Problem 19 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Each of the two triplets of numbers \(\log a, \log b, \log c\) and \(\log a-\log 2 b, \log 2 b-\log 3 c, \log 3 c-\log a\) are in A.P. Can the numbers \(a, b, c\) be the lengths of the sides of a triangle?

Step-by-Step Solution

Answer

The three numbers a, b, and c found from the given arithmetic progressions cannot form the lengths of the sides of a triangle since the triangle inequality theorem is not satisfied. We derived the expressions for the sides as \(a = 3c\), \(b = \frac{3}{2}c\), and \(c = c\), but the triangle inequality conditions resulted in contradictions, suggesting that the side lengths cannot form a triangle.

Step 1: Write down the given information.

We have two arithmetic progressions: 1) \(\log a, \log b, \log c\) 2) \(\log a- \log 2b, \log 2b- \log 3c, \log 3c- \log a\)

Step 2: Use the properties of logarithms to simplify the second progression.

Recall that \(\log a - \log b = \log(\frac{a}{b})\). Using this property, we can rewrite the second progression as follows: \(\log(\frac{a}{2b}), \log(\frac{2b}{3c}), \log(\frac{3c}{a})\)

Step 3: Use the arithmetic progression property to derive equations.

For both arithmetic progressions, let the common difference be d. Thus, for the first progression: \[\log b = \log a + d\] \[\log c = \log a + 2d\] And for the second progression: \[\log(\frac{2b}{3c}) = \log(\frac{a}{2b}) + d\] \[\log(\frac{3c}{a}) = \log(\frac{a}{2b}) + 2d\]

Step 4: Eliminate d from the second progression's equations.

Subtract the first equation from the second in the second progression: \[\log(\frac{3c}{a}) - \log(\frac{2b}{3c}) = d\] \[\log(\frac{3c}{a}) = \log(\frac{a}{2b}) + d\] Now we have: \[\log(\frac{2b}{3c}) = \log(\frac{a}{2b})\] \[\log(\frac{3c}{a}) = \log(\frac{a}{2b}) + d\]

Step 5: Simplify and solve for a, b, and c.

From the first equation of the second progression, we get: \[\frac{a}{2b} = \frac{2b}{3c}\] Cross-multiplying, we have: \[3ac = 4b^2\] Now using the first equation of the second progression and the equation we just found, we get: \[\frac{3c}{a} = 1\] \[\Rightarrow a = 3c\] Now substituting this back into the earlier equation: \[3(3c)c = 4b^2\] \[9c^2 = 4b^2\] \[b = \frac{3}{2}c\] Therefore, the lengths of the sides are: \[a = 3c\] \[b = \frac{3}{2}c\] \[c = c\]

Step 6: Check if the triangle inequality theorem is satisfied.

The triangle inequality theorem states that for any triangle with sides of length a, b, and c: \[a + b > c\] \[a + c > b\] \[b + c > a\] Substituting our expressions for a, b, and c: \[3c+\frac{3}{2}c > c\] \[3c + c > \frac{3}{2}c\] \[\frac{3}{2}c + c > 3c\] After simplifying all three inequalities: \[\frac{9}{2}c > c\] \[4c > \frac{3}{2}c\] \[\frac{5}{2}c > 3c\] From the first inequality, we can deduce: \[c(\frac{7}{2}) > 0\] \[c > 0\] From the second inequality, we get: \[2c > \frac{3}{2}c\] \[c > 0\] From the third inequality, we find: \[-\frac{c}{2} > 0\] \[c < 0\] This is a contradiction as the third inequality suggests that c must be negative, but the first and second inequalities require c to be positive for a triangle to exist.

Step 7: Conclusion:

The three numbers a, b, and c found from the given arithmetic progressions cannot form the lengths of the sides of a triangle since the triangle inequality theorem is not satisfied.

Key Concepts

Logarithmic Properties

Understanding logarithmic properties is crucial when solving problems involving logarithmic expressions. Logarithms are the inverse operations of exponentiation, similar to how subtraction is the inverse of addition. The properties of logarithms allow us to manipulate and simplify logarithmic expressions efficiently.
One of the most fundamental properties is the

• Product Property: \(\log(a \times b) = \log a + \log b\)
• Quotient Property: \(\log \left( \frac{a}{b} \right) = \log a - \log b\)
• Power Property: \(\log(a^b) = b \cdot \log a\)

In the context of our exercise, we used the quotient property to transform terms in a sequence from subtraction. For example, \(\log a - \log b\) becomes \(\log \left( \frac{a}{b} \right)\).
These properties allow us to rewrite and eventually solve the arithmetic progression by simplifying the terms involved.

Triangle Inequality Theorem

The triangle inequality theorem is a fundamental property of all triangles in Euclidean Geometry. It states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
This can be formulated as:

• \(a + b > c\)
• \(a + c > b\)
• \(b + c > a\)

In our problem, to determine if the numbers derived from the arithmetic progressions can form a triangle, we must ensure these inequalities all hold true. We substitute the expressions we found for \(a\), \(b\), and \(c\), attempting to validate these inequalities. Through this, it revealed a contradiction—suggesting the derived conditions violated the theorem as not all inequalities were satisfied, indicating the impossibility of forming a triangle from these values. This contradiction underscores the importance of the triangle inequality theorem when determining valid side lengths for triangles.

Cross-Multiplication

Cross-multiplication is a technique often used to solve equations involving fractions or ratios. It involves multiplying across the equality in a diagonal pattern. If you have two fractions \(\frac{a}{b} = \frac{c}{d}\), you can solve for one of the variables by cross-multiplying:

• \(a \cdot d = b \cdot c\)

This technique simplifies the equation to one without fractions, allowing easier solving steps.
In our exercise, cross-multiplication was pivotal after relating the logarithmic equations to simplify the expressions into a standard algebraic form. By rearranging the terms in our logarithmic equality and using cross-multiplication, we transformed complex expressions into something more manageable, ultimately leading us to assess the potential side lengths \(a\), \(b\), and \(c\). Cross-multiplication helps bridge the gap between abstract logarithmic manipulation and straightforward algebraic solutions.