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Well, I can't guarantee that I can make you happy, but atleast guess that I can. A simple problem: Determine the set of all points lying in the plane of a triangle ABC (say P), for which (PA)^2+(PB)^2+(PC)^2 is minimum. We all know that this set is the singleton set: {Centroid G of ABC}. Unfortunately, a previous knowledge of the answer makes it easy to prove the assertion, but more unfortunately, even after stating:"We will show that the centroid is the only such point", most books give a long, boring proof, involving non-trivial and non-motivating constructions and lengthy calculations. Right, I'm going to give a what-I-think beautiful proof, which is due to me! I solved it while preparing for the Indian National Mathematical Olympiad last year. Just join P with C1,A1,B1, the midpoints of AB,BC,CA,resp and form the triangle A1B1C1. Note that A1B1C1 is the image of ABC under a homothety of factor -0.5 about their common centroid. Next, applying the Apolonius' theorem on the 3 triangles APB, BPC & CPA and adding the 3 relations, note that PA^2+PB^2+PC^2 is minimum, if and only if PA1^2+PB1^2+PC1^2 is minimum. That the set above is singleton, is immediate from the extremal principle applied to 2 possible points having the same property and showing that their midpoint has the sum of squares less than them. Now, let P' be the image of P under the homothety -0.5. Then, by properties of homothety, P' and P both have minimum sum-of-squares with respect to A1B1C1. But the set is singleton, so P=P'. However, the only point that remains invariant under a homothety is the centre, which in this case, is the centroid!!:) Right, whenever one sees the sum-of-squares form one guesses to apply Apolonius' theorem and this proof doesn't even require a pre-knowledge of the answer! Please tell me if you've enjoyed it or not!