Maths Olympiad Prep

Library / /156 of 520

Geometry Difficulty 5.0 AIME Find the answer

12. Given nn vectors in the plane as OP1O P_{1}, OP2O P_{2}, ⋯\cdots, OPnO P_{n}, the vector OP\boldsymbol{O P} makes PP12+PP22+⋯\boldsymbol{P P}_{1}^{2}+\boldsymbol{P P}_{2}^{2}+\cdots +PPn2+P P_{n}^{2} minimal. Then the vector OP\boldsymbol{O P} is \qquad

A number or a short expression. Spacing and $ signs are ignored.

Solution

 12. 1n∑k=1nOPk. ∑k=1nPPk2=∑k=1n(OPPk−OP)2=∑k=1n(OPk2−2OPk⋅OP+OP2)=∑k=1nOPk2−2(∑k=1nOPk)⋅OP+nOP2=∑k=1nOPk2−(∑k=1nOPk)2n+n[OP−1n(∑k=1nOPk)]2⩾∑k=1nOPk2−(∑k=1nOPk)2n.  \begin{array}{l} \text { 12. } \frac{1}{n} \sum_{k=1}^{n} O P_{k} \text {. } \\ \sum_{k=1}^{n} \boldsymbol{P P}_{\boldsymbol{k}}^{2}=\sum_{k=1}^{n}\left(\boldsymbol{O P} \boldsymbol{P}_{\boldsymbol{k}}-\boldsymbol{O P}\right)^{2} \\ =\sum_{k=1}^{n}\left(O P_{k}^{2}-2 O P_{k} \cdot O P+O P^{2}\right) \\ =\sum_{k=1}^{n} O P_{k}^{2}-2\left(\sum_{k=1}^{n} O P_{k}\right) \cdot O P+n O P^{2} \\ =\sum_{k=1}^{n} O P_{k}^{2}-\frac{\left(\sum_{k=1}^{n} O P_{k}\right)^{2}}{n}+ \\ n\left[O P-\frac{1}{n}\left(\sum_{k=1}^{n} O P_{k}\right)\right]^{2} \\ \geqslant \sum_{k=1}^{n} O P_{k}^{2}-\frac{\left(\sum_{k=1}^{n} O P_{k}\right)^{2}}{n} \text {. } \\ \end{array}

When OP=1n∑k=1nOPkO P=\frac{1}{n} \sum_{k=1}^{n} O P_{k}, the equality holds in the above expression, hence
OP=1n∑k=1nOPk O P=\frac{1}{n} \sum_{k=1}^{n} O P_{k}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.