HMMT 十一月 2018 · GEN 赛 · 第 6 题
HMMT November 2018 — GEN Round — Problem 6
题目详情
- Call a polygon normal if it can be inscribed in a unit circle. How many non-congruent normal polygons are there such that the square of each side length is a positive integer? √
解析
- Call a polygon normal if it can be inscribed in a unit circle. How many non-congruent normal polygons are there such that the square of each side length is a positive integer? Proposed by: Yuan Yao Answer: 14 √ √ ◦ ◦ ◦ ◦ The side lengths of the polygon can only be from the set { 1 , 2 , 3 , 2 } , which take up 60 , 90 , 120 , 180 √ of the circle respectively. By working modulo 60 degrees we see that 2 must be used an even number of times. We now proceed to casework on the longest side of the polygon. Case 1: If the longest side has length 2, then the remaining sides must contribute the remaining 180 √ √ √ degrees. There are 3 possibilities: (1 , 1 , 1 , 2) , (1 , 3 , 2) , ( 2 , 2 , 2). √ ◦ ◦ Case 2: If the longest side has length 3, then it takes up either 120 or 240 of the circle. In √ √ √ √ √ √ √ √ √ the former case we have 6 possibilities: (1 , 1 , 1 , 1 , 3) , (1 , 2 , 2 , 3) , ( 2 , 1 , 2 , 3) , (1 , 1 , 3 , 3) , √ √ √ √ √ √ (1 , 3 , 1 , 3) , ( 3 , 3 , 3). In the latter case there is only 1 possibility: (1 , 1 , 3). √ Case 3: If the longest side has length 2, then it shows up either twice or four times. In the former case √ √ √ √ we have 2 possibilities: (1 , 1 , 1 , 2 , 2) , (1 , 1 , 2 , 1 , 2). In the latter case there is only 1 possibility: √ √ √ √ ( 2 , 2 , 2 , 2). Case 4: If all sides have length 1, then there is 1 possibility: (1 , 1 , 1 , 1 , 1 , 1). Adding up all cases, we have 3 + 6 + 1 + 2 + 1 + 1 = 14 polygons. √