Given a triangle ABC, how can you find a point P on AC and a point Q on BC, such that AP=PQ=QB?

N.B. A construction method is sought, and only compass and straightedge are allowed.

This is not a construction, but a solution to the value of the length of AP and of BA so that they can be measured off from A and B.

Let the base be AB.  The line PQ forms the base of a different triangle that also includes vertex (and angle) C.

Calling the lengths of the sides in the given (larger) triangle by the lower-case of the opposite angle, the law of cosines gives

c^2=a^2+b^2-2ab cos C
so
cos C = (a^2+b^2-c^2)/(2ab)

Then, letting x be the sought distance, also the base of the smaller triangle:

x^2 = (a-x)^2 + (b-x)^2 - 2 (a-x)(b-x) cos C

This then eventually produces the quadratic

(1 - 2 cos C) x^2 + (a+b)(2 cos C - 2) x + c^2  = 0

The solution of this in which 0 < x < min(a,b) is the distance to measure off from A and from B.

When a or b (that is the length of either BC or AC) is more than twice the other, there is no such solution.

The following program evaluates this:

DEFDBL A-Z
DO
  INPUT a, b, c
  cosC = (a * a + b * b - c * c) / (2 * a * b)
  coSq = 1 - cosC * 2
  coLin = (a + b) * (cosC * 2 - 2)
  con = c * c

  rad = coLin * coLin - 4 * coSq * con

  IF rad < 0 OR coSq = 0 THEN
    PRINT "no answer"
  ELSE
    root = SQR(rad)
    ans1 = (-coLin + root) / (2 * coSq)
    ans2 = (-coLin - root) / (2 * coSq)
    IF ans1 <= a AND ans1 <= b AND ans1 >= 0 THEN PRINT ans1
    IF ans2 <= a AND ans2 <= b AND ans2 >= 0 THEN PRINT ans2
    IF (ans1 > a OR ans1 > b OR ans1 < 0) AND (ans2 > a OR ans2 > b OR ans2 < 0) THEN
      PRINT "No conventional answer."
    END IF
  END IF
LOOP

For an equilateral triangle, the answer should be halfway along AC; however, by the formula used, this would result in division by zero, and so no answer is given:

? 1,1,1
no answer

That is an exceptional condition.  It can be gotten around by deviating slightly from the values:

? 1,1,1.00001
 .5000024999877068
 
Some other examples are:

? 1,1,.5
 .3333333333333333
? 1,1,1.5
 .6
? 1,1.8,2
 .8994546157594839
? 1,1.9,2
 .9464399043856192
? 1,1.99,2
 .9943228831942601
? 1,2,2
 1
? 1,2,2.1
 .9999999999999999
? 1,2.1,2
No conventional answer.

The last is a case when side b is more than twice the length of side a.  When it was exactly twice the length of side a, the solution was degenerate and P was coincident with C, and Q the midpoint of CB.

Posted by Charlie on 2005-04-27 19:45:53