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

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
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

@@ Line 1: / Line 1: @@
-Solve the following system of equations:
+== Problem ==
-<math>x + y + z = a</math>
+(''Hungary'')
+Solve the system of equations:
-<math>x^2</math><math>+y^2+z^2=b^2</math>
+<center>
+<math>
+\begin{matrix}
+\quad x + y + z \!\!\! &= a \; \, \\
+x^2 +y^2+z^2 \!\!\! &=b^2 \\
+\qquad \qquad xy \!\!\!  &= z^2
+\end{matrix}
+</math>
+</center>
-<math>xy = z^2</math>
+where <math>a </math> and <math>b </math> are constants.  Give the conditions that <math>a </math> and <math>b </math> must satisfy so that <math>x, y, z </math> (the solutions of the system) are distinct positive numbers.
-where ''a'' and ''b'' are given real numbers.  What conditions must hold on ''a'' and ''b'' for the solutions to be positive and distinct?
+== Solution 1==
+Note that <math>x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2 </math>, so the first two equations become
+<center>
+<math>
+\begin{matrix}
+\quad (x + y) + z \!\!\! &= a \; \; (*) \\
+(x+y)^2 - z^2 \!\!\! &=b^2 (**)
+\end{matrix}
+</math>.
+</center>
+We note that <math>(x+y)^2 - z^2 = \Big[ (x+y)+z \Big]\Big[ (x+y)-z\Big] </math>, so if <math>a </math> equals 0, then <math>b </math> must also equal 0.  We then have <math> x+y = -z </math>; <math>xy = (x+y)^2 </math>.  This gives us <math>x^2 + xy + y^2 = 0 </math>.  Mutiplying both sides by <math>(x-y) </math>, we have <math>x^3 - y^3 = 0 </math>.  Since we want <math>x,y </math> to be real, this implies <math>x = y </math>.  But <math>x^2 + x^2 + x^2 </math> can only equal 0 when <math>x=0 </math> (which, in this case, implies <math>y,z = 0 </math>).  Hence there are no positive solutions when <math>a = 0 </math>.
+When <math>a \neq 0 </math>, we divide <math>(**) </math> by <math>(*) </math> to obtain the system of equations
+<center>
+<math>
+\begin{matrix}
+(x+y)+z &= a \; \quad \\
+(x+y)-z &= b^2/a
+\end{matrix}
+</math>,
+</center>
+which clearly has solution <math> x+y = \frac{a^2 + b^2}{2a} </math>, <math> z = \frac{a^2 - b^2}{2a} </math>.  In order for these both to be positive, we must have positive <math>a </math> and <math>a^2 > b^2 </math>.  Now, we have <math> x+y = \frac{a^2 + b^2}{2a} </math>; <math> xy = \left(\frac{a^2 - b^2}{2a}\right)^2 </math>, so <math>x,y </math> are the roots of the quadratic <math> m^2 - \frac{a^2 + b^2}{2a}m + \left(\frac{a^2 - b^2}{2a}\right)^2 </math>.  The [[discriminant]] for this equation is
+<center>
+<math>
+\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}
+</math>.
+</center>
+If the expressions <math>(3a^2 - b^2), (3b^2 - a^2) </math> were simultaneously negative, then their sum, <math>2(a^2 + b^2) </math>, would also be negative, which cannot be.  Therefore our quadratic's discriminant is positive when <math>3a^2 > b^2 </math> and <math>3b^2 > a^2 </math>.  But we have already replaced the first inequality with the sharper bound <math>a^2 > b^2 </math>.  It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from <math> \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 </math> or from [[Polynomial#Descartes.27_Law_of_Signs | Descartes' Rule of Signs]]).  We have now found the solutions to the system, and determined that it has positive solutions if and only if <math>a </math> is positive and <math>3b^2 > a^2 > b^2 </math>.  Q.E.D.
+==Solution 2==
+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>
+and
+<cmath>\left(r^2+1+\frac{1}{r^2}\right)z^2=b^2~~~(2)</cmath>
+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>
+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===
+https://www.youtube.com/watch?v=_stCjNU0_M4&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=3
+- AMBRIGGS
+https://youtu.be/e5cuvmW0clk  [Video Solution by little fermat]
+{{alternate solutions}}
+{{IMO box|year=1961|before=First question|num-a=2}}
+[[Category:Olympiad Algebra Problems]]

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