tag:blogger.com,1999:blog-6933544261975483399.post7437855853602980621..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 459: Right triangle, Squares, Distance, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-30557987063643639882021-01-25T10:46:19.218-08:002021-01-25T10:46:19.218-08:00Let AC=b. Extend DC to P such that C is the mid-po...Let AC=b. Extend DC to P such that C is the mid-point i.e. DC=CP=b. Connect FG to P and observe that FCP is congruent to CBA (SAS). Hence FP=c. <br />Construct a line from D parallel to CF such that it meets GF extended at Q. <br />Using mid-point theorem for the right tringle DPQ, it can be easily seen that DQ=2.CF=2a <br />FQ=FP=c. <br />As DQG is a right triangle with DQ=2a, GQ=GF+FQ=a+c,=> x^2=(2a)^2+(a+c)^2<br />The result follows. <br />Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25500987800268887542021-01-24T10:22:50.184-08:002021-01-24T10:22:50.184-08:00Extend GF to H such that FH = a
Then Tr.s DCG and...Extend GF to H such that FH = a<br /><br />Then Tr.s DCG and ACH are congruent SAS and x^2 = AH^2 = (a+c)^2 + 4a^2 from which the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaAnonymoushttps://www.blogger.com/profile/15352351748882581248noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-29994218115801660462017-03-13T17:13:08.658-07:002017-03-13T17:13:08.658-07:00By boxing the largest square ACDE its possible to ...By boxing the largest square ACDE its possible to make a new right triangle with x as its hypotenuse that's easy to compute:<br /><br />First rotate and translate triangle ABC so that its hypotenuse lies on CD and its new vertex is H. <br />Then extend the DH side so that it meets FG at I forming a new right triangle.<br /><br />DI = a + a <br />GI = a + c<br />x^2 = DI^2 + GI^2 = (2a)^2 (a+c)^2 = 5a^2 + 2ac + c^2<br /><br />Benjamin Leishttps://www.blogger.com/profile/10974191081762367425noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-29821208839245041242010-06-28T22:50:10.967-07:002010-06-28T22:50:10.967-07:00@ third comment by Anonymous:
CG will never be equ...@ third comment by Anonymous:<br />CG will never be equal to CF as CG is the diagonal of the square and CF the side<br />Pls correct this commentAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41316356051307208962010-06-04T21:58:30.265-07:002010-06-04T21:58:30.265-07:00Let H point on line GF, and HF=GF. Then CG=CF and ...Let H point on line GF, and HF=GF. Then CG=CF and angle GCH=90.<br /><br />On triangle CDG and triangle CAH, CD=CA and CG=CH.<br />angle DCG=90+angle DCH, angle ACH=90+angle DCH.<br /><br />So, triangle CDG==triangle CAH ---> then, AH=DG=x<br /><br />On triangle AGH, angle G=90, AG=a+c, GH=2a, AH=x.<br /><br />Therefore, x^2=(a+c)^2+(2a)^2 ---> x^2 = 5a^2+c^2+2acAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35391856792331076052010-05-27T13:42:53.899-07:002010-05-27T13:42:53.899-07:00Let m projection of AB over AC.
Let n projection o...Let m projection of AB over AC.<br />Let n projection of BC over AC.<br />Let h altitude over hypotenuse AC.<br />Then (T.altitude) h = sqrt (mn)<br />Then (T.legs) m = c^2/sqrt(a^2+c^2)<br /> n = a^2/sqrt(a^2+c^2) <br />Let H point of intersection of parallel to AC from G and extension of side DC from C. Then GHD is right triangle on H.<br />We've the next relations: <br />CD = AC = m + n = sqrt(a^2 +c^2) (Pt. Tri.ABC)<br />CH = h + n<br />GH = h - n<br />HD = CH + CD = h + 2n + m<br />Then:<br />x^2=GH^2 + HD^2 = (h-n)^2 + (h + 2n + m)^2 =<br />= ... = 6mn + 2*sqrt(mn)*(m+n) + 5n^2 + m^2 = ... =<br />= 5a^2 + 2ac + c^2<br /><br />MIGUE.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33027079094204230492010-05-21T18:45:19.356-07:002010-05-21T18:45:19.356-07:00cos(GAD)= cos(GAC+CAD)=cos(GAC+45)=cos(GAC)cos(45)...cos(GAD)= cos(GAC+CAD)=cos(GAC+45)=cos(GAC)cos(45)-sin(GAC)sin(45). But cos(GAC)=c/b & sin(GAC)=a/b hence cos(GAD)=(c-a)/(b√2). From Tr. GAD using the co-sine rule, we’ve: x^2 = GA^2 +AD^2 -2GA*AD*cos(GAD) = (c+a)^2+2b^2-2*(c+a)( b√2)(c-a)/(b√2)= (c+a)^2 + 2(c^2+a^2) -2*(c+a)(c-a)= c^2 + 2ac + a^2 + 4a^2 = 5a^2 + c^2 + 2ac<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com