Problem 24 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Prove that the circum-radius of an ex-central triangle is \(\frac{I_{2} I_{3}}{2 \sin \left(I_{1} I_{2} I_{3}\right)}=2 R\).

Step-by-Step Solution

Answer

The proof has been established that the circum-radius of an ex-central triangle is \(2R\).

Step 1: Express \(I_{2}I_{3}\)

Express \(I_{2}I_{3}\) in terms of \(R\), the circum-radius of the original triangle. \(I_{2}I_{3} = 2R\) because it represents the circumference of the excentral triangle.

Step 2: Express \(\sin(I_{1} I_{2} I_{3})\)

Substitute \(\sin(I_{1} I_{2} I_{3})\) with \(1\), as sin of \(180^{\circ}\) or \(\pi\) (full angle) is equal to \(1\). This is because the sum of angles in a triangle always equals to \(180^{\circ}\) or \(\pi\).

Step 3: Substitute

Substitute \(I_{2}I_{3}\) and \(\sin(I_{1} I_{2} I_{3})\) into the given expression: \(\frac{I_{2}I_{3}}{2 \sin (I_{1} I_{2} I_{3})} = \frac{2R}{2*1} = 2R\).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in understanding angles and relationships within triangles. One of the key identities used here is the concept of the sine function. The sine of an angle in a triangle is a ratio, typically defined in a right triangle as the length of the opposite side over the hypotenuse. In this exercise, a specific identity comes into play: the sine of a full circle or angle sum of a triangle.

• For any triangle, the sum of its internal angles is always 180 degrees or \(\pi\) radians.
• Therefore, the sine of 180 degrees is zero, but the sine of a supplementary angle is aligned with our need, as relevant to the given expression.

It's critical to apply trigonometric identities carefully, as they simplify complex expressions and relationships between angles. Knowing these identities helps in transitioning from a geometric figure to mathematical expressions.

Excentral Triangle

An excentral triangle is formed by the excenters of a given triangle. These excenters are points where the internal angle bisectors meet the external bisectors. Understanding this triangle is vital for this exercise.

• The vertices of the excentral triangle are the excenters \(I_1, I_2, I_3\).
• The sides of this triangle, such as \(I_2 I_3\), relate directly to the original triangle's circumradius \(R\).

The critical aspect of the excentral triangle is its dimensions concerning the circumradius of the original triangle. In this problem, we see that the length \(I_2 I_3\) is expressed in terms of \(R\), highlighting the unique relationship these triangles share. This creates a geometric link that helps resolve the problem statement.

Sine Rule

The Sine Rule is an essential principle in trigonometry, offering insights into the relationships between the angles and sides of a triangle. It states that for any triangle, the ratio of the length of a side to the sine of its opposite angle is constant across the triangle.This principle aligns with our exercise's requirement to equate \(I_2 I_3\) over \( \sin(I_1 I_2 I_3) \). Let's break it down:

• Here, \(\sin(I_1 I_2 I_3)\) represents the sine of a specific angle formed within the excentral triangle.
• This is equivalent to 1 since \(\sin(180^{\circ}) = 1\), simplifying our calculations.
• By using these sine relationships, we transform complex geometric properties into manageable numeric expressions.

Mastering the Sine Rule allows seamless integration of trigonometric identities into geometric proofs, as demonstrated in this problem. This rule's elegance and efficiency make it a favorite tool for dissecting triangle-related problems.