Agreed.

Chatting with a co-worker about this problem on Friday afternoon, we came up with another method (slightly more algebraic) to determine the length of the line segment from the center of a circle to the apex of the triangle, and came up with another quiz.

Alternate calculation: From the center of the circle of radius 1, a series of infinite circles can be inscribed in the circle/angle of radius 1/K, 1/K2, 1/K3, etc. The sum of the diameters of this infinite series of circles is 2/(K-1). When added to the radius of the larger circle, you get (K-1+2)/(K-1) or (K+1)/(K-1).

The follow-on quiz is about the golden ratio. Geometrically speaking, a golden rectangle is a rectangle which is similar to the rectangle created when a square is appended on its long side, or equivalently when a square is excised on the short side.

That is, in the diagram uploaded, given that AB=CD, AC:AB = AB:AD.

The quiz:

- Develop a geometric construction which produces a golden rectangle (simple).
- prove geometrically (that is, not using any analytic tools, including the quadratic equation or even the pythagorean theorem) that the construction developed in 1 (or another valid geometric construction) produces rectangles meeting the description above.

A hint on constructing a golden rectangle:

The definition of the rectangle implies that if the ratio of the longer side to the shorter side is named Ф, the following equation holds:

Ф2 = Ф + 1

Using the quadratic formula and selecting the positive root yields:

Ф = (1 + √5) / 2

OK, no takers on the construction, so here it is, along with an **analytic **proof.

Construction:

- Construct square ABDC, extending sides AB and CD as shown.
- Bisect segment AB to find point E
- Draw circle centered at E with radius ED. Intersection with line AB is point F.
- Draw perpendicular to AF at F; intersection with CD is G

Tedious analytical proof that rectangles BFGD and AFGC are both golden rectangles:

- BFGD and AFGC differ by a square of their common length.
- Define λ to be the length of EB.
- By construction, BD = 2 λ
- By construction, EBD is a right angle
- By the Pythagorean theorem hypotenuse ED = √((2λ)2 + λ2) = √5 λ.
- As EF and ED are radii of the same circle, EF = √5 λ.
- BF = (√5-1) λ, AF = (√5+1)λ.
- AF/BD = (√5+1)λ/2λ = (√5+1)/2
- BD/BF = 2λ/(√5-1)λ = 2(√5+1) / (√5-1)(√5+1) = 2(√5+1) /(5-1) = 2(√5+1)/4 = (√5+1)/2
- As AF/BD = BD/BF, BFGD and AFGC are similar.
- As BFGD and AFGC are similar and differ by a square on their common side, they are both golden rectangles.

There’s still the matter of a **pure geometric proof** of this or another construction. There is at least one much more elegant than the analytic proof above, based on essentially this same construction.

To both bump this thread and provide a hint, here’s a proof of Thale’s Theorem, which I used in my golden rectangle proof. Thale’s Theorem states that **an angle inscribed on a circle which intersects a diameter of the circle is a right angle**. It is a special case of the general inscribed angle theorem, which states that an angle inscribed on a circle spans an arc of the circle which is twice the angle.

OBTW, is supposed to indicate an angle. It looks like CD has put in a different symbol.

- Begin with BAD inscribed on circle centered at C across diameter BD.
- Draw segment CA.
- The triangle BAD has angles:
- CBA + CAB + CAD +CDA
- Because CB, CA, and CD are all radii of circle C, they have the same lengths.
- Triangles BCA and ACD are isosceles. This implies that:
- CAB = CBA
- CAD = CDA
- The angles of triangle BAD therefore add up to: 2 CAB + 2 CAD.
- BAD =CAB + CAD, so the sum of the angles of triangle BAD is therefore 2 BAD.
- The sum of a triangle’s angles is two right angles, therefore BAD is a right angle. Q.E.D.

Additionally, Jason and I have not let Ether’s original question go; we spend a few minutes each day looking for alternate solutions. I have a derivation for the original quiz which does not require the Pythagorean Theorem, just similar triangles and a few odd (but easily proven) circle/angle theorems. As a clue, let me note that the vertical distance between the centers of the two circles equals the geometric mean of the diameters of the two circles. The constructions in my derivation are anything but obvious.

