tag:blogger.com,1999:blog-6933544261975483399.post7445206590609433185..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 881: Triangle, three Squares, Centers, AreasAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-6933544261975483399.post-33479381736970555972013-05-30T07:12:34.054-07:002013-05-30T07:12:34.054-07:00Please read my solution.S, is midpoint of the ACPlease read my solution.S, is midpoint of the ACMichael Tsourakakishttps://www.blogger.com/profile/01337344627083642570noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88737286114466238742013-05-29T19:19:14.986-07:002013-05-29T19:19:14.986-07:00What is S?. Kindly clarify.
If you mean S is the p...What is S?. Kindly clarify.<br />If you mean S is the point where diagonals of the square MNPQ meet, it is trivial to see that SP = SN = SM.<br />If you mean S is the point where a diagonal, say MP, meets AC, you need to prove that S is the midpoint of the side AC of the triangle ABC.<br />Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86489438411662253992013-05-29T09:51:42.078-07:002013-05-29T09:51:42.078-07:00this is very easy.
Because MSP, is, diagonal of th...this is very easy.<br />Because MSP, is, diagonal of the square MNPQ ,and angle MSN=90 ,angle SMN=angle MNS = 45 ,angle MNP=90 ,then, angle SNP=45 ,so, angle NPS =45 .Therefore SP=SN=SM so,,S midpoint of the MP<br />Michael Tsourakakishttps://www.blogger.com/profile/01337344627083642570noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19252959326146273832013-05-29T00:01:40.250-07:002013-05-29T00:01:40.250-07:00Prove that the center of the square MNPQ coincides...Prove that the center of the square MNPQ coincides with the midpoint of AC.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20543883783615491622013-05-27T15:07:47.325-07:002013-05-27T15:07:47.325-07:00Problem 881: solution by Michael Tsourakakis from ...Problem 881: solution by Michael Tsourakakis from Greece at <br /><a href="http://www.gogeometry.com/school-college/p881-geometry-mixalis-greece.pdf" rel="nofollow">www.gogeometry.com/school-college/p881-geometry-mixalis-greece.pdf</a><br /><br />Thanks, <br />MixalisAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75536885693365939982013-05-24T22:51:22.529-07:002013-05-24T22:51:22.529-07:00BD=AB=sqrt(32), BF=BC=sqrt(46), MN=sqrt(67)
BM=sqr...BD=AB=sqrt(32), BF=BC=sqrt(46), MN=sqrt(67)<br />BM=sqrt(16)=4, BN=sqrt(23)<br />By cosine law, <br />MN^2 = BM^2 + BN^2 - 2*BM*BN*cosMBN<br />MN^2 = BM^2 + BN^2 - 2*BM*BN*cos(90+ABC)<br />MN^2 = BM^2 + BN^2 + 2*BM*BN*sinABC<br />MN^2 = BM^2 + BN^2 + 2*1/sqrt2*1/sqrt2*(AB*BC*sinABC)<br />67 - 16 - 23 = AB*BC*sinABC<br />28 = AB*BC*sinABC<br />14 = Area nakahttps://www.blogger.com/profile/11277356476170372732noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30032279671512165242013-05-24T19:35:37.404-07:002013-05-24T19:35:37.404-07:00Join BM, BN.
BM^2 = BD^2 / 2 = 16
BN^2 = BF^2 / ...Join BM, BN. <br /><br />BM^2 = BD^2 / 2 = 16<br />BN^2 = BF^2 / 2 = 23<br />MN^2 = 67<br /><br />MN^2 = BM^2 + BN^2 − 2×BM×BN×cos(∠B+90°)<br />67 = 16 + 23 + 2×BM×BN×sin∠B<br />BM×BN×sin∠B = 14<br /><br />S(ABC) = 1/2×BA×BC×sin∠B<br />= 1/2×(√2×BM)×(√2×BN)×sin∠B<br />= BM×BN×sin∠B<br />= 14Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56841917749334642372013-05-24T14:30:52.500-07:002013-05-24T14:30:52.500-07:00BM=4 , BN= SQRT(23), MN=SQRT(67)
In triangle MBN w...BM=4 , BN= SQRT(23), MN=SQRT(67)<br />In triangle MBN we have MN^2=MB^2+NB^2- 2.MB.NB.cos(B+90)<br />But cos(B+90)=- sin(B)<br />So sin(B)=(MN^2-MB^2-NB^2)/2.MB.NB<br />=(67-16-23)/(2 * 4 .*SQRT(23)= 7/(2 . SQRT(23))<br />Area (ABC)=1/2 . AB* BC* Sin(B) = 14<br />BM=4 , BN= SQRT(23), MN=SQRT(67)<br />In triangle MBN we have MN^2=MB^2+NB^2- 2.MB.NB.cos(B+90)<br />But cos(B+90)=- sin(B)<br />So sin(B)=(MN^2-MB^2-NB^2)/2.MB.NB<br />=(67-16-23)/(2 * 4 .*SQRT(23)= 7/(2 . SQRT(23))<br />Area (ABC)=1/2 . AB* BC* Sin(B) = 14<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com