HMMT 二月 2001 · 团队赛 · 第 12 题
HMMT February 2001 — Team Round — Problem 12
题目详情
- All subscripts in this problem are to be considered modulo 6, that means for example that ω is the same as ω . Let ω , . . . ω be circles of radius r , whose centers lie on a regular 7 1 1 6 hexagon of side length 1. Let P be the intersection of ω and ω that lies further from the i i i +1 center of the hexagon, for i = 1 , . . . 6. Let Q , i = 1 . . . 6, lie on ω such that Q , P , Q i i i i i +1 are colinear. Find the number of possible values of r .
解析
- All subscripts in this problem are to be considered modulo 6, that means for example that ω is the same as ω . Let ω , . . . ω be circles of radius r , whose centers lie on a regular 7 1 1 6 hexagon of side length 1. Let P be the intersection of ω and ω that lies further from the i i i +1 center of the hexagon, for i = 1 , . . . 6. Let Q , i = 1 . . . 6, lie on ω such that Q , P , Q i i i i i +1 are colinear. Find the number of possible values of r . ′ Solution: Consider two consecutive circles ω and ω . Let Q , Q be two points on ω i i +1 i i i ′ ′ ′ and Q , Q on ω such that Q , P and Q are colinear and also Q , P and Q . i +1 i +1 i i i +1 i i +1 i i +1 ′ ′ ′ ′ Then Q Q = 2 ∠ Q P Q = 2 ∠ Q P Q = ∠ Q Q . Refer to the center of ω as O . i i i i +1 i i +1 i i i i i +1 i +1 The previous result shows that the lines O Q and O Q meet at the same angle as the i i i +1 i +1 ′ ′ lines O Q and O Q , call this angle ψ . ψ is a function solely of the circles ω and ω i i +1 i i i i +1 i i +1 ′ and the distance between them (we have just showed that any two points Q and Q on ω i i i give the same value of ψ , so ψ can’t depend on this.) Now, the geometry of ω and ω i i i i +1 is the same for every i , so ψ is simply a constant ψ which depends only on r . We know i 6 ψ = 0 mod 2 π because Q = Q . 7 1 We now compute ψ . It suffices to do the computaiton for some specific choice of Q . i Take Q to be the intersection of O O and ω which is further from O . We are to i i i +1 i i +1 compute the angle between O Q and O Q which is the same as ∠ O O Q . Note the i i i +1 i +1 i i +1 i +1 triangle 4 O P O is isosceles, call the base angle ξ . We have ∠ O O Q = ∠ O O P + i i i +1 i i +1 i +1 i i +1 i ∠ P O Q = ξ + ( π − 2 ∠ O P Q ) = ξ + ( π − 2( π − ∠ Q O P − ∠ P Q O )) = i i +1 i +1 i +1 i i +1 i i +1 i i i i +1 ξ − π + 2( ξ + (1 / 2) ∠ P O O ) = ξ − π + 2( ξ + (1 / 2) ξ ) = 4 ξ − π. i i i +1 So we get 6(4 ξ − π ) = 0 mod 2 π . Noting that ξ must be acute, ξ = π/ 12 , π/ 6 , π/ 4 , π/ 3 or 5 π/ 12. r is uniquely determined as (1 / 2) sec ξ so there are 5 possible values of r .