tag:blogger.com,1999:blog-6933544261975483399.post6446133669196965045..comments2024-03-19T00:02:30.728-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 65Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-86927154074228785482018-12-18T17:06:22.177-08:002018-12-18T17:06:22.177-08:00In the spirit of Sumith but doing it from the insi...In the spirit of Sumith but doing it from the inside:<br />Draw l1 and l2 respectively // to AB and BC passing through M.<br />Draw l3 and l4 // to DE respectively passing through C and A.<br />See graph here : https://drive.google.com/file/d/1BZMCKpBag3aYhfvj69zVX2-9bNxG4kKc/view?usp=sharing<br /><br />By construction, MF//BC and MG//AB : ΔFMG is rectangle in M.<br />By construction, ADMG and MECF are //ograms so AG=DM=ME=CF.<br />Look at AFCG : AG//CF and AG=CF so AC and FG intersect in their middle which is N.<br />So in ΔFMG rectangle in M, N is middle of FG, and FM=EC=e and MG=AD=d so d^2+e^2=(2*MN)^2 so 1/2*Sqr(d^2+e^2)=x. QED. <br />Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27140287073959169412015-10-04T10:13:43.673-07:002015-10-04T10:13:43.673-07:00Variation of this problem
If B = 60 and d = e the...Variation of this problem<br /><br />If B = 60 and d = e then d = 2xSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46369716796983413632015-10-04T10:04:35.974-07:002015-10-04T10:04:35.974-07:00Extend AM to P such that AM = MP so ADPE is a pare...Extend AM to P such that AM = MP so ADPE is a parellelogram and Tr. PEC is a right Tr. with EP = d<br /><br />Now extend MN to Q such that MN = NQ = x. So AMCQ is a //ogram with QS = AM = MP. Hence MPCQ is a //ogram and PC = MQ = 2x.<br /><br />Now apply Pythagoras to right Tr. CEP and d^2 + e^2 = 4x^2 and so the result follows. <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47015353997156776502009-09-24T00:43:05.341-07:002009-09-24T00:43:05.341-07:00I like the geometric solution Joe so I got this so...I like the geometric solution Joe so I got this solution : Let K and L be the orthogonal projections of A and C on (DE), take S the midpoint of [KL] such that SM=p now KBEA and DBLC are concyclic so the two triangles AKD and CLE are similar then LC*KA=LE*KD-------(1)<br />now p=(KM-LM)/2 and the median NS=(CL+AK)/2<br />Now x^2=(NS)^2+P^2<br /> =(KM^2-2KM*LM+LM^2+CL^2+2CL*AK+AK^2)/4<br /> Let DM=ME=y<br />so x^2=((KD+y)^2-2(KD+y)(LE+y)+(LE+y)^2+CL^2+2CL*AK+AK^2)/4<br /> = (CL^2+AK^2+KD^2+2yKD+y^2+LE^2+2yLE+y^2-2KD*LE-2yKD-2yLE-2y^2+2CL*AK)/4<br /> =(CL^2+AK^2+KD^2+LE^2-2KD*LE+CL*AK)/4<br /> =(e^2+d^2)/4 (from (1)<br /> but this solution is long Is there shorter Antonio????? why there is no post for problem 358 the solution is already prepared and I have suggest for you <br />make an area in your site to our problems if we have because I have many problems didn't mention in this with these problems thx a lotTouskanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-54661851092003069482009-03-25T05:02:00.000-07:002009-03-25T05:02:00.000-07:00Let B be (0,0) BE the x-axis & BA the y-axis. ...Let B be (0,0) BE the x-axis & BA the y-axis. Further, let BE=p and BD=q. This gives, E:(p,0), C:(p+e,0),D:(0,q), A:(0,d+q). Thus N:((p+e)/2,(q+d)/2)and M;(p/2,q/2). By the distance formula, MN^2 = x^2 = ((p+e)/2 -p/2)^2+((q+d)/2 -q/2)^2 = e^2/4 + d&2/4 0r x^2 =(1/4)*(d^2+e^2)<BR/>or x=(1/2)*(d^2+e^2)^(1/2)<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com