Solution of Triangles: Distance of incentre from the vertices

Solution of Triangles: Distance of Incentre from the Vertices

The solution of triangles is an important topic in trigonometry that deals with finding various elements of a triangle, such as its sides, angles, and area, given some initial information. One of the key points of interest in a triangle is the incentre, which is the point where the angle bisectors of the triangle intersect. The incentre is equidistant from all sides of the triangle and is the center of the inscribed circle (incircle), which touches all the sides of the triangle.

Understanding the Incentre

The incentre (I) of a triangle can be found by the intersection of the angle bisectors of the three angles. The distances from the incentre to the vertices of the triangle are not equal, unlike the distances from the incentre to the sides of the triangle (which are all equal to the radius of the incircle).

Formulas for Distance of Incentre from Vertices

The distance of the incentre from the vertices of the triangle can be calculated using the following formulas:

  1. Distance from Incentre to Vertex A (IA): $$ IA = \frac{2}{b+c-a} \cdot \sqrt{s(s-a)(s-b)(s-c)} $$

  2. Distance from Incentre to Vertex B (IB): $$ IB = \frac{2}{c+a-b} \cdot \sqrt{s(s-a)(s-b)(s-c)} $$

  3. Distance from Incentre to Vertex C (IC): $$ IC = \frac{2}{a+b-c} \cdot \sqrt{s(s-a)(s-b)(s-c)} $$

Where:

  • ( a, b, c ) are the lengths of the sides of the triangle opposite to vertices A, B, and C, respectively.
  • ( s ) is the semi-perimeter of the triangle, given by ( s = \frac{a+b+c}{2} ).

Table of Differences and Important Points

Property Distance from Incentre to Vertices Distance from Incentre to Sides
Equality The distances are not necessarily equal All distances are equal (radius of incircle)
Formula Depends on the semi-perimeter and sides Equal to the radius of the incircle
Affected by Changes in the side lengths Not affected by changes in the side lengths as long as the angles remain the same

Examples

Let's go through an example to understand how to calculate the distance of the incentre from the vertices of a triangle.

Example 1: Calculating Distance from Incentre to Vertices

Given a triangle ABC with sides ( a = 7 ) units, ( b = 8 ) units, and ( c = 9 ) units, find the distance of the incentre from each vertex.

Solution:

  1. Calculate the semi-perimeter ( s ): $$ s = \frac{a+b+c}{2} = \frac{7+8+9}{2} = 12 \text{ units} $$

  2. Use the formulas to find the distances:

    • Distance from Incentre to Vertex A (IA): $$ IA = \frac{2}{b+c-a} \cdot \sqrt{s(s-a)(s-b)(s-c)} $$ $$ IA = \frac{2}{8+9-7} \cdot \sqrt{12(12-7)(12-8)(12-9)} $$ $$ IA = \frac{2}{10} \cdot \sqrt{12 \cdot 5 \cdot 4 \cdot 3} $$ $$ IA = \frac{1}{5} \cdot \sqrt{720} $$ $$ IA = \frac{1}{5} \cdot 6\sqrt{20} $$ $$ IA = \frac{6\sqrt{20}}{5} \approx 5.385 \text{ units} $$

  • Distance from Incentre to Vertex B (IB): $$ IB = \frac{2}{c+a-b} \cdot \sqrt{s(s-a)(s-b)(s-c)} $$ $$ IB = \frac{2}{9+7-8} \cdot \sqrt{12 \cdot 5 \cdot 4 \cdot 3} $$ $$ IB = \frac{2}{8} \cdot \sqrt{720} $$ $$ IB = \frac{1}{4} \cdot 6\sqrt{20} $$ $$ IB = \frac{6\sqrt{20}}{4} \approx 6.708 \text{ units} $$

  • Distance from Incentre to Vertex C (IC): $$ IC = \frac{2}{a+b-c} \cdot \sqrt{s(s-a)(s-b)(s-c)} $$ $$ IC = \frac{2}{7+8-9} \cdot \sqrt{12 \cdot 5 \cdot 4 \cdot 3} $$ $$ IC = \frac{2}{6} \cdot \sqrt{720} $$ $$ IC = \frac{1}{3} \cdot 6\sqrt{20} $$ $$ IC = \frac{6\sqrt{20}}{3} \approx 8.944 \text{ units} $$

In this example, we can see that the distances from the incentre to the vertices are different and depend on the side lengths of the triangle.