Geometry

Incenter

A triangle has three sides. Is there a point the same distance from all three, and a circle inside the triangle that touches every side?

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CircumcenterGeometry Angle bisectors of a triangleGeometry

Unlocks

Tangent propertiesGeometry

Where you are 4 in Triangle centers

Incenter

Centroid Orthocenter and the Euler line

Why this exists

A carpenter has a triangular board and wants to cut the largest possible round tabletop from it. A town has a triangular park bounded by three straight roads and wants to put a fountain as far from the traffic as it can. Both are asking for a point deep inside a triangle that keeps its distance from all three sides, and the fair choice is a point the same distance from every side.

For two sides the answer is already known. By the angle bisector theorem and its converse, the points inside an angle that are equally far from its two sides are exactly the points of its bisector. A triangle has three angles and so three bisectors, and that lesson proved that they always cross at a single point. This lesson is about that point: the circle it carries, where that circle touches the sides, and how its radius is tied to the triangle’s area and perimeter.

It is one of the classic triangle centers. The Circumcenter is equally far from the three corners. The incenter is equally far from the three sides.

Definition

The incenter of a triangle is the point where its three angle bisectors meet. It is usually called I. (British texts often spell it incenter.)

The distance from a point to a side is the length of the perpendicular segment from the point to the line containing that side (Distance from a point to a line).

A line is tangent to a circle when it meets the circle at exactly one point, called the point of tangency.

The inscribed circle, or incircle, of a triangle is a circle inside the triangle that is tangent to all three sides. Its center is the incenter, and its radius, written r, is the inradius. The triangle is then said to be circumscribed about the circle.

With the usual names a = BC, b = CA and c = AB for the sides opposite A, B and C, the semiperimeter is half the perimeter:

s = \tfrac{1}{2}(a + b + c)

Throughout, X, Y and Z are the feet of the perpendiculars from I to \overline{BC}, \overline{CA} and \overline{AB}, which turn out to be the points where the incircle touches those sides, as in Figure 1. K is the area of the triangle.

Figure 1The bisectors meet at I, and the circle centered there touches each side once, at the foot of the perpendicular from I.

That the incircle exists, and that a triangle has only one, is a theorem rather than part of the definition. It is among the results below.

Intuition

Picture the three sides as walls that move inwards at the same steady speed, each staying parallel to where it started. The triangle they enclose shrinks but keeps its shape. Each corner of the shrinking triangle is always equally far from the two walls that meet there, so it slides inwards along the bisector of that angle. The three corners close in on one another, and at the moment the triangle shrinks to a single point, that point is equally far from all three walls. It is the incenter, and the distance each wall has traveled is r.

The picture also explains why the incenter never leaves the triangle, which the circumcenter can. Each bisector starts at a vertex and runs into the inside of the triangle, so the point where they meet is inside too, for every shape of triangle: acute, right or obtuse.

Results

Incenter theorem. The three angle bisectors of any triangle meet at one point I, inside the triangle. I is the same distance r from all three sides,

IX = IY = IZ = r

and it is the only point inside the triangle with that property.

The inscribed circle. The circle with center I and radius r is tangent to each side at the foot of the perpendicular from I, and it lies inside the triangle. It is the only circle inside the triangle that is tangent to all three sides.

Tangent lengths. From each vertex, the two pieces of the sides up to the points of tangency are equal, and each equals the semiperimeter minus the opposite side:

\begin{aligned} AY = AZ &= s - a\\ BZ = BX &= s - b\\ CX = CY &= s - c \end{aligned}

Figure 2From each vertex, the two pieces of the sides up to the points of tangency are equal.

Area. The area is the inradius times the semiperimeter:

K = rs, \qquad\text{so}\qquad r = \frac{K}{s}

Figure 3Three triangles, each with a side of \triangle ABC as its base and height r.

Right triangles. If \angle C is a right angle, so that a and b are the legs and c is the hypotenuse, then

r = \frac{a + b - c}{2}

Figure 4In a right triangle the radii to the two legs make a square with the corner, so each leg loses exactly r.

Angles at the incenter.

m\angle BIC = 90^\circ + \tfrac{1}{2}\,m\angle A

and in the same way m\angle CIA = 90^\circ + \tfrac{1}{2}\,m\angle B and m\angle AIB = 90^\circ + \tfrac{1}{2}\,m\angle C. Each of the three is obtuse.

