Proposed Problem

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

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Complete Problem 429

Level: High School, SAT Prep, College geometry

## Friday, March 5, 2010

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

Labels:
90,
circumscribed,
inscribed,
pentagon,
perpendicular,
Pythagoras,
regular polygon

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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)

ReplyDeleteThus: 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)?

in [0;90[ 18 is the only root of

ReplyDelete(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

.-.

rotate A'B'C'D'E' so that A' is on AB

ReplyDeletemE'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

The sweetest proof seems to be this...

ReplyDeleteLet 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!