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Diagonal Product (Posted on 2005-10-20) Difficulty: 4 of 5
Suppose we have the N vertices of a regular N-gon inscribed in a circle of radius 1. Select one vertex W and draw line segments from W to each of the other N-1 vertices. What is the total product of the lengths of these line segments? (old problem - original author unknown)

See The Solution Submitted by owl    
Rating: 4.3333 (3 votes)

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Hints/Tips No proof, but ... (spoiler) | Comment 1 of 3

Each line segment is a chord of the circumscribing circle, subtending 360*i/n degrees, where i goes from 1 to n-1. The length of such a chord is twice the sine of half the subtended arc. So, using Prod to signify what is usually denoted by a capital Pi, the formula would be Prod{i=1 to n-1} 2 sin(180 i / n deg). Tabulated for n = 3 to 25, this comes out to

3       3.00000000000000000000000000000000000000000000000000000000000001
4       4.00000000000000000000000000000000000000000000000000000000000007
5       5.00000000000000000000000000000000000000000000000000000000000008
6       6.0000000000000000000000000000000000000000000000000000000000002
7       7.00000000000000000000000000000000000000000000000000000000000063
8       8.00000000000000000000000000000000000000000000000000000000000064
9       9.00000000000000000000000000000000000000000000000000000000000084
10      10.0000000000000000000000000000000000000000000000000000000000008
11      11.00000000000000000000000000000000000000000000000000000000000158
12      12.00000000000000000000000000000000000000000000000000000000000224
13      13.00000000000000000000000000000000000000000000000000000000000275
14      14.00000000000000000000000000000000000000000000000000000000000305
15      15.00000000000000000000000000000000000000000000000000000000000361
16      16.0000000000000000000000000000000000000000000000000000000000048
17      17.00000000000000000000000000000000000000000000000000000000000476
18      18.00000000000000000000000000000000000000000000000000000000000467
19      19.00000000000000000000000000000000000000000000000000000000000578
20      20.00000000000000000000000000000000000000000000000000000000000718
21      21.0000000000000000000000000000000000000000000000000000000000102
22      22.00000000000000000000000000000000000000000000000000000000000912
23      23.00000000000000000000000000000000000000000000000000000000000929
24      24.00000000000000000000000000000000000000000000000000000000001086
25      25.00000000000000000000000000000000000000000000000000000000001551

indicating that the non-zeros near the ends of these numbers are the result of rounding errors, and that the simplified formula for the product is actually N.

For N=3, for example, the sides of the triangle are sqrt(3). Only two of the sides are multiplied together, so the result is 3.  For a square (N=4), the sides are sqrt(2) and the diagonal is 2. The product of two sides and the diagonal is 4.

 4   point 13
 5   Pi=atan(1)*4:Dr=Pi/180
10   for N=3 to 25
20     P=1
30     for I=1 to N-1
40      P=P*2*sin(180*I*Dr/N)
50     next I
60     print N,P
70   next N

Edited on October 20, 2005, 2:09 pm
  Posted by Charlie on 2005-10-20 13:54:01

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