Let $ABC$ be an equilateral triangle and let $P_0$ be a point outside this triangle, such that $\triangle{AP_0C}$ is an isoscele triangle with a right angle at $P_0$. A grasshopper starts from $P_0$ and turns around the triangle as follows. From $P_0$ the grasshopper jumps to $P_1$, which is the symmetric point of $P_0$ with respect to $A$. From $P_1$, the grasshopper jumps to $P_2$, which is the symmetric point of $P_1$ with respect to $B$. Then the grasshopper jumps to $P_3$ which is the symmetric point of $P_2$ with respect to $C$, and so on. Compare the distance $P_0P_1$ and $P_0P_n$. $n \in N$.
Problem
Source: Pan African 2001
Tags: geometry, geometric transformation
yetti
06.10.2005 04:25
Let a = AB be the side of the equilateral triangle $\triangle ABC$. Since $AP_0 = AP_1,\ BP_1 = BP_2$, AB is a midline of the triangle $\triangle P_0P_1P_2$. Hence, the point $P_2$ is obtained from $P_0$ by a translation in the direction $\vec{AB}$ by the distance 2a. Similarly, $P_4$ is obtained from $P_2$ by a translation in the direction $\vec{CA}$ by the distance 2a and $P_6$ from $P_4$ by a translation in the direction $\vec{BC}$ by the distance 2a. It follows that the points $P_0, P_2, P_4$ form an equilateral triangle with the side lengths 2a and the sides $P_0P_2 \parallel AB,\ P_2P_4 \parallel CA, P_4P_0 \parallel BC$ and that the point $P_6 \equiv P_0$ are identical. For similar reasons, $P_3$ is obtained from $P_1$ by a translation in the direction $\vec{BC}$ by the distance 2a, $P_5$ is obtained from $P_3$ by a translation in the direction $\vec{AB}$ by the distance 2a, and $P_7$ is obtained from $P_5$ by a translation in the direction $\vec{CA}$ by the distance 2a. It follows that the points $P_1, P_3, P_5$ also form an equilateral triangle with the side lengths 2a and the sides $P_1P_3 \parallel BC,\ P_3P_5 \parallel AB, P_5P_1 \parallel CA$ and that the point $P_7 \equiv P_1$ are identical. Thus the equilateral triangles $\triangle P_0P_2P_4 \cong \triangle P_3P_5P_1$ are congruent and have parallel sides and the second is obtained from the first by a translation in the direction $\vec{P_0P_3}$ by the distance $P_0P_3$. Up to now, we did not need the fact that the triangle $\triangle ABC$ is equilateral or that the triangle $\triangle AP_0C$ is isosceles, only the fact that $\vec{AB} + \vec{BC} + \vec{CA} = 0$, which holds for any triangle. Now we use these facts. Since $P_6 \equiv P_0$ are identical and $P_0A = P_0C, AP_0 = AP_1, CP_0 = CP_5$, the triangle $\triangle P_1P_0P_5$ is also isosceles. This means that the point $P_0$ lies on the altitude of the equilateral triangle $\triangle P_3P_5P_1$ from the vertex $P_3$. Since in addition, the isosceles triangle $\triangle P_1P_0P_5$ is right, $P_0P_3 = P_1P_5 \dfrac{\sqrt 3}{2} - \dfrac{P_1P_5}{5} = a (\sqrt 3 - 1)$ Obviously, $P_0P_5 = P_0P_1 = 2AP_0 = 2CA \dfrac{\sqrt 2}{2} = a \sqrt 2$ and $P_0P_2 = P_0P_4 = 2a$. As a result, $P_0P_2 = P_0P_4 = P_0P_1 \sqrt 2$ $P_0P_3 = P_0P_1 \dfrac{\sqrt 3 - 1}{\sqrt 2} = P_0P_1 \dfrac{\sqrt 6 - \sqrt 2}{2}$ $P_0P_5 = P_0P_1$ $P_0P_6 = 0$ $P_0P_n = P_0P_{(n \mod 6)}$
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