Diagonal of a Pentagon
In a regular pentagon of side 1, find the length of the line connecting
two non-adjacent corners. Can you do this with and without trigonometry
(using sines and cosines)?
Source: Based upon puzzle 124 in Ivan Moscovich's
1000 PlayThinks, though I've seen it elsewhere.
Solutions were received from
Steven M. Murray,
Odette De Meulemeester (and students),
and Alexey Vorobyov.
The length is the golden ratio: (1+sqrt(5))/2 = 2cos(36). Many good solutions
were received. A representative sampling is below.
Denis Borris sent an URL of a good solution:
Adrian Atanasiu sent a good algebraic solution:
Shall us consider AB=BC=CD=DE=EA=a and let be BD=EB=EC=x.
We denote by M = the intersection between BD and EC.
Because BD||AE and AB||EC (it is easy to verify that), we have MB=EM=a and the angles EMB=EAB.
Thus MC=MD=x-a and the triangles MCD and ABE are similar.
Now, we can write
CD MC a a-x
---- = ---- that is --- = -----
EB AB x a
Solving this equation, we obtain x=a(1+sqrt(5))/2 (the second solution is negative).
In peculiar, for a=1:
Radu Ionescu's short algebraic solution:
Let be ABCDE the pentagon.
In ABCD AD=DB=AC=x.
Ptolemeus theorem: AD*BC+AB*DC=AC*DB ==>
x+1=x^2 ==> x=(1+ sqrt(5))/2
Radu's short trigonometric solution:
There isn't a problem with trigonometry: x=2cos(36)= ( 1+ sqrt(5) )/2.
Philippe Fondanaiche's trigonometric solution:
[In Pentagon ABCDE, let BE have midpoint H.]
The line BE connecting two non-adjacent corners B and E is such as:
BE = 2*BH = 2*AB*cos(2*pi/5) = 2*cos(2*pi/5)
It is easy to check that sin(5*a) = 5*sin(a) - 20*sin^3(a) + 16*sin^5(a)
For a=pi/5, we get the equation 16*u^4 - 20*u^2 + 5 = 0 with u=sin(pi/5)
Then sin(pi/5) = sqrt(10-2*sqrt(5))/4 and cos(pi/5) = (sqrt(5)+1)/4
So BE = 2*(2*cos^2(pi/5)-1) = (sqrt(5)+1)/2 which is the golden ratio.
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