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.

(In reply to re(2): Two solutions for one triangle! by McWorter)

I think any triangle other than one where angle C is 60 degrees (which exception includes equilateral triangles) will have two solutions so long as you allow P and/or Q to lie outside the bounds of AC and BC respectively.  The discriminant in the quadratic is positive for all shapes of the triangle, as shown by the following table, where columns represent the angle at C and rows the relative size of one BC to AC (or vice versa):

       20      40      60      80     100     120     140     160
1.0   0.4825  1.8716  4.0000  6.6108  9.3892 12.0000 14.1284 15.5175
1.1   0.5660  2.0823  4.4100  7.2731 10.3293 13.2100 15.5647 17.1046
1.2   0.7202  2.3399  4.8400  7.9378 11.2718 14.4400 17.0479 18.7623
1.3   0.9451  2.6444  5.2900  8.6049 12.2168 15.6900 18.5781 20.4907
1.4   1.2406  2.9959  5.7600  9.2744 13.1642 16.9600 20.1553 22.2897
1.5   1.6067  3.3943  6.2500  9.9464 14.1139 18.2500 21.7794 24.1593
1.6   2.0435  3.8397  6.7600 10.6207 15.0661 19.5600 23.4504 26.0996
1.7   2.5509  4.3320  7.2900 11.2975 16.0207 20.8900 25.1684 28.1106
1.8   3.1290  4.8712  7.8400 11.9767 16.9777 22.2400 26.9333 30.1921
1.9   3.7777  5.4574  8.4100 12.6582 17.9371 23.6100 28.7452 32.3443
2.0   4.4970  6.0906  9.0000 13.3422 18.8990 25.0000 30.6040 34.5672
2.1   5.2870  6.7707  9.6100 14.0287 19.8632 26.4100 32.5098 36.8607
2.2   6.1476  7.4977 10.2400 14.7175 20.8299 27.8400 34.4625 39.2248
2.3   7.0789  8.2717 10.8900 15.4087 21.7990 29.2900 36.4621 41.6596
2.4   8.0808  9.0926 11.5600 16.1024 22.7705 30.7600 38.5087 44.1650
2.5   9.1533  9.9605 12.2500 16.7984 23.7443 32.2500 40.6023 46.7411
2.6  10.2965 10.8754 12.9600 17.4969 24.7207 33.7600 42.7428 49.3877
2.7  11.5104 11.8371 13.6900 18.1978 25.6994 35.2900 44.9302 52.1051
2.8  12.7948 12.8458 14.4400 18.9011 26.6805 36.8400 47.1646 54.8931
2.9  14.1500 13.9015 15.2100 19.6068 27.6640 38.4100 49.4460 57.7517
3.0  15.5757 15.0041 16.0000 20.3149 28.6500 40.0000 51.7742 60.6810

The only complication in the formula is division by zero when the angle at C is 60 degrees. But come to think about it, that's when the quadratic actually becomes a linear equation, as the thing that becomes zero is the coefficient of the square of x.

So whenever the angle at C is 60 degrees there is only one solution. Otherwise there are two.

Posted by Charlie on 2005-05-08 03:51:48