RMO 2019 Problem 4 Solution

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Consider the following \( 3 \times 2 \) array formed by using the numbers \( 1 , 2 , 3 ,4 ,5 , 6 \ : \)\( \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix} = \begin{pmatrix} 1 &6 \\ 2 & 5 \\ 3 & 4 \end{pmatrix} \) . Observe that all row sums are equal , but the sum of the squares is not the same for each row .Extend the above arrar to a \( 3 \times k \) array \( (a_{ij}){3 \times k} \) for a suitable k , adding more columns , using the numbers \( 7 , 8 , 9 , ....,3k \) such that \( \displaystyle \sum_{j=1}^{k} a_{1j} = \sum_{j=1}^{k} a_{2j}= \sum_{j=1}^{k}a_{3j} \ and \ \sum_{j=1}^{k} (a_{1j})^2= \sum_{j=1}^{k} (a_{2j})^2 = \sum_{j=1}^{k} (a_{3j})^2 \)

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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"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]

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 : \( \begin{pmatrix} 1 & 6 & 8 & 11 & 18 & 13 & 21 & 23 & 25 \\ 2 & 5 & 7 & 12 & 15 & 17 & 19 & 22 & 27 \\ 3 & 4 & 9 & 10 & 14 & 16 & 20 & 24 & 26\end{pmatrix} \) [/et_pb_tab][/et_pb_tabs][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"]

Watch video

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Connected Program at Cheenta

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][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="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]