## Monday, March 2, 2009

### Problem 261: Regular Pentagon inscribed in a circle, sum of distances

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.

1. 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.
Apply 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

2. Here is my solution...

1. Nice proof.

In the video, at 6:06, the "established fact that Φ∧2= Φ+1" is because of quadrilateral ABDE.

3. 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.

For 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

1. Nice... but i think it should be:
BFCD -> p.d = p.b + q.c
ABFE -> q.a = p.b + p.e

2. Thank you, cha12va, for your comment, which is correct, and welcome to this blog !