Library / Iberoamerican Mathematical Olympiad /25 of 57

Iberoamerican Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Ibero-American Mathematical Olympiad

Problem:

Points $A_1, A_2, \ldots, A_n$ are equally spaced on the side $BC$ of the triangle $ABC$ (so that $BA_1 = A_1A_2 = \ldots = A_{n-1}A_n = A_nC$ ). Similarly, points $B_1, B_2, \ldots, B_n$ are equally spaced on the side $CA$ , and points $C_1, C_2, \ldots, C_n$ are equally spaced on the side $AB$ . Show that
$(AA_1^2 + AA_2^2 + \ldots + AA_n^2 + BB_1^2 + BB_2^2 + \ldots + BB_n^2 + CC_1^2 + \ldots + CC_n^2)$
is a rational multiple of $(AB^2 + BC^2 + CA^2).$

Solution

Solution:

Using the cosine formula, $AA_k^2 = AB^2 + k^2 BC^2/(n+1)^2 - 2k AB \cdot BC/(n+1) \cos B$ . So
$\sum AA_k^2 = n AB^2 + BC^2/(n+1)^2 (1^2 + 2^2 + \ldots + n^2) - 2 AB \cdot BC \cos B (1 + 2 + \ldots + n)/(n+1).$
Similarly for the other two sides.

Thus the total sum is
$n(AB^2 + BC^2 + CA^2) + n(2n+1)/(6(n+1))(AB^2 + BC^2 + CA^2) - n(AB \cdot BC \cos B + BC \cdot CA \cos C + CA \cdot AB \cos A).$
But $AB \cdot BC \cos B = (AB^2 + BC^2 - CA^2)/2$ , so
$AB \cdot BC \cos B + BC \cdot CA \cos C + CA \cdot AB \cos A = (AB^2 + BC^2 + CA^2)/2.$
Thus the sum is a rational multiple of $(AB^2 + BC^2 + CA^2)$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.