A Problem from Iranian Mathematical Olympiad 1995
March 16, 2024
Problem 6 from 13th Iranian Mathematical Olympiad 1995
Let \(a, b, c\) be positive real numbers. Find all real numbers \(x, y, z\) such that \[ \begin{gathered} x+y+z=a+b+c \\ 4 x y z-\left(a^2 x+b^2 y+c^2 z\right)=a b c . \end{gathered} \]
This is not a real solution though. Please see section below for a solution in Crux Mathematicorum (CRUX) Vol. 28, No. 2, pp.72 by by Mohammed Aassila, Strasbourg, France.
How did the problem writer come up with this problem? Most likely they started from an acute triangle with side lengths \(\sqrt{x}\), \(\sqrt{y}\) and \(\sqrt{z}\). Let the corresponding angles be \(\alpha\), \(\beta\) and \(\gamma\). By the law of cosines, we have \[ \begin{aligned} & x = y + z - 2 \sqrt{yz} \cos \alpha \\ & y = x + z - 2 \sqrt{xz} \cos \beta \\ & z = x + y - 2 \sqrt{xy} \cos \gamma \\ \end{aligned} \] Let \(a = 2 \sqrt{yz} \cos \alpha\), \(b = 2 \sqrt{xz} \cos \beta\), \(c = 2 \sqrt{xy} \cos \gamma\), we have \(x = \dfrac{b+c}{2}\), \(y = \dfrac{a+c}{2}\), \(z = \dfrac{a+b}{2}\), and \[ \boxed{x+y+z = a + b + c}.\]
Also note that for \(\alpha + \beta + \gamma = \pi\), the following trignometry identity holds:
\[ 1 - 2 \cos\alpha \cos\beta \cos\gamma = \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma.\]
Substituting \(\cos \alpha\), \(\cos \beta\) and \(\cos \gamma\), we get
\[ 1- \frac{2abc}{8xyz} = (\frac{a}{2\sqrt{yz}})^2 + (\frac{b}{2\sqrt{xz}})^2 + (\frac{c}{2\sqrt{xy}})^2 \]
Multiplying \(4xyz\) on both sides yields \[ \boxed{4xyz - abc = a^2x + b^2y + c^2z}. \]
Finally we note that since \(\sqrt{x}\), \(\sqrt{y}\) and \(\sqrt{z}\) are sides of an acute triangle, \(2x\), \(2y\) and \(2z\) also make a triangle, meaning there exist three positive reals \(a\), \(b\), \(c\) such that \(2x=b+c\), \(2y=a+c\) and \(2z=a+b\), which is also known as Ravi substitution.
Source: Crux Mathematicorum (CRUX) Vol. 28, No. 2, pp.72 for a solution by Mohammed Aassila, Strasbourg, France.
The second equation is equival ent to \[ 4=\frac{a^2}{y z}+\frac{b^2}{z x}+\frac{c^2}{x y}+\frac{a b c}{x y z} \] and also to \[ 4=x_1^2+y_1^2+z_1^2+x_1 y_1 z_1 . \] where \[ \begin{gathered} 0<x_1=\frac{a}{\sqrt{y z}}<2, \quad 0<y_1=\frac{b}{\sqrt{z x}}<2, \\ 0<z_1=\frac{c}{\sqrt{x y}}<2 . \end{gathered} \] 73 Setting \(x_1=2 \sin u, 0<u<\frac{\pi}{2}\), and \(y_1=2 \sin v, 0<v<\frac{\pi}{2}\), we have \[ 4=4 \sin ^2 u+4 \sin ^2 v+z_1^2+4 \sin u \cdot \sin v \cdot z_1 . \]
Hence, \[ z_1+2 \sin u \cdot \sin v=2 \cos u \cdot \cos v, \] and then, \[ z_1=2(\cos u \cdot \cos v-\sin u \cdot \sin v) \quad(=2 \cos (u+v)) . \]
Thus, \[ \begin{aligned} & a=2 \sqrt{y z} \sin u \\ & b=2 \sqrt{z x} \sin v \\ & c=2 \sqrt{x y}(\cos u \cdot \cos v-\sin u \cdot \sin v) . \end{aligned} \]
From \(x+y+z=a+b+c\), we get \[ (\sqrt{x} \cos v-\sqrt{y} \cos u)^2+(\sqrt{x} \sin v+\sqrt{y} \sin u-\sqrt{z})^2=0 \] which implies \[ \sqrt{z}=\sqrt{x} \sin v+\sqrt{y} \sin u=\sqrt{x} \frac{y_1}{2}+\sqrt{y} \frac{x_1}{2} . \]
Therefore, \[ \sqrt{z}=\sqrt{x} \cdot \frac{b}{2 \sqrt{z x}}+\sqrt{y} \cdot \frac{a}{2 \sqrt{y z}} . \] and thus, \(z=\frac{a+b}{2}\). Similarly, \(x=\frac{a+b}{2}, y=\frac{c+a}{2}\). The triple \[ (x, y, z)=\left(\frac{b+c}{2}, \frac{c+a}{2}, \frac{a+b}{2}\right) \] is the unique solution. Comment by Pierre Bornsztein, Pontoise, France. The problem and a solution are in ” \(36^{\text {th }}\) International Mathematical Olympiad”, published by the Canadian Mathematical Society p. 122-123.