Geometry Problem

Click the figure below to see the complete problem 533 about the sides of the regular pentagon, regular hexagon and regular decagon inscribed in the same circle.

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Complete Problem 533

Level: High School, SAT Prep, College geometry

## Friday, October 29, 2010

### Problem 533: Euclid's Elements Book XIII, Proposition 10

Labels:
decagon,
Elements,
Euclid,
hexagon,
pentagon,
regular polygon,
right triangle

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Volker Poschel, from Germany, sent the following solution/comment:

ReplyDeleteA big THANK YOU for the very good site. HERE IS A SOLUTION for Euklids book XIII 10. A very clever inviting proof.

I think the german words are easy to understand. But its not a "classical" answer - have you one?

Volker Pöschel

Germany

Visit PoeGot's Math-Club at www.poegot.org/www_seite/www_Geometrie_o.htm

See attached picture in the hyperlink below

ReplyDeletehttp://yfrog.com/6wproblem533j

Let H is the intersection of OA to BD and E is the symmetric of A over BD

Let x=BE=BA=DA and y=BH=HD=1/2BD=1/2AC

Note that angle DBA=angle EBH=18 , angle EOB=angle OBE= 36 and BE is the angle bisector of angle OBA

Since BE is the angle bisector so we have x/R=(R-x)/x or R.x=R^2-x^2

In right triangle OHB we have BH^2=OB^2-OH^2 or y^2=R^2-(R+x)^2/4 or 4.y^2=3.R^2-2.R.x-x^2

Replace R.x=R^2-x^2 in above expression we have 4.y^2=R^2 +x^2

And AC^2=BC^2+AB^2 . Triangle ABC is a right triangle

Peter Tran

Thanx, a good idea.

ReplyDeleteYou can find the solution in german here: http://www.poegot.org/www_seite/Historie/Viereckring.pdf at page 2 und 3

ReplyDeleteTo Volker Pöschel

ReplyDeleteThank you for clear graphic and additional explanation solution in Germany .

Peter Tran

TO prove the proposition we need to see that in the given figure if we apply Pythagoras theorem then we shall get〖 4R〗^2 〖sin〗^2 18+R^(2 )=4R^2 〖sin〗^2 36.

ReplyDeleteThen we need to see whether the values of sin 18 and sin 36 satisfy the above relation.

For this we need to simplify the expression after which we shall get 〖sin〗^2 36-〖sin〗^2 18=1/4. to see really if it is true then we substitute θ in place of 18 and 36 respectively and then solve the equation〖 sin〗^2 2θ-〖sin〗^2 θ=1/4. By using identity we get sin3θ.sinθ= 1/4. which then simplifies to 3〖sin〗^2 θ-4〖sin〗^4 θ= 1/4 proceeding step by step we get the following:

12〖sin〗^2 θ-16〖sin〗^4 θ=1. ⇒ 12〖sin〗^2 θ=1+16〖sin〗^4 θ ⇒ 12〖sin〗^2 θ=〖(1+〖4sin〗^2 θ)〗^2- 8〖sin〗^2 θ ⇒20〖sin〗^2 θ=〖(1+〖4sin〗^2 θ)〗^(2 ) ⇒2√5 sinθ= 1+4〖sin〗^2 θ

⇒4〖sin〗^2 θ-2√5 sinθ+1=0. solving this equation as quadratic in sin θ we get sin θ = (√5 ±1)/4 now we get the value of θ as 18 degrees (by applying inverse function of sin) and that of 3θ as 54 degrees. So it shows that the expression 〖 4R〗^2 〖sin〗^2 θ+R^(2 )=4R^2 〖sin〗^2 θ is satisfied by the obtained values of θ. Thus the proposition is proved.

Q. E. D.

Let the centre of the pentagon be X and let the bisector of < AXC meet circle X at Y. Let AY and XC extended meet at Z. Draw CW // to AZ to meet XY at W.

ReplyDeleteAY = YC = CZ = CW = XW = l10

R^2 = l10(l10 + R) from Tr. XYZ .....(1)

Now l5^2/4 + (R-l10)^2/4 = l10^2

i.e. l5^2 = (l10^2 + R^2) - (2R^2 - 2l10^2 - 2Rl10^2)

From (1) the expression in the 2nd bracket of RHS = 0

So l5^2 = R^2 + l10^2 and hence

< ABC = 90

Sumith Peiris

Moratuwa

Sri Lanka