SMT 2013 Team Test Solutions

  • Published 2013

Abstract

2. In unit square ABCD, diagonals AC and BD intersect at E. Let M be the midpoint of CD, with AM intersecting BD at F and BM intersecting AC at G. Find the area of quadrilateral MFEG. Answer: 1 12 Solution: Let (ABC) denote the area of polygon ABC. Note that 4AFB ∼ 4MFD with AB/MD = 2, so we have DF = 13BD. This implies that (MFD) = 1 3(MBD) = 1 3( 1 2(CBD)) = 1 12 . By symmetry, (MGC) = 1 12 as well. Therefore, we have (MFEG) = (CED) − (MBD) − (MGC) = 1 4 − 1 12 − 1 12 = 1 12 .

Cite this paper

@inproceedings{2013SMT2T, title={SMT 2013 Team Test Solutions}, author={}, year={2013} }