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By Ernst Dieterich and Lars Lindberg

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The equations of the lines Γe , Γf and Γr/2 are x = −2α, x = 2β and y = 1/r, respectively. Both of the radii of the circles Γd and Γg are R, and their centres are (−α + β, 1/r) and (−α + β, 1/r + 2R), respectively. Let D be the distance between (0, 0) and the centre of Γg . Then we have 2g = 1 2R 1 − = 2 , D−R D+R D − R2 which shows g = R/(D2 − R2 ). What we have to show is g ≤ 1 − the following computation: √ √ 3/2 r, that is 4 + 2 3 g ≤ r. This is verified by √ √ r− 4+2 3 g =r− 4+2 3 = = D2 r − R2 √ 1 R r 2 2 (D − R ) − 4 + 2 3 R = D 2 − R2 D 2 − R2 r 2 √ 1 1 + 2R + (α − β)2 − R2 − 4 + 2 3 R r r r 1 √ 3 R − D 2 − R2 3r ≥ 0.

A2m−1 ) satisfies (∗) ⇐⇒ (a0 + k, . . , a2m−1 + k) satisfies (∗) (1) (a0 , . . , a2m−1 ) can be permuted so that l=0 It is easy to verify that for all integers k, and that (a0 , . . , a2m−1 ) satisfies (∗) ⇐⇒ (ka0 , . . , ka2m−1 ) satisfies (∗) (2) for all integers k ≡ 0 (mod 11). For an integer k, denote by k the nonnegative integer less than 11 congruent to k modulo 11. For a fixed j ∈ {0, 1, . . , 9}, let k be the unique integer such that k ∈ {1, 2, . . , 10} and (j + 1)k ≡ 1 (mod 11).

This completes the proof. Solution 2. Suppose that the bisectors of ∠ABC and ∠ADC meet AC at L and M , respectively. Since AL/CL = AB/CB and AM/CM = AD/CD, the bisectors in question 32 meet on AC if and only if AB/CB = AD/CD, that is, AB · CD = CB · AD. We will prove that AB · CD = CB · AD is equivalent to P Q = QR. Because DP ⊥ BC, DQ ⊥ AC, DR ⊥ AB, the circles with diameters DC and DA contain the pairs of points P , Q and Q, R, respectively. It follows that ∠P DQ is equal to γ or 180◦ − γ, where γ = ∠ACB.

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7D dissident maps by Ernst Dieterich and Lars Lindberg


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