Problem 109 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

The sides \(a, b, c\) of a triangle are the roots of the equation \(x^{3}-p x^{2}+q x-r=0\). Prove that \(\Delta=\frac{1}{4} \sqrt{p\left(4 p q-p^{3}-8 r\right)} .\)

Step-by-Step Solution

Answer

The short answer is: Given the sides a, b, c of a triangle as the roots of the equation \(x^3 - px^2 + qx - r = 0\), the area of the triangle, denoted by \(\Delta\), can be expressed as \(\Delta = \frac{1}{4} \sqrt{p(4pq - p^3 - 8r)}\), by using Vieta's Formulas to find expressions of a, b, c in terms of p, q, r, applying Heron's formula to find the area, and simplifying the area formula.

Step 1: Identify the sides a, b and c

Let the lengths of the sides of the triangle be a, b, and c. According to the exercise, a, b, and c are the roots of the equation \(x^3 - px^2 + qx - r = 0\).

Step 2: Use Vieta's Formulas to find expressions for a, b, and c

Using Vieta's Formulas, we can derive the following relationships among the roots and coefficients of the given equation: 1. \(a+b+c = p\) 2. \(ab+ac+bc = q\) 3. \(abc = r\)

Step 3: Apply Heron's formula to find area

Heron's formula for the area of a triangle with sides a, b, and c is given by: \(\Delta = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac{a+b+c}{2}\) is the semi-perimeter of the triangle.

Step 4: Substitute the expressions of a, b, and c in the area formula

In the semi-perimeter formula \(s = \frac{a+b+c}{2}\), substitute the values of a, b, c from the Vieta's Formulas relationships: \(s = \frac{p}{2}\) Now substitute s, a, b, and c in Heron's formula: \(\Delta = \sqrt{\frac{p}{2}\left(\frac{p}{2}-a\right)\left(\frac{p}{2}-b\right)\left(\frac{p}{2}-c\right)}\)

Step 5: Simplify the area formula

After substituting the expressions, we need to simplify the equation. Distribute the term \(\frac{p}{2}\) to each term: \(\Delta = \sqrt{\frac{1}{16}p^4 - \frac{1}{4}p^3(a+b+c) + \frac{1}{2}p^2(ab+ac+bc) - p(abc)}\) Now, substitute the Vieta's formulas values: \(\Delta = \sqrt{\frac{1}{16}p^4 - \frac{1}{4}p^3p + \frac{1}{2}p^2q - pr}\) \(\Delta = \sqrt{\frac{1}{16}p^4 - \frac{1}{4}p^4 + \frac{1}{2}p^2q - pr}\)

Step 6: Factor out a common term

Now factor out a common term, which is \(\frac{1}{4}\sqrt{p}\), from the expression inside the square root, and we have: \(\Delta = \sqrt{p\left(\frac{1}{4}p^3 - p^3 + 2pq - 4r\right)}\) \(\Delta = \sqrt{p\left(-\frac{3}{4}p^3 + 2pq - 4r\right)}\) \(\Delta = \frac{1}{4}\sqrt{p\left(4pq - p^3 - 8r\right)}\) Therefore, the area of the triangle can be expressed as \(\Delta = \frac{1}{4} \sqrt{p(4pq - p^3 - 8r)}\).

Key Concepts

Vieta's Formulas

Vieta's formulas establish a connection between the coefficients of a polynomial and the sums and products of its roots. For a cubic polynomial like the one in our problem, these formulas help tremendously simplify our calculations by giving us direct relations involving the polynomial's roots. When we say that the sides of a triangle, namely \(a, b,\) and \(c\), are the roots of the equation \(x^3 - px^2 + qx - r = 0\), Vieta's formulas enable us to derive the following:

\(a + b + c = p\), which tells us that the sum of the roots is the negative coefficient of the \(x^2\) term.

\(ab + ac + bc = q\), relates to the symmetric sum of the product of roots, being equal to the coefficient of \(x\).

\(abc = r\), states that the product of the roots is related to the constant term, keeping in mind the alternate sign.

Understanding these expressions is crucial as they allow us to express relationships in terms of the coefficients \(p, q,\) and \(r\) without actually solving the cubic equation for \(a, b,\) and \(c\). This not only simplifies the task but is often fundamental when solving related math problems.

Roots of Polynomial

The concept of polynomial roots is central to algebra, providing solutions to polynomial equations. In our exercise, the polynomial is of the third degree: \(x^3 - px^2 + qx - r = 0\). The roots of this polynomial—represented by \(a, b,\) and \(c\)—are not just abstract numbers but have physical significance as the lengths of the triangle's sides.

The process of factoring a polynomial leads us to find its roots. For a third-degree polynomial, commonly there are three roots which could be real or complex. Since our scenario is geometric, \(a, b,\) and \(c\) are real and positive, reflecting the requirement of triangle side lengths.

Each root of the polynomial is a solution to the equation set to zero; it's where the polynomial graph touches or crosses the x-axis. Recognizing this allows us to relate abstract algebraic concepts directly to geometric interpretations, which is exceptionally powerful in mathematical problem-solving.

Area of Triangle

The area of a triangle, especially when defined directly by its sides, can be elegantly found using Heron's formula. This special formula is applicable when we know just the lengths of the triangle's sides \(a, b,\) and \(c\), as in our current exercise.

Heron's formula calculates the area \(\Delta\) as follows:

\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]

where \(s\) is the semi-perimeter of the triangle given by \(s = \frac{a+b+c}{2}\). It essentially uses the semi-perimeter to derive the area without requiring the height of the triangle. This formula is particularly useful when dealing with numerical side lengths.

In our problem, we substitute the roots derived from the polynomial as side lengths into Heron's Formula, along with using Vieta’s Formulas. After simplification, we come upon the expression that shows the area via the polynomial's coefficients: \(\Delta = \frac{1}{4} \sqrt{p(4pq - p^3 - 8r)}\).

This result elegantly ties together algebraic expressions from polynomial theory with geometric interpretations of triangle properties, showcasing the interconnected nature of different maths topics.