1988 OIM Problems/Problem 4

Problem

Let $ABC$ be a triangle which sides are $a$ , $b$ , $c$ . We divide each side of the triangle in $n$ equal segments. Let $S$ be the sum of the squares of the distances from each vertex to each of the points dividing the opposite side different from the vertices.

Prove that $\frac{S}{a^2+b^2+c^2}$ is rational.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Drop a perpendicular from $A$ onto $BC$ , and let the foot be $M_a$ . Let $p_a=BM_a$ and $q_a=CM_a$ (so $p_a+q_a=a$ ), and define similar values for $B$ and $C$ . Furthermore, let $S_a$ represent the contributions of the distances from $A$ to the opposite side and $S_b$ and $S_c$ similarly; thus $S=S_a+S_b+S_c$ .

Notice that each segment on $BC$ has length $\frac{a}{n}$ ; let $h_a$ be the length of the altitude from $A$ (so $h_a^2+p_a^2=b^2$ and $h_a^2+q_a^2=c^2$ by the Pythagorean Theorem). Then, by multiple uses of the Pythagorean Theorem: $\begin{aligned} S_{a} & = (h_{a}^{2} + {(p_{a} - \frac{a}{n})}^{2}) + (h_{a}^{2} + {(p_{a} - \frac{2 a}{n})}^{2}) + \dots + (h_{a}^{2} + {(p_{a} - \frac{a (n - 1)}{n})}^{2}) \\ = (n - 1) h_{a}^{2} + \sum_{i = 1}^{n - 1} {(p_{a} - \frac{i a}{n})}^{2} \end{aligned}$ But notice that, due to symmetry, we can substitute $q_a$ in place of $p_a$ and yield the same result. Therefore, $\begin{aligned} S_{a} & = \frac{1}{2} S_{a} + \frac{1}{2} S_{a} \\ = \frac{1}{2} ((n - 1) h_{a}^{2} + \sum_{i = 1}^{n - 1} {(p_{a} - \frac{i a}{n})}^{2}) + \frac{1}{2} ((n - 1) h_{a}^{2} + \sum_{i = 1}^{n - 1} {(q_{a} - \frac{i a}{n})}^{2}) \\ = (n - 1) h_{a}^{2} + \frac{1}{2} \sum_{i = 1}^{n - 1} ({(p_{a} - \frac{i a}{n})}^{2} + {(q_{a} - \frac{i a}{n})}^{2}) \\ = (n - 1) h_{a}^{2} + \frac{1}{2} \sum_{i = 1}^{n - 1} (p_{a}^{2} - \frac{2 i a p_{a}}{n} + \frac{i^{2} a^{2}}{n^{2}} + q_{a}^{2} - \frac{2 i a q_{a}}{n} + \frac{i^{2} a^{2}}{n^{2}}) \\ = \frac{1}{2} (n - 1) h_{a}^{2} + \frac{1}{2} (n - 1) p_{a}^{2} + \frac{1}{2} (n - 1) h_{a}^{2} + \frac{1}{2} (n - 1) q_{a}^{2} + \frac{1}{2} \sum_{i = 1}^{n - 1} (\frac{2 i^{2} a^{2}}{n^{2}} - \frac{2 i a (p_{a} + q_{a})}{n}) \\ = \frac{1}{2} (n - 1) (b^{2} + c^{2}) + \frac{1}{2} \sum_{i = 1}^{n - 1} (\frac{2 i^{2} a^{2}}{n^{2}} - \frac{2 i a^{2}}{n}) \\ = \frac{1}{2} (n - 1) (b^{2} + c^{2}) + a^{2} \sum_{i = 1}^{n - 1} (\frac{i^{2}}{n^{2}} - \frac{i}{n}) \end{aligned}$ We can similarly find $S_b$ and $S_c$ . Then, $\begin{aligned} S & = S_{a} + S_{b} + S_{c} \\ = (n - 1) (a^{2} + b^{2} + c^{2}) + (a^{2} + b^{2} + c^{2}) \sum_{i = 1}^{n - 1} (\frac{i^{2}}{n^{2}} - \frac{i}{n}) \end{aligned}$ Thus, $\frac{S}{a^2+b^2+c^2}=n-1+\sum_{i=1}^{n-1}\left(\frac{i^2}{n^2}-\frac{i}{n}\right)$ which is rational.

~eevee9406

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1988 OIM Problems/Problem 4

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