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Difference between revisions of "1961 IMO Problems/Problem 1"

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==Solution 2==
 
==Solution 2==
  
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The third equation implies that <math>x,z,y</math> is a geometric sequence. Then let <math>x=\frac{z}{r}</math> and <math>y=rz</math>, with <math>r\neq1,z>0</math>. Then the first two equations become:
+
Obviously, <math>a=x+y+z>0</math>. The third equation implies that <math>x,z,y</math> is a geometric sequence. Then let <math>x=\frac{z}{r}</math> and <math>y=rz</math>, with <math>r,z>0</math> and <math>r\neq1</math>. Then the first two equations become:
−
<cmath>\left(r+1+\frac{1}{r}\right)z=a~~(1)</cmath>
+
<cmath>\left(r+1+\frac{1}{r}\right)z=a~~~(1)</cmath>
 
and
 
and
−
<cmath>(r^2+1+\frac{1}{r^2})z^2=b^2~(2)</cmath>
+
<cmath>\left(r^2+1+\frac{1}{r^2}\right)z^2=b^2~~~(2)</cmath>
−
Taking <math>\frac{(2)}{(1)}</math>, we get:
+
Taking <math>\frac{(2)}{(1)}</math> (since <math>z>0</math>), we get:
−
<cmath>\frac{(r^2+1+\frac{1}{r^2})z^2}{(r+1+\frac{1}{r})z}=\left(r-1+\frac{1}{r}\right)z=\frac{b^2}{a}~(3)</cmath>
+
<cmath>\frac{(r^2+1+\frac{1}{r^2})z^2}{(r+1+\frac{1}{r})z}=\left(r-1+\frac{1}{r}\right)z=\frac{b^2}{a}~~~(3)</cmath>
 
We can then take <math>(1)^2-(2)</math> and <math>(2)-(3)^2</math> to get:
 
We can then take <math>(1)^2-(2)</math> and <math>(2)-(3)^2</math> to get:
 +
<cmath>\left(r^2+2r+3+\frac{2}{r}+\frac{1}{r^2}\right)z^2-\left(r^2+1+\frac{1}{r^2}\right)z^2=2z^2\left(r+1+\frac{1}{r}\right)=a^2-b^2~~~(4)</cmath>
 +
and
 +
<cmath>\left(r^2+1+\frac{1}{r^2}\right)z^2-\left(r^2-2r+3-\frac{2}{r}+\frac{1}{r^2}\right)z^2=2z^2\left(r-1+\frac{1}{r}\right)=b^2-\frac{b^4}{a^2}=\frac{b^2}{a^2}(a^2-b^2)~~~(5)</cmath>
 +
Let <math>k=r+\frac{1}{r}</math>. By AM-GM, <math>k\ge2</math> with equality at <math>r=\frac{1}{r}\implies r=1</math>, which is impossible. Hence, <math>k>2</math>. Then, <math>\frac{(5)}{(4)}</math> becomes:
 +
<cmath>\frac{k-1}{k+1}=\frac{b^2}{a^2}~(6)\implies a^2+b^2=(a^2-b^2)k>2(a^2-b^2)\implies3b^2>a^2</cmath>
 +
From the above restrictions on <math>a</math> and <math>b</math>, we see that there must exist some <math>k>2</math> satisfying <math>(6)</math>, and hence, some <math>r>0\neq1</math> satisfying <math>(6)</math>. From <math>(4)</math>, if <math>a^2-b^2>0</math>, then there must exist some positive <math>z</math> satisfying <math>(4)</math>, and consequently since <math>(4)</math> and <math>(6)</math> are equivalent to the remaining equations, they satisfy <math>(1)</math> and <math>(2)</math>. Hence, <math>x,y,z</math> satisfy the original system, and from the restrictions on <math>r</math> and <math>z</math>, they are distinct positive reals. Hence, <math>\boxed{a>0\text{ and }3b^2>a^2>b^2}</math>. <math>\blacksquare</math>
 +
 +
~rhydon516
 +
 
===Video Solution===
 
===Video Solution===
 
https://www.youtube.com/watch?v=_stCjNU0_M4&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=3
 
https://www.youtube.com/watch?v=_stCjNU0_M4&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=3

Latest revision as of 17:15, 26 March 2024

Problem

(Hungary) Solve the system of equations:

$\begin{matrix} \quad x + y + z \!\!\! &= a \; \, \\ x^2 +y^2+z^2 \!\!\! &=b^2 \\ \qquad \qquad xy \!\!\! &= z^2 \end{matrix}$

where $a$ and $b$ are constants. Give the conditions that $a$ and $b$ must satisfy so that $x, y, z$ (the solutions of the system) are distinct positive numbers.

Solution 1

Note that $x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2$, so the first two equations become

$\begin{matrix} \quad (x + y) + z \!\!\! &= a \; \; (*) \\ (x+y)^2 - z^2 \!\!\! &=b^2 (**) \end{matrix}$.

