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Incircle, n.

(Geometry) A circle inscribed within a triangle, i.e., a circle tangent to all three sides of a triangle. Usually mentioned in conjunction with a triangle, such as: "the incircle I of triangle ABC..."

Related words:

What is the largest circle that can fit inside a triangle? There is only one, it turns out: the incircle. The incircle's center is called the incenter I, and its radius, r, sometimes also called the inradius.

```      K      2K
r  =  -  = -------                                   (1a)
s    (a+b+c)
```

where:

```a      = side length of the triangle's side opposite to vertex A
b      = side length opposite to B
c      = side length opposite to C
s      = semiperimeter of a triangle, equal to (a+b+c)/2
K      = area of the triangle, equal to sqrt(s*(s-a)*(s-b)*(s-c)) by Heron's Formula
```

Alternative expressions for the inradius are:

```      (a*b*c)
r  = ---------                                      (1b)
4*s*R

K^2
r  = ---------                                      (1c)
rA*rB*rC

r  = R*( cos(alpha) + cos(beta) + cos(gamma) -1 )   (1d)

r  = 0.5*sqrt((b+c-a)*(c+a-b)*(a+b-c)/(a+b+c))      (1e)
```

where:

```alpha  = triangle's interior angle at vertex A
beta   = interior angle at B
gamma  = interior angle at C
rA     = Radius of the excircle located opposite to the vertex A
rB     = Radius of the excircle located opposite to the vertex B
rC     = Radius of the excircle located opposite to the vertex C
```

SOURCE: MathWorld, Inradius. CAVEAT: I haven't used these personally so cannot vouch for their veracity, but they're so beautiful and tie in with other radii (of circles associated with a triangle) that I just had to include them. Equation (1d) is so pretty that I couldn't not include it, even if it's potentially wrong. In fact, some of the formulae for the inradius of common and interesting triangles seem to be either wrong or expressed very poorly/unclearly, so the truth of the expressions on that page are somewhat suspect.

Equation (1e) can be used to solve for the unknown length of side c, if sides a and b and the circumradius r is known. It will result in a cubic equation, which is a solvable one, even if it is a bit grotesque.

Assume you choose three vertices for a triangle, A, B, and C, and you assign them Cartesian coordinates (Ax, Ay), (Bx, By), and (Cx, Cy). Then you can find side lengths a, b, and c, as well as the semiperimeter s and area K:

```I      = (Ix, Iy)... Cartesian coordinates of the incenter
A      = (Ax, Ay)... the coordinates of vertex A of triangle ABC
B      = (Bx, By)... the coordinates of vertex B of triangle ABC
C      = (Cx, Cy)... the coordinates of vertex C of triangle ABC
a      = length of the side opposite vertex A
= length of the side BC
= sqrt((Cx-Bx)^2 + (Cy-By)^2)                (2a)
b      = length of the side opposite vertex B
= length of the side CA
= sqrt ((Ax-Cx)^2 + (Ay-Cy)^2)               (2b)
c      = length of the side opposite vertex C
= length of the side AB
= sqrt ((Bx-Ax)^2 + (By-Ay)^2)               (2c)
s      = (a+b+c)/2                                  (3)
K      = sqrt(s*(s-a)*(s-b)*(s-c))                  (4)
```

Example 1: An acute triangle has vertices A, B, and C at A = (-2,-2), B = (5,3), and C = (1,4). The first things we calculate are the lengths of the sides, a, b, and c, the semiperimeter s, and then the area of the triangle, K.

```A = (Ax, Ay) = (-2,-2)
B = (Bx, By) = ( 5, 3)
C = (Cx, Cy) = ( 1, 4)

a = length of the side opposite vertex A
= length of the side BC
= sqrt((Cx-Bx)^2 + (Cy-By)^2)
= sqrt((1 - 5)^2 + (4 - 3)^2)
= sqrt(16 + 1) = sqrt(17)
= 4.1
b = length of the side opposite vertex B
= length of the side CA
= sqrt ((Ax-Cx)^2 + (Ay-Cy)^2)
= sqrt ((-2 - 1)^2 + (-2 - 4)^2)
= sqrt(9 + 36) = sqrt(45)
= 6.7
c = length of the side opposite vertex C
= length of the side AB
= sqrt ((Bx-Ax)^2 + (By-Ay)^2)
= sqrt ((5 - -2)^2 + (3 - -2)^2)
= sqrt(49 + 25) = sqrt(74)
= 8.6
s = (a+b+c)/2
= (4.1 + 6.7 + 8.6)/2
= 9.7
K = sqrt(s*(s-a)*(s-b)*(s-c))
= sqrt(9.7*(9.7-4.1)*(9.7-6.7)*(9.7-8.6))
= 13.5
```

