MathDB

1987 Iran MO (2nd round)

Part of Iran MO (2nd Round)

Subcontests

(3)
3
2

Prove that 5 S{A'B'C'D'} =S{ABCD} - [Iran Second Round 1987]

In the following diagram, let ABCDABCD be a square and let M,N,PM,N,P and QQ be the midpoints of its sides. Prove that SABCD=15SABCD.S_{A'B'C'D'} = \frac 15 S_{ABCD}. [asy] import graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttzz = rgb(0,0.2,0.6); pen qqzzff = rgb(0,0.6,1); draw((0,4)--(4,4),qqttzz+linewidth(1.6pt)); draw((4,4)--(4,0),qqttzz+linewidth(1.6pt)); draw((4,0)--(0,0),qqttzz+linewidth(1.6pt)); draw((0,0)--(0,4),qqttzz+linewidth(1.6pt)); draw((0,4)--(2,0),qqzzff+linewidth(1.2pt)); draw((2,4)--(4,0),qqzzff+linewidth(1.2pt)); draw((0,2)--(4,4),qqzzff+linewidth(1.2pt)); draw((0,0)--(4,2),qqzzff+linewidth(1.2pt)); dot((0,4),ds); label("AA", (0.07,4.12), NE*lsf); dot((0,0),ds); label("DD", (-0.27,-0.37), NE*lsf); dot((4,0),ds); label("CC", (4.14,-0.39), NE*lsf); dot((4,4),ds); label("BB", (4.08,4.12), NE*lsf); dot((2,4),ds); label("MM", (2.08,4.12), NE*lsf); dot((4,2),ds); label("NN", (4.2,1.98), NE*lsf); dot((2,0),ds); label("PP", (1.99,-0.49), NE*lsf); dot((0,2),ds); label("QQ", (-0.48,1.9), NE*lsf); dot((0.8,2.4),ds); label("AA'", (0.81,2.61), NE*lsf); dot((2.4,3.2),ds); label("BB'", (2.46,3.47), NE*lsf); dot((3.2,1.6),ds); label("CC'", (3.22,1.9), NE*lsf); dot((1.6,0.8),ds); label("DD'", (1.14,0.79), NE*lsf); clip((-4.44,-11.2)--(-4.44,6.41)--(16.48,6.41)--(16.48,-11.2)--cycle); [/asy]
[SXS_{X} denotes area of the X.X.]