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In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is called the triangle's incenter.

An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two.

Every triangle has three distinct excircles, each tangent to one of the triangle's sides.

The center of the incircle can be found as the intersection of the three internal angle bisectors

The center of an excircle is the intersection of the internal bisector of one angle and the external bisectors of the other two. Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system.

Relation to area of the triangle

The radii of the incircles and excircles are closely related to the area of the triangle. Suppose  $\triangle ABC$ has an incircle with radius r and center I.

Let a be the length of BC, b the length of AC, and c the length of AB.

Now, the incircle is tangent to AB at some point C′, and so

$\angle AC'I$ is right.

Thus the radius C'I is an altitude of

$\triangle IAB$

Therefore

$\triangle IAB$

has base length c and height r, and so has area

$\tfrac{1}{2}cr$.

Similarly,

$\triangle IAC$

has area

$\tfrac{1}{2}br$

and

$\triangle IBC$

has area $\tfrac{1}{2}ar$.

Since these three triangles decompose $\triangle ABC$, we see that

$\Delta = \frac{1}{2} (a+b+c) r = s r,$

where $\Delta$ is the area of $\triangle ABC$ and $s= \frac{1}{2}(a+b+c)$ is its semiperimeter.

The radii in the excircles are called the exradii. Let the excircle at side AB touch at side AC extended at G, and let this excircle's

radius be $r_c$ and its center be $I_c$. Then $I_c G$ is an altitude of $\triangle ACI_c$,

so $\triangle ACI_c$ has area $\tfrac{1}{2}br_c$. By a similar argument,

$\triangle BCI_c$

has area

$\tfrac{1}{2}ar_c$

and $\triangle ABI_c$

has area

$\tfrac{1}{2}cr_c$.

Thus

$\Delta = \frac{1}{2}(a+b-c)r_c = (s-c)r_c$.

So, by symmetry,

$\Delta = sr = (s-a)r_a = (s-b)r_b = (s-c)r_c$.

By the law of cosines, we have

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$

Combining this with the identity $\sin^2 A + \cos^2 A = 1$, we have

$\sin A = \frac{\sqrt{-a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2 c^2 + 2 a^2 c^2}}{2bc}$

But $\Delta = \tfrac{1}{2}bc \sin A$, and so

\begin{align} \Delta &= \frac{1}{4} \sqrt{-a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2 c^2 + 2 a^2 c^2} \\ &= \frac{1}{4} \sqrt{ (a+b+c) (-a+b+c) (a-b+c) (a+b-c) }\\ & = \sqrt{s(s-a)(s-b)(s-c)}, \end{align}

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which is Heron's formula.

Combining this with $sr=\Delta$, we have

$r^2 = \frac{\Delta^2}{s^2} = \frac{(s-a)(s-b)(s-c)}{s}$.

Similarly, $(s-a)r_a = \Delta$ gives

$r_a^2 = \frac{s(s-b)(s-c)}{s-a}$.

From these formulas one can see that the excircles are always larger than the incircle and that the largest excircle is the one tangent to the longest side and the smallest excircle is tangent to the shortest side. Further, combining these formulas  formula yields:

$\Delta=\sqrt{rr_ar_br_c}.$

The ratio of the area of the incircle to the area of the triangle is less than or equal to $\frac{\pi}{3\sqrt{3}}$, with equality holding only for equilateral triangles.

Nine-point circle and Feuerbach point

The circle tangent to all three of the excircles as well as the incircle is known as the nine-point circle. The point where the nine-point circle touches the incircle is known as the Feuerbach point.

Gergonne triangle and point

The Gergonne triangle(of ABC) is defined by the 3 touchpoints of the incircle on the 3 sides.

Those vertices are denoted as TA, etc.

The point that TA denotes, lies opposite to A.

This Gergonne triangle TATBTC is also known as the contact triangle or intouch triangle of ABC.

