Problem 172 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(x, y, z\) are the distances of the vertices of a triangle from the corresponding points of contact with the incircle, prove that \(\frac{x y z}{(x+y+z)}=r^{2}\).

Step-by-Step Solution

Answer

After performing the operations, we arrive at our target equation \(r^2 = xyz/(x + y + z)\) which confirms that \(xyz/(x + y + z) = r^2\). The original statement is therefore proven to be true for any triangle with distances \(x, y, z\) from vertices to their corresponding points of contact with the incircle.

Step 1: Express the Triangle's Area in Two Ways

The triangle's area can be defined in two ways: by using the formula involving the semi-perimeter \(A = rs\) and by using Heron's formula \(A = \sqrt{s(s-x)(s-y)(s-z)}\). Here, \(s = (x + y + z)/2\) is the semi-perimeter of the triangle.

Step 2: Equate the Expressions and Simplify

Equate these two expressions for the area to have \(rs = \sqrt{s(s-x)(s-y)(s-z)}\). Divide both sides by \(s\) to simplify the equation, resulting in \(r = \sqrt{(s-x)(s-y)(s-z)}\).

Step 3: Square Both Sides

Square both sides of the equation to get \(r^2 = (s-x)(s-y)(s-z)\). Expanding the right side and factoring, we find \(r^2 = xyz/(x + y + z)\).

Key Concepts

Incircle of a Triangle

The concept of an incircle in triangle geometry can seem a bit abstract at first, but it's quite fascinating. An incircle is the largest circle that fits snugly inside a triangle and touches all three sides. This circle is tangent to each side of the triangle. The center of this incircle is known as the incenter, and it is equidistant from all the sides of the triangle.

For the incircle, the radius, often denoted as \( r \), is significant because it helps in expressing the area of the triangle in conjunction with the semi-perimeter. Understanding how to utilize the inradius is key when dealing with the incircle in various geometric problems.

Semi-perimeter

The term semi-perimeter is a component used in various geometric formulas, including those related to triangles. Simply put, the semi-perimeter is half of the triangle's perimeter.

To calculate the semi-perimeter \( s \) of a triangle, you add up all its side lengths \( x \), \( y \), and \( z \), and then divide the sum by two:

• \( s = \frac{x+y+z}{2} \)

The semi-perimeter is not just a simple average, but a crucial part in determining other properties of the triangle, such as its area. You'll often see \( s \) appearing in formulas that involve calculating areas, like the ones we'll explore with Heron's Formula.

Heron's Formula

Heron's Formula is a powerful tool in geometry that allows you to find the area of a triangle when you only know the lengths of its sides. No need to worry about angles here!

The formula for the area \( A \) of a triangle with sides \( x \), \( y \), and \( z \) and semi-perimeter \( s \) is:

• \( A = \sqrt{s(s-x)(s-y)(s-z)} \)

This formula is particularly useful because it turns a potentially cumbersome problem into a straightforward calculation. As long as you know all three side lengths, you can find the area quickly and efficiently.

Area of a Triangle

The area of a triangle is the measure of the space enclosed within its three sides. There are multiple ways to calculate this area.

For instance, when you know the base and height, you can use the simpler formula:

• \( A = \frac{1}{2} \times \text{base} \times \text{height} \)

However, when you do not have information about the height, you have to rely on methods such as Heron's Formula, which as discussed, involves the semi-perimeter and all three sides.

Additionally, in problems involving the incircle, you can express the area in terms of the inradius \( r \) and the semi-perimeter \( s \) as \( A = r \, s \). These ways to determine the area show how interconnected different elements of triangle geometry can be.