Why it is true

The bisectors meet inside the triangle. This is proved in full in the angle bisector lesson. In short: the bisectors from A and B run from their vertices into the triangle and cross at a point I inside it. Being on the bisector of \angle A, I is as far from \overline{AB} as from \overline{CA}; being on the bisector of \angle B, it is as far from \overline{AB} as from \overline{BC}. So it is equally far from all three sides, and since it is inside \angle C, the converse puts it on the bisector of \angle C as well. Any point inside the triangle that is equally far from all three sides lies on the bisectors of \angle A and \angle B for the same reason, and those two lines cross only once, so I is the only such point.

The circle touches each side line once. X is the foot of the perpendicular from I to line BC, so IX = r. For any other point P of line BC, triangle IXP has a right angle at X, and the Pythagorean theorem gives

IP^2 = r^2 + XP^2 > r^2

So P is outside the circle, and line BC meets the circle at X and nowhere else: it is tangent there. The same argument works for the other two sides.

The circle lies inside the triangle. Take any point Q on the circle. If Q were on the far side of line BC from I, the segment \overline{IQ} would cross line BC at a point P with IP < IQ = r, but every point of line BC is at least r from I. So Q is on the same side of line BC as I, or on the line itself, and the same is true for the other two side lines. The points on the inner side of all three side lines make up the triangle, so the circle is inside it. In particular X is a point of the circle on line BC, so it lies on the side \overline{BC} itself and not on its extension; the same holds for Y and Z.

There is only one such circle. Suppose a circle with center P and radius \rho is inside the triangle and tangent to all three sides. Its center is exactly \rho from each side line: if the line were closer than \rho it would pass through the inside of the circle and cut it twice, and if it were farther it would miss the circle altogether. Tangent properties treat this in full. So P is a point inside the triangle equally far from all three sides, which makes it I, and then \rho = r.

Tangent lengths. Triangles AZI and AYI have right angles at Z and Y, share the hypotenuse \overline{AI}, and have legs IZ = IY = r. They are congruent by HL, so AZ = AY. (This is the equality BF = BG from the angle bisector lesson’s proof.) The same holds at B and at C. Call the three equal pairs x, y and z, as in Figure 2. The points of tangency lie on the sides themselves, so each side is made of two of the pieces:

\begin{aligned} x + y &= c\\ y + z &= a\\ z + x &= b \end{aligned}

Adding all three equations gives 2(x + y + z) = a + b + c, so x + y + z = s. Then x = s - (y + z) = s - a, and in the same way y = s - b and z = s - c.

Area. Join I to the three vertices, as in Figure 3. Since I is inside the triangle, the triangles IBC, ICA and IAB fill \triangle ABC without overlapping. Each has a side of \triangle ABC as its base, and its height is the perpendicular from I to that side, which is r. Adding the three areas:

\begin{aligned} K &= \tfrac{1}{2}ar + \tfrac{1}{2}br + \tfrac{1}{2}cr\\ &= \tfrac{1}{2}(a + b + c)\,r = rs \end{aligned}

Right triangles. In Figure 4 the angle at C is right, and \angle IXC and \angle IYC are right because \overline{IX} and \overline{IY} are perpendiculars to the sides. The angles of a quadrilateral add to 360^\circ, so the fourth angle, at I, is right too, and CXIY is a rectangle with two neighboring sides IX = IY = r: a square. So CX = CY = r, which leaves BX = a - r and AY = b - r. By the tangent lengths, BZ = BX and AZ = AY, so

c = AZ + ZB = (b - r) + (a - r)

and solving for r gives 2r = a + b - c. This is the tangent-length result CX = s - c read another way.

Angles at the incenter. In \triangle BIC the angles at B and C are half of \angle B and half of \angle C, and the triangle angle sum gives

\begin{aligned} m\angle BIC &= 180^\circ - \tfrac{1}{2}(m\angle B + m\angle C)\\ &= 180^\circ - \tfrac{1}{2}(180^\circ - m\angle A)\\ &= 90^\circ + \tfrac{1}{2}\,m\angle A \end{aligned}

Since m\angle A is between 0^\circ and 180^\circ, m\angle BIC is between 90^\circ and 180^\circ.