The three lines ATA, BTB and CTC intersect in a single point called Gergonne point, denoted as Ge - X(7).

Interestingly, the Gergonne point of a triangle is the symmedian point of the Gergonne triangle.

The touchpoints of the three excircles with segments BC,CA and AB are the vertices of the extouch triangle.  The points of intersection of the interior angle bisectors of ABC with the segments BC,CA,AB are the vertices of the incentral triangle.

Nagel triangle and point

The Nagel triangle of ABC is denoted by the vertices XA, XB and XC that are the three points where the excircles touch the reference triangle ABC and where XA is opposite of A, etc. This triangle XAXBXC is also known as the extouch triangle of ABC.  The circumcircle of the extouch triangle XAXBXC is called the Mandart circle.  The three lines AXA, BXB and CXC are called the splitters of the triangle; they each bisect the perimeter of the triangle, and they intersect in a single point, the triangle's Nagel point Na - X(8).

Trilinear coordinates for the vertices of the intouch triangle are given by

• $A-\text{vertex}= 0 : \sec^2 \left(\frac{B}{2}\right) :\sec^2\left(\frac{C}{2}\right)$
• $B-\text{vertex}= \sec^2 \left(\frac{A}{2}\right):0:\sec^2\left(\frac{C}{2}\right)$
• $C-\text{vertex}= \sec^2 \left(\frac{A}{2}\right) :\sec^2\left(\frac{B}{2}\right):0$

Trilinear coordinates for the vertices of the extouch triangle are given by

• $A-\text{vertex} = 0 : \csc^2\left(\frac{B}{2}\right) : \csc^2\left(\frac{C}{2}\right)$
• $B-\text{vertex} = \csc^2\left(\frac{A}{2}\right) : 0 : \csc^2\left(\frac{C}{2}\right)$
• $C-\text{vertex} = \csc^2\left(\frac{A}{2}\right) : \csc^2\left(\frac{B}{2}\right) : 0$

Trilinear coordinates for the vertices of the incentral triangle are given by

• $\ A-\text{vertex} = 0 : 1 : 1$
• $\ B-\text{vertex} = 1 : 0 : 1$
• $\ C-\text{vertex} = 1 : 1 : 0$

Trilinear coordinates for the vertices of the excentral triangle are given by

• $\ A-\text{vertex}= -1 : 1 : 1$
• $\ B-\text{vertex}= 1 : -1 : 1$
• $\ C-\text{vertex}= 1 : -1 : -1$

Trilinear coordinates for the Gergonne point are given by

$\sec^2\left(\frac{A}{2}\right) : \sec^2 \left(\frac{B}{2}\right) : \sec^2\left(\frac{C}{2}\right)$,

or, equivalently, by the Law of Sines,

$\frac{bc}{b+ c - a} : \frac{ca}{c + a-b} : \frac{ab}{a+b-c}$.

Trilinear coordinates for the Nagel point are given by

$\csc^2\left(\frac{A}{2}\right) : \csc^2 \left(\frac{B}{2}\right) : \csc^2\left(\frac{C}{2}\right)$,

or, equivalently, by the Law of Sines,

$\frac{b+ c - a}{a} : \frac{c + a-b}{b} : \frac{a+b-c}{c}$.

It is the isotomic conjugate of the Gergonne point.

Coordinates of the incenter

The Cartesian coordinates of the incenter are a weighted average of the coordinates of the three vertices using the side lengths of the triangle as weights.  (The weights are positive so the incenter lies inside the triangle as stated above.)  If the three vertices are located at $(x_a,y_a)$, $(x_b,y_b)$, and $(x_c,y_c)$, and the sides opposite these vertices have corresponding lengths $a$, $b$, and $c$, then the incenter is at

