## Sunday, March 14, 2010

### Problem 430: Circumscribed and Inscribed Regular Pentagon, Perpendicular, Areas

Proposed Problem
Click the figure below to see the complete problem 430 about Circumscribed and Inscribed Regular Pentagon, Perpendicular, Areas. Complete Problem 430

Level: High School, SAT Prep, College geometry

1. S/7=OAB-OA'B', S1/7=OA''B''.
OAB:OA'B':OA''B''=AB^2:A'B'^2:A''B''^2
AB^2-A'B'^2=A''B''2
Then, S/7=S1/7
q.e.d.

2. To anonymous:
7 or 5?

3. name h altitude of A'B'C'D'E'
from similarity of AOE and A'OE'
a/R = b/h =>

R = ah/b (1)

join E' to B' and name G E'B' meet A"B" we get
right triangle E'A'G with A'E' = b, A'G = c , GE' =a)
( B'B"G is isoceles B'B" = B"G = ...A'A" )

from similarity of A'OM (M midpoint of A'E')
and E'A'G ( right tr and angle 36°)
(b/2)/R = c/a
substitute R from (1)

h = b²/2c (2)
=>
R = ab/2c (3)

from similarity of A"ON ( N midpoint of A"E") and GA'E' (c/2)/h1 = c/b ( h1 altitude of pentagon C)

h1 = b/2 (4)

Sa = (5∙a ∙R)/2

Sa = 5a²b/4c ( R from 3)

Sb = ( 5∙b∙h )/2

Sb = 5b³/4c ( h from 2)
=>
Sj = 5a²b/4c - 5b³/4c = 5b(a² - b²)/4c ( a²-b²=c²)

Sj = 5bc/4 (5)

Sc = 5∙c∙h1/2

Sc = 5bc/4 (6) ( h1 from 4)

comparing (5) and (6) give the result

4. the area of a regular pentagon with side s is:
A=s²(5/4)cotan36
and we know (previous problem) that c²=a²-b²....
.-.

5. Masterstream said…

The sweetest proof seems to be this:-

Let a,b,c be the side-lengths of pentagons ABCDE, A'B'C'D'E', A''B''C''D''E'' respectively, as in Problem 429(q.v.); and let S2,S3 be the areas of the inscribed regular pentagon A'B'C'D'E' and the circumscribed regular pentagon ABCDE respectively.

Since all regular pentagons are similar,
S3:S2:S1 = a^2:b^2:c^2.

So S:S1 = (S3–S2):S1 = (a^2–b^2):c^2 = 1 ,
(since a^2–b^2 = c2, by Problem 429 [q.v.])

i.e. S = S1 . QED