tag:blogger.com,1999:blog-6933544261975483399.post3050221008041381548..comments2023-12-05T12:24:14.189-08:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1230: Quadrilateral, 60 Degrees, Congruence, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-6933544261975483399.post-39941298350278745472016-11-10T09:08:47.909-08:002016-11-10T09:08:47.909-08:00Extend AB to meet DC at P and form the equlateral ...Extend AB to meet DC at P and form the equlateral triangle APD<br />Since given AD = 3a and AB = a<br />we have BP = 2a, PC = a, CD = 2a and since m(BPC) = 60<br />=> BPC is a 30-60-90 triangle <br />Hence BC = √3a<br /><br />Since m(BCP) = 90 => m(BCD) = 90 (Hence BC _|_ CD) <br />Applying pythogrous to triangle BCD <br />=> BD^2 = BC^2+CD^2 <br />=> BD^2 = 3a^2+4a^2<br />=> BD = √7a = AC<br /><br />Observe that the triangles PAC and ADB are congruent per SSA (since PA = AD, AC=BD and m(APC) = m(BAD) = 60)<br />Let m(ADB)=m(PAC)= α<br />=> m(PCA) = 120-α => m(BCA) = 30-α<br />and m(BDC) = 60-α => m(DBC) = 30+α (Corrected this from m(DBA) to m(DBC))<br />∴ in the triangle BEC, m(BEA) = 120<br />=> m(BEA) = 60<br />So triangle EAB is similar to ADB <br />∴ AB/BE = BD/AB<br />=> BE = a^2/√7a<br />Hence BE = a/√7<br /><br />Similarly AE/BE = AD/AB = a/3a = 1/3<br />Hence AE = 3BE<br /><br />Finally, Area of Equilateral triangle APD = √3(9a^2)/4<br />Area of right triangle PBC = √3(a^2)/2<br />∴ Area ABCD = Area APD - Area PBC = √3(a^2)[9/4-1/2] = 7√3(a^2)/4Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11586738399768777052016-11-06T07:15:07.225-08:002016-11-06T07:15:07.225-08:00No need of Trigonometry to prove CD = 2a.
Since A...No need of Trigonometry to prove CD = 2a.<br /><br />Since AC = AF, triangles AKC and ABD are congruent ASA.<br />Hence KC = AB = a. <br />So CD = 2aSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-84996308889027152152016-09-03T17:26:55.467-07:002016-09-03T17:26:55.467-07:00whoops, at the end, I meant to say "AE has le...whoops, at the end, I meant to say "AE has length of 3*a/sqrt(7)" which is really thrice of BE.Anonymoushttps://www.blogger.com/profile/14379158334268338516noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79533530027191174442016-09-03T00:29:03.769-07:002016-09-03T00:29:03.769-07:00I said in an earlier post that F has mass 3, D is ...I said in an earlier post that F has mass 3, D is mass 1 and A is mass 3. This indicates that FD has total mass of 3.<br />So, mass ratio A:C = 4:3, which means that AE:EC = 3:4<br />Hence, AE is 3/7 of AC.<br /><br />I also found in that other post that AC is sqrt(7) a. This means at AE has length of 3*sqrt(7)*a, which is thrice of BE.Anonymoushttps://www.blogger.com/profile/14379158334268338516noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56244358051648066752016-09-03T00:21:12.586-07:002016-09-03T00:21:12.586-07:00Oops, you used trigonometry. Ah well.Oops, you used trigonometry. Ah well.Anonymoushttps://www.blogger.com/profile/14379158334268338516noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57342548582999062802016-09-03T00:15:25.993-07:002016-09-03T00:15:25.993-07:00Make F such that AFD is equilateral (true as angle...Make F such that AFD is equilateral (true as angle BAD and ADC are 60 degrees).<br />Make H on FD such that AH = BD, BH is parallel to AD. (Feasible as triangle AFD is equilateral which are symmetrical about their altitudes).<br />Make G on FD such that AG is perpendicular to FD. As AFD and ACH show properties of isosceles triangles, G is bisector of CH and FD.<br />Hence, CD = 2a.<br /><br />Mass points: Let E be the center of mass if fixed masses are placed on A, F, D (lines are frames which are strong and massless).<br />Let D have mass 1. As E is center of mass, B (which is concurrent with E and D) must be center of mass of AF. Similarly, C is center of mass of FD.<br />As CD = 2a, FC must be a, so the mass on F must be 2.<br />As AB = a, BF must be 2a and therefore the mass on A is 4.<br />So the total mass of AF is 6. This means that mass ratio of B:D is 6:1, so the length ratio of BE:ED = 6:1<br />This indicates that BE has 1/7 of length of BD.<br /><br />BD = AC, AC^2 = AG^2 + CG^2 = AD^2 - GD^2 + CG^2 = 7*a^2<br />Hence BD = sqrt(7) a<br />That means that BE = a/sqrt(7)Anonymoushttps://www.blogger.com/profile/14379158334268338516noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10950272001916812412016-07-31T01:09:34.529-07:002016-07-31T01:09:34.529-07:00Note that
- triangle AKD is equilateral
- FD= a
- ...Note that<br />- triangle AKD is equilateral<br />- FD= a<br />- triangle FAC is isosceles ( FA=BD=AC) <br />from above information we can calculate CD= 2a Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11715279363536802672016-07-29T05:25:05.708-07:002016-07-29T05:25:05.708-07:00There is no explanation why cd=2aThere is no explanation why cd=2aAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57925151341731859172016-07-06T17:54:57.869-07:002016-07-06T17:54:57.869-07:00https://goo.gl/photos/baBUwbeKwDLUEN1V9
(1) KBF ...https://goo.gl/photos/baBUwbeKwDLUEN1V9<br /><br />(1) KBF is equilateral and AC=AF=BF= a√7<br />Apply Van Aubel theorem for point E in triangle KAD<br />AE/EC=BA/BK+XA/XD= ¾<br />So AE/3=EC/4=AC/7=a/√7<br />And AE= 3.a/√7 => BE/AE=1/3<br />(2) .. In equilateral triangle KBF, the median BC perpendicular to KF<br />(3)…BC= KB. √3/2= a√3<br />(4).. S( ABCD)=S(KAD)- S(KBC)<br />S(KAD)= AD^2. √3/4<br />S(KBC)=SB^2. √3/8<br />Replace AD= 3.a and SB=2.a we get S(ABCD)=7/4. a^2. √3<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28969031016012817592016-07-05T21:01:01.116-07:002016-07-05T21:01:01.116-07:00https://goo.gl/photos/TSKinFA2FzwTcMNU6
Let AB me...https://goo.gl/photos/TSKinFA2FzwTcMNU6<br /><br />Let AB meet DC at K<br />AKD is an equilateral triangle <br />Draw BF // AD and let KE meet AD at X<br />We have KB=2.a and CD=2.a<br />We have BD^2= a^2+9a^2- 2.3a.ạ cos(60)= 7.a^2 => BD= a√7<br />Apply Ceva’s theorem in triangle AKD we have<br />BA/BK x CK/CD x XD/XA = 1 => XD/XA= 4<br />We also have ED/EB= XD/XA + CD/CK= 4+ 2= 6<br />So ED/6=EB/1=BD/7 => EB= BD/7= a/√7<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com