HMMT 十一月 2012 · GEN 赛 · 第 6 题

HMMT November 2012 — GEN Round — Problem 6

[ 5 ] ABCD is a parallelogram satisfying AB = 7, BC = 2, and ∠ DAB = 120 . Parallelogram ECF A is contained in ABCD and is similar to it. Find the ratio of the area of ECF A to the area of ABCD .

解析

[ 5 ] ABCD is a parallelogram satisfying AB = 7, BC = 2, and ∠ DAB = 120 . Parallelogram ECF A is contained in ABCD and is similar to it. Find the ratio of the area of ECF A to the area of ABCD . 39 Answer: First, note that BD is the long diagonal of ABCD , and AC is the long diagonal of 67 ECF A . Because the ratio of the areas of similar figures is equal to the square of the ratio of their side 2 AC lengths, we know that the ratio of the area of ECF A to the area of ABCD is equal to the ratio . 2 BD 2 2 2 ◦ 2 Using law of cosines on triangle ABD , we have BD = AD + AB − 2( AD )( AB ) cos(120 ) = 2 + 1 2 7 − 2(2)(7)( − ) = 67. 2 2 2 2 ◦ 2 2 Using law of cosines on triangle ABC , we have AC = AB + BC − 2( AB )( BC ) cos(60 ) = 7 + 2 − 1 2(7)(2)( ) = 39. 2 2 AC 39 Finally, = . 2 BD 67 General Test