Finding the Relationship Between Numbers Using HM and GM

This question asks us to find the relationship between two positive numbers, m and n, given a specific relationship between their Harmonic Mean (HM) and Geometric Mean (GM). Let's denote the HM as x and the GM as y.

Understanding Harmonic Mean (HM) and Geometric Mean (GM)

For two positive numbers, m and n:

• The Harmonic Mean (HM), x, is defined as the reciprocal of the arithmetic mean of the reciprocals of the numbers.
  \( x = \frac{2}{\frac{1}{m} + \frac{1}{n}} = \frac{2mn}{m+n} \)
• The Geometric Mean (GM), y, is defined as the square root of the product of the numbers.
  \( y = \sqrt{mn} \)

Using the Given Relationship 5x = 4y

We are given the relationship 5x = 4y. We can substitute the formulas for x and y into this equation:

\( 5 \left( \frac{2mn}{m+n} \right) = 4 \left( \sqrt{mn} \right) \)

This simplifies to:

\( \frac{10mn}{m+n} = 4\sqrt{mn} \)

Solving for the Relationship Between m and n

Since m and n are positive numbers, \(\sqrt{mn}\) is positive and non-zero. We can divide both sides of the equation by \(\sqrt{mn}\):

\( \frac{10mn}{(m+n)\sqrt{mn}} = \frac{4\sqrt{mn}}{\sqrt{mn}} \)

\( \frac{10\sqrt{mn}}{m+n} = 4 \)

Divide both sides by 2:

\( \frac{5\sqrt{mn}}{m+n} = 2 \)

Now, square both sides of the equation to eliminate the square root:

\( \left( \frac{5\sqrt{mn}}{m+n} \right)^2 = (2)^2 \)

\( \frac{25mn}{(m+n)^2} = 4 \)

Multiply both sides by \((m+n)^2\):

\( 25mn = 4(m+n)^2 \)

Expand the right side:

\( 25mn = 4(m^2 + 2mn + n^2) \)

\( 25mn = 4m^2 + 8mn + 4n^2 \)

Rearrange the terms to form a quadratic equation:

\( 0 = 4m^2 + 8mn + 4n^2 - 25mn \)

\( 4m^2 - 17mn + 4n^2 = 0 \)

Solving the Quadratic Equation for m and n

We can solve this quadratic equation by treating it as a quadratic in terms of \(\frac{m}{n}\). Divide the entire equation by \(n^2\) (assuming \(n \neq 0\)):

\( 4\frac{m^2}{n^2} - 17\frac{mn}{n^2} + 4\frac{n^2}{n^2} = 0 \)

\( 4\left(\frac{m}{n}\right)^2 - 17\left(\frac{m}{n}\right) + 4 = 0 \)

Let \(z = \frac{m}{n}\). The equation becomes a standard quadratic equation in z:

\( 4z^2 - 17z + 4 = 0 \)

We can solve this by factoring. We look for two numbers that multiply to \(4 \times 4 = 16\) and add up to -17. These numbers are -1 and -16.

\( 4z^2 - 16z - z + 4 = 0 \)

Factor by grouping:

\( 4z(z - 4) - 1(z - 4) = 0 \)

\( (4z - 1)(z - 4) = 0 \)

This gives two possible solutions for z:

• Case 1: \(4z - 1 = 0 \implies 4z = 1 \implies z = \frac{1}{4}\)
• Case 2: \(z - 4 = 0 \implies z = 4\)

Relating the Solution back to m and n

Recall that \(z = \frac{m}{n}\).

• Case 1: \(\frac{m}{n} = \frac{1}{4} \implies 4m = n\)
• Case 2: \(\frac{m}{n} = 4 \implies m = 4n\)

Both \(4m=n\) and \(m=4n\) are possible relationships between m and n based on the given condition.

Checking the Options

Let's compare our derived relationships with the given options:

• Option 1: 5m = 4n (Does not match either derived relationship)
• Option 2: 2m = n (Does not match either derived relationship)
• Option 3: 4m = 5n (Does not match either derived relationship)
• Option 4: m = 4n (Matches one of the derived relationships)

Therefore, the correct relationship between m and n from the given options is m = 4n.

Revision Table: Key Concepts

Additional Information: AM-GM-HM Inequality

For any set of positive numbers, the Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM) are related by the inequality: AM \(\ge\) GM \(\ge\) HM.

Equality holds if and only if all the numbers are equal. In our case, for two positive numbers m and n:

\( \frac{m+n}{2} \ge \sqrt{mn} \ge \frac{2mn}{m+n} \)

The problem states that \(5x = 4y\), which means \(5 \times HM = 4 \times GM\). Since both HM and GM are positive for positive m and n, this implies \(GM = \frac{5}{4} HM\). Since \(\frac{5}{4} > 1\), this relation \(GM > HM\) is consistent with the AM-GM-HM inequality when m and n are not equal. If m and n were equal, then \(HM=GM=AM\), and \(5x=4y\) would mean \(5x=4x\), which implies \(x=0\), which is not possible for positive numbers.

We solved the equation \(4m^2 - 17mn + 4n^2 = 0\), which arose from \(5HM = 4GM\). The roots gave us \(\frac{m}{n} = 4\) or \(\frac{m}{n} = \frac{1}{4}\). These represent the two scenarios where the ratio of the numbers satisfies the given condition. For instance, if \(m=4\) and \(n=1\), \(HM = \frac{2(4)(1)}{4+1} = \frac{8}{5}\) and \(GM = \sqrt{4 \times 1} = 2\). \(5 \times HM = 5 \times \frac{8}{5} = 8\). \(4 \times GM = 4 \times 2 = 8\). So, \(5HM=4GM\) holds. If \(m=1\) and \(n=4\), \(HM = \frac{2(1)(4)}{1+4} = \frac{8}{5}\) and \(GM = \sqrt{1 \times 4} = 2\). \(5 \times HM = 5 \times \frac{8}{5} = 8\). \(4 \times GM = 4 \times 2 = 8\). So, \(5HM=4GM\) holds.

The option m = 4n corresponds to one of the two possible ratios found from the quadratic equation.