What is In-Circle and In-Centre?
The incircle of a triangle is the largest circle that can fit inside the triangle and is tangent to all three sides. The centre of the incircle is called the incentre (I) of the triangle. The radius of the incircle, known as the inradius (r), is the distance from the incentre to any side of the triangle:
$r = \text{distance from } I \text{ to any side of the triangle}$
The incentre serves as a central point for constructing the incircle and is crucial for solving triangle geometry problems and applying incircle-related formulas.
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Incentre of a Triangle: Properties
The incentre ($I$) is the point where the internal angle bisectors of a triangle intersect. It is also the centre of the incircle, which touches all three sides of the triangle internally. Key properties of the incentre are:
If $I$ is the incentre of triangle $ABC$, then the line segments from the vertices to the points of tangency are equal:
$AE = AG, \quad CG = CF, \quad BF = BE$
Using the angle bisector theorem, the angles formed at the incentre are equal:
$\angle BAI = \angle CAI, \quad \angle BCI = \angle ACI, \quad \angle ABI = \angle CBI$
The sides of the triangle are tangents to the incircle, and the distances from the incentre to the points of tangency are equal to the inradius $r$:
$EI = FI = GI = r$
If $s$ is the semi-perimeter and $r$ is the inradius, the area of the triangle is:
$\Delta = A = s \cdot r$
The incentre always lies inside the triangle.
A circle that touches all sides of a triangle internally is called an incircle, and its centre is the incentre. The incentre of a triangle is generally denoted by $I$.
How to construct the incenter of a triangle
The construction of the incenter of a triangle is possible with the help of a compass. Here are the steps to construct the incenter of a triangle:
- Step 1: Place one of the compass's ends at one of the triangle's vertices. The other side of the compass is on one side of the triangle.
- Step 2: Draw two arcs on two sides of the triangle using the compass.
- Step 3: By using the same width as before, draw two arcs inside the triangle so that they cross each other from the point where each arc crosses the side.
- Step 4: Draw a line from the vertex of the triangle to where the two arcs inside the triangle cross.
- Step 5: Repeat the same process from the other vertex of the triangle.
- Step 6: The point at which the two lines meet or intersect is the incenter of a triangle.
In-Radius Formula
The radius of the in-circle of a triangle is called the in-radius and is denoted by $r$.
In-radius in terms of area ($\Delta$) and semi-perimeter ($s$):
$r = \frac{\Delta}{s}$
In-radius in terms of sides and angles:
$r = (s-a) \tan \frac{A}{2} = (s-b) \tan \frac{B}{2} = (s-c) \tan \frac{C}{2}$
In-radius in terms of circumradius ($R$) and angles:
$r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
Derivation of In-Radius and Related Formulas
Area in Terms of In-Radius
Consider triangle $ABC$. The area of the triangle can be written as the sum of areas of triangles formed with the in-centre $I$:
$\Delta = \Delta IBC + \Delta ICA + \Delta IAB$
Using the in-radius $r$ and sides $a$, $b$, $c$:
$\Delta = \frac{1}{2} a r + \frac{1}{2} b r + \frac{1}{2} c r = \frac{1}{2} r (a + b + c) = \frac{1}{2} r (2s) \ [\text{since } s = \frac{a+b+c}{2}] = r s$
Using Half-Angle Tangent Formula
From the half-angle formula:
$\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
Multiplying both sides by $(s-a)$:
$(s-a) \tan \frac{A}{2} = \sqrt{(s-a)(s-b)(s-c)/s} = \frac{\Delta}{s} = r \ [\text{since } \Delta = \sqrt{s(s-a)(s-b)(s-c)}]$
Similarly, other angles can be used to prove related formulas.
Using Half-Angle Sine Formulas
$\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$
$\sin \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}$
$\sin \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}}$
Using $\Delta = \frac{abc}{4R}$ and $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, the in-radius can also be expressed as:
$4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4R \frac{(s-b)(s-c)}{bc} \frac{(s-a)(s-c)}{ac} \frac{(s-a)(s-b)}{ab} = \frac{4R (s-a)(s-b)(s-c)}{abc} = 4R \frac{\Delta}{abc} = r$
Length of Tangents from Vertices to the In-Circle
If $AF = AE = x$, $BF = BD = y$, and $CE = CD = z$, then:
$a + b + c = 2x + 2y + 2z \Rightarrow s = x + y + z$
Hence, tangent lengths are:
$AE = s - a, \quad BF = s - b, \quad CE = s - c$
Distance of In-Centre from Vertices
$AI = \frac{r}{\sin \frac{A}{2}}, \quad BI = \frac{r}{\sin \frac{B}{2}}, \quad CI = \frac{r}{\sin \frac{C}{2}}$
Length of Angle Bisectors
$AP = \frac{2bc}{b+c} \cos \frac{A}{2}, \quad BQ = \frac{2ac}{a+c} \cos \frac{B}{2}, \quad CR = \frac{2ab}{a+b} \cos \frac{C}{2}$
Relation Between In-Centre and Other Triangle Centres
The in-centre of a triangle is closely related to other triangle centres like the circumcentre, centroid, and orthocentre. While each centre has a unique geometric property, comparing them helps in understanding the triangle’s internal and external structure. In this section, we explore the relation between in-centre and other triangle centres with clear differences and formulas.