Now we can calculate the incircle coordinates.

```      (4.1*(-2) + 6.7*5 + 8.6*1)    33.9
Ix = ---------------------------- = ---- = 1.7
(4.1 + 6.7 + 8.6)         19.4

(4.1*(-2) + 6.7*3 + 8.6*4)    46.3
Iy = ---------------------------- = ---- = 2.4
(4.1 + 6.7 + 8.6)         19.4

I = (Ix, Iy) = (1.7, 2.4)
```

Point of Concurrency of Angle Bisectors: The incenter is the point of concurrency of the angle bisectors. If you bisect every angle, and you draw the ray from the vertex to the opposite side, then, amazingly, every one of these line segments intersects at one point!

Now you would suspect that two of these lines would intersect at a point, but when the third line segment intersects at exactly the same point, you know something's up.

Euclid proved that the angle bisectors all meet at the same point, always, for every possible triangle (which is the definition for the point of concurrency). He did this in Book 4, Proposition 4 of his magnum opus, The Elements.

There's a wonderful Web site devoted to Euclid's Elements here, and the wonderfulness is enhanced by its illustrations, which turn out to be Java applets. (You must permit a Java applet to run on your browser.) You can move the vertices of a triangle around and see how the incircle moves with the moving vertices. The incircle is always the intersection of the angle bisectors.

(I can see this a million times and never get bored. The sheer audacity of mathematics to have proved that this is always the case!)

Touch Points: The incircle touches the triangle at three contact points, commonly called touch points. Let's label them TA, TB, and TC.

If you draw a line segment connecting each touch point to the incenter, each line segment will make a right angle to the side it touches. (It has to - the radius of a circle is always perpendicular to the tangent line at the point of contact.)

The line segments are all radii of the incircle, so their lengths are all equal to r.

The distances of the touch points to the vertices is quite interesting. The distance from vertex A is the same to the two closest touchpoints. Let's label this quantity x. Similarly, let's label distances y as the distance between the two closest touchpoints to B, and z to C. It turns out that the distances are as follows:

```
x = s - a                                      (5a)
y = s - b                                      (5b)
z = s - c                                      (5c)
```

which means that now we can compute the point coordinates:

```  T_Ax = Bx + (y/a)*(Cx - Bx)                       (6a)
T_Ay = By + (y/a)*(Cy - By)                       (6b)
T_Bx = Cx + (z/b)*(Ax - Cx)                       (6c)
T_By = Cy + (z/b)*(Ay - Cy)                       (6d)
T_Cx = Ax + (x/c)*(Bx - Ax)                       (6e)
T_Cy = Ay + (x/c)*(By - Ay)                       (6f)
```

Example 2: Using the same acute triangle as in Example 1, find the contact points. First, we compute the distances x, y, and z, then we compute the touchpoint coordinates.

```A = (Ax, Ay) = (-2,-2)
B = (Bx, By) = ( 5, 3)
C = (Cx, Cy) = ( 1, 4)
a = 4.1
b = 6.7
c = 8.6
s = 9.7

x = 9.7 - 4.1 = 5.6
y = 9.7 - 6.7 = 3.0
z = 9.7 - 8.6 = 1.1

T_Ax =  5 + (3.0/4.1)*(1-5)    = 2.1
T_Ay =  3 +     0.73 *(4-3)    = 3.7
T_Bx =  1 + (1.1/6.7)*(-2 - 1) = 0.5
T_By =  4 +     0.17 *(-2 - 4) = 3.0
T_Cx = -2 + (5.6/8.6)*(5 - -2) = 2.6
T_Cy = -2 +     0.65 *(3 - -2) = 1.3

T_A = (T_Ax, T_Ay) = (2.1, 3.7)
T_B = (T_Bx, T_By) = (0.5, 3.0)
T_C = (T_Cx, T_Cy) = (2.6, 1.3)
```
The results are that TA = (2.1, 3.7), which resides on side BC, TB = (0.5, 3.0), on side CA, and TC = (2.6, 1.3) on side AB.

I created a little spreadsheet in Excel that would graph the triangle, the incircle, the incenter, and the touch points just to verify that these formulas yielded the right values. A picture of this is on my homenode, and will stay there for a brief time.

Each side of the triangle is split by its touch point. Each segment has a length that is of the form (s - xx), where xx is either a, b, or c.