We note that $(x+y)^2 - z^2 = \Big[ (x+y)+z \Big]\Big[ (x+y)-z\Big]$, so if $a$ equals 0, then $b$ must also equal 0. We then have $x+y = -z$; $xy = (x+y)^2$. This gives us $x^2 + xy + y^2 = 0$. Mutiplying both sides by $(x-y)$, we have $x^3 - y^3 = 0$. Since we want $x,y$ to be real, this implies $x = y$. But $x^2 + x^2 + x^2$ can only equal 0 when $x=0$ (which, in this case, implies $y,z = 0$). Hence there are no positive solutions when $a = 0$.

When $a \neq 0$, we divide $(**)$ by $(*)$ to obtain the system of equations

$\begin{matrix} (x+y)+z &= a \; \quad \\ (x+y)-z &= b^2/a \end{matrix}$,

which clearly has solution $x+y = \frac{a^2 + b^2}{2a}$, $z = \frac{a^2 - b^2}{2a}$. In order for these both to be positive, we must have positive $a$ and $a^2 > b^2$. Now, we have $x+y = \frac{a^2 + b^2}{2a}$; $xy = \left(\frac{a^2 - b^2}{2a}\right)^2$, so $x,y$ are the roots of the quadratic $m^2 - \frac{a^2 + b^2}{2a}m + \left(\frac{a^2 - b^2}{2a}\right)^2$. The discriminant for this equation is

$\left(\frac{a^2 + b^2}{2a}\right)^2 - \left(2\frac{a^2 -b^2}{2a}\right)^2 = \frac{ (3a^2 - b^2)(3b^2 - a^2) }{4a^2}$.

If the expressions $(3a^2 - b^2), (3b^2 - a^2)$ were simultaneously negative, then their sum, $2(a^2 + b^2)$, would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when $3a^2 > b^2$ and $3b^2 > a^2$. But we have already replaced the first inequality with the sharper bound $a^2 > b^2$. It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from $\left(\frac{a^2 + b^2}{2a}\right)^2 > \left(\frac{a^2 + b^2}{2a}\right)^2 - \left(2\frac{a^2 -b^2}{2a}\right)^2$ or from Descartes' Rule of Signs). We have now found the solutions to the system, and determined that it has positive solutions if and only if $a$ is positive and $3b^2 > a^2 > b^2$. Q.E.D.

Solution 2

Obviously, $a=x+y+z>0$. The third equation implies that $x,z,y$ is a geometric sequence. Then let $x=\frac{z}{r}$ and $y=rz$, with $r,z>0$ and $r\neq1$. Then the first two equations become: \[\left(r+1+\frac{1}{r}\right)z=a~~~(1)\] and \[\left(r^2+1+\frac{1}{r^2}\right)z^2=b^2~~~(2)\] Taking $\frac{(2)}{(1)}$ (since $z>0$), we get: \[\frac{(r^2+1+\frac{1}{r^2})z^2}{(r+1+\frac{1}{r})z}=\left(r-1+\frac{1}{r}\right)z=\frac{b^2}{a}~~~(3)\] We can then take $(1)^2-(2)$ and $(2)-(3)^2$ to get: \[\left(r^2+2r+3+\frac{2}{r}+\frac{1}{r^2}\right)z^2-\left(r^2+1+\frac{1}{r^2}\right)z^2=2z^2\left(r+1+\frac{1}{r}\right)=a^2-b^2~~~(4)\] and \[\left(r^2+1+\frac{1}{r^2}\right)z^2-\left(r^2-2r+3-\frac{2}{r}+\frac{1}{r^2}\right)z^2=2z^2\left(r-1+\frac{1}{r}\right)=b^2-\frac{b^4}{a^2}=\frac{b^2}{a^2}(a^2-b^2)~~~(5)\] Let $k=r+\frac{1}{r}$. By AM-GM, $k\ge2$ with equality at $r=\frac{1}{r}\implies r=1$, which is impossible. Hence, $k>2$. Then, $\frac{(5)}{(4)}$ becomes: \[\frac{k-1}{k+1}=\frac{b^2}{a^2}~(6)\implies a^2+b^2=(a^2-b^2)k>2(a^2-b^2)\implies3b^2>a^2\] From the above restrictions on $a$ and $b$, we see that there must exist some $k>2$ satisfying $(6)$, and hence, some $r>0\neq1$ satisfying $(6)$. From $(4)$, if $a^2-b^2>0$, then there must exist some positive $z$ satisfying $(4)$, and consequently since $(4)$ and $(6)$ are equivalent to the remaining equations, they satisfy $(1)$ and $(2)$. Hence, $x,y,z$ satisfy the original system, and from the restrictions on $r$ and $z$, they are distinct positive reals. Hence, $\boxed{a>0\text{ and }3b^2>a^2>b^2}$. $\blacksquare$

~rhydon516

Video Solution

https://www.youtube.com/watch?v=_stCjNU0_M4&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=3 - AMBRIGGS

https://youtu.be/e5cuvmW0clk [Video Solution by little fermat]


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1961 IMO (Problems) • Resources
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First question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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