PUMaC 2010 · 组合（A 组） · 第 6 题
PUMaC 2010 — Combinatorics (Division A) — Problem 6
题目详情
- All the diagonals of a regular decagon are drawn. A regular decagon satisfies the property that if three diagonals concur, then one of the three diagonals is a diameter of the circumcircle of the decagon. How many distinct intersection points of diagonals are in the interior of the decagon? 1
解析
- All the diagonals of a regular decagon are drawn. A regular decagon satisfies the property that if three diagonals concur, then one of the three diagonals is a diameter of the circumcircle of the decagon. How many distinct intersection points of diagonals are in the interior of the decagon? Answer: 161 4 Solution: First we classify the intersection points. There are points on exactly 2 lines, points on exactly 3 lines, and the center of the decagon, which is on all 5 diameters. To prove there are no other points, suppose a point other than the center is on at least 4 lines. The point is on a diameter by the given property. Let the vertices of the decagon be P , · · · , P clockwise 1 10 around the decagon, such that the point is on P P . Now consider all the intersection points 1 6 on P P . Given a diagonal P P that intersects P P , let f ( P P ) be the intersection point. 1 6 i j 1 6 i j Then the intersection points on P P between P and the center of the decagon are 1 6 1 f ( P P ) , f ( P P ) = f ( P P ) , f ( P P ) = f ( P P ) , f ( P P ) 2 10 2 9 3 10 2 8 4 10 3 9 Of these points, f ( P P ) is the closest to P , and f ( P P ) = f ( P P ) is the next closest to 2 10 1 2 9 3 10 P . The other two points are clearly further from P . Therefore, our point can only exist if 1 1 these other two points are the same point, so that f ( P P ) = f ( P P ) = f ( P P ). But by 2 8 4 10 3 9 symmetry, P P and P P intersect on the diameter of the circumcircle of the decagon that 2 8 3 9 is perpendicular to P P and P P . Therefore, if our point exists, then it is on two different 1 10 5 6 diameters of the circumcircle of the decagon, so our point is the center of the decagon, a con- tradiction. Therefore our original classification of points was correct. Now we take a PIE-like approach to count the intersection points. Each set of four points ( ) 10 corresponds to exactly one intersection point, contributing = 210 intersection points. 4 However, we have overcounted, since some of these intersection points are actually the same point. Consider intersection points of exactly three diagonals. By the given property, one diagonal is a diameter, and the other two diagonals are reflections of each other over the diameter, since otherwise we could reflect the diagonals over the diameter to get more diagonals that pass through the same point. We can produce such an intersection point by choosing a diameter and two of the four points on one side of the diameter, taking the corresponding two points on ( ) 4 the other side of the diameter, and drawing the resulting diagonals. There are 5 = 30 such 2 ( ) 3 diagonals. Each of the 3-line diagonals was counted = 3 times, so we need to subtract 60. 2 ( ) 5 At this point, we’ve only miscounted the center. We counted it = 10 times in the first step, 2 and we counted it 5 · 2 = 10 times in the second step, which we actually subtracted twice, so we’ve counted it − 10 times total. Therefore, we need to add 11. Therefore, the answer is 210 − 60 + 11 = 161 .