Maths Olympiad Prep

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Problem 512

AMC 10/12, early questions
Geometry Difficulty 3.1 Find the answer Japan Junior Mathematical Olympiad · Japan

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

Figure 1

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

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Official solution

Let EE be the point of intersection of the lines ABAB and CDCD. Then, since EBC=ECB=60\angle EBC = \angle ECB = 60^\circ, the triangle EBCEBC is an equilateral triangle. If we let x=EAx = EA, then the triangle ADEADE is a right triangle with EAD=90\angle EAD = 90^\circ and since AED=60\angle AED = 60^\circ, we have DE=2xDE = 2x. From EB=ECEB = EC we obtain 5+x=4+2x5 + x = 4 + 2x. Solving this equation we get x=1x = 1, and therefore, we conclude that BC=EB=1+5=6BC = EB = 1 + 5 = 6.

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