Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.8 AMC 10/12 Find the answer South Africa

$\triangle ABC$ is right-angled at $A$ , and $\triangle ABP$ is equilateral with $AB = 2$ . The length of $AC$ is

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Solution

$BAP = 60^\circ$ while $BAC = 90^\circ$ , so $PAC = 30^\circ$ . With $BPA = 60^\circ$ that means $C = 30^\circ$ , and so $\angle PAC$ is isosceles. Then $BC$ has length $4$ , and so by Pythagoras the length of $AC$ is $\sqrt{4^2 - 2^2}$

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