Worked examples

Example 1

Tangent pieces as expressions

The incircle of \triangle ABC touches \overline{AB} at Z and \overline{AC} at Y. You know AZ = 4x - 3, AY = 2x + 9 and AI = 29. Find x, the length AZ, and the inradius.

Solution

The two tangent pieces from the same vertex are equal, so AZ = AY:

\begin{aligned} 4x - 3 &= 2x + 9\\ 2x &= 12\\ x &= 6 \end{aligned}

Then AZ = 4(6) - 3 = 21, and AY = 2(6) + 9 = 21 agrees.

The radius \overline{IZ} is perpendicular to \overline{AB}, so triangle AZI has a right angle at Z and hypotenuse AI = 29. By the Pythagorean theorem,

r^2 = 29^2 - 21^2 = 841 - 441 = 400

so the inradius is r = 20. That is also the distance from I to each of the three sides.

Example 2

Working back from an angle at the incenter

I is the incenter of \triangle ABC, with m\angle BIC = 118^\circ and m\angle ABC = 72^\circ. Find the other two angles of the triangle, and m\angle AIB and m\angle AIC.

Solution

\overline{BI} bisects \angle B, so m\angle IBC = 36^\circ. The angle sum of \triangle BIC then gives

m\angle ICB = 180^\circ - 118^\circ - 36^\circ = 26^\circ

\overline{CI} bisects \angle C, so m\angle ACB = 2(26^\circ) = 52^\circ, and the angle sum of \triangle ABC gives

m\angle BAC = 180^\circ - 72^\circ - 52^\circ = 56^\circ

The result m\angle BIC = 90^\circ + \tfrac{1}{2}m\angle A agrees: 90^\circ + 28^\circ = 118^\circ.

The angles at A, B and C inside the small triangles are the halves 28^\circ, 36^\circ and 26^\circ, so

\begin{aligned} m\angle AIB &= 180^\circ - 28^\circ - 36^\circ = 116^\circ\\ m\angle AIC &= 180^\circ - 28^\circ - 26^\circ = 126^\circ \end{aligned}

Check. The three angles at I go once around the point: 118 + 116 + 126 = 360.

Example 3

Where the incircle touches the sides

In \triangle ABC, AB = 7, BC = 9 and CA = 8. The incircle touches \overline{BC} at X, \overline{CA} at Y and \overline{AB} at Z. Find all six tangent pieces.

Solution

The semiperimeter is s = \tfrac{1}{2}(7 + 9 + 8) = 12. Each tangent piece is s minus the side opposite its vertex:

\begin{aligned} AZ = AY &= 12 - 9 = 3\\ BZ = BX &= 12 - 8 = 4\\ CX = CY &= 12 - 7 = 5 \end{aligned}

Check. Each side should be the sum of its two pieces: AZ + ZB = 3 + 4 = 7, BX + XC = 4 + 5 = 9 and CY + YA = 5 + 3 = 8. All three match.

The formula is a shortcut, not a necessity. With x = AZ, y = BZ and z = CX, the sides give x + y = 7, y + z = 9 and z + x = 8, and solving those three equations gives the same answer.

Example 4

The inradius of a right triangle

A right triangle has legs 8 and 15. Find its inradius two ways, and place its incenter when the right angle is at the origin and the legs lie along the positive axes.

Solution

The hypotenuse is \displaystyle \sqrt{8^2 + 15^2} = \sqrt{289} = 17.

From the right-triangle formula. r = \tfrac{1}{2}(8 + 15 - 17) = \tfrac{1}{2}(6) = 3.

From the area. The legs are a base and height, so K = \tfrac{1}{2}(8)(15) = 60. The semiperimeter is s = \tfrac{1}{2}(8 + 15 + 17) = 20, so r = \tfrac{60}{20} = 3. The two methods agree.

With the right angle at the origin, the legs lie along the axes, and I is 3 from each of them. It is inside the triangle, so both its coordinates are positive: I = (3, 3).

Example 5

The incenter on a coordinate grid

Find the incenter and inradius of the triangle with vertices A(0, 0), B(14, 0) and C(5, 12).

