tag:blogger.com,1999:blog-6933544261975483399.post457038068051463866..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1261: Cyclic Quadrilateral, Circle, Diameter, Midpoint, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger16125tag:blogger.com,1999:blog-6933544261975483399.post-57104107902533540132019-03-01T09:51:28.566-08:002019-03-01T09:51:28.566-08:00Ok thanksOk thanksc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67846249667847117812019-03-01T05:22:52.198-08:002019-03-01T05:22:52.198-08:00From: Sumith Peiris. March 1, 2019 at 3:07 AM
To c...From: Sumith Peiris. March 1, 2019 at 3:07 AM<br />To c.t.e.o<br /><br />Problem 1261<br /><br />Mark X on AB such that BX = BC<br />Mark Y on AD such that DC = DY<br /><br />Use Pythagoras and show CX = AY and AX = CY that is AXCD is a parallelogram.<br />X mid point of AB so C midpoint of BEAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4205689673085603982019-02-08T07:43:24.281-08:002019-02-08T07:43:24.281-08:00Draw EP ꓕ AC, from similarity CP = a, PE = 2a, AC ...Draw EP ꓕ AC, from similarity CP = a, PE = 2a, AC = 5a<br />From ∆ABC => a = x/√5, from ∆PEC => EC = xc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30065022452737061472019-02-03T09:09:42.029-08:002019-02-03T09:09:42.029-08:00Let CD = x and BC = y
then AD = 3x and AB = 2y
(...Let CD = x and BC = y <br />then AD = 3x and AB = 2y<br /><br />(2y)^2 + y^2 = (3x)^2 + x^2<br />5y^2 = 10x^2<br />y = sqrt(2)x<br /><br />Let CE = t and DE = p<br />then x^2 + p^2 = t^2<br />t = sqrt(x^2 + p^2)<br /><br /><br />(2y)^2 + (y + t)^2 = (3x + p)^2<br />4y^2 + y^2 + 2yt + t^2 = 9x^2 + 6xp + p^2<br /><br />4(sqrt(2)x)^2 + (sqrt(2)x)^2 + 2(sqrt(2)x)(sqrt(x^2 + p^2)) + (sqrt(x^2 + p^2))^2 = 9x^2 + 6xp + p^2<br />8x^2 + 2x^2 + 2x(sqrt(2))(sqrt(x^2 + p^2)) + x^2 + p^2 = 9x^2 + 6xp + p^2<br />2x(sqrt(2))(sqrt(x^2 + p^2)) = 6xp - 2x^2<br />(sqrt(2x^2 + 2p^2)) = 2x(3p - x)/2x<br />2x^2 + 2p^2 = 9p^2 - 6xp + x^2<br />x^2 + 6xp - 7p^2 = 0 <br />x^2 + 7xp - xp - 7p^2 = 0<br />x(x + 7p) - p(x + 7p) = 0<br />(x - p)(x + 7p) = 0<br />x - p = 0<br />x = p<br />CD = DE<br /><DCE = <DEC = 45 (Isosceles tri theorem)<br /><DEC = 45 = <EAB (Sum of angles in a tri)<br />BA = BE<br />But BA = 2BC<br />thus BE = 2BC<br />BE = BC + CE<br />2BC = BC + CE <br />BC = CE <br />therefore C is the midpoint of BE<br /><br /><br /><br /><br /><br /><br /> <br /><br /><br /><br /><br /><br /><br /><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31024051765962136412018-11-24T18:55:31.617-08:002018-11-24T18:55:31.617-08:00Let BC = x
then AB = 2x
AC^2 = x^2 + (2x)^2
