Finding an Incenter via Formulas

Terms

For every polygon, there is a largest circle that fits entirely within that polygon. If that circle touches all sides of the polygon, then it is said to be inscribed; it is called the incircle, and its center is called the incenter (which is then also called the polygon’s incenter). Every triangle has an incenter and an incircle.

An angle bisector is the ray that divides any angle into two congruent smaller angles. Geometrically, a triangle’s incenter can be located by drawing any two of its three angle bisectors and finding where they intersect, which is called the point of concurrency. This point is the same distance from all three sides of the triangle.

In this article, I will derive how to calculate the incenter and the radius of the incircle based only on the lengths of the sides of the triangle; I’ll also discuss some other related information.

Finding the Incenter Geometrically

Let’s consider a basic triangle: triangle

The angle bisector at ∠A splits it into two 29° angles; likewise, ∠B’s angle bisector creates two 21° angles and ∠C’s creates two 40° angles. These three bisectors intersect at a single point, H:
incenter

The distance from H to each side of the triangle is the segment perpendicular to the side that goes to H:incenter

Because H is the incenter, these three perpendicular segments are spokes (radii) of the incircle, and are hence the same length.

Finding the tangent points algebraically

The first step to finding the location of H using formulas is finding the location of D, E, and F. This is the easiest of the two steps.

Notice that we have divided our original triangle into six right triangles. What do we know about these triangles?

Consider △ADH and △AFH. ∠HAD ≅ ∠HAF because AH bisects ∠BAC; HA ≅ HA; ∠HDA ≅ ∠HFA because they’re both right angles. Thus, △ADH ≅ △AFH by AAS, and AD ≅ AF by CPCTC.

We can likewise demonstrate that BD ≅ BE and CF ≅ CE. In casual terms, this means that the two segments between each corner and each point along the side are the same length.

Let’s call these segment lengths a, b, and c, so that a = AD = AF, b = BD = BE, and c = CE = CF. And let’s call the three sides of the triangle s1, s2, and s3, as shown: labels

We can set up these equations: \[ a + b = s_1 \\ a + c = s_2 \\ b + c = s_3 \]

We can then solve these in terms of \(a\): \[ b = s_3 – c \\ a + s_3 – c = s_1 \\ c = s_2 – a \\ a + s_3 – (s_2 – a) = s_1 \\ a + a + s_3 – s_2 = s_1 \\ 2a = s_1 + s_2 – s_3 \\ a = \frac{s_1 + s_2 – s_3}{2} \]

This provides a way of finding the incenter of a triangle using a ruler with a square end: First find two of these tangent points based on the length of the sides of the triangle, then draw lines perpendicular to the sides of the triangle. The intersection point will be the incenter.

The formula for the radius

The radius of the incircle is the length of DH, FH, and EH. Consider △ADH. We now have a formula for AD. We need a formula for DH. Start with \[ \tan(\angle HAD) = \frac{DH}{AD} \\ \implies DH = AD\cdot \tan(\angle HAD) \]
Since ∠HAD is an angle bisector, m∠HAD is half of m∠BAC. Because we know the sides of the triangle, we can use the Law of Cosines to calculate m∠BAC: \[c^2 = a^2 + b^2 – 2ab \cos(C) \\ \implies \cos(C) = \frac{a^2 + b^2 – c^2}{2ab} \\ \implies C = \cos^{-1}\left(\frac{a^2 + b^2 – c^2}{2ab}\right) \\ m\angle BAC = \cos^{-1}\left(\frac{s_1^2 + s_2^2 – s_3^2}{2s_1s_2}\right) \\ m\angle HAD = \frac{\cos^{-1}\left(\frac{s_1^2 + s_2^2 – s_3^2}{2s_1s_2}\right)}{2} \]

This gives a fairly messy formula for the radius of the incircle, given only the side lengths:\[r = \left(\frac{s_1 + s_2 – s_3}{2}\right) \tan\left(\frac{\cos^{-1}\left(\frac{s_1^2 + s_2^2 – s_3^2}{2s_1s_2}\right)}{2}\right)\]

Coordinates of the Incenter

We can now calculate the coordinates of the incenter if we know the coordinates of the three vertices. Given a triangle △ABC where A is at the origin and B is on the x-axis, with side lengths \(s_1, s_2, s_3\), the incenter of △ABC is located at \[\left(\left(\frac{s_1 + s_2 – s_3}{2}\right), \left(\frac{s_1 + s_2 – s_3}{2}\right) \tan\left(\frac{\cos^{-1}\left(\frac{s_1^2 + s_2^2 – s_3^2}{2s_1s_2}\right)}{2}\right)\right)\]Note that we can calculate \(s_1, s_2,\) and \(s_3\) with the coordinates of A, B, and C using the distance formula: \[s_1 = \sqrt{(x_A – x_B)^2 + (y_A – y_B)^2} \\ s_2 = \sqrt{(x_A – x_C)^2 + (y_A – y_C)^2} \\ s_3 = \sqrt{(x_B – x_C)^2 + (y_B – y_C)^2} \]If A is not at the origin or B is not on the x-axis, the coordinate formula can be adjusted accordingly (with a bit of work).

