Test of Mathematics Solution Subjective 81 - Cyclic and Symmetric Simultaneous Equations

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 81 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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Problem

Find all possible real numbers \(a,b,c,d,e\) which satisfy the following set of equations:

\( \left\{ \begin{array}{ccc} 3a & = & ( b + c+ d)^3 \\ 3b & = & ( c + d +e ) ^3 \\ 3c & = & ( d + e +a )^3 \\ 3d & = & ( e + a +b )^3 \\ 3e &=& ( a + b +c)^3. \end{array}\right.\)

 

Solution

Observing the symmetry and the cyclicity of the given set of equations it can be easily inferred that the real numbers \(a,b,c,d\) and \(e\) cannot be ordered i.e. one number cannot be greater or smaller than the other number else the system of equations will be inconsistent. This can be shown easily with the help of inequality.

Without loss of generality, we can say  that \(a\) is greater or equal to them all, i.e., \(a\geq b,c,d,e\). Thus we have

\((d+e+a) \geq ( c+d+e)\)

\(=> (d+e+a)^3 \geq (c+d+e)^3\)

\(=>  3c \geq 3b\)

\(=> c \geq b\)

As \(a \geq d\), we also have,

\((a+b+c) \geq ( b+c+d)\)

\(=> (a+b+c)^3 \geq (b+c+d)^3\)

\(=>  3e \geq 3a\)

\(=> e \geq a\)

Thus we have \(a=e\)

Further calculations will show that \(a=b=c=d=e\) is the only possible solution to this set of equations.

Thus we are left to solve just one equation, i.e., \(3a = (3a)^3\)

which gives \(a= -\frac{1}{3},0,\frac{1}{3}\)

Thus the possible values of the 5-tuple \((a,b,c,d,e)\) are:

\(( -\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3})\)

\((0,0,0,0,0)\)

\((\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})\)

Algebra, Germany MO 2019, Problem 6

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Understand the problem

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Suppose that real numbers $x,y$ and $z$ satisfy the following equations: \begin{align*} x+\frac{y}{z} &=2,\\ y+\frac{z}{x} &=2,\\ z+\frac{x}{y} &=2. \end{align*}
Show that $s=x+y+z$ must be equal to $3$ or $7$. Note: It is not required to show the existence of such numbers $x,y,z$.
[/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="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0" hover_enabled="0"]

Germany MO 2019, Problem 6 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Algebra, Simultaneous Equations [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]6/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Challenges and Thrills of 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"]Observe that x = y = z = 1 gives a valid solution of the set of equations. In this case s = x+y+z = 3. Now, observe one thing that this set of equations is symmetric in (x,y,z). Observe that we are required to comment on (x+y+z).  Rewriting the equations as: $$xz+y = 2z, \qquad (1)$$
$$xy + z = 2x, \qquad (2)$$
$$yz + x = 2y \qquad (3)$$
and then summing gives us that $x+y+z = xy + yz + zx = s.$ Our aim will be to reduce all the equations into a single variable. ( maybe a polynomial ). Let's consider the case, where all of x,y,z is not 1. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]From now on we consider $x,y,z \neq 1$. This also gives $x \neq y \neq z \neq x$ Solving the first expression  $x=\frac{2z-y}{z}$  then plugging this into the second two gives:
$$y+\frac{z^2}{2z-y}=2 \Rightarrow (2z-y)y+z^2=2(2z-y)$$$$z+\frac{2z-y}{yz}=2 \Rightarrow yz^2+2z-y=2yz \Rightarrow y=-\frac{2z}{z^2-2z-1}$$ as z is not equal to 1. Plugging the latter into the former and simplifying gives:
$$\frac{z^2 (z^4-8z^3+14z^2-7)}{(z^2-2z-1)^2}=0 \Rightarrow z^4-8z^3+14z^2-7=0$$ [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Plugging the latter into the former and simplifying gives:
$$\frac{z^2 (z^4-8z^3+14z^2-7)}{(z^2-2z-1)^2}=0 \Rightarrow z^4-8z^3+14z^2-7=0$$ Now, observe that we already know z = 1 is a solution. This gives rise to  $$0=z^4-8z^3+14z^2-7=(z-1)(z^3-7z^2+7z+7) \Rightarrow z^3-7z^2+7z+7=0$$   [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Observe that the polynomial we have got in terms of z is also satisfied by x,y,z as the equations are symmetric in x,y,z. Hence we can claim that \( t^3 - 7t^2 + 7t + 7 = 0 \) has three solutions x,y,z.  Hence, \( t^3 - 7t^2 + 7t + 7 = (t-x)(t-y)(t-z)\). Therefore, by Vieta's formula, x+y+z = 7. QED. [/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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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"]

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