Share your proof or solution in the comments below.
Target Audience: K-12, Honors Geometry, and College Mathematics Education.
In Problem 1627, we explore three mutually tangent circles with collinear centers and a common external tangent line. The challenge is to prove using synthetic similarity that the radii satisfy a geometric progression, and to establish the elegant radical and harmonic difference relations involving the radii of their curvilinear inscribed circles.
Explore the full theorem and illustrated diagram by clicking the image below.
Target Audience: K-12, Honors Geometry, and College Mathematics Education.
In Problem 1627, we explore three mutually tangent circles with collinear centers and a common external tangent line. The challenge is to prove using synthetic similarity that the radii satisfy a geometric progression, and to establish the elegant radical and harmonic difference relations involving the radii of their curvilinear inscribed circles.
Explore the full theorem and illustrated diagram by clicking the image below.
Click for additional details and full diagram.
How to contribute:
Post your step-by-step proof in the comments below. Feel free to:
Post your step-by-step proof in the comments below. Feel free to:
- Describe the theorems applied.
- Share a link to your digital proofs (GeoGebra, Desmos).
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Please use the box below to Enter your Comment or Solution. You can use plain text or provide links to your digital proofs.
Please use the box below to Enter your Comment or Solution. You can use plain text or provide links to your digital proofs.
Let P₁, P₂, P₃, Q₁, and Q₂ be the points of tangency of circles ω₁, ω₂, ω₃, and the two inscribed circles with the common circumscribed line, respectively. Furthermore, let R₁ be the foot of the perpendicular from O₁ to O₂P₂, and let R₂ be the foot of the perpendicular from O₂ to O₃P₃.
ReplyDelete(1) Since O₁O₂ : O₂R₂ = O₂O₃ : O₃R₂,
a + b : b − a = b + c : c − b
Hence, we obtain b² = ac.
(2)(3) Since (O₁R₁)² + (O₂R₂)² = (a + b)² and from the result of (1), O₁R₁ = 2√ab
Let S₁ be the center of the incircle with radius r₁, T₁ the foot of the perpendicular from S₁ to O₁P₁, and T₂ the foot of the perpendicular from S₂ to O₂P₂. Applying the Pythagorean theorem to the right triangles O₁S₁T₁ and O₂S₁T₂ respectively,
we obtain P₁Q₁ = 2√(ar₁) and Q₁P₂ = 2√(br₁).
Therefore, since 2√(ab) = 2√(ar₁) + 2√(br₁),
√r₁ = √(ab)/(√a + √b)
Similarly, for the incircle with radius r₂,
√r₂ = √(bc)/(√b + √c)
Equations (2) and (3) follow from these two equations and the result in (1).
What a beautiful challenge — the diagram makes the tangent circles and their curvilinear incircles so clear. Thanks for sharing such an elegant problem and a clear, step-by-step solution.
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