Proposed Problem

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

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Complete Problem 430

Level: High School, SAT Prep, College geometry

## Sunday, March 14, 2010

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

Labels:
area,
circle,
circumscribed,
inscribed,
pentagon,
perpendicular,
regular polygon

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S/7=OAB-OA'B', S1/7=OA''B''.

ReplyDeleteOAB: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.

To anonymous:

ReplyDelete7 or 5?

name h altitude of A'B'C'D'E'

ReplyDeletefrom 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

the area of a regular pentagon with side s is:

ReplyDeleteA=s²(5/4)cotan36

and we know (previous problem) that c²=a²-b²....

.-.

Masterstream said…

ReplyDeleteThe 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