$\bigg(\frac{a x_a+b x_b+c x_c}{P},\frac{a y_a+b y_b+c y_c}{P}\bigg) = \frac{a(x_a,y_a)+b(x_b,y_b)+c(x_c,y_c)}{P}$

where $\ P = a + b + c.$

Trilinear coordinates for the incenter are given by

$\ 1 : 1 : 1.$

Barycentric coordinates for the incenter are given by

$\ a : b : c$

or equivalently

$\sin(A):\sin(B):\sin(C).$

Equations for four circles

Let x : y : z be a variable point in trilinear coordinates, and let u = cos2(A/2), v = cos2(B/2), w = cos2(C/2).  The four circles described above are given by these equations:

• Incircle:
$\ u^2x^2+v^2y^2+w^2z^2-2vwyz-2wuzx-2uvxy=0$
• A-excircle:
$\ u^2x^2+v^2y^2+w^2z^2-2vwyz+2wuzx+2uvxy=0$
• B-excircle:
$\ u^2x^2+v^2y^2+w^2z^2+2vwyz-2wuzx+2uvxy=0$
• C-excircle:
$\ u^2x^2+v^2y^2+w^2z^2+2vwyz+2wuzx-2uvxy=0$

Euler's theorem

Euler's theorem states that in a triangle:

$(R-r_{in})^2=d^2+r_{in}^2,$

where R and rin are the circumradius and inradius respectively, and d is the distance between the circumcenter and the incenter.

For excircles the equation is similar:

$(R+r_{ex})^2=d^2+r_{ex}^2,$

where rex is the radius of one of the excircles, and d is the distance between the circumcenter and this excircle's center.

Other incircle properties

Suppose the tangency points of the incircle divide the sides into lengths of x and y, y and z, and z and x.  Then the incircle has the radius

$r = \sqrt{\frac{xyz}{x+y+z}}$

and the area of the triangle is

$K=\sqrt{xyz(x+y+z)}.$

If the altitudes from sides of lengths a, b, and c are ha, hb, and hc then the inradius r is one-third of the harmonic mean of these altitudes, i.e.

$r = \frac{1}{h_a^{-1}+h_b^{-1}+h_c^{-1}}.$

The product of the incircle radius r and the circumcircle radius R of a triangle with sides a, b, and c is

$rR=\frac{abc}{2(a+b+c)}.$

$ab+bc+ca=s^2+(4R+r)r,$
$a^2+b^2+c^2=2s^2-2(4R+r)r.$

Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle.

The distance from the incenter to the centroid is less than one third the length of the longest median of the triangle.

Denoting the distance from the incenter to the Euler line as d, the length of the longest median as v, the length of the longest side as u, and the semiperimeter as s, the following inequalities hold:

$\frac{d}{s} < \frac{d}{u} < \frac{d}{v} < \frac{1}{3}.$

Denoting the center of the incircle of triangle ABC as I, we have

$\frac{\overline{IA} \cdot \overline{IA}}{\overline{CA} \cdot \overline{AB}} + \frac{\overline{IB} \cdot \overline{IB}}{\overline{AB} \cdot \overline{BC}} + \frac{\overline{IC} \cdot \overline{IC}}{\overline{BC} \cdot \overline{CA}} = 1.$

Other excircle properties

The circular hull of the excircles is internally tangent to each of the excircles, and thus is an Apollonius circle. The radius of this Apollonius circle is $\frac{r^2+s^2}{4r}$ where r is the incircle radius and s is the semiperimeter  of the triangle.

The following relations hold among the inradius r, the circumradius R, the semiperimeter s, and the excircle radii r'a, rb, rc:

$r_a+r_b+r_c=4R+r,$
$r_a r_b+r_br_c+r_cr_a = s^2,$
$r_a^2 + r_b^2 + r_c^2 = (4R+r)^2 -2s^2,$

The circle through the centers of the three excircles has radius 2R.

If H is the orthocenter of triangle ABC, then

$r_a+r_b+r_c+r=AH+BH+CH+2R,$
$r_a^2+r_b^2+r_c^2+r^2=AH^2+BH^2+CH^2+(2R)^2.$