Maximising the area of a triangle when the 3 sides are at most 2, 3, 4 This is question 5 from the 2012 BMO1, please can someone provide a full solution? There is an online video solution but is incomplete as it assumes that the area of a triangle is maximised when two side lengths are maximised without justification. Thank you.
 A: At least one of the triangle sides has to be maximized, equal to 2, 3 or 4. If all sides are shorter, it is always possible to scale up the triangle until one side reaches the maximum value. And that triangle certainly has a bigger area. So we have three cases to check: the maximized side can be 2, 3 or 4.
Case AB=2: Draw circles with radii 3 and 4 with points A and B as centers. The third vertex of the triangle has to be somewhere in the common area of these two circles. The most distant point from AB is the optimal solution (because it gives the maximum triangle height). And that's point C' (giving triangle ABC' with sides 2, 3, $\sqrt{13}$), not point C (giving triangle ABC with sides 2, 3 and 4). 

Case AB=3: Same as above, just a slightly different picture. Draw circles with radii 2 and 4. Optimal triangle ABC' is actually the same as in the previous case with sides 2, 3, $\sqrt{13}$. 

Case AB=4: In this case, the most distant point is actually the intersection point of circles with radii 2 and 3 and centers A,B. The otimal triangle is the triangle ABC with sides 2, 3 and 4. But in the first case we already saw that that solution is inferior to triangle with sides $2, 3, \sqrt{13}$

So the soluton to this problem is $2,3,\sqrt{13}$
A: The online video solution (0:40 to 1:00) does provide justification that two of the sides must have maximum length. The reasoning is effectively as follows:
A triangle's area is determined by its base and height. If the base is fixed and the other two sides can be made longer, then we can increase the area by pulling the vertex away from the base so as to increase the height.
