tag:blogger.com,1999:blog-6933544261975483399.post6564942921810481703..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1345: Three Equilateral Triangles, Center, Area, Common Vertex, 60 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-30624960752584990032017-09-22T04:15:18.220-07:002017-09-22T04:15:18.220-07:00Do u have a pure geometry solution Antonio?Do u have a pure geometry solution Antonio?Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51967937278111954812017-09-22T03:51:59.967-07:002017-09-22T03:51:59.967-07:00Trigonometry Solution
Let AB = a, OE = b and CE ...Trigonometry Solution<br /> <br />Let AB = a, OE = b and CE = c and <br />let < ACE = α.<br /> <br />S1 = √3b2/4 …(1) and<br />S2 = a2/(4√3)…(2) <br />being 1/3rd of S(ABC) = (√3a2/4)/3<br /> <br />S3 = S(ADC) – S(ACE) – S(CDE)<br />S3 = ½ ac.sin(α+60) - √3c2/4 – ½ ac.sinα<br />S3 = ½ ac.(1/2 sinα + √3/2 cosα – sinα) - √3c2/4<br />S3 = ½ ac(cos30.cosα – sin30.sinα) - √3c2/4<br />S3 = ½ ac cos(α+30) - √3c2/4…(3)<br /> <br />Now from ∆OCE, cos(α+30) = (a2/3 + c2 – b2)/(2ac/√3) …(4)<br /> <br />Substituting in (3),<br />S3 = √3/4. (a2/3 + c2 – b2) - √3c2/4<br />S3 = a2/(4√3) – √3b2/4<br />So S3 = S2 – S1 from (1) and (2).<br /> <br />If S1 > S2, then S3 = S1 – S2.<br /> <br />So generally S3 = IS2 – S1I<br /> <br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com