Proposed Problem
See complete Problem 261 at:
gogeometry.com/problem/p261_regular_pentagon_inscribed_circle_distance.htm
Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, March 2, 2009
Problem 261: Regular Pentagon inscribed in a circle, sum of distances
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ABCDE is a regular pentagon;hence all its sides are equal. Let it's side = x. Hence BD = 2a*sin(54). Let 2sin(54) = s. Thus, BD =AC=AD= x*s.
ReplyDeleteApply Ptolemy's Theorem to BFCD,AFCD and FCDE to obtain: b*x+sc*x=d*x or b+sc=d --(1)Similarly, a+sc=sd --(2)and c + e = sd--(3) . Add (1) and (3) to get b+c+e+sc = sd + d or b+c+e = sd-sc+d = a + d by equation (2). QED
Vihaan: vihaanup@gmail.com
Here is my solution...
ReplyDeletehttps://www.youtube.com/watch?v=VBia9qMI-Lw
Nice proof.
DeleteIn the video, at 6:06, the "established fact that Φ∧2= Φ+1" is because of quadrilateral ABDE.
One combination which avoids substitutions in the equations which are hard to "see" is to look at 4 different quadrilaterals and just add the resulting equations.
ReplyDeleteFor instance, if p = side of the pentagon and q length of the chord intercepting 2 sides :
ABFC -> p.a = q.b + p.c
BFCD -> q.a = p.b + q.c
ABFE -> p.d = p.b + p.e
BFDE -> q.d = q.e - p.b
Sum -> (p+q).(a+d) = (p+q).(b+c+e) then divide by (p+q) QED
Nice... but i think it should be:
DeleteBFCD -> p.d = p.b + q.c
ABFE -> q.a = p.b + p.e
Thank you, cha12va, for your comment, which is correct, and welcome to this blog !
Delete