Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.5 AMC 10/12 Find the answer South Africa

Points $A_1$ , $A_2$ , $A_3$ ... are constructed as follows: the length $OA_1$ is $4$ , $\angle OA_1A_2 = 90^\circ$ and the length $A_1A_2 = 1$ ; then a right angle is constructed at $A_2$ to find $A_3$ , and so on as shown in the diagram.
The length of $OA_{21}$ is

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

6 By Pythagoras, $OA_2 = \sqrt{17}$ . Then $OA_3 = \sqrt{18}$ , $OA_4 = \sqrt{19}$ and so on, with $OA_n = \sqrt{n+15}$ , and thus $OA_{21} = \sqrt{36} = 6$ .

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