Regional Math Olympiad, 2019 Problem 4
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Inequality[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]8/10
[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" open="on"]Challenges and Thrills in Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]1+2+3+ ... 3k = \( \frac{3k(3k+1)}{2}\)
So, \( \sum_{j=1}^{k} a_{1j} = \sum_{j=1}^{k} a_{2j}= \sum_{j=1}^{k}a_{3j} = \frac{k(3k+1)}{2}\)
\( 1^2 + 2^2 +... (3k)^2 = \frac{k(3k+1)(6k+1)}{2}\)
So, \(\sum_{j=1}^{k} (a_{1j})^2= \sum_{j=1}^{k} (a_{2j})^2 = \sum_{j=1}^{k} (a_{3j})^2 = \frac{k(3k+1)(6k+1)}{6}\).
This means that 3 | k.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]Step 1: Try out with k = 3. Prove that it is not possible to arrange them in the desired order as already some numbers are fixed.
We will now try for k = 6.
Claim: If 3|k and k > 3 then it is always possible. (This is our conjecture too)
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]n + (n + 5) + (m + 1) + (m + 4) + (l + 2) + (l + 3) = 2n + 2m + 2l + 15 = (n + 1) + (n + 4) + (m + 2) + (m + 3) + (l) + (l+ 5)
\(n^2 + (n + 5)^2 - (n + 1)^2 - (n + 4)^2 = 8\)
\( (m + 1)^2 + (m + 4)^2 - (m + 2)^2 - (m + 3)^2 = 4\)
\( (l + 2)^2 + (l + 3)^2 - (l)^2 - (l+ 5)^2 = - 12 \)
So, we get \( n^2 + (n + 5)^2 + (m + 1)^2 + (m + 4)^2 + (l + 2)^2 + (l + 3)^2 = (n + 1)^2 + (n + 4)^2 + (m + 2)^2 +(m + 3)^2 + (l)^2 + (l+ 5)^2 \).
Also,
n + (n + 5) + (m + 1) + (m + 4) + (l + 2) + (l + 3) = 2n + 2m + 2l + 15 = (n + 1) + (n + 4) + (m + 2) + (m + 3) + (l) + (l+ 5)
Hence putting suitable values of l, m, and n, we get an array like the one below:
\( \begin{pmatrix} 1 & 6 & 8 & 11 & 15 & 16 \\ 2 & 5 & 9 & 10 & 13 & 18 \\ 3 & 4 & 7 & 12 & 14 & 17\end{pmatrix} \)
\( (n + 1)^2 + (n + 6)^2 + (n + 8)^2 + (n + 11)^2 + (n + 15)^2 + (n + 16)^2
= (n + 2)^2 + (n + 5)^2+ (n + 9)^2 + (n + 10)^2+ (n + 13)^2+ (n + 18)^2
= (n + 3)^2 + (n + 4)^2 + (n + 7)^2+ (n + 12)^2 + (n+14)^2 + (n+17)^2 \).
Using this and the above two matrices, you can prove by induction that the claim holds!
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Using the array found similarly :
\( \begin{pmatrix} 1 & 6 & 8 & 11 & 18 & 13 & 21 & 23 & 25 \\ 2 & 5 & 7 & 12 & 15 & 17 & 19 & 22 & 27 \\ 3 & 4 & 9 & 10 & 14 & 16 & 20 & 24 & 26\end{pmatrix} \)
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