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Sunday, September 17, 2017
Geometry Problem 1346: Equilateral Triangle, Point on the Incircle, Altitude, Perpendicular, Sum of Squares, Distance
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An attempt at the solution, but feel like not a complete one
ReplyDeleteJoin BD, AD and CD and consider the individual triangles, BDA, ADC, CDB and let the side of the eqilateral triangle be 'a'
S(BDA) = 0.5*a*d
S(ADC) = 0.5*a*e
S(CDB) = 0.5*a*f
As we know S(ABC) =S(BDA)+S(ADC)+S(CDB)
=> 0.5*a*h = 0.5*a*(d+e+f)
=> h = d+e+f
From the above, we can conclude that D is independent and can be any point on the incircle.
Assuming D on the altitude and at the tip of the incircle, we have
r = h/3 ( r-Radius of incircle)
f = 2r = 2h/3
d = e = h/6 ( Since BD = h/3 )
Therefore d^2+e^2+f^2 = h^2/36+h^2/36+4h^2/9 = 18h^2/36 = h^2/2
h = d+e+f holds true for any point inside the triangle
DeleteI too deduced that h = d+e+f using the same reasoning but could not proceed further
ReplyDeleteSo we need to prove that de+ef+fd = h^2/4 to arrive at the result.
Only issue is the above proof does not use the fact that D can be any point on the incircle
However I'm sure Antonio has a different straightforward proof. Antonio will u confirm do that we may continue to try?
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
ReplyDeleteLet 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
ReplyDeleteSince = S(ABC) = S(ABD) + S(ACD) + S(BCD), we infer easily that
h = d+e+f …(1)
Now O is also the centroid of ∆ABC,
so OX = h/3 = OD.
Also DY = f – h/3
It is also not difficult to see that,
OY= f/√3 + 2d/√3 - h/√3
Hence applying Pythagoras to ∆ODY,
h2/9 = (f-h/3)2 + (f+2d-h)2/3 which simplifies to
4(f2 + d2) + h2 = 4(dh + fh – df)…..(2)
Similarly we can show that
4(f2 + e2) + h2 = 4(eh + fh – ef)…..(3)
(2) + (3) gives us,
4(d2 + e2 + f2) + 4f2 + 2h2
= 4h(d+e+f) + 4f(h-d-e)
= 4h2 + 4f2 from (1)
Upon simplification,
d2 + e2 + f2 = h2/2
Sumith Peiris
Moratuwa
Sri Lanka