tag:blogger.com,1999:blog-6933544261975483399.post3039382270971602654..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 103Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-88411286443485682042011-05-23T00:21:41.690-07:002011-05-23T00:21:41.690-07:00http://img198.imageshack.us/img198/538/problem103....http://img198.imageshack.us/img198/538/problem103.png<br />Draw equilateral triangles CDE, DBG and ADF and we have a hexagon CEBGAF ( see picture)<br />We have triangles (CDB) congruence to (AGB) case SAS<br />similarly triangles (AFC) congruence to triangles (ADB) and triangles (CDA) congruence to triangles (CEB) <br />Area(CEBGAF)= 2.* Area(ABC)=2S<br />We also have congruence triangles EDB, GAD, CDF ( case SSS) with sides d, e, f .<br />Area(CEBGAF)= area(CDE)+Area(DBG)+Area(ADF)+3.Area(EDB)=2S<br />= f^2*SQRT(3)/4+ e^2*SQRT(3)/4+ d^2*SQRT(3)/4+3SQRT(s(s-e)(s-f)(s-d))<br />Divide both sides by 2 we will get the result<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77892883121316888342011-05-21T20:42:42.845-07:002011-05-21T20:42:42.845-07:00Some errors have crept into my earlier proof. Plea...Some errors have crept into my earlier proof. Please ignore the same. It must read as follows: ABC is the given equilateral triangle with a point D inside which is d, e & f units away from A, B & C resply. Rotate the triangle anti-clockwise thru' 60 deg. so that C goes to B, B to F and D to E. Now we've: AD=AE=d and BE=BD=e while FE=CD=f. It's easy to see that triangles DAC and EAB are congruent (corresponding sides equal) and, therefore, /_DAC = /_EAB which makes /_DAE=60 deg and since AD=AE, we conclude that triangle ADE is equilateral.<br />Now Tr. DAC + Tr. DAB = Tr. EAB + Tr. DAB = Tr. ADE + Tr. EDB and Tr. EDB has sides of d, e & f while Tr ADE is equilateral with side d.<br />Hence, Tr. DAC + Tr. DAB=(3)^(1/2)*(d^2)/4 +(s(s-a)(s-b)(s-c))^(1/2)/4 using Heron's Formula where 2s=d+e+f.<br />We can likewise prove that, Tr, DAB + Tr DBC=(3)^(1/2)*(e^2)/4 + (s(s-a)(s-b)(s-c))^(1/2)/4 and Tr. DBC + DCA =(3)^(1/2)*(f^2)/4+(s(s-a)(s-b)(s-c))^(1/2)/4<br />Add LHS and RHS of the three equations to get: <br />2*[Tr. ABC]=(3)^(1/2)*(d^2)/4+(3)^(1/2)*(e^2)/4 +(3)^(1/2)*(f^2)/4+3(s(s-a)(s-b)(s-c))^(1/2)/4 Or [Tr. ABC] = S =(3)^(1/2)(d^2)/8 +(3)^(1/2)*(e^2)/8+( 3)^(1/2)(f^2)/8+(3/8)( s(s-a)(s-b)(s-c))^(1/2) Or S=(1/2){(3)^(1/2)(d^2)/4 +(3)^(1/2)(e^2)/4+(3)^(1/2)(f^2)/4+3(s(s-a)(s-b)(s-c))^(1/2)} whics was to be proven.<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-64231605376823495092008-11-17T06:18:00.000-08:002008-11-17T06:18:00.000-08:00E, F, and G are exterior points in the plane of th...E, F, and G are exterior points in the plane of the triangle ABC so that CEAFBG is a hexagon.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-64000381220167688802008-11-16T19:54:00.000-08:002008-11-16T19:54:00.000-08:00where are points E,F,G?thankswhere are points E,F,G?<BR/>thanksssriv97589https://www.blogger.com/profile/03735851879095373783noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32683462559299522172008-11-16T17:27:00.000-08:002008-11-16T17:27:00.000-08:00HINT: Draw equilateral triangles CDE, ADF, and BDG...<B>HINT:</B> Draw equilateral triangles CDE, ADF, and BDG. Apply SAS congruent triangles.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.com