Difference between revisions of "1988 OIM Problems/Problem 4"
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Prove that <math>\frac{S}{a^2+b^2+c^2}</math> is rational. | Prove that <math>\frac{S}{a^2+b^2+c^2}</math> is rational. | ||
| + | |||
| + | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
== Solution == | == Solution == | ||
| − | {{ | + | Drop a perpendicular from <math>A</math> onto <math>BC</math>, and let the foot be <math>M_a</math>. Let <math>p_a=BM_a</math> and <math>q_a=CM_a</math> (so <math>p_a+q_a=a</math>), and define similar values for <math>B</math> and <math>C</math>. Furthermore, let <math>S_a</math> represent the contributions of the distances from <math>A</math> to the opposite side and <math>S_b</math> and <math>S_c</math> similarly; thus <math>S=S_a+S_b+S_c</math>. |
| + | |||
| + | Notice that each segment on <math>BC</math> has length <math>\frac{a}{n}</math>; let <math>h_a</math> be the length of the altitude from <math>A</math> (so <math>h_a^2+p_a^2=b^2</math> and <math>h_a^2+q_a^2=c^2</math> 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 <math>q_a</math> in place of <math>p_a</math> 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 <math>S_b</math> and <math>S_c</math>. 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, | ||
| + | <cmath>\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)</cmath> | ||
| + | which is rational. | ||
| + | |||
| + | ~eevee9406 | ||
| + | |||
| + | == See also == | ||
| + | https://www.oma.org.ar/enunciados/ibe3.htm | ||
Latest revision as of 01:53, 26 March 2025
Problem
Let 
be a triangle which sides are 
, 
, 
. We divide each side of the triangle in 
equal segments. Let 
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 
is rational.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Drop a perpendicular from 
onto 
, and let the foot be 
. Let 
and 
(so 
), and define similar values for 
and 
. Furthermore, let 
represent the contributions of the distances from to the opposite side and 
and 
similarly; thus 
.
Notice that each segment on has length 
; let 
be the length of the altitude from (so 
and 
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 
in place of 
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 and . 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,
![\[\frac{S}{a^2+b^2+c^2}=n-1+\sum_{i=1}^{n-1}\left(\frac{i^2}{n^2}-\frac{i}{n}\right)\]](http://latex.artofproblemsolving.com/6/3/8/638951679fd0e7546f04c5ef6cea7238ca7b280b.png)
which is rational.
~eevee9406
