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 Equilateral Triangle (Posted on 2003-12-18)
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.

 No Solution Yet Submitted by Ravi Raja Rating: 4.0000 (5 votes)

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 If ... (an alternate question) | Comment 3 of 19 |
(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

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