Maths Olympiad Prep

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Geometry Difficulty 6.8 National Olympiad Prove it Romania

Consider a circle centered at OO with radius rr and a line \ell not passing through OO. A grasshopper is jumping to and fro between the points of the circle and the line, the length of each jump being rr. Prove that there are at most 8 points for the grasshopper to reach.

Figure 1

Solution

We assume that, when having the choice between only two places to jump to, the grasshopper never jumps back to the point from which he got to that place. Let us denote by P1P_1 the starting point of the grasshopper, with P2P_2 the point on the line on which he has jumped from P1P_1, and so on. As the length of the jumps are all equal to rr, OP1P2P3OP_1P_2P_3 is a rhombus (possibly a degenerate one). Similarly, OP3P4P5OP_3P_4P_5 is also a rhombus. It follows that the triangles P1OP5P_1OP_5 and P2P3P4P_2P_3P_4 are congruent (SAS), and from here we obtain that P1P5P_1P_5 is parallel to \ell. We deduce that P5P_5 is the reflection of P1P_1 across the perpendicular line from OO onto \ell. (This fact remains true even in the degenerate cases.) From P5P_5, the grasshopper can get to P9P_9 which, as above, is the reflection of P5P_5 across the perpendicular line from OO onto \ell, i.e. P1P_1. In conclusion, the grasshopper can reach only the points PkP_k, k=1,8k = \overline{1,8} (which are not necessarily distinct).

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.