MathDB

497

Part of 1989 All Soviet Union Mathematical Olympiad

Problems(1)

ASU 497 All Soviet Union MO 1989 area XQYP/area ABCD < mn/(m^2 + mn + n^2)

Source:

8/13/2019
ABCDABCD is a convex quadrilateral. XX lies on the segment ABAB with AXXB=mn\frac{AX}{XB} = \frac{m}{n}. YY lies on the segment CDCD with CYYD=mn\frac{CY}{YD} = \frac{m}{n}. AYAY and DXDX intersect at PP, and BYBY and CXCX intersect at QQ. Show that SXQYPSABCD<mnm2+mn+n2\frac{S_{XQYP}}{S_{ABCD}} < \frac{mn}{m^2 + mn + n^2}.
equal ratiosgeometryareasgeometric inequality