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

Equal incircles in two adjacent triangles

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 tan(∠ADC/2)tan(∠BDC/2) = 1.

Lemma 2'

Assume points Mi, I = 1, 2, ..., N (N > 3) lie on side AB of ΔABC, A = M1, B = MN. Assume further that the incircles of triangles M1CM2, M2CM3, ..., MN-1CMN all have equal radii, say ρ. If, as before, h is the altitude from C and r is the inradius of ΔABC, then

(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, j = 2, ..., N-1, all equal to 1, leaving only the product tan(α/2)tan(β/2) which is exactly (and still) 1 - 2r/h.

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

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