I.S.I. and C.M.I. Entrance USA Math Olympiad

AMC 10 (2013) Solutions

12. In (\triangle ABC, AB=AC=28) and BC=20. Points D,E, and F are on sides (\overline{AB}, \overline{BC}), and (\overline{AC}), respectively, such that (\overline{DE}) and (\overline{EF}) are parallel to (\overline{AC}) and (\overline{AB}), respectively. What is the perimeter of parallelogram ADEF?

(\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }72\qquad )

Solution: Perimeter = 2(AD + AF). But AD = EF (since ABCD is a parallelogram).
Hence perimeter = 2(AF + EF).
Now ABC is isosceles (AB = AC = 28). Thus angle B = angle C. But EF is parallel to AB. Thus angle FEC = angle B which in turn is equal to angle C.
Hence triangle CEF is isosceles. Thus EF = CF.
Perimeter = 2(AF + EF) = 2(AF + EF) =2AC = (2 \times 28) = 56.

Ans. (C) 56

By Ashani Dasgupta

Ph.D. in Mathematics from University of Wisconsin Milwaukee (USA)
Founder - Faculty at Cheenta

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