tag:blogger.com,1999:blog-6933544261975483399.post4275213858762263862..comments2024-10-11T10:08:49.545-07:00Comments on GoGeometry.com (Problem Solutions): Geometry Problem 1438: Triangle, Incenter, Centroid, Circumcenter, Parallel, 90 Degree, Art, PosterAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-6933544261975483399.post-57208046563580475802019-06-09T21:27:19.125-07:002019-06-09T21:27:19.125-07:00
HiPeter,
Here is proof for b+c=2a
Let altitude A...<br />HiPeter, <br />Here is proof for b+c=2a<br />Let altitude AH to BC is equal to h, If we draw perpendicular to BC from G, then it is equal to (1/3)h, since IG is parallel to BC, perpendicular to BC from I is r=h/3, <br />r=S/s<br />h/3=(1/2)ah/s, we get 2s=3a or a+b+c=3a<br />Or b+c=2a. <br />Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36615730939198605142019-06-08T19:11:47.027-07:002019-06-08T19:11:47.027-07:00Thanks, Pradyumna.Thanks, Pradyumna.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4419560773720891002019-06-07T21:01:21.080-07:002019-06-07T21:01:21.080-07:00dear Pradyumna Agashe.
referring to line 3 of you...dear Pradyumna Agashe.<br /><br />referring to line 3 of your comment, I wonder how do you get b+c= 2a.<br />please explain<br />Peter Tran<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58410333496453088522019-06-07T17:03:20.971-07:002019-06-07T17:03:20.971-07:00Yes, thanks Pradyumna. AT =a/2; that is the crux o...Yes, thanks Pradyumna. AT =a/2; that is the crux of the matter. But where and how have we utilised the fact that IG//BCAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53498413772048107572019-06-07T11:38:44.055-07:002019-06-07T11:38:44.055-07:00Pradyumna indeed as I have mentioned above b+c = 2...Pradyumna indeed as I have mentioned above b+c = 2a but it must first be provedSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9513862664394983722019-06-07T11:36:37.429-07:002019-06-07T11:36:37.429-07:00Let AI meet BC at U and the circle at M.Let MO cut...Let AI meet BC at U and the circle at M.Let MO cut BC at P, the midpoint of BC<br /><br />Let BM = BI = p (since < MBI = < MIB = (A+B)/2)<br />Let AI = 2q so that IU = q since GP = AG/2 and IG//UP<br /><br />Now AB/BU = AI/IU = 2q/q = 2 so that BU = c/2<br /><br />Triangles MBU & ABM are similar so AM/BM = AB/BU hence (p+2q)/p = c/(c/2) = 2<br /><br />Therefore p + 2q = 2p and so p = 2q<br /><br />Hence AI = MI and I is the midpoint of AM and OI is perpendicular to AI<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86676332319704307672019-06-06T22:01:37.260-07:002019-06-06T22:01:37.260-07:00Adding to Ajit's proof,
Let touch point of inc...Adding to Ajit's proof,<br />Let touch point of incircle at at AC is T. <br />Then AT=s-a=(b+c-a)/2, since b+c=2a, AT=a/2<br />AT=BP=a/2,Angle IAT=Angle MBP=A/2, and Angle ATI =Angle BPM=90<br />Hence Triangle ATI is congruent to Triangle BPM. And AI=BM=CM=MI,<br />Hence I is midpoint of chord AM and Angle AIO=90. <br />Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4002099577673364642019-06-04T20:40:07.031-07:002019-06-04T20:40:07.031-07:00Sumith,
I'm afraid my proof is incomplete. Mus...Sumith,<br />I'm afraid my proof is incomplete. Must figure out a way to prove my assumption. I hope Antonio will help.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33315819184452054502019-06-04T14:43:16.913-07:002019-06-04T14:43:16.913-07:00In this triangle, 2a = b + cIn this triangle, 2a = b + cSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3069896428006497422019-06-04T14:39:53.629-07:002019-06-04T14:39:53.629-07:00Borrowing from Ajit’s proof
When we show that MP ...Borrowing from Ajit’s proof<br /><br />When we show that MP = r (not clear how this is so) AI = MI = BM<br /><br />So I is the mid point of chord AM and so OI is perpendicular to AM<br /><br />Ajit - I would like u to clarify how MP = rSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75112931021205252622019-06-03T01:51:23.485-07:002019-06-03T01:51:23.485-07:00Extend AI to meet the circle in M, the midpoint of...Extend AI to meet the circle in M, the midpoint of smaller arc BC. Further, Join M to O and produce it to meet the circumcircle in N. Let the in-circle touch AB in L. From M draw a perpendicular to BC meeting the latter in P. Now we can see that Tr. AIL///Tr. NBM; AI*MB=IL*NM=2r*R. We further observe that MP=r since GI//BC and thus r/MB=sin(A/2) as also r/AI=sin(A/2) which implies that AI=MB & therefore AI²=2r*R. We now recall by Euler's Formula : OI²=R² -2r*R or OI²=R²-AI² or AI²+OI²=R² or /_OIA =90° by the converse of Pythagoras Theorem.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com