Incircle of a Right Triangle

In a right triangle, \(s_1^2 + s_2^2 = s_3^2\), so \(\cos^{-1}\left(\frac{s_1^2 + s_2^2 – s_3^2}{2s_1s_2}\right) = \cos^{-1}\left(\frac{0}{2s_1s_2}\right) = \cos^{-1}(0)\), and \(\tan(\frac{\cos^{-1}(0)}{2}) = 1\). This means that the radius of the incircle of a right triangle is simply \(\frac{s_1 + s_2 – s_3}{2}\). For instance, the incircle of the standard 3, 4, 5 right triangle is \(\frac{3 + 4 – 5}{2} = \frac{2}{2} = 1\). In other words, the incircle of the standard 3, 4, 5 right triangle is the unit circle: Another reason to love that particular right triangle.

However, this is not the only right triangle with a unit incircle. Indeed, any right triangle where \(a + b – c = 2\) will have a unit incircle. We can calculate the leg length of an isosceles right triangle with a unit incircle two ways. Let’s use the fact that an isosceles triangle has two legs the same length, so \(a = b\): \[a + a – c = 2 \\ a^2 + a^2 = c^2 \implies 2a^2 = c^2 \\ \implies \sqrt{2a^2} = \sqrt{2}a = c \\ 2a – \sqrt{2}a = (2 – \sqrt{2})a = 2 \\ a = \frac{2}{2 – \sqrt{2}}\]From here, we rationalize by using the complement \(2 + \sqrt{2}\): \[\frac{2}{2 – \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{2(2 + \sqrt{2})}{(2 – \sqrt{2})(2 + \sqrt{2})} \\ = \frac{2(2 + \sqrt{2})}{4 – 2} = \frac{2(2 + \sqrt{2})}{2} \\ = 2 + \sqrt{2}\]
We can solve this using a different method, without the complement but a bit longer. First, as stated, any right triangle that satisfies these conditions will have a unit incircle:\[a^2 + b^2 = c^2 \\ a + b – c = 2\]The second condition implies \(a + b – 2 = c\). Since \(c = \sqrt{a^2 + b^2}\), \(a + b – 2 = \sqrt{a^2 + b^2}\). Square both sides and solve for a:\[(a + b – 2)^2 = a^2 + ab – 2a + ab + b^2 – 2b – 2a – 2b + 4 \\ = a^2 + 2ab + b^2 – 4a – 4b + 4 \\ = a^2 + b^2 + 2ab – 4a – 4b + 4 = a^2 + b^2 \\ \implies 2ab – 4a – 4b + 4 = 0 \\ \implies ab – 2a – 2b + 2 = 0 \\ \implies (b – 2)a = (2b – 2)\]Hence, any right triangle where the legs satisfy \(a = \frac{2b – 2}{b – 2}\) will have a unit incircle. We can check this for \((a, b) = (3, 4)\): \[3 \stackrel{?}{=} \frac{2\cdot4 – 2}{4 – 2} \\ = \frac{8-2}{2} = \frac{6}{2} = 3\] as well as for \((a, b) = (4, 3)\): \[4 \stackrel{?}{=} \frac{2\cdot3 – 2}{3 – 2} \\ = \frac{6-2}{1} = \frac{4}{1} = 4\]As expected, this relationship is true for the 3, 4, 5 right triangle.

Note that the formula means that b (and a) must be greater than 2. Setting \(b = a\) (for an isosceles right triangle) gives us: \[(a – 2)a = (2a – 2) \\ \implies a^2 – 2a = 2a – 2 \\ \implies a^2 – 4a + 2 = 0\] which we can then solve using the Quadratic Formula, \[x = \frac{-b + \sqrt{b^2 – 4ac}}{2a}\]In this case, it yields \(x = 2\pm\sqrt{2}\); \(x = 2-\sqrt{2}\) is less than 2, so the only viable solution is \(x = 2+\sqrt{2}\).

Conclusion

In this article, I discussed how to find the tangent points and the radius of an incircle when you know the lengths of the sides of a triangle (\(s_1, s_2, s_3\)). The tangents create segments of each permutation of: \[a = \frac{s_1 + s_2 – s_3}{2}\] The radius is: \[r = \left(\frac{s_1 + s_2 – s_3}{2}\right) \tan\left(\frac{\cos^{-1}\left(\frac{s_1^2 + s_2^2 – s_3^2}{2s_1s_2}\right)}{2}\right)\].

I also pointed out that the radius of an incircle of a right triangle is simply \[r = \frac{a + b – c}{2}\] and provided some commentary on unit circles and isosceles right triangles.

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