AC^2...Let BC = x <br />then AB = 2x<br /><br />AC^2 = x^2 + (2x)^2<br />AC^2 = x^2 + 4x^2<br />AC = sqrt(5)x<br /><br />Let CD = y<br />then AD = 3y<br />AC^2 = (3y)^2 + y^2<br />AC^2 = 9y^2 + y^2 <br />AC = sqrt(10)y<br /><br />sqrt(5)x = sqrt(10)y<br /> y = sqrt(5)x/sqrt(10)<br /> = sqrt(5)x*sqrt(10)/sqrt(10)*sqrt(10)<br /> = sqrt(50)x/10<br /> = 5*sqrt(2)x/10<br /> = sqrt(2)x/2<br /><br />Let CE = b<br />then y^2 + DE^2 = b^2<br /> DE = sqrt(b^2 - y^2)<br /><br />In tri CDE and tri ABE<br /><ABC = <CDE (diameter AC)<br /><AEB = <DEC (common angle)<br /><BAC = <DCE (remaining angles of a tri)<br />therefore triangle ABE and triangle CDE are equiangular triangles<br />BE/DE = AB/CD<br />(x + b)/sqrt(b^2 - y^2) = 2x/x*sqrt(2)/2 = sqrt(8)<br />x+b = sqrt(8)*sqrt{b^2 - (x*sqrt(2)/2)^2}<br />(x+b)^2 = 8(b^2 - x^2/2)<br />x^2 + 2xb + b^2 -8b^2 +4x^2 = 0<br />5x^2 + 2xb - 7b^2 = 0 <br />5x^2 + 7xb -5xb - 7b^2 = 0<br />x(5x+7b) -b(5x+7b) = 0<br />(x-b)(5x+7b) = 0<br />x-b = 0<br />x = b<br />CE = b = x<br />CE = BC = x<br />therefore C is the midpoint of BE<br /> <br /> <br /><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74577574678658477982016-09-19T13:39:18.011-07:002016-09-19T13:39:18.011-07:00Let CD = 1
=> AD = 3
=> AC^2 = CD^2 + AD^2...Let CD = 1 <br />=> AD = 3 <br />=> AC^2 = CD^2 + AD^2 = Sqrt(10)<br /><br />Similarly solving for AB & AC in the right triangle ABC, <br />we get BC = sqrt(2) --------(1)<br />and AB = 2*Sqrt(2) <br /><br />Let CE = x and DE = y<br />As CE^2 = DE^2 + CD^2<br />=> 1+y^2 = x^2 ----------- (2)<br /><br />As ED.EA = EC.EB (From Secant-Secant Rule)<br />=> y.(3+y) = x.(x+Sqrt(2))<br />=> y^2+3y = x^2+Sqrt(2).x<br />=> 3y = (x^2-y^2) + Sqrt(2).x<br />=> 3y = 1 + Sqrt(2).x (from (1) )<br />=> y = [1+Sqrt(2).x]/3 ----------(3)<br /><br />Substituting the value of y in Eq(2)<br /><br />=> 9 + (1+2x+2Sqrt(2)x) = 9x^2<br />=> 7x^2 - 2Sqrt(2)x -10 = 0 <br /><br />Therefore x = (2Sqrt(2) + Sqrt(8+288))/14 = 14.Sqrt(2)/14 = Sqrt(2) = BC (from (1) )<br /><br />Hence C is midpoint of BE<br /> Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46399912926047989492016-09-18T19:21:40.337-07:002016-09-18T19:21:40.337-07:00Good pure geometry solutionGood pure geometry solutionSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27816777143277587472016-09-18T17:29:18.169-07:002016-09-18T17:29:18.169-07:00Thank u. Good pure geometry proofThank u. Good pure geometry proofSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47582826130369861062016-09-18T15:50:24.575-07:002016-09-18T15:50:24.575-07:00Since AB=2BC=> BC=BE=AE ( see sketch)
Triangles...Since AB=2BC=> BC=BE=AE ( see sketch)<br />Triangles AFE, EGB and BGC are 45-45-90 triangles and are congruent ( case ASA)<br />so FA=FE=EG=GC => AF/FC= 1/3Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-49691812472607101282016-09-18T10:08:00.959-07:002016-09-18T10:08:00.959-07:00Palai Renato,
Reply to Sumith Peiris to request t...Palai Renato,<br />Reply to Sumith Peiris to request to motivate why angle.. is 45°.<br />If we calla <BAC= and <CAD= , <BAD= + , bat<br />tang ( + )=(tang + tang )/(1 - tang tang ) and introducing the values of and , we have: (2+3)//1-2x3)= -1, tang 1 = 45°<br />Reply<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16004603139315133882016-09-18T07:33:46.721-07:002016-09-18T07:33:46.721-07:00Problem 1261