Radials and Radial Lengths: The radials are line segments that connect the incenter to each of the vertices. The radial lengths are easily computed:

```    a' = x sec(α/2)                           (7a)
b' = y sec(β/2)                           (7b)
c' = z sec(γ/2)                           (7c)
```

Central Angles: The central angles measured from the incenter are measured counterclockwise from the first ray to the second one. Central angle φC is the angle swept counterclockwise from ray IA to IB. The sum of the central angles equals 360°.

There may be better ways of computing this, but the way I do it is not the law of cosines, which does not work for angles greater than 180 degrees but by measuring the angle of rays IA and IB using the four-quadrant arctan2 function (available for most computer languages and computational applications) and then subtracting angle IA from angle IB. If this angle is negative, then 360° is added.

Procedurally:

• Find the angle θA of line segment IA
• θA = atan2(Ay-Iy, Ax-Ix).................NOTE 1
• Find the angle θB of line segment IB
• θB = atan2(By-Iy, Bx-Ix)
• Find the angle θC of line segment IC
• θC = atan2(Cy-Iy, Cx-Ix)
• Compute the central angle φA:
• φA = θC - θB
• If φA <0, then add 2π
• Compute the central angle φB:
• φB = θA - θC
• If φB <0, then add 2π
• Compute the central angle φC:
• φC = θB - θA
• If φC <0, then add 2π

The coordinates of a triangle when the origin is the center of the incircle: Suppose we orient the triangle ΔABC so that A and B are horizontal (i.e., share the same y coordinate), and C is the top vertex. Then the vertices are found by:

```    A = (Ax, Ay) = ( -(s-a), -r)                                          (8a)
B = (Bx, By) = (   s-b , -r)                                          (8b)
C = (Cx, Cy) = (Ax + sqrt(b^2 - hc^2), hc - r)    if a^2 < b^2 + c^2  (8ca)
= (       Ax            , hc - r)    if a^2 = b^2 + c^2  (8cb)
= (Ax - sqrt(b^2 - hc^2), hc - r)    if a^2 > b^2 + c^2  (8cc)
```

where hc is the altitude of the triangle, which is equal to hc = 2K/c. Equation (8ca) is for when the angle at A is less than 90°, and (8cc) for when it's greater than 90 ° When the side b is equal to the altitude, the triangle is right.

This allows you to create a triangle without having to specify vertex coordinates. I find this useful when graphing triangles in Excel or Matlab.

Barycentric Coordinates: The barycentric coordinates of the incenter are a:b:c.

Trilinear Coordinates: The trilinear coordinates of the incenter are 1:1:1.

NOTES

1. Microsoft Excel has, for reasons known only but to its corporate self, defined the 4-quadrant arc-tangent function ATAN2 to reverse the order of input parameters. The arctangent function is the inverse tangent of (Δy/Δx), the ratio of the difference in y values to the the difference in x values, and most people, I daresay, see the Δy parameter as preceding the Δx parameter. (I spent many headscratching minutes tracing this error down. Silly me. I should have expected that Microsoft might have done things in a nonstandard way.)
2. I truly truly hate HTML and how it doesn't permit pretty equations.

Everything2 Writeups: Articles on (topic)

1. tongpoo, incenter, Feb. 7, 2002
2. IWhoSawTheFace, incenter, Nov. 3, 2011
3. tongpoo, incircle, Sep. 19, 2002
4. tongpoo, triangle, Feb. 8, 2002

