tag:blogger.com,1999:blog-6933544261975483399.post1210652112051615917..comments2024-09-10T22:02:29.582-07:00Comments on GoGeometry.com (Problem Solutions): Problem 864: Parallelogram, Diagonal, Congruence, Similarity, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-42280474498721499702016-09-12T02:47:14.699-07:002016-09-12T02:47:14.699-07:00We use Barycentric Coordinates.
Definition: Capita...We use Barycentric Coordinates.<br />Definition: Capital letters will denote vectors. P = x A + y B + z C will be written as P = (x, y, z). By definition, x + y + z = 1<br />Let A = (1, 0, 0), B = (0, 1, 0) and C = (0, 0, 1).<br />ABCD is a parallelogram, hence D + B = A + C, which results in D = (1, -1, 1)<br />Let F = q B + (1 – q) G. As FB = GE, we have G = (1 – q) F + q E<br />Let F = (p, 0, 1 – p). This leads to G = (p, -q, 1 – p) / (1 – q)<br />Consider the fact that G is the intersection of CD and BF. If (x, y, z) is on a specified straight line, then 0 = ux + vy + wz for some constant u, v, w. After solving the systems of equations, it’s clear that line BF is x (1 – p) – z p = 0 and CD is x + y = 0. To find G, we also need x + y + z = 1. With this, it’s clear that G = (p / (1 – p), -p / (1 – p), 1).<br />Using the two expressions of G, we get that p = q. Using these facts leads to finding E = (p – 1 + 1/(1-p), -1/(1-p), -p + 2).<br />But we also know that E is the intersection of AD [z + y = 0] and BF [x(1-p) – zp = 0]. So E = (1, -(1 – p)/p, (1 – p)/p). Combining the two expressions for E, we get p = 0.5 (3 + sqrt(5)) or 0.5(3 – sqrt(5)). This gives two possible coordinates for E: 0.5 (2, sqrt(5) – 1, 1 – sqrt(5)) or 0.5 (2, -sqrt(5) – 1, 1 + sqrt(5)).<br />D is between A and E. Thus, must be possible to express D as mA + nE, such that m + n = 1 for positive m and n. From the formula of the division of an interval, AD : DE = n : m.<br />Straight away, it is clear that E = 0.5 (2, -sqrt(5) – 1, 1 + sqrt(5)) only, as the alternative would produce negative m or n. Simple arithmetic tells us that (sqrt(5) – 1) A / (sqrt(5) + 1) + 2 E / (1 + sqrt(5)) = D.<br />Hence, AD : DE = 2 : sqrt(5) – 1, which means that AD:AE = 2 : 1 + sqrt(5)<br />It’s given that AD = 3<br />So quite obviously, AE = 1.5 (1 + sqrt(5)).Anonymoushttps://www.blogger.com/profile/14379158334268338516noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55725775895953894292013-03-30T07:46:04.422-07:002013-03-30T07:46:04.422-07:00From ΔFBC~ΔFEA
3/x = BF/FE := a/b
From ΔGBC~ΔGED
...From ΔFBC~ΔFEA<br />3/x = BF/FE := a/b<br /><br />From ΔGBC~ΔGED<br />3/(x-3) = BG/GE = b/a<br /><br />⇒ x(x-3) = 9<br />⇒ x^2 - 3x - 9 = 0<br />⇒ x = 3(1 + √5)/2Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35721257586625623592013-03-30T06:21:13.916-07:002013-03-30T06:21:13.916-07:00Constrct a line PQ through F parrallel to AB,CD su...Constrct a line PQ through F parrallel to AB,CD such that P lies on BC and Q lies on AD.<br />FP:FQ = 3:x<br />Since triangle(BFP) = triangle(EGD), FP = GD, so CG = FQ.<br />By triangle(BCG) ~ triangle(EDG), <br /> GD/CG = DE/BC<br />=> FP/FQ = DE/BC<br />=> 3/x = (x - 3)/3 <br />=> x = 3*(1 + sqrt5)/2<br />W Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-6041699343277672102013-03-29T18:34:46.322-07:002013-03-29T18:34:46.322-07:00if i worng, please do not sanction my post
but if ...if i worng, please do not sanction my post<br />but if my solution is true:<br />BF=Y=GE<br />FG=Z<br />DE=X-3<br />3/X=y/(z+y)<br />y/(z+y)=(x-3)/3<br />3/x=(x-3)/3<br />9=x^2-3x<br />x^2-3x-9=0<br />x1,2=(3+-sqrt(9-4*1*(-9)))/2=<br />=1.5+sqrt(45)/2<br />Anonymoushttps://www.blogger.com/profile/11694885671410014993noreply@blogger.com