Solution

First the sides, by the The distance formula:

\begin{aligned} c = AB &= 14\\ b = CA &= \sqrt{5^2 + 12^2} = 13\\ a = BC &= \sqrt{9^2 + 12^2} = 15 \end{aligned}

\overline{AB} lies along the x-axis and C is 12 above it, so K = \tfrac{1}{2}(14)(12) = 84. The semiperimeter is s = \tfrac{1}{2}(13 + 14 + 15) = 21, so r = \tfrac{84}{21} = 4.

That fixes the height of I: it is 4 from the x-axis and inside the triangle, so I has y-coordinate 4. To find its x-coordinate, use the point Z where the incircle touches \overline{AB}. The tangent length from A is AZ = s - a = 21 - 15 = 6, so Z = (6, 0). The radius \overline{IZ} is perpendicular to the x-axis, so it is vertical and I is directly above Z:

I = (6, 4)

Figure 5The incircle of A(0, 0), B(14, 0), C(5, 12). Its center sits straight above the point where it touches \overline{AB}.

Check. Triangle CYI has a right angle at Y, with legs CY = s - c = 7 and IY = 4, so CI^2 should be 7^2 + 4^2 = 65. From the coordinates, CI^2 = (5 - 6)^2 + (12 - 4)^2 = 65. In the same way BI^2 should be (s - b)^2 + 4^2 = 8^2 + 4^2 = 80, and from the coordinates it is (14 - 6)^2 + 4^2 = 80.

Example 6

Incenter and circumcenter of one triangle

In \triangle ABC, AB = AC = 10 and BC = 12. Find the inradius and the position of the incenter, and compare it with the circumcenter.

Solution

Let M be the midpoint of \overline{BC}. Triangles ABM and ACM are congruent by SSS (SSS, SAS, ASA, AAS), so \angle BAM \cong \angle CAM and \overline{AM} is the bisector of \angle A. The same congruence makes the two angles at M equal, and they add to 180^\circ, so \overline{AM} \perp \overline{BC}. With BM = 6, the Pythagorean theorem gives \displaystyle AM = \sqrt{10^2 - 6^2} = 8.

The area is K = \tfrac{1}{2}(12)(8) = 48 and the semiperimeter is s = \tfrac{1}{2}(10 + 10 + 12) = 16, so

r = \frac{48}{16} = 3

I lies on the bisector \overline{AM}. Since \overline{AM} \perp \overline{BC}, the perpendicular from I to \overline{BC} runs along \overline{AM} and lands at M, so the incircle touches \overline{BC} at its midpoint and IM = r = 3. The incenter is 3 above M, and AI = 8 - 3 = 5.

Check. The tangent length from A is s - a = 16 - 12 = 4, and the right triangle AZI should have AI^2 = 4^2 + 3^2 = 25. It does: AI = 5.

The Circumcenter lesson worked with this same triangle and found its circumcenter O on the same line \overline{AM}, but only \tfrac{7}{4} above M. Both centers sit on the line of symmetry, as they must, yet they are different points, 3 - \tfrac{7}{4} = \tfrac{5}{4} apart. Being equally far from the corners and being equally far from the sides are different conditions.

Example 7

The largest circle in a triangle

Show that no circle inside a triangle has a radius larger than the inradius. Then find the radius of the largest round fountain that fits in a triangular lawn with sides 10 m, 17 m and 21 m, whose corner opposite the 21 m side is 8 m from that side.

Solution

Let a circle with center P and radius \rho lie inside \triangle ABC, and let d_a, d_b and d_c be the distances from P to the lines BC, CA and AB. Each of them is at least \rho: if d_a were smaller than \rho, the point of the circle that lies \rho from P in the direction of the perpendicular to BC would be past line BC, outside the triangle.

Now split the triangle at P, exactly as in the proof of K = rs. The three pieces have bases a, b, c and heights d_a, d_b, d_c, so

\begin{aligned} K &= \tfrac{1}{2}(a\,d_a + b\,d_b + c\,d_c)\\ &\ge \tfrac{1}{2}(a + b + c)\,\rho = s\rho \end{aligned}

So \rho \le \tfrac{K}{s} = r. The incircle itself has radius exactly r, so it is the largest circle that fits.

For the lawn, the 21 m side is a base with height 8 m, so K = \tfrac{1}{2}(21)(8) = 84 square meters. The semiperimeter is s = \tfrac{1}{2}(10 + 17 + 21) = 24 m, so

r = \frac{84}{24} = \frac{7}{2}

The largest fountain has radius 3.5 m, and its center goes at the incenter.

