tag:blogger.com,1999:blog-6933544261975483399.post5239029505407588413..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1286 Triangle, Quadrilateral, 60, 135, 105 Degree Angles, Congruence, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-75063645279872331742016-11-10T14:32:09.278-08:002016-11-10T14:32:09.278-08:00https://goo.gl/photos/YmHy2K7uN38Yp5xY8
Let AB mee...https://goo.gl/photos/YmHy2K7uN38Yp5xY8<br />Let AB meet CD at E<br />Draw BF//AD , AH⊥CD, CG⊥BE ( see sketch)<br />We have AED and BEF are 60-60-60 triangles<br />AHD , AHE and CGE are 30-60-90 triangles<br />And CCG is 45-45-90 triangle<br />Observe that H is the midpoint of CF and DE ;and AF=BD=AC <br />Let u= CG<br />So BG= u and BC= u.sqrt(2)<br />EC=2u/sqrt(3)<br />EG= u/sqrt(3)<br />EF=BE=u(1+1/sqrt(3))<br />CF= EF-EC=u(1-1/sqrt(3))<br />AB=FD=CE=2u/sqrt(3)<br />CD=EF=u(1+1/sqrt(3))<br />AD=2.DH= 2.DF+CF= u(1+sqrt(3))<br />Verify that BC/AB=sqrt(6)/2<br />CD/AB=(sqrt(3)+1)/2<br />And AD/AB=(sqrt(3)+3)/2<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34487079358855146642016-11-10T13:24:57.631-08:002016-11-10T13:24:57.631-08:00Problem 1286
Let AB=a, BC=b, CD=c,AD=d. The sides...Problem 1286<br />Let AB=a, BC=b, CD=c,AD=d. The sides AB and CD intersects at point K then triangle KAD is equilateral (<ADC=360-60-135-105=60). Draw BE//AD( the point E lies on the CD).Then ABED is isosceles trapezoid (AB=ED).So AC=BD=AE or triamgle ACE is isosceles .Then <AED=<ACK and <DAE=<KAC.But triangles KAC=DAE and AB=ED=KC, KB=CD. Draw CF<br />perpendicular in KB, then KF=KC/2=a/2,CF=√(3 )a/2=BF(<KCF=30,<FCB=<FBC=45).But b^2=2CF^2 or b/a=√(6 )/2.<br />c=CD=KB=KF+FB=a/2+√(3 )a/2 =a(√(3 ) +1)/2 or c/a=(√(3 ) +1)/2.<br />And d=AD=KA=AB+BK=a+c=a+ a(√(3 ) +1)/2=a (√(3 ) +3)/2 or d/a=(√(3 ) +3)/2 .<br /> APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE <br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.com