RMO 2019 Problem 4 Solution
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 \( \displaystyle \sum_{j=1}^{k} a_{1j} = \sum_{j=1}^{k} a_{2j}= \sum_{j=1}^{k}a_{3j} \ and \ \sum_{j=1}^{k} (a_{1j})^2= \sum_{j=1}^{k} (a_{2j})^2 = \sum_{j=1}^{k} (a_{3j})^2 \)[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.0"]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
[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first![/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 :
\( \begin{pmatrix} 1 & 6 & 8 & 11 & 18 & 13 & 21 & 23 & 25 \\ 2 & 5 & 7 & 12 & 15 & 17 & 19 & 22 & 27 \\ 3 & 4 & 9 & 10 & 14 & 16 & 20 & 24 & 26\end{pmatrix} \)
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[/et_pb_text][et_pb_blurb title="Math Olympiad Program" url="https://cheenta.com/matholympiad/" url_new_window="on" image="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" link_option_url="https://cheenta.com/matholympiad/" link_option_url_new_window="on"]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"]
Construction: Can you find a circle?
Nine-point circle passes through feet of altitudes, the midpoint of sides and midpoints of AH, BH, CH (where H is orthocenter) of a triangle. This is a well-known theorem from geometry.
In this case, it passes through L, E, D, N, and F.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]
Chase three cyclic quadrilaterals to understand DH is the internal angle bisector of \( \angle EDF \).
Since \( \angle BFE = \angle BEC = 90 ^o \) hence BFEC is cyclic hence \(\angle FBE = \angle FCE \) (angle subtended by the segment FE).
Since \( \angle ADC = \angle AFC = 90 ^o \) hence AFDC is cyclic hence \(\angle FDA = \angle FCA or \angle FCE \) (angle subtended by the segment FA).
Since \( \angle BDA = \angle BEA = 90 ^o \) hence BDEA is cyclic hence \(\angle ADE = \angle ABE \) (angle subtended by the segment AE).
Combining we have \( \angle FDA = \angle EDA \) implying DA bisects \( \angle FDE \)
Can you show L-M-N are collinear and perpendicular to EF?
We show that \( \Delta LMF \) is congruent to \( \Delta LME \).
Clearly LM = LM, ME = MF (as M is the midpoint). We will show that LE = LF. Why? Since L is the midpoint of LH and AFH is right-angled, hence L being the midpoint of hypotenuse is the circumcenter of AFH.
This implies LF = LA. Similarly, L is the circumcenter of AEH implying LE = LA.
Thus LF = LE.
Hence we showed that \( \Delta LMF \) is congruent to \( \Delta LME \) making \( \angle LMF = \angle LME = 90^o \)
Similarly NM is perpendicular to EF.
Now can you find two cyclic quadrilaterals? [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]
Construction: Complete the hexagon.
Can you find some parallelograms in the picture? Particularly, can you prove \( BGCA_1 \) is a parallelogram? [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]
(We will prove that the marked blue angles are equal)
Notice that \( \angle ABC = A_1 C_1 B_1 \) (given)
Also \( \angle A_1 C_1 B_1 = \angle A_1 B B_1 \) (subtended by the chord \(A_1 B_1\) at the circumference.
Hence \( \angle ABC = \angle A_1 B B_1 \)
Now substracting \( \angle CBB_1 \) from both side we have
IMPORTANT: \( \angle A_1 B C = \angle ABB_1 \)
Similarly notice that \( \angle BAC =\ C_1 B_1 A_1 \) (given)
Also \( \angle C_1 B_1 A_1 = \angle C_1 A A_1 \) (subtended by the chord \(C_1 A_1 \) at the circumference.
Hence \( \angle BAC = \angle C_1 A A_1 \)
Now substracting \( \angle BAA_1 \) from both side we have
IMPORTANT: \( \angle C_1 A B = \angle A_1AC\)
Now \( \angle A_1 B C = \angle A_1 A C \) (subtended by the same segment \(A_1 C \) at the circumference.)
And \( \angle C_1 A B = \angle C_1 C B \) ( subtended by the same segment \( C_1 B \) )
Thus \( \angle C_1 C B = \angle A_1 B C \)
Thus alternate angles are equal making \( BA_1 \) parallel to GC.
Thus in \( \Delta A_1MB \) and \( \Delta GMC \) we have BM = CM, \( \angle BMA_1 = \angle GMC \) and \( \angle C_1 C B = \angle A_1 B C \)
Hence the triangles are congruent, making \( BA_1 = GC \)