Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it Ireland

Let $AOB$ be a triangle with $\angle AOB = 90^\circ$ . Define a sequence of points $X_1, X_2, X_3, \dots, X_{2022}$ as follows:
* $X_1$ lies on the ray $\vec{OA}$ with $|OX_1| = 1$ .
* $X_2$ lies on the ray $\vec{OB}$ with $|X_1X_2| = 2$
* For $3 \le n \le 2022$ , the $X_n$ lie alternately on $\vec{OA}$ ( $n$ odd) or $\vec{OB}$ ( $n$ even), with $|X_{n-1}X_n| = n$ .
Find the distance $|X_{2020}X_{2022}|$ .

Solution

Let $X_0 = O$ and, for $0 \le n \le 2022$ , let $d_n = |OX_n|$ . Then $d_0 = 0$ and by Pythagoras' theorem, for $n \ge 1$

$d_{2n-1}^2 + d_{2n}^2 = (2n)^2$
$d_{2n-1}^2 + d_{2n-2}^2 = (2n-1)^2.$
Subtracting these equations we get $d_{2n}^2 - d_{2n-2}^2 = (2n)^2 - (2n-1)^2 = 4n-1$ .
Since $d_0^2 = 0$ we then have
$d_{2n}^2 = \sum_{k=1}^{n} d_{2k}^2 - d_{2k-2}^2 = \sum_{k=1}^{n} 4k - 1 = 4 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 \\ = 2n(n+1) - n = n(2n+1).$
This implies that:
$d_{2020}^2 = 1010 \cdot 2021 = 2041210$
$d_{2022}^2 = 1011 \cdot 2023 = 2045253$
and so $|X_{2020}X_{2022}| = \sqrt{2045253} - \sqrt{2041210}$ .

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