Animated Solution for Mathematics - Trigonometry: Three forces $\vec{P},\vec{Q}$ and $\vec{R}$ acting along IA, IB and IC, where I is the incentre of a $△ABC$ are in equilibrium. Then $P:Q:R$ is

Imagine you are standing at the heart of a triangle, at the very point where the three angle bisectors meet—the incenter $I$. You are holding three forces, $\vec{P}$, $\vec{Q}$, and $\vec{R}$, pulling outwards along the lines $IA$, $IB$, and $IC$.

The problem states that these forces are in perfect equilibrium. When three concurrent forces are in equilibrium, we apply Lami's Theorem.

This theorem states that each force is proportional to the sine of the angle between the other two:

Let us focus on the triangle $△BIC$. Since $I$ is the incenter, the lines $IB$ and $IC$ are angle bisectors, meaning $\angle IBC=\frac{B}{2}$ and $\angle ICB=\frac{C}{2}$.

The sum of angles in any triangle is $180^{\circ }$, so $\angle BIC+\frac{B}{2}+\frac{C}{2}=180^{\circ }$. This leads to the expression:

In the main triangle $ABC$, we know $A+B+C=180^{\circ }$, which implies $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^{\circ }$. Therefore, $\frac{B}{2}+\frac{C}{2}=90^{\circ }-\frac{A}{2}$.

Substituting this back, we find:

By the elegance of symmetry, we can deduce the remaining angles: $\angle CIA=90^{\circ }+\frac{B}{2}$ and $\angle AIB=90^{\circ }+\frac{C}{2}$.

We now bridge geometry and trigonometry by substituting these angles into Lami's Theorem:

Using the trigonometric identity $\sin (90^{\circ }+\theta )=\cos \theta$, the denominators transform into $\cos \frac{A}{2}$, $\cos \frac{B}{2}$, and $\cos \frac{C}{2}$.

The final ratio of the forces is:

This result demonstrates how the internal geometric structure of a triangle dictates the balance of forces at its incenter.