Check. The data agree with each other: the foot of the 8 m height splits the 21 m side into 6 m and 15 m, and 6^2 + 8^2 = 10^2 and 15^2 + 8^2 = 17^2.

Boundaries

It needs a real triangle. Three points on one line make no angles to bisect and no region to put a circle in. As a triangle is flattened toward a line, its area shrinks toward 0 while its semiperimeter does not, so r = \tfrac{K}{s} shrinks to 0 as well.

Equally far from three sides, or from three lines? Inside the triangle there is exactly one point equally far from all three sides. If the sides are extended to full lines, there are four such points. The other three are the excenters, each outside the triangle, where the bisector of one angle meets the bisectors of the exterior angles at the other two vertices. Each is the center of an excircle, tangent to one side and to the extensions of the other two. That is why the results above keep saying “inside”.

Tangent to the side, not just to its line. The incircle touches each side at a point of the side itself. A circle can be tangent to all three side lines and still not be inside the triangle: the excircles are.

More sides, no guarantee. The angle bisectors of a quadrilateral need not meet at one point, and many quadrilaterals have no circle tangent to all four sides. A rhombus has one. A 2 by 4 rectangle has none: a circle touching both long sides has diameter 2 and cannot reach both short sides, which are 4 apart.

One center among several. The incenter is generally not the circumcenter, as Example 6 showed, and not the Centroid either. In an isosceles triangle all of them lie on the line of symmetry, and in an equilateral triangle they are the same point (Exercises 21 and 22).

r = \tfrac{K}{s} needs the area. When only the three sides are known, the area has to come first, from a height or from Heron’s formula.

Where this goes next

The incircle is the first circle you meet that is defined by tangent lines rather than by points on it. Tangent properties take up the two facts used here again in general: a radius to a point of tangency is perpendicular to the tangent, and the two tangent segments from an outside point to a circle are equal.

The splitting argument behind K = rs works for any polygon that has a circle tangent to all its sides. For a regular polygon the inradius is called the apothem, and the area of a regular polygon is half the apothem times the perimeter for exactly this reason.

The bisector from A does more than pass through I: it cuts \overline{BC} in the ratio AB : AC, which is the triangle angle bisector theorem. And the incenter takes its place beside the Circumcenter, the Centroid and the Orthocenter and the Euler line as one of the four classic triangle centers.

Exercises

Distances and angles at the incenter

I is the incenter of \triangle ABC. Its distance to \overline{AB} is 3x - 4 and its distance to \overline{BC} is x + 6. Find x and the distance from I to \overline{CA}.
I is the incenter of \triangle DEF. Its distance to \overline{DE} is x^2 - 8 and its distance to \overline{EF} is 2x. Find x and the inradius.
I is the incenter of \triangle ABC, AI = 17, and the distance from I to \overline{AB} is 8. The incircle touches \overline{AB} at Z and \overline{AC} at Y. Find AZ and AY.
In \triangle ABC, m\angle A = 40^\circ and m\angle B = 80^\circ, and I is the incenter. Find m\angle IAB, m\angle ICB and m\angle BIC.
I is the incenter of \triangle ABC and m\angle BIC = 135^\circ. Find m\angle A.
I is the incenter of \triangle ABC and m\angle AIB = 100^\circ. Find m\angle C.

Tangent lengths

In \triangle ABC, AB = 10, BC = 12 and CA = 14. The incircle touches \overline{BC} at X, \overline{CA} at Y and \overline{AB} at Z. Find AZ, BX and CY.
The incircle of \triangle ABC touches \overline{BC} at X and \overline{AB} at Z, with BX = 5, CX = 7 and AZ = 3. Find the three sides and the perimeter.
In \triangle ABC, AB = AC = 13 and BC = 10. Where does the incircle touch \overline{BC}, and how far from A does it touch \overline{AB}?

Inradius and area

Find the inradius of a right triangle with legs 6 and 8.
Find the inradius of a right triangle with legs 9 and 40.
A triangle has area 30 and perimeter 30. Find its inradius.
A triangle has sides 25, 29 and 36 and area 360. Find its inradius.
Find the inradius of an equilateral triangle with side 12.
A triangle has inradius 3 and perimeter 40. Find its area.
In \triangle ABC, AB = AC = 17 and BC = 16. Find the inradius.

