HMMT 二月 2001 · 团队赛 · 第 12 题

英文解析

12. All subscripts in this problem are to be considered modulo 6, that means for examplethat 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 thei i i +1
    center of the hexagon, for i = 1 , . . . 6. Let Q , i = 1 . . . 6, lie on ω such that Q , P , Qi 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 ωi i +1 i i
    ′ ′ ′iand 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 ii +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 ii i i +1 i +1
    The previous result shows that the lines O Q and O Q meet at the same angle as thei 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 ωi 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 ωi igive the same value of ψ , so ψ can’t depend on this.) Now, the geometry of ω and ωii i i i +1
    is the same for every i , so ψ is simply a constant ψ which depends only on r . We knowi
    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 .
    Take Q to be the intersection of O O and ω which is further from O . We are toii i i +1 i i +1
    compute the angle between O Q and O Q which is the same as ∠ O O Q . Note thei 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 .