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A Point in a Triangle (Posted on 2007-04-12) Difficulty: 4 of 5
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?

See The Solution Submitted by Brian Smith    
Rating: 3.0000 (1 votes)

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Solution solution (spoiler) | Comment 1 of 8

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.


  Posted by Charlie on 2007-04-12 15:40:14
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