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{align*} S_a&=\left(h_a^2+\left(p_a-\frac{a}{n}\right)^2\right)+ \left(h_a^2+\left(p_a-\frac{2a}{n}\right)^2\right)+\cdots+ \left(h_a^2+\left(p_a-\frac{a(n-1)}{n}\right)^2\right)\\ &=(n-1)h_a^2+\sum_{i=1}^{n-1}\left(p_a-\frac{ia}{n}\right)^2\\ \end{align*} But notice that, due to symmetry, we can substitute $q_a$ in place of $p_a$ and yield the same result. Therefore, \begin{align*} S_a&=\frac{1}{2}S_a+\frac{1}{2}S_a\\ &=\frac{1}{2}\left((n-1)h_a^2+\sum_{i=1}^{n-1}\left(p_a-\frac{ia}{n}\right)^2\right)+\frac{1}{2}\left((n-1)h_a^2+\sum_{i=1}^{n-1}\left(q_a-\frac{ia}{n}\right)^2\right)\\ &=(n-1)h_a^2+\frac{1}{2}\sum_{i=1}^{n-1}\left(\left(p_a-\frac{ia}{n}\right)^2+\left(q_a-\frac{ia}{n}\right)^2\right)\\ &=(n-1)h_a^2+\frac{1}{2}\sum_{i=1}^{n-1}\left(p_a^2-\frac{2iap_a}{n}+\frac{i^2a^2}{n^2}+q_a^2-\frac{2iaq_a}{n}+\frac{i^2a^2}{n^2}\right)\\ &=\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}\left(\frac{2i^2a^2}{n^2}-\frac{2ia(p_a+q_a)}{n}\right)\\ &=\frac{1}{2}(n-1)(b^2+c^2)+\frac{1}{2}\sum_{i=1}^{n-1}\left(\frac{2i^2a^2}{n^2}-\frac{2ia^2}{n}\right)\\ &=\frac{1}{2}(n-1)(b^2+c^2)+a^2\sum_{i=1}^{n-1}\left(\frac{i^2}{n^2}-\frac{i}{n}\right)\\ \end{align*} We can similarly find $S_b$ and $S_c$ . Then, \begin{align*} 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}\left(\frac{i^2}{n^2}-\frac{i}{n}\right)\\ \end{align*} 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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