Circles Connected with Triangles

Introduction

Several important circles are associated with any given triangle, each with unique properties and formulas. These include the circumcircle, the incircle, and the excircles. Understanding their properties is crucial for solving many problems in trigonometry and geometry. ⭕

1. Circumcircle and Circumradius (R)

• Circumcircle: The unique circle that passes through all three vertices (A, B, C) of a triangle.

• Circumcenter: The center of the circumcircle. It is the point where the perpendicular bisectors of the sides intersect.

• Circumradius (R): The radius of the circumcircle. It is given by the following key formulas:

$$R = \frac{a}{2\sin A} = \frac{b}{2\sin B} = \frac{c}{2\sin C} \quad \text{(Extended Sine Rule)}$$

$$R = \frac{abc}{4\Delta} \quad \text{(where } \Delta \text{ is the area)}$$

2. Incircle and Inradius (r)

• Incircle: The unique circle that is tangent to all three sides of the triangle internally.

• Incenter: The center of the incircle. It is the point where the angle bisectors of the triangle intersect.

• Inradius (r): The radius of the incircle. It is related to the area ($\Delta$) and semi-perimeter (s) by:

$$r = \frac{\Delta}{s}$$

• Other Formulas for Inradius:

$$r = (s-a)\tan(A/2) = (s-b)\tan(B/2) = (s-c)\tan(C/2)$$

$$r = 4R\sin(A/2)\sin(B/2)\sin(C/2)$$

3. Excircles and Exradii ($r_1, r_2, r_3$)

• Excircle: A circle that lies outside the triangle and is tangent to one side and the extensions of the other two sides. Every triangle has three excircles.

• Excenter: The center of an excircle. It is the intersection of one internal angle bisector and two external angle bisectors.

• Exradii ($r_1, r_2, r_3$): The radii of the excircles opposite to vertices A, B, and C, respectively.

$$r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c}$$

$$r_1 = s\tan(A/2)$$

$$r_1 = 4R\sin(A/2)\cos(B/2)\cos(C/2)$$

Example 1: Finding Circumradius (R)

Question: For a triangle with sides a=13, b=14, c=15, find the circumradius R.

Solution:

First, find the area. The semi-perimeter is $s = (13+14+15)/2=21$. The area is $\Delta = \sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21(8)(7)(6)}=84$. Using the formula $R = \frac{abc}{4\Delta}$: $R = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{65}{8}$.

Example 2: Finding Inradius (r)

Question: For the same triangle, find the inradius r.

Solution:

We have $s=21$ and $\Delta=84$. Using the formula $r = \Delta/s$: $r = \frac{84}{21} = 4$.

Example 3: Finding an Exradius ($r_1$)

Question: For the same triangle, find the exradius $r_1$ opposite to side a.

Solution:

We have $s=21$ and $\Delta=84$. Also, $s-a = 21-13=8$. Using the formula $r_1 = \frac{\Delta}{s-a}$: $r_1 = \frac{84}{8} = \frac{21}{2} = 10.5$.

Example 4: Right-Angled Triangle Radii

Question: For a right-angled triangle with sides 5, 12, 13, find R and r.

Solution:

• Circumradius (R): For a right-angled triangle, the circumradius is half the hypotenuse. $R = \frac{13}{2} = 6.5$.

• Inradius (r): The area is $\Delta = \frac{1}{2}(5)(12)=30$. The semi-perimeter is $s=(5+12+13)/2=15$. $r = \Delta/s = 30/15=2$. A useful shortcut for a right triangle is $r=(a+b-c)/2 = (5+12-13)/2=2$.

Example 5: Identity with All Radii

Question: In a triangle ABC, prove that $rr_1r_2r_3 = \Delta^2$.

Solution:

LHS = $(\frac{\Delta}{s})(\frac{\Delta}{s-a})(\frac{\Delta}{s-b})(\frac{\Delta}{s-c}) = \frac{\Delta^4}{s(s-a)(s-b)(s-c)}$.

By Heron's formula, the denominator $s(s-a)(s-b)(s-c)$ is equal to $\Delta^2$. So, LHS = $\frac{\Delta^4}{\Delta^2} = \Delta^2$ = RHS.

Example 6: Identity with Reciprocal Radii

Question: Prove that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$.

Solution:

LHS = $\frac{s-a}{\Delta} + \frac{s-b}{\Delta} + \frac{s-c}{\Delta} = \frac{(s-a)+(s-b)+(s-c)}{\Delta} = \frac{3s-(a+b+c)}{\Delta}$. Since $a+b+c=2s$, the numerator is $3s-2s=s$. So, LHS = $\frac{s}{\Delta}$. We know $r=\Delta/s$, so $1/r = s/\Delta$. Thus, LHS = RHS.

Example 7: Identifying Triangle Type from Radii

Question: In a triangle ABC, if $r_1 = r+r_2+r_3$, what type of triangle is it?

Solution:

This is a known property. The condition $r_1 = r+r_2+r_3$ simplifies to $b^2+c^2=a^2$, which means the triangle is right-angled at vertex A.

Example 8: Proving Identity with Half-Angles

Question: In $\triangle ABC$, prove that $r_1 = s\tan(A/2)$.

Solution:

RHS = $s \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \sqrt{\frac{s^2(s-b)(s-c)}{s(s-a)}} = \sqrt{\frac{s(s-a)(s-b)(s-c)}{(s-a)^2}} = \frac{\sqrt{\Delta^2}}{(s-a)} = \frac{\Delta}{s-a} = r_1$ = LHS.

Example 9: Equilateral Triangle Radii

Question: For an equilateral triangle of side 'a', find the ratio R:r.

Solution:

For an equilateral triangle, $\Delta=\frac{\sqrt{3}}{4}a^2$ and $s=3a/2$.

$R = \frac{abc}{4\Delta} = \frac{a^3}{4(\sqrt{3}/4)a^2} = \frac{a}{\sqrt{3}}$.

$r = \frac{\Delta}{s} = \frac{(\sqrt{3}/4) a^2}{3a/2} = \frac{a}{2\sqrt{3}}$.

The ratio $R:r = \frac{a}{\sqrt{3}} : \frac{a}{2\sqrt{3}} = 1 : 1/2$, which is 2:1.

Example 10: Finding Sides from Radii

Question: In a $\triangle ABC$, if $r=1, r_1=2, r_2=3$, find the sides a and b.

Solution:

We use the identity $1/r = 1/r_1+1/r_2+1/r_3$. So $1 = 1/2+1/3+1/r_3 \implies 1/r_3=1-5/6=1/6 \implies r_3=6$.

Now, use the identity $\Delta = \sqrt{rr_1r_2r_3} = \sqrt{1 \cdot 2 \cdot 3 \cdot 6} = \sqrt{36} = 6$.

From $r_1=\Delta/(s-a) \implies 2=6/(s-a) \implies s-a=3$.

From $r_2=\Delta/(s-b) \implies 3=6/(s-b) \implies s-b=2$.

From $r_3=\Delta/(s-c) \implies 6=6/(s-c) \implies s-c=1$.

Adding these gives $3s-(a+b+c)=6 \implies 3s-2s=6 \implies s=6$.

Finally, $a = s-(s-a) = 6-3=3$.

And $b = s-(s-b) = 6-2=4$.