

(since the two arcs are symmetrical and meet.)
Arcs diagram

By Brad Rodgers (P1930) on Thursday, January 25, 2001 - 12:33 am :

θ = 60

r = 36 / π

. You can see this also by seeing that triangle ABC is equilateral. Where do I go from here though?
Thanks,

Brad

By Tom Hardcastle (P2477) on Thursday, January 25, 2001 - 12:49 pm :

Here's a possible way to do it that also introduces a useful idea for similar problems.
Say that the origin is at A.
Then the cartesan equation for the arc BC is

x^{2} + y^{2} = (36 / π)^{2}

The cartesan equation for the inscribed circle is
(x - 0.5r)² + (y - q)² = q²
where q is the radius of the inscribed circle (can you see why?).
Then you use these equations to eliminate y and form a quadratic in x (exercise for the reader).
Because the inscribed circle is at a tangent to the arc, it touches the arc at only one point.
Therefore, there is only one solution to the quadratic.
Therefore, the determinant of the quadratic is zero, (b² - 4ac = 0). Use this to determine q, the radius of the inscribed circle, and its circumference turns out to be 27.

By Brad Rodgers (P1930) on Friday, January 26, 2001 - 01:06 am :

That's a good approach, Tom.

Brad

By Michael Doré (Md285) on Saturday, January 27, 2001 - 01:16 pm :

This problem can be solved very simply, using Mobius maps. In particular we can use the map $z \to 1 / z$ (where $z$ is a complex number representing the position vector of a point on the diagram, relative to a fixed point). Apparently this map is sometimes called inversion (for obvious reasons). We take our origin as the midpoint of $B C$ (let's call this $O$ ), and then replace each point $T$ in the diagram with a point $T'$ such that $O T \times O T' = 1$ , and so that the vectors $O T$ and $O T'$ are in the same direction. (And technically you should flip the diagram over too because if $z \to 1 / z$ then the argument of the complex number is negated but we don't need to bother about that.)

Now remember that in another discussion we showed that if you apply this inversion Mobius map then the following changes occur:

- circles and lines which pass through the origin turn into lines.
- circles and lines which don't pass through the origin turn into circles.

OK, let's apply this at

O

then. Forget completely the inner circle for the moment. Let

L

be the line through

A O B

,extending to infinity in each direction.

Now imagine drawing out the whole circle, which coincides with the arc $A C$ . Call this circle $P$ . First note that the circumference of circle $P$ intersects $L$ at $A$ and one other point, call it $K$ (this one is to the right of $B$ ). You can see $B A = B K = 36 π$ (as $36 / π$ is the radius of $P$ , as Tom and you worked out above). So $O A = 1 / 2 B A = 18 / π$ and $O K = A K - AO = 72 / π - 18 / π = 54 / π$ .

Now $A'$ and $K'$ (those are the images of $A$ and $K$ under the Mobius map) must lie on $L$ , and the image of $P$ (which we'll call $P'$ ) is going to be a circle with diameter on $L$ (by symmetry about $L$ ). It follows that $A'$ and $K'$ are on the diameter of $P'$ . We know that $O A' = π / 18$ and $O K' = π / 54$ and clearly $A'$ , $O$ , $K'$ appear in that order, so $A' K' = π / 18 + π / 54 = 2 π / 27$ . So $P'$ is a circlewith diameter along $L$ , and of radius $π / 27$ .

Now by symmetry, the image of $Q$ (called $Q'$ ) also is a circle on the line $L$ of radius $π / 27$ - it is simply the reflection of $P$ in the line $O C$ .

Now what is the image of $R$ , the ''inscribed circle''? $R$ passes through $O$ and $C$ and its image must be a line (as it passes through $O$ ). This line $R'$ is horizontal (by symmetry in the line $O C$ ). It passes through circles $P'$ and $Q'$ (each at only one point; each point is the image of the intersection points of circles $R$ and $P$ , and of circles $R$ and $Q$ respectively). But circles $P'$ and $Q'$ both have diameters on $L$ and are of equal radius therefore $R'$ must be tangential to both $P'$ , $Q'$ , and a distance of $π / 27$ from $L$ (because the radius of $P'$ and $Q'$ is $π / 27$ ). Obviously $R'$ is parallel to $L$ . So if we draw a perpendicular (vertical) line from $O$ till the intersection point with $R'$ , it would be of length $π / 27$ and the intersection point is the image of the highest point of the original inscribed circle $R$ , therefore the diameter of $R$ is the reciprocal of $π / 27$ , so circumference of $R = π \times diameter = π / (π / 27) = 27$ as Tom obtained.

Obviously that's not how you're supposed to do it, but I think it is a fun approach. I think that the inversion approach could be quite useful generally. The reason it works here is because we are able to change the inscribed circle $R$ , which intersects with theother two circles, into a line. It is much, much easier to see how lines intersect with things than circles, so the proof is then relatively straightforward.