## Re: fermat point (more)

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• In Hyacinthos message #7961, I proved the following result by Antoine Verroken: If a point P lies in the interior of a circle M, and three lines intersecting
Message 1 of 4 , Jan 22, 2004
In Hyacinthos message #7961, I proved the following
result by Antoine Verroken:

If a point P lies in the interior of a circle M, and
three lines intersecting at P at angles of 60 degrees
meet the circle M at the points A, B, C, D, E, F such
that A and D lie on the first line, B and E on the
second line, and C and F on the third line, and these
points A, B, C, D, E, F form a convex hexagon, then
AP + CP + EP = BP + DP + FP.

In Hyacinthos message #7969, I conjectured that the sum
AP + CP + EP depends on the radius of M and the distance
from P to the center of M only, and not of the concrete
position of the three lines through P. I asked if
anybody could prove this. In fact, nobody could or can:
the conjecture turned out to be wrong!

Since the falseness of the conjecture can hardly be
observed on any geometric sketch, here is an algebraic
way to disprove it:

If XY is any chord of the circle M such that XY passes
through P, and O and R are the center and the radius of
M, then

XY = sqrt(R^2 - OP^2 * sin^2 phi)/2,

where phi is the angle between OP and XY.

Now, if t is the angle between OP and AD, then t + 60 is
the angle between OP and BE, and t - 60 is the angle
between OP and CF. [Here, 60 means 60 degrees.]

Hence,

AD + BE + CF = sqrt(R^2 - OP^2 * sin^2 t)/2
+ sqrt(R^2 - OP^2 * sin^2 (t + 60))/2
+ sqrt(R^2 - OP^2 * sin^2 (t - 60))/2.

Since AP + CP + EP = BP + DP + FP and obviously
AP + CP + EP + BP + DP + FP = AD + BE + CF, we have
AP + CP + EP = (AD + BE + CF)/2, and

AP + CP + EP = 1/4 * (sqrt(R^2 - OP^2 * sin^2 t)
+ sqrt(R^2 - OP^2 * sin^2 (t + 60))
+ sqrt(R^2 - OP^2 * sin^2 (t - 60))).

If we abbreviate x = R^2 / OP^2, then

AP + CP + EP = OP/4 * (sqrt(x - sin^2 t)
+ sqrt(x - sin^2 (t + 60))
+ sqrt(x - sin^2 (t - 60))).

Now, the conjecture we wanted to disprove stated that
this term is independent of t. It is enough to show that

sqrt(x - sin^2 t) + sqrt(x - sin^2 (t + 60))
+ sqrt(x - sin^2 (t - 60))

depends on t.

The number x can take any value >1; for instance, we can
take x = 50. Then, for t = 10 degrees, the above sum is
21.1067364..., while for t = 30 degrees, it is
21.1067359... (calculated by Maple). Qed..

Darij Grinberg
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