PUMaC 2015 · 团队赛 · 第 12 题
PUMaC 2015 — Team Round — Problem 12
题目详情
- [ 8 ] Alice is stacking balls on the ground in three layers using two sizes of balls: small and large. All small balls are the same size, as are all large balls. For the first layer, she uses 6 identical large balls A , B , C , D , E , and F all touching the ground and so that D, E, F touch each other, A touches E and F , B touches D and F , and C touches D and E . For the second layer, she uses 3 identical small balls, G , H , and I ; G touches A , E , and F , H touches B , D , and F , and I touches C , D , and E . Obviously, the small balls do not intersect the ground. Finally, for the top layer, she uses one large ball that touches D , E , F , G , H , and I . If the large balls have volume 2015, the sum of the volumes of all the balls in the pyramid can be √ written in the form a b + c for integers a, b, c where no integer square larger than 1 divides b . What is a + b + c ? (This diagram may not have the correct scaling, but just serves to clarify the layout of the problem.) Figure 1: The projection of the balls onto the ground.
解析
- [ 8 ] Alice is stacking balls on the ground in three layers using two sizes of balls: small and large. All small balls are the same size, as are all large balls. For the first layer, she uses 6 identical large balls A , B , C , D , E , and F all touching the ground and so that D, E, F touch each other, A touches E and F , B touches D and F , and C touches D and E . For the second layer, she uses 3 identical small balls, G , H , and I ; G touches A , E , and F , H touches B , D , and F , and I touches C , D , and E . Obviously, the small balls do not intersect the ground. Finally, for the top layer, she uses one large ball that touches D , E , F , G , H , and I . If the large balls have volume 2015, the sum of the volumes of all the balls in the pyramid can be √ written in the form a b + c for integers a, b, c where no integer square larger than 1 divides b . What is a + b + c ? Solution: Let R be the radius of the large balls and r of the small balls. The difference in height between the centers of the balls in the second layer and the first is simply the height of the triangular pyramid with base lengths 2 R and lateral lengths R + r . Since the small balls in the second layer are still 2 R away from one another, the difference in height between the center of the ball in the third layer and the second is equal to the height of the same pyramid. Then, the difference between the heights of the first and third layer is: 5 √ ( ) √ √ √ 2 √ 2 3 1 √ 2 2 2 h = 2 · ( R + r ) − · · 2 R = 2 · r + 2 Rr − R 3 2 3 Since the large ball in the third layer also touches the large balls in the first layer, the difference between the heights of the first and third layer must equal the height of the triangular pyramid with base lengths 2 R and lateral lengths 2 R , so: √ ( ) √ √ √ 2 √ 2 3 24 √ 2 2 h = (2 R ) − · · 2 R = R 3 2 9 Setting the two equations equal and solving for r in terms of R , we have: ( )( ) √ √ ( ) ( ) 36 2 2 4 r − 8 Rr − R = 4 · r − 2 − 1 R r − − 2 − 1 R = 0 9 √ ( ) √ 3 r Since r > 0, we have r = ( 2 − 1) R . Then, V = · V = 2015 · (5 2 − 7), so the total r R R ( √ ) √ √ volume is V = 2015 · 7 · 1 + 5 · (5 2 − 7) = 2015 · (15 2 − 14) = 30225 2 − 28210, so a + b + c = 2017 . Author: Bill Huang