## Friday, March 5, 2010

### Problem 429: Circumscribed and Inscribed Regular Pentagon, Perpendicular, Measurement

Proposed Problem
Click the figure below to see the complete problem 429 about Circumscribed and Inscribed Regular Pentagon, Perpendicular, Measurement. Complete Problem 429

Level: High School, SAT Prep, College geometry

1. From figure geometry: (b/2)/sin(36) =(a/2)/tan(36) or b=acos(36). Moreover, (c+btan(18))cos(18)=b which gives us: c =b(1-sin(18))/cos(18)=btan(36) or c = a*cos(36)*tan(36)=a*sin(36)
Thus: b^2+ c^2 = a^2[(cos(36))^2 +sin(36))^2]=a^2 QED
PS: Will someone pl. explain why (1-sin(18))/cos(18)= tan(36)?

2. in [0;90[ 18 is the only root of
(1-sinx)-cosx.tan2x=0
4sin18=sqr(5)-1
4cos18=sqr(10+2sqr5)
4tan36=(sqr(5)-1).sqr(10-2sqr5)
this numbers are related with the golden ratio
.-.

3. rotate A'B'C'D'E' so that A' is on AB
mE'A'O=mE"A"O => O,A",A',E' are concyclic => mA"OE'=pi/2=mEE'O => A"O//E'E, note that mE'A"O=mOEE' so A"OEE' is a //gram and A"E'=OE
consider OA"E' we have OE^2=A"E'^2=OE'^2+OA"^2
as three regular pentagons are similar to each others, it yields a^2=b^2+c^2

4. The sweetest proof seems to be this...
Let x, y, z be the perpendicular distances from O to AE, A'E', A''B'’ respectively.
Note that x = the radius of circle O, so x=OA', and also that z=b/2.
Since all regular pentagons are similar,
x:y:z= a:b:c.
But x^2 = y^2 + z^2 (Pythagoras),
so a^2 = b^2 + c^2. QED

Corollary: In fact, z=b/2 =(b/2c)×c;
so y=(b/2c)×b=b^2/2c and x=(b/2c)×a = ab/2c.
So if d=diameter of circle O,
then d=2(ab/2c)=ab/c. ∴ ab=cd. How elegant!