Is <BAC+<CAD=45, so <CED=...Problem 1261<br />Is <BAC+<CAD=45, so <CED=45 then CD=DE. Let the points K, L on the side AD such that<br />AK=KL=LD=AD/3=DC=DE. So AL=LE and BL is perpendicular at AE so BL//CD. Therefore <br />BC=CE (LD=DE).<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37128746312533184502016-09-17T18:02:14.732-07:002016-09-17T18:02:14.732-07:00ABCD is cyclic, with AC being diameter. Hence, Ang...ABCD is cyclic, with AC being diameter. Hence, Angle ABC and Angle ADC are right angles, by Thales' theorem.<br /><br />By pythagoras’ theorem, it’s evident that BC = AC/sqrt(5), AB = 2AC/sqrt(5), CD = AC/sqrt(10), AD = 3AC/sqrt(10). We need the sine and cosines of angle BAC, angle CAD.<br />If we let O be the circumcenter of ABCD, then angle BOC = 2 angle BAC, angle COD = 2 angle CAD. This is obvious high school geometry.<br /><br />From here, we use complex numbers.<br />Let the circle on ABCD be the unit circle and c = 1, a = -1.<br />With a bit of trig (especially double angle formula), b = cis(angle BOC) = cis(2 angle BAC) = (3/5) + i(4/5). Similarly, d = cis(-angle COD) = (4/5) – i(3/5).<br /><br />Now let’s make a point F such that C is the midpoint of B and F. This indicates (f + b)/2 = c, which rearranges to f = 2c – b. We compute f = (4/5) – i(4/5).<br /><br />We now assert that A, D and F are collinear. This is proved by checking that (a – f)/(a – d) is a real number (justification is De Moivre theorem; (a - d) multiplied by a real number would make (a-f) iff a-f and a-d make an angle of 0 or 180 degrees with each other). <br /><br />(a – f)/(a – d) = (12/5 – i(4/5))/((-9/5) – i(3/5)) = (4/5) / (-3/5) = -4/3 which is definitely a real number. So we conclude that F is on line AD and BC (as C is midpoint of BF).<br /><br />But we know that E is the intersection of AD and BC. This means E = F, and the proof is complete.<br />Anonymoushttps://www.blogger.com/profile/14379158334268338516noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-6888221422368476902016-09-17T17:07:10.129-07:002016-09-17T17:07:10.129-07:00Dear Peter - How is AF/FC = CD/AD?Dear Peter - How is AF/FC = CD/AD?Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75995685459785462372016-09-17T17:04:19.363-07:002016-09-17T17:04:19.363-07:00How is < BAC = 45?How is < BAC = 45?Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31830408452794311642016-09-17T12:48:16.891-07:002016-09-17T12:48:16.891-07:00Palai Renato
We put AB=1, BC = 2 AC= √5,
We put ...Palai Renato<br />We put AB=1, BC = 2 AC= √5, <br />We put CD=x we have x+(3x)2=√5, and therefore x= 1/√2<br /><BAC =45 (regle of sum of tangent), CD=DE and CE=1<br />CE=CB and C is midpoint of BE<br />Reply<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10695455594642692972016-09-17T10:03:45.812-07:002016-09-17T10:03:45.812-07:00https://goo.gl/photos/tDgKutmrL4Bn9Lur8
Draw poin...https://goo.gl/photos/tDgKutmrL4Bn9Lur8<br /><br />Draw points O, E, F, G per sketch<br />Note that OE ⊥AB , AF⊥ EF and triangles EBC, AFE are 45-45- 90 triangles <br />We also have AF/FC= CD/AD= 1/3 => EC//AE<br />And BE/BA=BC/BE= 1/2<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com