Before I start posting my incredibly long and unpractical (probably far from optimal) solution, is the answer 5pi over 6?
It is an old problem , my idea is to reflect across the three sides , let the reflection be respectively .
I find that are triangles . Then has the sides ratio so it is right-angled .
which it is also the answer given by RaisinBread .
The second solution is more elegant , but not by me .
The idea is to rotate by , by sending to , to . Then , is equilateral and so . Therefore ,
It's all good, I'm happy to have found the answer! I see the solution is really simple, but believe it or not, I solved the whole thing algebraically with functions and intersections.
In short, I did this:
1. I represented the triangle in the x-y plane, such that one side is resting on the x axis, and that the summit opposed to that side is on the y axis. I can thus make any equilateral triangle with the following equations:
, and , where a is any real number.
2. Let the left summit be A, the top be B and the right be C. I centered 3 circles of respective radii 3, 4 and 5 on these summits.
I know I have to find the point a (upon which all my equations depend) at which the three circles intersect at a single point, that point being P.
3. I knew that three non concentric circles could only intersect at the intersection of their radical axis, so i isolated the radical axis of the circles with center A and C, using trigonometry, which gave me the x coordinate of where the circles would intersect.
I then proceeded to solve quadratic equations to find at which y did the circles around A and B intersected, and then the same thing with the circles around B and C. I got two solutions each time, but I could see that in both case, the lowest intersection would be the one where the three circles intersected.
(These are the x and y coordinates of the point P, provided the two y are equal)
I then found the value of a by conditioning that the two y coordinates be equal, and isolated
4. Putting these results in the graph I had a confirmation that I was on the right track; . Now all I needed to do was find the equations of the line from A to P, and the line from P to B, which was easy to do now since I had the value of a. I then changed these slopes into polar coordinates to find what angle they make with the y axis, subtracted the angle values and eureka, got 5pi over 6 radians!
Very long and impractical solution, but I can't say I didn't have fun doing it!