References: Useful books and references on geometry

1. H.S.M. Coxeter, Introduction to Geometry, 2nd Ed., (c) 1969
*SIGH* What a magnificent book.
§ 1.4, “The Medians and the Centroid,” p. 10
§ 1.5, “The Incircle and the Circumcircle,” pp. 11-16
§ 1.6, “The Euler Line and the Orthocenter,” p. 17
§ 1.7, “The Nine Point Circle,” pp. 18-20
§ 1.9, “Morley’s Theorem,” pp. 23-25
§ 1.6, “The Euler Line and the Orthocenter,” p. 17
2. Dan Pedoe, Geometry: A Comprehensive Course
3. C. Stanley Ogilvy, Excursions in Geometry, (c) 1969
An elegant, thin discourse on geometry. Surprises in every chapter, even for the mathematically astute geometer.
Ch. 8, Some Euclidean Topics,
§ 8.1, “A navigation Problem,” p. 111
§ 8.2, “A three-circle Problem,” p. 115
§ 8.3, “The Euler line,” p. 117
§ 8.4, “The nine-point circle,” p. 119
§ 8.5, “A triangle problem,” p. 120
4. J.L. Heilbron, Geometry Civilized, ©2000
6. David Wells, Ed., The Penguin Dictionary of Curious and Interesting Geometry, ©1991
7. Hans Schwerdtfeger, Geometry of Complex Numbers, ©1962
8. Roland Deaux, Introduction to the Geometry of Complex Numbers, Howard Eves, Tr., Originally published in 1956
§ 38, “Centroid of a triangle,” p. 60
§ 39, “Algebraic value for the area of a triangle,” p. 60
9. Bruce Meserve, Fundamental Concepts of Geometry, ©1983
10. Will Dunham, Journey Through Genius, ©1990
Ch 5, “Heron’s formula for triangular area,” p. 113
11. Melvin Hausner, A Vector Space Approach to Geometry, ©1965
12. M. de Berg, M. van Kreveld, M. Overmans, and O. Schwartzkopf, Computational Geometry, 2nd Ed, ©2000
13. Gerald Farin and Dianne Hansford, The Geometry Toolbox, ©1998
Ch. 3, 2D Lines
§ 3.6, “Distance of a point to a line,” p. 40
§ 3.7, “The foot of a point,” p. 44
§ 3.8, “Computing intersections,” p. 45
Ch. 8, Breaking it up: Triangles
§ 8.1, “Barycentric coordinates,” p. 126
§ 8.2, “Affine invariance,” p. 128
§ 8.3, “Some special points,” p. 128
14. Clifford Pickover, The Math Book, ©2009
c.600 BC, “Pythagorean Theorem and Triangles,” p. 40
1639, “Projective Geometry,” p. 142
1899, “Pick’s Theorem,” p. 294
1899, “Morley’s Trisector Theorem,” p. 296
15. Keith Devlin, The Language of Mathematics, ©2000
16. Daniel Zwillinger, Ed., The CRC Standard Mathematical Tables and Formulae, 30th Ed, ©1996
Ch. 4, Geometry,
§ 4.5.1, “Triangles,” p. 271
§ 4.6, “Circles,” p. 277
17. Siobhan Roberts, King of Infinite Space: Donald Coxeter, the Man Who Saved Geometry, ©2006
18. Alfred S. Posamentier and Charles T. Salkind, Challenging Problems in Geometry, ©1988
19. Edna E. Kramer, The Nature and Growth of Modern Mathematics, ©1970, 1981
Figure 2.7, the square law spiral, p. 28
20. Hans Rademacher and Otto Toeplitz, The Enjoyment of Mathematics, Published in 1957 by the Princeton University Press
§ 5, “A minimum property of the pedal triangle,” p. 27 (minimum path length of light when crossing a boundary)
§ 14, “Pythagorean numbers and Fermat’s Theorem,” p. 88
§ 16, “The spanning circle of a finite set of points,” p. 108 (H.W.E. Jung’s theorem)
§ 26, “A characteristic property of the circle,” p. 160
§ 28, “The indispensability of the compass for the constructions of elementary geometry,” p. 177

Internet References

1. David E. Joyce, Euclid's Elements, Book IV, Prop. 4. Euclid's proof and Java applets showing the incenter of a triangle being the point of concurrency of the three angle bisectors.
2. David E. Joyce, Homepage at Clark University. Joyce is a professor of Mathematics and Computer Science at Clark University, Worcester, MA. He has rendered Euclid's Elements for the computer and added numerous Java applets in the next reference. In my opinion, his site is one of the greatest reasons that geometers should use the Internet.
3. David E. Joyce, Euclid's Elements. Euclid's Elements on steroids. The java applets make geometry and geometric proofs come alive. I defy you to read a few of his propositions, try out the applets, and then not be amazed at Euclid's genius.
4. P. Ballew, “Orthocenter of a triangle
5. Wikipedia, "Triangle"
6. Wikipedia, "Incircle and Excircles of a Triangle"
7. Weisstein, Eric W. "Incenter" From MathWorld--A Wolfram Web Resource.
8. Weisstein, Eric W. "Inradius" From MathWorld--A Wolfram Web Resource.
9. Weisstein, Eric W. "Triangle" From MathWorld--A Wolfram Web Resource.
10. Alexander Bogomolny, "Incircle and Excircles of a Triangle" From Cut The Knot--mathematical topics. Cut the Knot has a full range of geometric topics. It is a reference I use quite frequently, as the language is clean and to the point, without being unnecessarily over-mathematical, as Mathworld's information tends to be.

In*cir"cle (?), v. t.

See Encircle.

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