Question

If \(y^2+z^2=3yz,\ z^2+x^2=8zx,\ x^2+y^2=4xy\), then the value of \(\dfrac{y^2}{xz}+\dfrac{xz}{y^2}\) is:

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

When equations are of the form \(a^2+b^2=kab\), divide by \(ab\) to convert them into reciprocal expressions like \(\frac{a}{b}+\frac{b}{a}=k\).

Answer 1

The Correct Option is C

Step 1: Divide each equation by the product of two variables.
From \(y^2+z^2=3yz\): divide by \(yz\) to get \(\tfrac{y}{z}+\tfrac{z}{y}=3\). From \(z^2+x^2=8zx\): divide by \(zx\) to get \(\tfrac{z}{x}+\tfrac{x}{z}=8\). From \(x^2+y^2=4xy\): divide by \(xy\) to get \(\tfrac{x}{y}+\tfrac{y}{x}=4\).

Step 2: Compute the required expression.
Let \(u=\tfrac{y}{z}, v=\tfrac{z}{x}, w=\tfrac{x}{y}\). Then \(\tfrac{y^2}{xz} = \tfrac{y}{z}\cdot\tfrac{y}{x} = u\cdot\tfrac{1}{w}\) and \(uvw=1\). Note \(\tfrac{y^2}{xz}\cdot\tfrac{xz}{y^2}=1\), so the expression \(t+\tfrac{1}{t}\) where \(t=\tfrac{y^2}{xz}\). From the three equations, \((\tfrac{y}{z}+\tfrac{z}{y})(\tfrac{x}{y}+\tfrac{y}{x}) - (\tfrac{z}{x}+\tfrac{x}{z}) = 3\cdot4-8=4\). One can verify \(t+\tfrac{1}{t}=4\), giving the answer 4.
\[\boxed{4}\]