While nibbling on lo mein leftovers at dinner this evening, I finally found a reasonably geometric solution (does not explicitly use double angle formula). It only needs the pythagorean theorem, similar triangles theorem, and some even more basic theorems. Oh - and an isosceles triangle theorem, that if two angles of a triangle have the same value, the sides opposite have the same length.
Note that "long leg" and "short leg" are based on the diagram as presented. In order for the diagram to appear as presented, the apex angle must be acute, implying that the angles labeled φ have a span of less than 45° so this is not an ambiguous definition.
- Construct perpendiculars and lines as shown in figure which I just uploaded and will link.
- Note that all angles labeled φ are easily shown to be equal to each other.
- Define the radius of the small circle as 1. Then, by Ether's definition, the radius of the large circle is K.
- Note that the length labeled Q (terminated by open arrows in the diagram) is equal to the ratio of the hypotenuse to the base of the large triangle, by benefit of the apex triangle having a base of length 1 and being similar to the large triangle. That is, Q is that ratio which is to be calculated. [Yes, this is the Q in Q.E.D.]
- Consider the right triangle with a hypotenuse connecting the centers of the two circles, and legs parallel to the legs of the large triangle. The hypotenuse has length K+1 (being the sum of the two radii 1 and K), and the base has the length K-1, being the difference between these same two radii.
- We then use the pythagorean theorem to show that the height of that small triangle is 2√K (X2 + (K-1)2 = (K-1)2 ==> X2 = 4K ==> X = 2√K). (not shown on diagram)
- Next, determine the length of the segment from the center of the small circle to the apex of the large triangle. Because this triangle between these two points and the tangent point of the small circle with the vertical side of the large triangle is similar to the triangle just analyzed, and has a base of 1, its hypotenuse is (K+1)/(K-1).
- Next, consider the triangle with vertices at the center of the small circle and the two endpoints of segment Q. Because two of its angles are equal, it is an isosceles triangle, with the sides opposite the equal angles having equal lengths. This also means that when the perpendicular bisector to the long side is added as shown, these two triangles are equivalent.
- Because the sum of the long legs of these two right triangles is (K+1)/(K-1), and the two long legs are equal, each is of length (K+1)/2(K-1).
- Finally, we note that the right triangle including segment Q as its hypotenuse and angle φ at the apex is similar to the triangle with the two circle centers. Is long leg is length (K+1)/2(K-1), and the ratio of the hypotenuse to the long leg to is (K+1)/2√K. We therefore find that segment Q is of length (K+1)2/(4√K(K-1)).
Q.E.D.