Limits of iterated geometric constructions

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In triangle ABC1A B C_{1}, the bisector C1C2C_{1} C_{2} is drawn; in triangle AC1C2A C_{1} C_{2}, the bisector C2C3C_{2} C_{3} is drawn; in triangle AC2C3A C_{2} C_{3}, the bisector C3C4C_{3} C_{4}, and so on. Prove that the sequence of measures of the angles Cn+1CnA,n∈NC_{n+1} C_{n} A, n \in \mathbb {N}, has a limit, and find it, if the angle BACB A C equals α\alpha.

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The circle inscribed in triangle A1B1C1A_{1} B_{1} C_{1} touches its sides B1C1,C1A1B_{1} C_{1}, C_{1} A_{1}, and A1B1A_{1} B_{1} at the points A2,B2,C2A_{2}, B_{2}, C_{2}, respectively; the circle inscribed in triangle A2B2C2A_{2} B_{2} C_{2} touches its sides B2C2,C2A2,A2B2B_{2} C_{2}, C_{2} A_{2}, A_{2} B_{2} at the points A3,B3,C3A_{3}, B_{3}, C_{3}, respectively, and so on. Find the limit of the sequence of measures of the angles BnAnCnB_{n} A_{n} C_{n} as n→∞n \rightarrow \infty.

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In trapezoid ABCDA B C D (AD∥BC)(A D \parallel B C), the points B1B_{1} and C1C_{1} are the midpoints of the diagonals ACA C and BDB D; in trapezoid AB1C1DA B_{1} C_{1} D, the points B2B_{2} and C2C_{2} are the midpoints of the diagonals AC1A C_{1} and B1DB_{1} D, and so on. Find lim⁡n→∞∣BnCn∣\lim_{n \rightarrow \infty }\left|B_{n} C_{n}\right|, if ∣AD∣=a\left|A D\right| = a, in the cases: а) ∣AD∣>∣BC∣\left|A D\right| > \left|B C\right|; б) ∣AD∣<∣BC∣\left|A D\right| < \left|B C\right|.

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In triangle ABCA B C, medians AD,BEA D, B E, and CFC F are drawn. On the side ACA C, a point M0M_{0} is taken, M0≠A,M0≠CM_{0} \neq A, M_{0} \neq C, and from it a point M1M_{1} on ACA C is found as follows: segments [M0N0]∥[CF](N0∈AB)\left[M_{0} N_{0}\right] \parallel [C F] \left(N_{0} \in A B\right) are drawn, then [N0P0]∥[AD](P0∈BC)\left[N_{0} P_{0}\right] \parallel [A D] \quad \left(P_{0} \in B C\right), and finally [P0M1]∥[BE]\left[P_{0} M_{1}\right] \parallel [B E]. Starting from the point M1M_{1}, the point M2M_{2} is found analogously, and so on.

Prove that the sequence of points {Mn}\left\{ M_{n}\right\} converges, and find its limit.

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Let P0P1P2P_0 P_1 P_2 be a triangle right-angled at P2P_2, P2P3P_2 P_3 the perpendicular from P2P_2 to P0P1P_0 P_1, similarly P3P4⊥P1P2P_3 P_4 \perp P_1 P_2, P4P5⊥P2P3P_4 P_5 \perp P_2 P_3, etc. Find the point lim⁡n→∞Pn\lim_{n \to \infty } P_n.

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Remark.
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E. Steinitz, problem: Arch. Math. Phys. Ser. 3, 19, p. 361 (1912). Solution by W. Gaedecke: Arch. Math. Phys. Ser. 3, 21, p. 290 (1913).

References

[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Предел. Непрерывность. Дифференцируемость. 2-е изд., ФИЗМАТЛИТ, 2003.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.