Each of the sides PQ, QR, RS, ST, TU, UV and VP of a regular heptagon
PQRSTUV have length 2. The respective lengths of two of the diagonals PR and PS of the heptagon are denoted by X and Y.
Determine 1/X + 1/Y.
Each of the seven angles of a regular heptagon is equal to (5*pi/7) radians or approximately 128.57143 degrees.
Using the law of cosines, PR2 = PQ2 + QR2 - 2*PQ* QR*cos(5*pi/7)
Therefore, given that PQ = QR = 2,
X = PR = [22 + 22 - 2*2*2*cos(5/7 pi)]1/2 ~= 3.60388
Let us identify two points on PS. A, such that PA is perpendicular to PS, and B, such that RA is also perpendicular to PS. Triangles PAQ and QBS would both be right triangles with equal angles and side lengths. As angle AQR would also be a 90 degree angle, angle PQA would then be 3*pi/14 radians or approximately 38.57143 degrees.
Using the law of sines, sin(Angle PQA)/PA = sin(Angle PAQ)/PQ,
therefore PA = sin(3*pi/14)*sin(pi/2)/2 ~= 1.346397
PA = BS and QR = AB, therefore
Y = PS = ( PA + AB + BS) ~= (1.346397 + 2 + 1.345396) ~= 4.692794
1/X + 1/Y ~= (1/3.60388) + (1/4.692794) ~= 0.5 (though I suspect it equals 0.5)
Posted by Dej Mar
on 2007-11-17 19:44:41