Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.3 AMC 10/12 Find the answer

Quadrilateral $ABCD$ is inscribed in a circle with side $AD$ , a diameter of length $4$ . If sides $AB$ and $BC$ each have length $1$ ,
then side $CD$ has length

Pick one

Solution

Note that the length 4 forms a semicircle. We can then use the Law of Cosines. Take the center and form a line segment with the other two points.
Let's find the cosine of angle $AOB$ . The cosine of that angle is 7/8. We use the double cosine angle to find angle $AOC$ :
$cos(2x) = 2 cos^2 (x)- 1$
Therefore, $cos(2x) = 17/32$
Applying the Law of Cosines to the triangle $DOC$ , we get
$CD^2 = 2^2 + 2^2 - 2(4)(-17/32)$ (because $cos(x) = -cos(180-x)$ )
Therefore,
$CD^2 = 8 + 17/4$
and we find that is equal to $7/2$ , or option A!

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.