PUMaC 2022 · 团队赛 · 第 2 题
PUMaC 2022 — Team Round — Problem 2
题目详情
- A triangle △ A A A in the plane has sidelengths A A = 7 , A A = 8 , A A = 9. For 0 1 2 0 1 1 2 2 0 i ≥ 0, given △ A A A , let A be the midpoint of A A and let G be the centroid of i i +1 i +2 i +3 i i +1 i ∞ △ A A A . Let point G be the limit of the sequence of points { G } . If the distance i i +1 i +2 i i =0 √ a b between G and G can be written as , where a, b, c are positive integers such that a and c 0 c 2 2 2 are relatively prime and b is not divisible by the square of any prime, find a + b + c . 28 28 27
解析
- A triangle △ A A A in the plane has sidelengths A A = 7 , A A = 8 , A A = 9. For 0 1 2 0 1 1 2 2 0 i ≥ 0, given △ A A A , let A be the midpoint of A A and let G be the centroid of i i +1 i +2 i +3 i i +1 i ∞ △ A A A . Let point G be the limit of the sequence of points { G } . If the distance i i +1 i +2 i i =0 √ a b between G and G can be written as , where a, b, c are positive integers such that a and c 0 c 2 2 2 are relatively prime and b is not divisible by the square of any prime, find a + b + c . Proposed by Frank Lu Answer: 422 To do this, we work with vectors. Let r ⃗ be the vector between G and G . Then, notice i i i +1 1 1 that, by definition, we have that G = ( A + A + A ) , meaning that r ⃗ = ( A − A ) = i i i +1 i +2 i i +3 i 3 3 1 1 1 1 ( A − A ) . However, notice that we have that r ⃗ = ( A − A ) = ( ( A + A ) − A ) i +1 i i i +1 i i − 1 i − 2 i 6 6 6 2 1 1 1 = − ( A − A ) − ( A − A ) = r ⃗ − r ⃗ . From here, we explicitly consider one i i − 1 i − 1 i − 2 i − 1 i − 2 6 12 2 coordinate: notice then that we have the characteristic equation for, say, the x − coordinate, 1 2 i i r + r + = 0 , with the resulting solution for x = Ar + Br . But from here, notice that i 1 2 2 − 1+ i − 1 − i the solutions for r here are and . Hence, we see that the solutions for both x, y are 2 2 − 1+ i − 1 − i k k ⃗ of this form. In particular, we see that r ⃗ = a ⃗ ( ) + b ( ) . Therefore, we see that the k 2 2 ∞ P − 1+ i − 1 − i k k ⃗ vector between G and G is equal to a ⃗ ( ) + b ( ) . But using geometric series, we 0 2 2 k =0 1 1 2 2 3+ i 3 − i ⃗ ⃗ ⃗ see that this is just equal to a ⃗ + b = a ⃗ + b = a ⃗ + b . We just need − 1+ i − 1 − i 3 − i 3+ i 5 5 1 − 1 − 2 2 ⃗ to find what a ⃗ and b are. Returning to our original triangle, position our triangle such that A = (0 , 0) , A = (0 , 9) , and A has positive y − coordinate. Then, notice that we see that, if 0 2 1 2 2 2 2 A = ( x, y ) , we have that x + y = 49 , (9 − x ) + y = 64 means that − 18 x + 81 = 15 , or that 1 √ 11 8 5 − 1+ i − 1 − i ⃗ ⃗ x = , and y = . But notice then that we have that a ⃗ + b = r ⃗ and a ⃗ + b = r ⃗ . 0 1 3 3 2 2 3+ i 3 − i ⃗ Notice therefore that a ⃗ + b = 2 / 5 r ⃗ + 4 / r 5 ⃗ Simplifying this we see that this is equal 1 0 5 5 1 √ 2 1 1 1 38 8 5 to ( A − A + ( A − A ) = ( A + A − 2 A ) . But this is then equal to ( , ) . Our 1 0 2 1 2 1 0 15 15 15 15 3 3 √ √ √ √ 1 1 1 42 14 14 1 2 final answer is therefore 38 + 320 = 1444 + 320 = 1764 = = = , or 45 45 45 45 15 15 that we have 196 + 1 + 225 = 422 . 28 28 27