Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.1 AMC 10/12 Find the answer Japan

Suppose for a quadrilateral $ABCD$ , $\angle DAB = 90^\circ$ , $\angle ABC = \angle BCD = 60^\circ$ . If $AB = 5$ and $CD = 4$ , what is the value of $BC$ ? Here for a line segment $XY$ , its length is also denoted by $XY$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let $E$ be the point of intersection of the lines $AB$ and $CD$ . Then, since $\angle EBC = \angle ECB = 60^\circ$ , the triangle $EBC$ is an equilateral triangle. If we let $x = EA$ , then the triangle $ADE$ is a right triangle with $\angle EAD = 90^\circ$ and since $\angle AED = 60^\circ$ , we have $DE = 2x$ . From $EB = EC$ we obtain $5 + x = 4 + 2x$ . Solving this equation we get $x = 1$ , and therefore, we conclude that $BC = EB = 1 + 5 = 6$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.