Rather than leaving this hanging, here’s the geometric proof of a golden rectangle construction:

Construction is as in my preceding post, with the addition of finding the other intersection of circle ED with line AB (point H), and a perpendicular at H to find point I on line CD.

Geometric Proof that the construction produces golden rectangles:

- HACI and BFGD are congruent by the symmetry of construction.
- Likewise, HBDI and AFGC are congruent by the symmetry of construction.
- It is clear that these pairs of rectangles differ by a square (ABDC); It remains to be shown that these two pairs of rectangles are similar.
- Draw segments HC and CF.
- Because HCF is an inscribed angle on diameter HF of circle E, HCF is a right angle (Thale’s Theorem)
- Angles HAC and CAF are right angles because of the initial construction of square ABCD.
- As triangles AFC and HFC share the angle at F and both have a right angle, they are similar to each other.
- As triangles AHC and HFC share the angle at H and both have a right angle, they are similar to each other.
- Because triangles AFC and AHC are both similar to triangle HFC, they are similar to each other.
- Because of this, rectangles AFGC and AHIC are similar to each other.
- GACI, BHJD, GBDI, and AHJC are all golden rectangles.

Here’s the alternate proof of the original quiz which does not use the Pythagorean Theorem, though it does use the geometric mean theorem. The geometric mean theorem is essentially proven in steps 5 through 8 of the Golden proof above, adding that because AH/AC = AF/AH, AH * AH = AF * AC, or AH is the geometric mean of AF and AC.

- Bisect angle BAC. Note that because AB and AC are both tangent to both circles, this line passes through their centers, designated E and H.
- Identify points D, F, G, and I being the intersections of perpendiculars of AB and AC through E and H, and therefore the points of tangency of those circles with the lines.
- Note that ADE and AFD are congruent, AGH and AIH are congruent, and these pairs of triangles are similar by virtue of having two angles (at A and a right angle) equal.
- Identify point J, being the intersection of AH with the circles. Construct a perpendicular to AH at J, extending past AB and AC. The intersections are points K and L. Because AJK and AJL are congruent (ASA), KJ = JL.
- Note that line KL is tangent to both circles, as are DG and FI. Therefore, DK and KG are both equal to KJ and FL and LI are equal to JL, so all are equal to each other, meaning that K bisects DG and L bisects FI.
- Designate the radius of the small circle as r.
- By definition of k, the large circle has radius rk. EH therefore has length r(k+1)
- Bisect segment EH to identify point M. Note that EM and MH are both length r(k+1)/2.
- Note that AMK and AML are similar to ADE and AHG so that length KM and ML are the average of DE and GH, or r(k+1)/2.
- Draw the circle centered on M passing through E (and therefore H, K, and L).
- Angle EKH is inscribed on circle M and spans diameter EH, so EKH is a right angle.
- By the geometric mean theorem, KJ is the geometric mean of EJ and JH. As EJ is r and JH is rk, KJ is r√k, as are JL, DK, KG, IL, and LF. KL is 2 r√k.
- Construct a perpendicular to AB through L. Designate the intersection with AB as N.
- Because triangles ADE and AGH are similar triangles:
- AD/r = (AD+2r√k)/rk, or AD k = AD + 2r√k, or AD = 2r√k/(k-1), and
- AE/r = (AE + r + rk)/rk, or AE k = AE + r + rk, or AE = r(k+1)/(k-1).
- Because NL is perpendicular to AD, and KL is perpendicular to AE, angle NLK equals angle DAE. Because angle KNL and ADE are both right angles, triangles DAE and NLK are similar. This means:
- NL = KL * AD / AE = 2r√k * 2r√k/(k-1) / (r(k+1)/(k-1)) = 4 r k / (k+1)
- AL = AK = AD + DK = 2r√k/(k-1) + r√k = r√k(k+1)/(k-1)
- As triangles ABC and ANL share the angle at A and have right angles at N and B, they are similar.
- So AC/BC = AL/NL = (r√k(k+1)/(k-1)) / (4 r k / (k+1)) = (k+1)2 /4 √k (k-1).