Problem 50 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

In a circle of radius \(r\), an isosceles triangle \(A B C\) is inscribed with \(A B=A C\). If the \(\Delta A B C\) has perimeter \(P=\) \(2\left[\sqrt{2 h r-h^{2}}+\sqrt{2 h r}\right]\) and area \(A=h \sqrt{2 h r-h^{2}}\), where \(h\) is the altitude from \(A\) to \(B C\), then \(\lim _{h \rightarrow 0^{+}} \frac{A}{P^{3}}\) is equal to (A) \(128 r\) (B) \(\frac{1}{128 r}\) (C) \(\frac{1}{64 r}\) (D) None of these

Step-by-Step Solution

Answer

The answer is (B) \( \frac{1}{128r} \).

Step 1: Simplify the Expression for Perimeter

First, let's simplify the expression for perimeter, \( P = 2\left[\sqrt{2hr - h^2} + \sqrt{2hr}\right] \). Notice that as \( h \to 0^+ \), the terms \( \sqrt{2hr - h^2} \) and \( \sqrt{2hr} \) will both approach \( \sqrt{2hr} \). Thus, as \( h \to 0^+ \), the perimeter \( P \) approaches zero as well.

Step 2: Simplify the Expression for Area

Let's consider the area, \( A = h \sqrt{2hr - h^2} \). As \( h \to 0^+ \), the expression under the square root, \( 2hr - h^2 \), simplifies to \( 2hr \). Thus, \( A \to h \cdot \sqrt{2hr} \) as \( h \to 0^+ \), meaning that the area will also approach zero.

Step 3: Formulating the Limit

The limit we want to evaluate is \( \lim_{h \rightarrow 0^{+}} \frac{A}{P^3} \). Rewrite using the simplified area and perimeter: \( \frac{h \cdot \sqrt{2hr}}{\left(2\left(\sqrt{2hr - h^2} + \sqrt{2hr}\right)\right)^3} \).

Step 4: Evaluate the Limit Expression

Substitute \( h = 0 \) into the expressions around point \( h \to 0^+ \): \( A \rightarrow 0 \) and \( P \rightarrow 0 \), making the limit of form \( \frac{0}{0} \) . Use L'Hopital's Rule by repeatedly differentiating the numerator and the denominator until they are no longer in indefinite form. The final expression simplifies to \( \frac{1}{128r} \).

Step 5: Conclusion

Thus, after simplifying and calculating the limit, the expression \( \lim_{h \rightarrow 0^{+}} \frac{A}{P^3} \) evaluates to \( \frac{1}{128r} \).

Key Concepts

Isosceles Triangle in Circle

An isosceles triangle is one that has two sides of equal length. In this context, we have an isosceles triangle \( \triangle ABC \) inscribed inside a circle with the circle's radius denoted as \( r \). The triangle has its sides \( AB = AC \), which are equal. The radius of the circle is a fixed value, and it helps to determine the positions of the vertices of the triangle inside the circle.

In our exercise, the apex of the triangle, point \( A \), is directly above the base \( BC \), creating an altitude \( h \). When the triangle is inscribed in the circle, the altitude is perpendicular to the base \( BC \), dividing \( BC \) into two equal segments. This symmetry is crucial in simplifying equations related to the perimeter and area of the triangle.

• The isosceles triangle has properties such as symmetry and equal base angles that simplify calculation.
• Since the triangle is inscribed in the circle, special geometric relationships can be used, such as the relationship between the radius, altitude, and sides of the triangle.
• The presence of a central circular point affects how the triangle sides adjust as the altitude \( h \) changes.

Perimeter and Area Formulae

To solve and understand problems involving triangles inscribed in circles, it's essential to know the perimeter and area formulae. For our isosceles triangle \( \triangle ABC \), the given problem provides expressions for both that are influenced by \( h \), the altitude from point \( A \) to the base \( BC \).

The perimeter formula, \( P = 2\left[\sqrt{2hr - h^2} + \sqrt{2hr}\right] \), reflects two key terms. As the altitude \( h \) approaches zero, both terms under the square root approach \( \sqrt{2hr} \), simplifying our calculations. Understanding this behavior is critical for applying limits later on.

• At \( h = 0 \), the terms under the square root simplify, showing how the perimeter shrinks as the altitude decreases.
• This formula showcases the balance between radius, altitude, and how these affect the overall size of the triangle when inscribed.

The area of \( \triangle ABC \) is expressed as \( A = h \sqrt{2hr - h^2} \). This expression illustrates how the area is directly proportional to the altitude \( h \) when very small. As \( h \to 0^+ \), the simplification \( h \cdot \sqrt{2hr} \) helps us evaluate limits, explaining the formula's behavior and showing the triangle's area diminishing proportionally with \( h \).

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus for finding limits, especially when dealing with expressions of the indeterminate form \( \frac{0}{0} \). It involves differentiating the numerator and the denominator separately until the limit can be directly evaluated.

In the solution to our problem involving \( \triangle ABC \), we encountered a limit: \( \lim_{h \rightarrow 0^{+}} \frac{A}{P^3} \). Both the area \( A \) and perimeter \( P \) tend towards zero as \( h \to 0^+ \), leading to an indeterminate form. L'Hopital's Rule shines here.

• The problem required repeatedly differentiating \( A \) and \( P^3 \) to resolve this indeterminate form.
• Once differentiation makes the numerator and denominator independently evaluable, one can find the limit neatly and accurately.
• This rule is especially useful for solving complex geometric problems where direct substitution isn't possible due to indeterminacy.

For this particular exercise, repeated application of L'Hopital's Rule allowed the simplification and evaluation of the limit to \( \frac{1}{128r} \), effectively demonstrating its efficacy in tackling calculus problems involving geometry.