tag:blogger.com,1999:blog-6933544261975483399.post5231667204659472767..comments2024-04-22T04:55:16.794-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1558: Three Equilateral TrianglesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-62932275931064156082023-11-13T22:11:13.479-08:002023-11-13T22:11:13.479-08:00Perpendiculars from C1, C and C2 to B1B2 (extended...Perpendiculars from C1, C and C2 to B1B2 (extended to R) meet B1B2 at P, Q and R respectively. <br />Let C1B1= a and CB2 = b. <br />Let B1P = t1, PB = t2, BQ = t3, QB2 = t4 and B2R = t5. <br />Let C1P = u and C2R = v<br /><br />C1PRC2 is a trapezoid with C1P//CQ//C2R, so by connecting C2P and applying the midpoint theorem, <br />CQ = (u+v)/2 ..........(1) and <br />t2 + t3 = t4 + t5 ......(2)<br />Now it is given that t1 + t2 = t3 + t4....... (3)<br />From (2) and (3), t3 = (t1 + t5) / 2..........(4)<br /><br />Now consider Right Triangles B1C1P, BCQ and B2C2R<br />CQ = (u + v)/2<br />BQ = t3 = (t1+ t5)/2<br />and the included angle is 90<br /><br />So BC = (a + b)/2<br />Similarly we can show that CA = (a+b)/2 and AB = (a+b)/2<br /><br />Therefore Triangle ABC has BC = CA = AB and is thus equilateral<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com