HMMT 二月 2020 · 冲刺赛 · 第 30 题
HMMT February 2020 — Guts Round — Problem 30
题目详情
- [18] Let S = { ( x, y ) | x > 0, y > 0, x + y < 200, and x, y ∈ Z } . Find the number of parabolas P with vertex V that satisfy the following conditions: • P goes through both (100 , 100) and at least one point in S , • V has integer coordinates, and • P is tangent to the line x + y = 0 at V .
解析
- [18] Let S = { ( x, y ) | x > 0, y > 0, x + y < 200, and x, y ∈ Z } . Find the number of parabolas P with vertex V that satisfy the following conditions: • P goes through both (100 , 100) and at least one point in S , • V has integer coordinates, and • P is tangent to the line x + y = 0 at V . Proposed by: James Lin Answer: 264 Solution: We perform the linear transformation ( x, y ) → ( x − y, x + y ), which has the reverse trans- ( ) a + b b − a formation ( a, b ) → , . Then the equivalent problem has a parabola has a vertical axis of 2 2 symmetry, goes through A = (0 , 200), a point B = ( u, v ) in ′ S = { ( x, y ) | x + y > 0, x > y , y < 200, x, y ∈ Z , and x ≡ y mod 2 } , ( ) 2 u v and a new vertex W = ( w, 0) on y = 0 with w even. Then 1 − = . The only way the RHS w 200 ′ u v ′ 2 can be the square of a rational number is if = where v = 2(10 − v ) . Since v is even, we can find w 10 conditions so that u, w are both even: ′ ′ v ∈ { 1 , 3 , 7 , 9 } = ⇒ (2 v ) | u, 20 | w ′ ′ v ∈ { 2 , 4 , 6 , 8 } = ⇒ v | u, 10 | w ′ v = 5 = ⇒ 2 | u, 4 | w ′ ′ It follows that any parabola that goes through v ∈ { 3 , 7 , 9 } has a point with v = 1, and any parabola ′ ′ that goes through v ∈ { 4 , 6 , 8 } has a point with v = 2. We then count the following parabolas: • The number of parabolas going through (2 k, 162), where k is a nonzero integer with | 2 k | < 162. • The number of parabolas going through (2 k, 128) not already counted, where k is a nonzero integer with | 2 k | < 128. (Note that this passes through ( k, 162).) • The number of parabolas going through (2 k, 50) not already counted, where k is a nonzero integer ( ) 2 k with | 2 k | < 50. (Note that this passes through , 162 , and any overlap must have been counted 5 in the first case.) The number of solutions is then ( ) 1 4 2 80 + · 64 + · 25 = 264 . 2 5