tag:blogger.com,1999:blog-6933544261975483399.post8462562372661112624..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 112Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-13421520563222006182018-12-10T02:14:20.544-08:002018-12-10T02:14:20.544-08:002nd Solution, much shorter.
Let S(AFD) = S2 &...2nd Solution, much shorter.<br /><br />Let S(AFD) = S2 & S(AFB) = S3<br /><br />Since E is the mid point of BD, S(BFE) = S1.<br /><br />Tr.s BFE & AFD are similar, so S2 = 2S1<br />Tr.s ABF & BFD are similar so S3/2S1 = 1/2 so S3 = S1<br /><br />2S1 + S2 + S3 = S/2<br />So 2S1 + 2S1 + S1 = S/2 and hence <br />S1 = S/10<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16718184991486586122017-07-26T11:58:09.797-07:002017-07-26T11:58:09.797-07:00Let the length of the square be 2 units
Join BF an...Let the length of the square be 2 units<br />Join BF and AF and let m(CFE) = x<br />Hence m(FDE) = x (external segment theorem)<br /><br />Since E is the center of the square, m(EAD) = m(EFD) = 45 -------(1)<br />Upon further angle chasing, we can deduce that C is the circumcenter of the triangle BFD<br />Hence m(BFE) = 135 and from (1) , m(BFE) = 90 <br />In the right triangle BFE, we have<br />BF^2+FE^2 = BE^2<br />=> BF^2+FE^2=2 ---------(2)<br /><br />Observe that the triangles, AFB and BFD are similar (AAA)<br />=> AB.FD=BD.BF<br />=> BF = FD/Sqrt(2) -------(3)<br /><br />Now extend FE to a point P such that FP = 2.FE. Join DP and form the right isosceles triangle DPF (since BF_|_FE, by symmetry DP_|_FP and DP = BF)<br /><br />Hence FE = FD/2.Sqrt(2) -----------(4)<br />Substitute (3) and (4) in (2)<br />=> FD^2=16/5 ---------(5)<br /><br />Area of triangle FED = 0.5*(FE)*(DP)<br />=> S1 = 0.5*(FE)*(BF)<br />-> S1 = 0.5*(FD/2.Sqrt(2))*(FD/Sqrt(2)) [Substitute value of FD^2 from (5)]<br />=> S1 = 2/5 = 4/10 = 1/10(Area of the square ABCD)<br />Hence S1 = S/10Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9037788243862405802016-04-29T23:24:50.273-07:002016-04-29T23:24:50.273-07:00Let FD, OC meet at X
Then AFD, FXC,DXC are all co...Let FD, OC meet at X<br /><br />Then AFD, FXC,DXC are all congruent triangles and since each are similar to Tr.ODC, OC = sqrt5 a/ 2 where a = side of the square<br /><br />Hence XD = a^2/4 /((sqrt5 a/2)/2) = a/sqrt5 = AF<br /><br />S(AFD) = S(ODC) AF^2/OD ^2 = a^2/5<br /><br />S(FCD) = 2a^2/5<br /><br />So d the altitude from F in Tr. AFD = a^/5/(a/2) = 2a/5<br /><br />So S(BFC) = (1/2)(3a/5)a = 3a^2/10<br /><br />Hence S(BFD) = 2a^2/5+3a^2/10-a^2/2 = a^2/5<br /><br />Therefore S(FED) = S(BFD)/2 = a^2/10 = S/10<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52790928561637923782012-05-13T18:27:12.219-07:002012-05-13T18:27:12.219-07:00I tried a solution without coordinates in problem ...I tried a solution without coordinates in problem 112.<br />As triangles DEF and BEF have equal bases BE and DE and same altitude relative to that bases, then both have area S1. Being O the center of the semicircle, and s the side of the square ABCD, let’s write the altitude of triangle BAF as FG = p, the altitude of triangle AFD as FH = q, and ang(ADF) = a.<br />DOF is isosceles then ang(DOF) = 180º - 2a. Quadrilateral CDOF is cyclic, so ang(DCF) = 2a. We have CD = CF (tangents from C to the circle) thus CDOF is a kite, OC bisects the angle DCF and ang(DCO) = a.<br />We can see now that FHD and ODC are similar, with q/(s-p) = (s/2)/s so p = s - 2q (1).<br />Also AHF and AFD are similar, because ang(AFD) = 90º, so AHF and ODC are similar, with p/q = (s/2)/s and q = 2p (2).<br />From (1) and (2) we get p = s/5 and q = 2s/5.<br />Let S2 and S3 be the areas of AFD and AFB respectively. We have, successively <br />2S1 + S2 + S3 = S/2, then 2S1 + sq/2 + sp/2 = S/2, then 2S1 + (s^2)/5 + (s^2)/10 = S/2, then 2S1 + (3S)/10 = S/2, then S1 = S/10.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11066605834980338052009-04-12T19:49:00.000-07:002009-04-12T19:49:00.000-07:00If we take the midpoint of AC as the origin and la...If we take the midpoint of AC as the origin and label A:(-a,0),B:(-a,2a),C:(a,2a)& D:(a,0) then E is clearly (0,a). Now CF & CD are both tangents to the semicircle so CF = CD = 2a which means F is one of the intersections of the two circles given by x^2+y^2=a^2 &(x-a)^2+(y-2a)^2= 4a^2. This gives F as (-3a/5,4a/5). <BR/>Therefore, Tr. DEF =S1=(1/2)[a(a-4a/5)-(3a/5(-a)]<BR/> = 4a^2/10<BR/>But S = 2a*2a = 4a^2; hence S1 = S/10<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78934375613682482182008-06-20T22:12:00.000-07:002008-06-20T22:12:00.000-07:00Seehttp://homepage2.nifty.com/tangoh/tan080616.htm...See<BR/><BR/>http://homepage2.nifty.com/tangoh/tan080616.html<BR/><BR/>Nice problem!!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33672207101317702852008-06-06T01:41:00.000-07:002008-06-06T01:41:00.000-07:00You need additional points to prove this.G - the c...You need additional points to prove this.<BR/><BR/>G - the center of semicircle<BR/>Then AG = DG = EG = FG = ½√S<BR/>And FG is perpendicular to CF<BR/>So knowing that CG = ½√(5S) lets us compute CF = √S = AB = AD = BC = CD<BR/><BR/>Using analytic coordinates, we can use the equations for CF = √S and FG = ½√S to prove that F is 1/5 of the way from A to D and 2/5 the way from A to B. Alternately, we can add the point H which is the closest point on AD to F, and use the similar 1:2:√5 triangles to prove the same idea.<BR/><BR/>Thus DE = √(S/2), DF = √(4S/5), and EF = √(S/10) which gives us the area of S/10Richardhttps://www.blogger.com/profile/12170713115050940479noreply@blogger.com