KöMaL 2013 A586 Two convex quadrilaterals have side lengths a1a_1a1, a2a_2a2, a3a_3a3, a4a_4a4, and A1A_1A1, A2A_2A2, A3A_3A3, A4A_4A4; their areas are ttt and TTT, respectively. Show that 16tT≤∑i=14ai2(Ai+12+Ai+22+Ai+32−Ai2)+4a1A2a3A4+4A1a2A3a4.16tT \le \sum_{i=1}^4 a_i^2 \big(A_{i+1}^2+A_{i+2}^2+A_{i+3}^2-A_i^2\big) + 4a_1A_2a_3A_4 + 4A_1a_2A_3a_4.16tT≤∑i=14ai2(Ai+12+Ai+22+Ai+32−Ai2)+4a1A2a3A4+4A1a2A3a4. (The indices are used modulo 4.) Official text KöMaL. Quoted with source. Solutions No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label. Add a solution Similar problems These problems have the most words in common with this statement. RU-ALLRUSSIAN 1974 Problem 13CMO 1988 Problem 2SMO 1997 Problem 4CMO 1987 Problem 5CMO 1969 Problem 1 PreviousA585 NextA587 Locator hu-komal/2013/a586 · Tag K8W8PYMNVG · Edit this page on GitHub · Keys: j next, k previous