PUMaC 2019 · 个人决赛（A 组） · 第 3 题
PUMaC 2019 — Individual Finals (Division A) — Problem 3
题目详情
- Let ABCDEF be a convex hexagon with area S such that AB ‖ DE, BC ‖ EF, CD ‖ F A holds, and whose all angles are obtuse and opposite sides are not the same length. Prove that the following inequality holds: A + A + A + A + A + A < S , where ABC BCD CDE DEF EF A F AB A is the area of triangle XY Z . XY Z 1
解析
- Let ABCDEF be a convex hexagon with area S such that AB ‖ DE, BC ‖ EF, CD ‖ F A holds, and whose all angles are obtuse and opposite sides are not the same length. Prove that the following inequality holds: A + A + A + A + A + A < S , where ABC BCD CDE DEF EF A F AB A is the area of triangle XY Z . XY Z Solution : Notice that out of two opposite sides one is always longer than the other one. We can label each side ”red” or ”blue” based on whether it’s longer or shorter than the opposite side, respectively. We claim that there are no two adjacent red sides. For the sake of contradiction, assume the contrary that AB and BC are both red. Exactly one of the sides AF and CD is red (the two are opposite sides so one is blue and one is red). Without loss of generality ′ ′ assume that AF is red. Denote F and C the orthogonal projections of F and C to AB . ′′ ′′ Similarly, denote F and C the orthogonal projections of F and C to DE . Notice that ′ ′ ′′ ′′ ′ ′′ F C = F C (these four points form a rectangle). ∆ AF F ∼ DCC since the corresponding ′ ′ sides are parallel, hence AF > DF ” and AF > DC . Similarly, BC > EF ”, so we have ′ ′ ′ ′ ′ ′ F C = F A + AB + BC > F ” D + DE + EF ” = F ” C ” = F C , which is absurd. Hence the red and blue sides alternate. Without loss of generality let AB be a blue side. Let k, l, m be lines through A, C, E parallel to BC, DE, F A respectively. Let k ∩ l = X, l ∩ m = Y, , m ∩ k = Z . Since AZEF and ABCX are parallelograms we have that AZ = EF < BC = AX , so the point Z is in between A and X . Similarly, X is in between C, Y and Y is in between E, Z . Then the following holds S = 1 A + A + A + A > A + A + A = 2( A + A + A ). ABCX CDEY EF AZ XY Z ABCX CDEY EF AZ ABC CDE EF A Similarly we can prove that S > 2( A + A + A ). Adding up the last two inequalities, BCD DEF F AB we get that 2 S > 2( A + A + A + A + A + A ), which means that ABC BCD CDE DEF EF A F AB S > A + A + A + A + A + A . ABC BCD CDE DEF EF A F AB 2