Suppose ABC is an equilateral triangle and P is a point inside the triangle, such that PA = 3 cms., PB = 4 cms., and PC = 5 cms.
Then find the length of the side of the equilateral triangle.

(In reply to Excel solution by Charlie)

In the queue someone brought up the possibility of the point lying outside the triangle rather than within it. That also has a solution:

Consider a triangle formed by the point and the nearest two points on the equilateral triangle. Its distances from the nearest vertices are 3 and 4 cm. If one side of the original triangle (the remaining side of the new triangle) is length s, then we can find angle a, adjacent to the length-3 side by the law of cosines: 4^2 = 3^2 + s^2 -2*3*s*cos(a). If we assume s is in Excel cell A10, we can put angle a in B10 with the formula =ACOS((9+A10^2-16)/(6*A10)).

Then the distance that should be 5 can be found via the law of cosines applied to the point in question and the points of the equilateral triangle that are 3 and (supposed to be) 5 cm away. The square of the side that is supposed to be length 5 is then 3^2 + s^2 - 2*3*s*cos(60+a), using degree measure. In Excel, in radian measure, we can place in cell C10 =SQRT(9+A10^2-6*A10*COS(PI()/3+B10)), and ask Solver that C10 be made into 5 by changing cell A10.

This results in a side length of 2.053141571 cm--the length the side would have if the point that is 3, 4 and 5 cm away from the points in the equilateral triangle had been specified to lie outside the triangle.

This methodology would also have been possible for the original question, where angle a would have been subtracted from 60 instead of added, and the calculation would have been simpler, as it was here, with one fewer steps.

Posted by Charlie on 2003-12-18 10:18:46