Coordinates

Find the incenter and inradius of the triangle with vertices (0, 0), (12, 0) and (0, 5).
Find the incenter and inradius of the triangle with vertices A(0, 0), B(21, 0) and C(5, 12).

Reasoning

Explain why \angle BIC is obtuse in every triangle, so that it can never be 80^\circ or 90^\circ.
The circumcenter of a triangle can lie outside it. Explain why the incenter never can.
Show that in an equilateral triangle the incenter and the circumcenter are the same point, and that the circumradius is twice the inradius.
Show that if the incenter and the circumcenter of a triangle are the same point, the triangle is equilateral.
Show that in a right triangle with hypotenuse c, the inradius is r = s - c.
The incircle of \triangle ABC touches \overline{BC} at X. Show that BX - CX = AB - AC.

Answers

x = 5, and the distance is 11. I is equally far from all three sides, so 3x - 4 = x + 6, giving 2x = 10. Then 3(5) - 4 = 11 and 5 + 6 = 11, and the distance to \overline{CA} is the same 11.
x = 4 and r = 8. Equal distances give x^2 - 8 = 2x, so x^2 - 2x - 8 = 0 and (x - 4)(x + 2) = 0. The root x = -2 would make the distance 2x = -4, which no distance can be, so x = 4, and 2x = 8 and x^2 - 8 = 8 agree.
AZ = AY = 15. Triangle AZI has a right angle at Z and hypotenuse 17, so \displaystyle AZ = \sqrt{17^2 - 8^2} = \sqrt{225} = 15. The two tangent pieces from A are equal, so AY = 15 too.
m\angle IAB = 20^\circ, m\angle ICB = 30^\circ, m\angle BIC = 110^\circ. First m\angle C = 180^\circ - 40^\circ - 80^\circ = 60^\circ. The bisectors halve the angles, so m\angle IAB = 20^\circ, m\angle IBC = 40^\circ and m\angle ICB = 30^\circ, and m\angle BIC = 180^\circ - 40^\circ - 30^\circ = 110^\circ, which is 90^\circ + 20^\circ.
m\angle A = 90^\circ. In \triangle BIC the halves of \angle B and \angle C add to 180^\circ - 135^\circ = 45^\circ, so m\angle B + m\angle C = 90^\circ and m\angle A = 90^\circ. Equivalently, 90^\circ + \tfrac{1}{2}m\angle A = 135^\circ.
m\angle C = 20^\circ. The halves of \angle A and \angle B add to 180^\circ - 100^\circ = 80^\circ, so m\angle A + m\angle B = 160^\circ and m\angle C = 180^\circ - 160^\circ = 20^\circ.
AZ = 6, BX = 4, CY = 8. Here a = 12, b = 14, c = 10 and s = 18, so AZ = s - a, BX = s - b and CY = s - c. Check: AZ + ZB = 6 + 4 = 10, BX + XC = 4 + 8 = 12, CY + YA = 8 + 6 = 14.
BC = 12, AB = 8, CA = 10, and the perimeter is 30. Equal tangent pieces give BZ = BX = 5, CY = CX = 7 and AY = AZ = 3. So BC = 5 + 7, AB = 3 + 5 and CA = 3 + 7.
At the midpoint of \overline{BC}, and 8 from A. Here s = \tfrac{1}{2}(13 + 13 + 10) = 18, so BX = s - b = 18 - 13 = 5, which is half of BC. The tangent piece from A is s - a = 18 - 10 = 8.
r = 2. The hypotenuse is 10, so r = \tfrac{1}{2}(6 + 8 - 10) = 2. By area: K = 24, s = 12, r = \tfrac{24}{12} = 2.
r = 4. The hypotenuse is \displaystyle \sqrt{9^2 + 40^2} = \sqrt{1681} = 41, so r = \tfrac{1}{2}(9 + 40 - 41) = 4. By area: K = 180, s = 45, and \tfrac{180}{45} = 4.
r = 2. The semiperimeter is 15, so r = \tfrac{K}{s} = \tfrac{30}{15} = 2.
r = 8. The semiperimeter is \tfrac{1}{2}(25 + 29 + 36) = 45, so r = \tfrac{360}{45} = 8.
\displaystyle r = 2\sqrt{3}. The height splits the base into halves of 6, so it is \displaystyle \sqrt{12^2 - 6^2} = 6\sqrt{3}, and \displaystyle K = \tfrac{1}{2}(12)(6\sqrt{3}) = 36\sqrt{3}. With s = 18, \displaystyle r = \tfrac{36\sqrt{3}}{18} = 2\sqrt{3}.
