# A circle intersectsWatch

#1
A circle intersects an equilateral triangle ABC in
six points, D, E, F, G, H, J. In traversing the perimeter
of the triangle, these points occur in the order A,
D, E, B, F, G, C, H, J. Prove that AD+BF +CH =
AJ + BE + CG.
0
10 years ago
#2
The best I can do is to say:

If you assume the circle and the triangle share the same centre, AD = AJ, BF = BE and CH = CG, because the combined shape is symmetrical in 3 axes (the same as the axes of symmetry of the triangle). Therefore AD+BF+CH = AJ+BE +CG

If you displace the circle in either of these 3 axes, this still holds true. For example, if you displace it towards the point A, then AD = AJ, BF = CG, BE = CH. This is because the whole shape still has one axis of symmetry.

The entire plane we're working in can be defined by these three axes. So shifting the circle around in this place is just the same as displacing it along the axes.

Ok, so it' not that good, but i dont really know what else to do
0
10 years ago
#3
(Original post by drmath)
A circle intersects an equilateral triangle ABC in
six points, D, E, F, G, H, J. In traversing the perimeter
of the triangle, these points occur in the order A,
D, E, B, F, G, C, H, J. Prove that AD+BF +CH =
AJ + BE + CG.
Let AD = a, DE = x, EB = b*, BF = b, FG = y, GC = c*, CH = c, HJ = z and JA = a*.
We know that a + x + b* = b + y + c* = c + z + a* = L, say.
Circle properties give us a[a + x] = a*[a* + z] so that a[L - b*] = a*[L - c]. Similarly b[L - c*] = b*[L - a] and c[L - a*] = c*[L - b].
Add these last 3 identities to get L[a + b + c] = L[a* + b* + c*] so that
a + b + c = a* + b* + c*, the required result.
0
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