tag:blogger.com,1999:blog-6933544261975483399.post4168442644195959822..comments2022-09-27T03:11:10.165-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1346: Equilateral Triangle, Point on the Incircle, Altitude, Perpendicular, Sum of Squares, DistanceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-20830853015590535652017-09-26T07:12:17.885-07:002017-09-26T07:12:17.885-07:00Let O be the centre of the incircle. Let BO meet A...Let O be the centre of the incircle. Let BO meet AC at X and let Y be the foot of the perpendicular from O to DG<br /> <br />Since = S(ABC) = S(ABD) + S(ACD) + S(BCD), we infer easily that<br />h = d+e+f …(1)<br /> <br />Now O is also the centroid of ∆ABC, <br />so OX = h/3 = OD.<br />Also DY = f – h/3<br />It is also not difficult to see that,<br />OY= f/√3 + 2d/√3 - h/√3<br /> <br />Hence applying Pythagoras to ∆ODY,<br /> h2/9 = (f-h/3)2 + (f+2d-h)2/3 which simplifies to<br /> <br />4(f2 + d2) + h2 = 4(dh + fh – df)…..(2)<br /> <br />Similarly we can show that<br />4(f2 + e2) + h2 = 4(eh + fh – ef)…..(3)<br /> <br />(2) + (3) gives us,<br /> <br />4(d2 + e2 + f2) + 4f2 + 2h2 <br />= 4h(d+e+f) + 4f(h-d-e)<br /> = 4h2 + 4f2 from (1)<br /> <br />Upon simplification,<br /> <br />d2 + e2 + f2 = h2/2<br /> <br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76085563220553682362017-09-19T05:36:07.694-07:002017-09-19T05:36:07.694-07:00h = d+e+f holds true for any point inside the tria...h = d+e+f holds true for any point inside the triangleAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-22349800917402486062017-09-18T23:22:10.083-07:002017-09-18T23:22:10.083-07:00In other words h = d+e+f is true for any point ins...In other words h = d+e+f is true for any point inside the equilateral triangle not just for a point on the in circle Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77092355967110370712017-09-18T23:08:42.396-07:002017-09-18T23:08:42.396-07:00I too deduced that h = d+e+f using the same reason...I too deduced that h = d+e+f using the same reasoning but could not proceed further<br /><br />So we need to prove that de+ef+fd = h^2/4 to arrive at the result. <br /><br />Only issue is the above proof does not use the fact that D can be any point on the incircle<br /><br />However I&#39;m sure Antonio has a different straightforward proof. Antonio will u confirm do that we may continue to try?Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60816524516543019742017-09-18T07:36:32.436-07:002017-09-18T07:36:32.436-07:00An attempt at the solution, but feel like not a co...An attempt at the solution, but feel like not a complete one<br /><br />Join BD, AD and CD and consider the individual triangles, BDA, ADC, CDB and let the side of the eqilateral triangle be &#39;a&#39;<br />S(BDA) = 0.5*a*d<br />S(ADC) = 0.5*a*e<br />S(CDB) = 0.5*a*f<br /><br />As we know S(ABC) =S(BDA)+S(ADC)+S(CDB)<br />=&gt; 0.5*a*h = 0.5*a*(d+e+f)<br />=&gt; h = d+e+f<br /><br />From the above, we can conclude that D is independent and can be any point on the incircle.<br />Assuming D on the altitude and at the tip of the incircle, we have<br />r = h/3 ( r-Radius of incircle)<br />f = 2r = 2h/3<br />d = e = h/6 ( Since BD = h/3 )<br /><br />Therefore d^2+e^2+f^2 = h^2/36+h^2/36+4h^2/9 = 18h^2/36 = h^2/2Anonymousnoreply@blogger.com