In-Centre vs Circumcentre
The in-centre (I) is the intersection of angle bisectors, while the circumcentre (O) is the intersection of perpendicular bisectors. In an equilateral triangle, both coincide, but in other triangles, they are distinct points.
In-Centre vs Orthocentre
The orthocentre (H) is the intersection of altitudes, whereas the in-centre is always inside the triangle. The in-centre’s location depends on the triangle type, but the orthocentre may lie outside in obtuse-angled triangles.
In-Centre vs Centroid
The centroid (G) is the intersection of medians, representing the triangle’s center of mass. The in-centre is related to the sides and angles, while the centroid depends solely on vertex coordinates:
$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$
Important Formulas Related to In-Circle and In-Centre
This section covers the important formulas related to the in-circle and in-centre of a triangle, including the in-centre coordinates, in-circle radius, and trigonometric relations. These formulas are essential for solving triangle geometry problems and practicing exam questions efficiently.
Radius Formula in Terms of Area and Semiperimeter
For a triangle with area $A$ and semi-perimeter $s$, the in-circle radius $r$ is:
$r = \frac{A}{s}, \quad s = \frac{a + b + c}{2}$
where $a$, $b$, and $c$ are the sides of the triangle.
Trigonometric Formulas for In-Circle
The in-circle radius can also be expressed in terms of angles:
$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
where $R$ is the circumradius, and $A$, $B$, $C$ are the angles of the triangle.
Coordinates of the in-centre $(I)$ in terms of vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$ and side lengths $a$, $b$, $c$ are:
$I = \left(\frac{a x_1 + b x_2 + c x_3}{a + b + c}, \frac{a y_1 + b y_2 + c y_3}{a + b + c}\right)$
Solved Examples Based on In-Circle and In-Centre
Example 1: To find the coordinates of the incenter of a triangle whose vertices are given as A(−36,15)B(0,0), and C(20,15).
Solution: Given that,
$A(-36, 15) = (x_1, y_1)$, $B(0, 0) = (x_2, y_2)$, $C(20, 15) = (x_3, y_3)$
To find the length of the sides using the distance formula:
$AB = c = \sqrt{(-36 - 0)^2 + (15 - 0)^2} = 39$
$BC = a = \sqrt{(0 - 20)^2 + (0 - 15)^2} = 25$
$CA = b = \sqrt{(20 + 36)^2 + (15 - 15)^2} = 56$
Substituting the values in the incenter formula:
$\left(\frac{a x_1 + b x_2 + c x_3}{a + b + c}, \frac{a y_1 + b y_2 + c y_3}{a + b + c}\right) = \left(\frac{-900 + 0 + 780}{120}, \frac{375 + 0 + 585}{120}\right) = (-1, 8)$
Hence, the answer is $(-1, 8)$.
Example 2: Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x+y=3. If R and r be the radius of circumcircle and incircle respectively of △ABC then (R+r) is equal to:
Solution:
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$r = OM = 32 \sin 30^\circ = 12$
$rR \Rightarrow R = 62$
$\therefore r + R = 92$
Hence, the answer is 92.
Example 3: Let a,b, and c be the length of sides of a triangle ABC such that a+b7=b+c8=c+a9 If r and R are the radius of the incircle and radius of the circumcircle of the triangle ABC, respectively, then the value of Rr is equal to:
[JEE MAINS 2022]
Solution:
Let $a + b7 = b + c8 = c + a9 = k$
So $a + b = 7k$, $b + c = 8k$, $c + a = 9k$
To solve this, we get $a = 4k$, $b = 3k$, $c = 5k$ and calculate semi-perimeter $s = \frac{a+b+c}{2} \Rightarrow s = 6k$
We know that $rR = 4 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}$
Hence, the answer is $5/2$.
Example 4: In triangle ABC in the ratio of a:b:c=3:4:5 then find the ratio of the radius of the circumcircle to that of the inner circle.
Solution:
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$AB \perp BC$
So AC is the radius of the circumcircle
$2R = AC = 2R = 5$
$EC = FC = 4 - r$
$AD = AF = 3 - r$
$AC = AF + FC$
$5 = 3 - r + 4 - r$
$r = 1$
$\frac{R}{r} = \frac{5}{2}$
Hence, the correct answer is 2.5.
