tag:blogger.com,1999:blog-6933544261975483399.post6930968555653395088..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1092. Equilateral Triangle, Square, Circle, Tangent, 90 Degree, SangakuAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-7804361996774966102021-01-18T11:29:04.472-08:002021-01-18T11:29:04.472-08:00https://photos.app.goo.gl/rJTz5amj6vkcWJQj6https://photos.app.goo.gl/rJTz5amj6vkcWJQj6c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33196411651606174832018-06-16T07:13:20.234-07:002018-06-16T07:13:20.234-07:00https://1drv.ms/u/s!AuFUZHYD5UUnzWied8htEcHx2GL8
S...https://1drv.ms/u/s!AuFUZHYD5UUnzWied8htEcHx2GL8<br />Solved by Fernando Rogério Gonçalves!<br />BrazilAnonymoushttps://www.blogger.com/profile/01559088151133874233noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33223258391453686952015-08-30T01:23:59.096-07:002015-08-30T01:23:59.096-07:00Let BA meet FE in N. Let the side of the square be...Let BA meet FE in N. Let the side of the square be a and let AG = b. Easily Tr. EFC is isoceles with FC = a+b. <br /><br />Now H, O, D are collinear and by congruent Tr.s easily we can prove that HB = a and BD = b. So BC = 2a+3b. <br /><br />Now since < BNF = 30, Tr. AEN is isoceles and AN = a+b, AH = 2b and HB = a. So BN = 2a+3b = BC.<br /><br />So AN + AB = BC hence BC - AB = AN = AE.<br /><br />(Note that a = sqrt3 X b which I have avoided using to maintain clarity)<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52704841571177972152015-03-05T07:23:19.654-08:002015-03-05T07:23:19.654-08:00Also: angle CAF = 15 degreeAlso: angle CAF = 15 degreeAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62384795953085770932015-03-05T02:25:51.615-08:002015-03-05T02:25:51.615-08:00Join the collinear points H, O, D.
The three 60-30...Join the collinear points H, O, D.<br />The three 60-30-90 Right Triangles AHG, HDB, HDM are congruent to one another.<br />Let each side of the equilateral triangle be "a" and each side of the square be "b".<br />So BD = DM = a - b. Also HB = HM = b, AH = 2AG = 2BD = 2DM = 2a - 2b.<br />Follows AB = AH + HB = (2a - 2b) + b = 2a - b, AE = AG + GE = (a - b) + b = a<br />while BC - AB = (BD + DC) - AB = [(a - b) + 2a] - [2a - b] = a.<br />Hence AE = BC – AB.<br />(Incidentally AB = BF and AE = FC)Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17550416371246868382015-03-03T18:22:14.423-08:002015-03-03T18:22:14.423-08:00Observe that ΔABC is a 30°-60°-90° triangle, and s...Observe that ΔABC is a 30°-60°-90° triangle, and so are ΔAHB, ΔHBD, ΔHMD. <br />Also observe that HBDM is a kite. <br /><br />Let BN⊥AC at N which is the mid-point of GE. <br />Let EP⊥BC at P which is the mid-point of DF. <br /><br />Let AG=a. <br />Then<br />AB = AH + HB = 2 AG + GH = (2+√3)a<br />BC = 2 ED + DM = 3 AG + 2 GH = (3+2√3)a<br />AE = AG + GH = (1+√3)a<br /><br />Hence, AE = BC − ABJacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com