103 Trigonometry Problems

implying that n = 23.
4. Solutions to Introductory Problems 105
Second Solution: Note that
Hence
(1 + tan k ◦ )(1 + tan(45 − k) ◦ )
= 1 +[tan k ◦ + tan(45 − k) ◦ ]+tan k ◦ tan(45 − k) ◦
= 1 + tan 45 ◦ [1 − tan k ◦ tan(45 − k) ◦ ]+tan k ◦ tan(45 − k) ◦
= 2.
(1 + tan 1 ◦ )(1 + tan 2 ◦ ) ···(1 + tan 45 ◦ )
= (1 + tan 1 ◦ )(1 + tan 44 ◦ )(1 + tan 2 ◦ )(1 + tan 43 ◦ )
···(1 + tan 22 ◦ )(1 + tan 23 ◦ )(1 + tan 45 ◦ )
= 2 23 ,
implying that n = 23.
31. [AIME 2003] Let A = (0, 0) and B = (b, 2) be points in the coordinate plane.
Let ABCDEF be a convex equilateral hexagon such that ̸ FAB = 120 ◦ ,
AB ‖ DE, BC ‖ EF, and CD ‖ FA, and the y coordinates of its vertices
are distinct elements of the set {0, 2, 4, 6, 8, 10}. The area of the hexagon can
be written in the form m √ n, where m and n are positive integers and n is not
divisible by the square of any prime. Find m + n.
Note: Without loss of generality, we assume that b>0. (Otherwise, we can
reflect the hexagon across the y axis.) Let the x coordinates of C, D, E, and F
be c, d, e, and f , respectively. Note that the y coordinate of C is not 4, since
if it were, the fact |AB| =|BC| would imply that A, B, and C are collinear or
that c = 0, implying that ABCDEF is concave. Therefore, F = (f, 4). Since
−→
AF = −→ CD, C = (c, 6) and D = (d, 10), and so E = (e, 8). Because the y
coordinates of B,C, and D are 2, 6, and 10, respectively, and |BC|=|CD|,
we conclude that b = d. Since −→ AB = −→ ED, e = 0. Let a denote the side length
of the hexagon. Then f