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Some interesting considerations
• In fig. 1, we can see that triangles A'BC' and ABC are similar.
• In fig. 2, we can see that AC' and AE' are tangent to the circle with center O₃. The segment O₃A represents then the symmetry axis along which the right triangle AO₃D is reflected on AE'.
What we have to prove
If we prove that angle γ (gamma, see fig. 2) of the right triangle AO₃D is complementary to angle α (alpha) of the isosceles triangle AC'A', then we prove that points E and E' are identical, and consecutively the segment O₃A is perpendicular to the diameter AB.


a) It is given:
radius O₁B = R (see fig. 3)
radius O₂B = r
radius O₃D = p
segment O₃A = t
β = 90° - α

b) According to the drawing:
A’C’ = AC’
MA = MA' = R - r
h ⊥ AA'
p ⊥ AC’
O₁O₃ = R - p
O₁A = 2r - R
O₁M = O₂B = r
O₂O₃ = r + p

c) According to the 'Secant-Tangent' theorem (see further below):
t² = p(p + 2r)

d) According to the Pythagorean theorem:
t² + O₁A² = O₁O₃² →
t² + (2r - R)² = (R - p)²

e) Solving the equations in c) and d) together, we obtain:
t = 2r[2R(R - r)] / (R + r)
and
p = 2r(R - r) / (R + r)

f) According to the Pythagorean theorem:
1) h = (O₁C'² - O₁M²) = (O₁B² - O₁M²) = (R² - r²)
and
2) AC’² = h² + MA² = (R² - r²) + (R - r)² →
AC’ = [2R(R + r)]

g) By solving the following equations:
1) (p / t)² = p² / t² = [2r(R - r)]²/ 4r²[2R(R - r)] =
= [2r(R - r)]² / 4rR[2r(R - r)] = (R - r) / 2R
and
2) (MA / AC')² = MA² / AC'² = (R - r)² / 2R(R + r) = (R - r) / 2R
We proved that p / t = MA / AC', then the triangles AO₃D and AC'M are similar, hence angle DÂO₃ = angle β = 90° - α.
In conclusion, angle γ (or DÂO₃) being complementary to angle α, the segment O₃A is perpendicular to the diameter AB.

(You will find here a more geometric way to solve this puzzle)

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More Math Facts behind the puzzle

Secant-tangent theorem: fig. 1) c² = a(a + b) and fig. 2) c² = a(a + 2r)

The angle, formed between a tangent line and a chord, is congruent to the inscribed angle on the other side of the chord and subtended by the chord (angle α' = angle α, see fig. 3). Then, with similar triangles:
c / (a + b) = a / c