HMMT 十一月 2021 · 冲刺赛 · 第 27 题
HMMT November 2021 — Guts Round — Problem 27
题目详情
- [13] O is the center of square ABCD , and M and N are the midpoints of BC and AD , respectively. ′ ′ ′ ′ ′ ′ ′ ′ Points A , B , C , D are chosen on AO, BO, CO, DO , respectively, so that A B M C D N is an equiangular √ ′ ′ ′ ′ [ A B M C D N ] a + b c hexagon. The ratio can be written as , where a, b, c, d are integers, d is positive, c is [ ABCD ] d square-free, and gcd( a, b, d ) = 1. Find 1000 a + 100 b + 10 c + d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2021, November 13, 2021 — GUTS ROUND Organization Team Team ID#
解析
- [13] O is the center of square ABCD , and M and N are the midpoints of BC and AD , respectively. ′ ′ ′ ′ ′ ′ ′ ′ Points A , B , C , D are chosen on AO, BO, CO, DO , respectively, so that A B M C D N is an equian- √ ′ ′ ′ ′ [ A B M C D N ] a + b c gular hexagon. The ratio can be written as , where a, b, c, d are integers, d is [ ABCD ] d positive, c is square-free, and gcd( a, b, d ) = 1. Find 1000 a + 100 b + 10 c + d . Proposed by: Joseph Heerens Answer: 8634 Solution: Assume without loss of generality that the side length of ABCD is 1 so that the area of the 1 ′ ′ ′ ′ square is also 1 . This also means that OM = ON = . As A B M C D N is equiangular, it can be seen 2 ′ ◦ ′ ′ ′ ′ ◦ ′ ◦ that ∠ A N O = 60 , and also by symmetry, that A B ‖ AB, so ∠ OA B = 45 and ∠ OA N = 75 . √ √ ′ Therefore, A N O is a 45 − 60 − 75 triangle, which has sides in ratio 2 : 1 + 3 : 6 , so we may compute √ √ √ 6 1 3 2 − 6 ′ ′ √ that A O = · = . Further, the area of A N O can be found by taking the altitude to 2 4 1+ 3 √ √ √ √ 1 3 3 − 3 1 1 3 − 3 3 − 3 ′ ′ √ N O, which has length of · = , so the area is · · = . The area of OA B is 2 4 2 2 4 16 1+ 3 ( ) √ √ √ 2 1 3 2 − 6 6 − 3 3 = . 2 4 8 √ ′ ′ ′ ′ ′ ′ ′ 3 − 3 Combining everything together, we can find that [ A B M C D N ] = 4[ A N O ] + 2[ OA B ] = + 4 √ √ 6 − 3 3 9 − 4 3 = . Therefore, our answer is 9000 − 400 + 30 + 4 = 8634 . 4 4