Let P be a point in triangle ABC such that angles APB, BPC, and CPA are all 120 degrees. Can lines AB, AC, BC, PA, PB, and PC all have integral lengths?
Call the sides opposite angles A, B and C, a, b and c. Call PB length d; PC, length e and PA, length F.
a^2 = d^2 + e^2 - 2de cos 120, or
a^2 = d^2 + e^2 - 2de (-.5)
a^2 = d^2 + e^2 + de
and similarly
b^2 = e^2 + f^2 + ef
c^2 = d^2 + f^2 + df
The following program goes through all possible combinations of d, e and f, with d<=e<=f in order of increasing total d+e+f up to 1,000,000 or until stopped.
DEFDBL A-Z
FOR t = 1 TO 1000000
FOR d = 1 TO t / 3
FOR e = d TO (t - d) / 2
f = t - d - e
asq = d * d + e * e + d * e
bsq = e * e + f * f + e * f
csq = d * d + f * f + d * f
a = INT(SQR(asq) + .5)
b = INT(SQR(bsq) + .5)
c = INT(SQR(csq) + .5)
IF a * a = asq AND b * b = bsq AND c * c = csq THEN
PRINT d; e; f; a; b; c
END IF
NEXT
NEXT
NEXT
It finds
d e f a b c
195 264 325 399 511 455
264 325 440 511 665 616
390 528 650 798 1022 910
528 650 880 1022 1330 1232
before the program was stopped at t = 2127 (i.e., when d+e+f would add up to 2127). The third and fourth solutions are merely the first and second doubled, but that doesn't preclude other, non-similar, solutions from existing at higher totals.
So all these lengths can be integral.
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Posted by Charlie
on 2007-04-12 15:40:14 |
solution (spoiler)
