## Sunday, September 17, 2017

### Geometry Problem 1346: Equilateral Triangle, Point on the Incircle, Altitude, Perpendicular, Sum of Squares, Distance

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. An attempt at the solution, but feel like not a complete one

Join 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(CDB) = 0.5*a*f

=> 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

1. h = d+e+f holds true for any point inside the triangle

2. I too deduced that h = d+e+f using the same reasoning but could not proceed further

So 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?

3. 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

4. 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

Since = 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