Equal Incircles Theorem, Angela Drei's Proof
The Equal Incircle Theorem is a generalization of a sangaku problem:
Let C be a point. Assume points Mi, I = 1, 2, ..., N (N > 3) lie on a line not through C. Assume further that the incircles of triangles M1CM2, M2CM3, ..., MN-1CMN all have equal radii. Then the same is true of triangles M1CM3, M2CM4, ..., MN-2CMN, and also of triangles M1CM4, M2CM5, ..., MN-3CMN, and so on.
The proof below is by Angela Drei, an Italian mathematics teacher. The proof first appeared in an Italian magazine Archimede. It depends on two lemmas.
Lemma 1
In any ΔABC,
1 - 2r/h = tan(α/2)tan(β/2),
where r is the inradius, h the altitude to AB, α = ∠BAC, β = ∠ABC.
Proof
Let a, b, c be the lengths sides BC, AC, AB and p = (a + b + c)/2 the semiperimeter of ΔABC. As we know,
tan²(α/2) = (p - b)(p - c) / p(p - a) and tan²(β/2) = (p - a)(p - c) / p(p - b)
Taking the product gives
tan²(α/2)tan²(β/2) = (p - c)² / p².
Next, letting S be the area of ΔABC,
(p - c) / p | = 1 - c/p | |
= 1 - ch·r / pr·h | ||
= 1 - 2Sr / Sh | ||
= 1 - 2r / h, |
which proves the lemma.
Lemma 2
Given a triangle ABC, D lie on AB, r1 and r2 are the radii of the circles inscribed in ACD and BCD, r is the inradius, h is the length of the altitude to the side AB. Then the following relation holds:
(1 - 2r1/h)(1 - 2r2/h) = 1 - 2r/h.
In particular, if r1 = r2, then (1 - 2r1/h)² = 1 - 2r/h.
Proof
Apply Lemma 1 to triangles ACD, BCD, and ABC and observe that ∠ADC/2 = 90° - ∠BDC/2 so that
Lemma 2'
Assume points Mi, I = 1, 2, ..., N (N > 3) lie on side AB of ΔABC,
(1 - 2ρ/h)N-1 = 1 - 2r/h.
Proof
Similar to the proof of Lemma 2, apply Lemma 1 to each of the triangles MiCMi+1 and multiply the identities. The products of the tangents due to pairs of supplementary angles at points Mj,
Proof of the Equal Incircle Theorem
This is a direct consequence of Lemma 2 applied to pairs of adjacent triangles Mi-1CMi and MiCMi+1.
Sangaku
- Sangaku: Reflections on the Phenomenon
- Critique of My View and a Response
- 1 + 27 = 12 + 16 Sangaku
- 3-4-5 Triangle by a Kid
- 7 = 2 + 5 Sangaku
- A 49th Degree Challenge
- A Geometric Mean Sangaku
- A Hard but Important Sangaku
- A Restored Sangaku Problem
- A Sangaku: Two Unrelated Circles
- A Sangaku by a Teen
- A Sangaku Follow-Up on an Archimedes' Lemma
- A Sangaku with an Egyptian Attachment
- A Sangaku with Many Circles and Some
- A Sushi Morsel
- An Old Japanese Theorem
- Archimedes Twins in the Edo Period
- Arithmetic Mean Sangaku
- Bottema Shatters Japan's Seclusion
- Chain of Circles on a Chord
- Circles and Semicircles in Rectangle
- Circles in a Circular Segment
- Circles Lined on the Legs of a Right Triangle
- Equal Incircles Theorem
- Equal Incircle Theorem, Angela Drei's Proof
- Equilateral Triangle, Straight Line and Tangent Circles
- Equilateral Triangles and Incircles in a Square
- Five Incircles in a Square
- Four Hinged Squares
- Four Incircles in Equilateral Triangle
- Gion Shrine Problem
- Harmonic Mean Sangaku
- Heron's Problem
- In the Wasan Spirit
- Incenters in Cyclic Quadrilateral
- Japanese Art and Mathematics
- Malfatti's Problem
- Maximal Properties of the Pythagorean Relation
- Neuberg Sangaku
- Out of Pentagon Sangaku
- Peacock Tail Sangaku
- Pentagon Proportions Sangaku
- Proportions in Square
- Pythagoras and Vecten Break Japan's Isolation
- Radius of a Circle by Paper Folding
- Review of Sacred Mathematics
- Sangaku à la V. Thebault
- Sangaku and The Egyptian Triangle
- Sangaku in a Square
- Sangaku Iterations, Is it Wasan?
- Sangaku with 8 Circles
- Sangaku with Angle between a Tangent and a Chord
- Sangaku with Quadratic Optimization
- Sangaku with Three Mixtilinear Circles
- Sangaku with Versines
- Sangakus with a Mixtilinear Circle
- Sequences of Touching Circles
- Square and Circle in a Gothic Cupola
- Steiner's Sangaku
- Tangent Circles and an Isosceles Triangle
- The Squinting Eyes Theorem
- Three Incircles In a Right Triangle
- Three Squares and Two Ellipses
- Three Tangent Circles Sangaku
- Triangles, Squares and Areas from Temple Geometry
- Two Arbelos, Two Chains
- Two Circles in an Angle
- Two Sangaku with Equal Incircles
- Another Sangaku in Square
- Sangaku via Peru
- FJG Capitan's Sangaku
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