60. K = rs = 3 \times 20 = 60.
r = \tfrac{24}{5}. The bisector from A meets \overline{BC} at its midpoint at a right angle (as in Example 6), so the height is \displaystyle \sqrt{17^2 - 8^2} = 15 and K = \tfrac{1}{2}(16)(15) = 120. With s = \tfrac{1}{2}(17 + 17 + 16) = 25, r = \tfrac{120}{25} = \tfrac{24}{5}.
I = (2, 2) and r = 2. The right angle is at the origin, the legs are 12 and 5, and the hypotenuse is 13, so r = \tfrac{1}{2}(12 + 5 - 13) = 2. I is 2 from each axis and inside the triangle.
I = \left(7, \tfrac{14}{3}\right) and r = \tfrac{14}{3}. The sides are AB = 21, \displaystyle CA = \sqrt{5^2 + 12^2} = 13 and \displaystyle BC = \sqrt{16^2 + 12^2} = 20, so s = 27. The area is \tfrac{1}{2}(21)(12) = 126, so r = \tfrac{126}{27} = \tfrac{14}{3}. The tangent piece from A is s - a = 27 - 20 = 7, so the incircle touches the x-axis at (7, 0) and I is directly above it.
In \triangle BIC, m\angle BIC = 180^\circ - \tfrac{1}{2}(m\angle B + m\angle C), which equals 90^\circ + \tfrac{1}{2}m\angle A. Since m\angle A > 0^\circ, this is more than 90^\circ, and since m\angle A < 180^\circ, it is less than 180^\circ. So \angle BIC is always obtuse, and 80^\circ and 90^\circ are impossible.
The bisector of \angle A runs from A into the inside of the angle and ends on \overline{BC}, so apart from its ends the whole segment is inside the triangle; the same is true of the bisector from B. Two segments inside the triangle can only cross inside it, so I is inside. Perpendicular bisectors carry no such guarantee: they do not pass through the vertices, and in an obtuse triangle two of them cross outside.
Take the bisector from A, meeting \overline{BC} at M. Since AB = AC, triangles ABM and ACM are congruent by SAS (equal sides, equal halves of \angle A, shared \overline{AM}), so BM = CM and the angles at M are equal and add to 180^\circ, making them right angles. So the bisector from A lies on the perpendicular bisector of \overline{BC}, and in the same way for the other two vertices. The three bisectors lie on the three perpendicular bisectors, so I is the circumcenter. Now R = IB and r = IM, and triangle IMB has a right angle at M and m\angle IBM = 30^\circ (half of 60^\circ). Reflect it across line BM to a triangle I'MB: then IB = I'B and m\angle IBI' = 60^\circ, so \triangle IBI' is isosceles with a 60^\circ apex and therefore equilateral. Hence IB = II' = 2\,IM, that is, R = 2r.
Call the common point P. The perpendicular bisector of \overline{BC} passes through P and is perpendicular to \overline{BC}, so the foot of the perpendicular from P to \overline{BC} is the midpoint of \overline{BC}. But that foot is also X, where the incircle touches. So BX = CX, that is, s - b = s - c, and b = c. In the same way, Y is the midpoint of \overline{CA}, so s - c = s - a and c = a. All three sides are equal.
With the right angle at C, r = \tfrac{1}{2}(a + b - c), and s - c = \tfrac{1}{2}(a + b + c) - c = \tfrac{1}{2}(a + b - c). They are the same. Geometrically, s - c is the tangent piece CX, which is a side of the square CXIY, so it equals r.
BX = s - b and CX = s - c, so BX - CX = (s - b) - (s - c) = c - b, which is AB - AC. For example, with AB = 10, AC = 14, BC = 12 (Exercise 7), BX - CX = 4 - 8 = -4, and AB - AC = 10 - 14 = -4.