tag:blogger.com,1999:blog-6933544261975483399.post8778887285472356748..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 533: Euclid's Elements Book XIII, Proposition 10Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-58174469802608808052019-05-08T11:52:40.864-07:002019-05-08T11:52:40.864-07:00Let the centre of the pentagon be X and let the bi...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.<br /><br />AY = YC = CZ = CW = XW = l10<br /><br />R^2 = l10(l10 + R) from Tr. XYZ .....(1)<br /><br />Now l5^2/4 + (R-l10)^2/4 = l10^2<br />i.e. l5^2 = (l10^2 + R^2) - (2R^2 - 2l10^2 - 2Rl10^2)<br /><br />From (1) the expression in the 2nd bracket of RHS = 0<br />So l5^2 = R^2 + l10^2 and hence<br /><br />< ABC = 90<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55357327923788940502011-03-03T00:55:32.992-08:002011-03-03T00:55:32.992-08:00TO prove the proposition we need to see that in th...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.<br />Then we need to see whether the values of sin 18 and sin 36 satisfy the above relation.<br />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:<br />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 <br />⇒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.<br />Q. E. D.sourav mishranoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18834919714543347452010-11-04T11:07:55.214-07:002010-11-04T11:07:55.214-07:00To Volker Pöschel
Thank you for clear graphic and...To Volker Pöschel<br /><br />Thank you for clear graphic and additional explanation solution in Germany .<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35609768786297534342010-11-03T02:59:56.986-07:002010-11-03T02:59:56.986-07:00You can find the solution in german here: http://w...You can find the solution in german here: http://www.poegot.org/www_seite/Historie/Viereckring.pdf at page 2 und 3Volker Pöschelhttp://www.poegot.orgnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-763657926292437342010-10-30T02:16:24.573-07:002010-10-30T02:16:24.573-07:00Thanx, a good idea.Thanx, a good idea.Volker Pöschelhttp://www.poegot.orgnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15103254064128918692010-10-29T17:51:32.846-07:002010-10-29T17:51:32.846-07:00See attached picture in the hyperlink below
http:/...See attached picture in the hyperlink below<br />http://yfrog.com/6wproblem533j<br />Let H is the intersection of OA to BD and E is the symmetric of A over BD<br />Let x=BE=BA=DA and y=BH=HD=1/2BD=1/2AC<br />Note that angle DBA=angle EBH=18 , angle EOB=angle OBE= 36 and BE is the angle bisector of angle OBA<br /><br />Since BE is the angle bisector so we have x/R=(R-x)/x or R.x=R^2-x^2<br />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<br />Replace R.x=R^2-x^2 in above expression we have 4.y^2=R^2 +x^2 <br />And AC^2=BC^2+AB^2 . Triangle ABC is a right triangle<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90122032909162421902010-10-29T07:44:02.823-07:002010-10-29T07:44:02.823-07:00Volker Poschel, from Germany, sent the following s...Volker Poschel, from Germany, sent the following solution/comment: <br /><br />A big THANK YOU for the very good site. <a href="http://www.gogeometry.com/problem/p533_volker_poschel_polygon.pdf" rel="nofollow">HERE IS A SOLUTION</a> for Euklids book XIII 10. A very clever inviting proof. <br />I think the german words are easy to understand. But its not a "classical" answer - have you one?<br /><br />Volker Pöschel<br />Germany <br />Visit PoeGot's Math-Club at <a href="http://www.poegot.org/www_seite/www_Geometrie_o.htm" rel="nofollow">www.poegot.org/www_seite/www_Geometrie_o.htm</a>Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.com