Maths Olympiad Prep

Library / /313 of 520

Geometry Difficulty 6.6 National olympiad Prove it

Points A1, A2,…,An\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{A}_{\mathrm{n}} are equally spaced on the side BC\mathrm{BC} of the triangle ABC\mathrm{ABC} (so that BA1=\mathrm{BA}_{1}= A1A2=…=An−1An=AnC)\left.A_{1} A_{2}=\ldots=A_{n-1} A_{n}=A_{n} C\right). Similarly, points B1,B2,…,BnB_{1}, B_{2}, \ldots, B_{n} are equally spaced on the side CA, and points C1,C2,…,Cn\mathrm{C}_{1}, \mathrm{C}_{2}, \ldots, \mathrm{C}_{\mathrm{n}} are equally spaced on the side AB\mathrm{AB}. Show that (AA12+AA22+…+\left(\mathrm{AA}_{1}{ }^{2}+\mathrm{AA}_{2}{ }^{2}+\ldots+\right. AAn2+BB12+BB22+…+BBn2+C12+…+CCn2)\left.\mathrm{AA}_{\mathrm{n}}{ }^{2}+\mathrm{BB}_{1}{ }^{2}+\mathrm{BB}_{2}{ }^{2}+\ldots+\mathrm{BB}_{\mathrm{n}}{ }^{2}+\mathrm{C}_{1}{ }^{2}+\ldots+\mathrm{CC}_{\mathrm{n}}{ }^{2}\right) is a rational multiple of (AB2+BC2+\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\right. CA2)\left.\mathrm{CA}^{2}\right).

Solution

Using the cosine formula, AAk2=AB2+k2BC2/(n+1)2−2kAB⋅BC/(n+1)cos⁡B\mathrm{AA}_{\mathrm{k}}{ }^{2}=\mathrm{AB}^{2}+\mathrm{k}^{2} \mathrm{BC}^{2} /(\mathrm{n}+1)^{2}-2 \mathrm{k} \mathrm{AB} \cdot \mathrm{BC} /(\mathrm{n}+1) \cos \mathrm{B}. So ∑AAk2=\sum \mathrm{AA}_{\mathrm{k}}{ }^{2}= nAB2+BC2/(n+1)2(12+22+…+n2)−2AB⋅BCcos⁡B(1+2+…+n)/(n+1)n \mathrm{AB}^{2}+\mathrm{BC}^{2} /(n+1)^{2}\left(1^{2}+2^{2}+\ldots+n^{2}\right)-2 \mathrm{AB} \cdot \mathrm{BC} \cos \mathrm{B}(1+2+\ldots+n) /(n+1). Similarly for the other two sides.

Thus the total sum is n(AB2+BC2+CA2)+n(2n+1)/(6(n+1))(AB2+BC2+CA2)−nn\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)+\mathrm{n}(2 \mathrm{n}+1) /(6(\mathrm{n}+1))\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)-\mathrm{n} (AB⋅BCcos⁡B+BC⋅CAcos⁡C+CA⋅ABcos⁡A)(\mathrm{AB} \cdot \mathrm{BC} \cos \mathrm{B}+\mathrm{BC} \cdot \mathrm{CA} \cos \mathrm{C}+\mathrm{CA} \cdot \mathrm{AB} \cos \mathrm{A}). But AB⋅BCcos⁡B=(AB2+BC2−CA2)/2\mathrm{AB} \cdot \mathrm{BC} \cos \mathrm{B}=\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}-\mathrm{CA}^{2}\right) / 2, so AB⋅BCcos⁡B+BC⋅CAcos⁡C+CA⋅ABcos⁡A=(AB2+BC2+CA2)/2\mathrm{AB} \cdot \mathrm{BC} \cos \mathrm{B}+\mathrm{BC} \cdot \mathrm{CA} \cos \mathrm{C}+\mathrm{CA} \cdot \mathrm{AB} \cos \mathrm{A}=\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right) / 2. Thus the sum is a rational multiple of (AB2+BC2+CA2)\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right).

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.