The incircle and excirles

There are many nice properties of the incirce and excircles of a given triangle, but we will have to restrict our attention to only few of these. Let first introduce our notations - see Figure 1. The incircle with radius r (inradius) and center I touches the sides BC, CA and AB at X, Y and Z. We know that AY=AZ, BZ=BX and CX=CY. We have accordingly labeled these segments x, y and z, so that:

  

Figure 1: The incircle and excircles

y+z=a, z+x=b, x+y=c.

Adding these equations we find that:

x+y+z=p.

(We are using the standard notation: a, b, c and p for the sides BC, AC, AB and the semi-perimeter of $\Delta ABC$) and:

Theorem 1. x=p-a, y=p-b, z=p-c

Problem 1. Prove that the area of the triangle ABC is given by the formula: S_ABC=pr.

Proof. Since we have that $S_{IBC}= \frac{1}{2}ar$, $S_{IAC}=\frac{1}{2}br$ and $S_{IAB}=\frac{1}{2}cr$, after adding them we can easily obtain the required identity.

Denote by I_a, I_b and I_c the centers of the excircles of $\Delta ABC$. Remind you that I_a is the intersecting point of the external bisectors of the angles B an C and internal bisector of A, and similarly for I_b and I_c. Denote with r_a, r_b and r_c the corresponding radii (exradii). Each circle touches one side of the triangle internally and the other two externally. Marking the internal touching points with X_a, Y_b and Z_c and the external with Y_a, Z_a; X_b, Z_b; X_c, Y_c we have that:

Theorem 2.

BX_c=BZ_c=CX_b=CY_b=p-a,

CY_a=CX_a=AY_c=AZ_c=p-b,

AZ_b=AY_b=BZ_a=BX_a=p-c.

Proof. Two tangents from a point are equal in length, so BX_b=BZ_b. Also we have that BX_b+BZ_b=BC+CX_b+Z_bA+AB=BC+CY_b+Y_bA+AB=a+b+c. So:

AY_a=AZ_a=BZ_b=BX_b=CX_c=CY_c=p,

which proves the theorem.

Theorem 1 and 2 are the main tool for proving identities connected with the incircle and excircles of a given triangle. A large number of formulas can be obtained from them. See for example the next two problems.

Problem 2. Prove that: S_ABC=r_a(p-a)=r_b(p-b)=r_c(p-c) (see Problem 1.)

Proof. Have that $S_{ABC}=S_{ABI_a}+S_{ACI_a}-S_{BCI_a}=\frac{1}{2} (cr_a+br_a-ar_a)=r_a(p-a)$ and similarly for the rest.

Problem 3. $\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}=\frac{1}{r}$.

Proof. Using problem 1 and problem 2 we have that:

$$\frac{S_{ABC}}{r_a}+\frac{S_{ABC}}{r_b}+\frac{S_{ABC}}{r_c}=p-a+p-b+p-c=p=\frac{S_{ABC}}{r}$$

Problem 4. $\Delta ABC$ is the orthic triangle of $\Delta I_a I_b I_c$, i.e. AI_a, BI_b and CI_c are the altitudes in the triangle I_a I_b I_c. The incenter I is exactly the orthocenter of $\Delta ABC$.

Proof. Lets prove that AI_a is perpendicular to I_b I_c, the rest will follow easily. Have that $\angle I_aAI_c=\angle I_aAB+ \angle BAI_c= \frac{1}{2} \angle A+\frac{1}{2} (180^{\circ}-\angle A)=90^{\circ}$

The next three problems require more knowledge about the Geometry of the plane. For the teachers it is recommended to discuss these problems with the advanced students.

Problem 5. Let MY be the diameter of the incircle, perpendicular to the base AC (see Figure 2). Let L be the intersection of BM and AC. Prove that AY=LC.

  

Figure 2: Problem 5

Proof. Draw the tangent line EF to the circle at the point M. $EF \parallel AC$. Consider the homotethic transformation T with the center in B mapping M into L. T will map EF onto AC, and the incircle into the circle tangent to BA, BC and AC outside of the $\Delta ABC$, i.e. the excircle. In particular we have that L=Y_b. Using Theorem 1 and Theorem 2 we have that AY=CY_b=p-a.

Note. Although not so attractive, this problem can be solved purely algebraically. For this purpose you need to prove first that EM.AY=FM.CN (use that the triangles AOE and COF are right). After that use the two pairs of similar triangles: $\Delta ALB \sim \Delta MEB$ and $\Delta CLB \sim \Delta MFB$ obtain that EM.LC=AL.MF, so combining the two equalities we get AN=LC.

Problem 6. Triangle ABC is right-angled at C. Let CD is the altitude and r, r₁ and r₂ are the radii of the circles inscribed in the triangles ABC, ACD and BCD. Prove that r+r₁+r₂=CD.

Proof. Using Theorem 1 for the triangles ABC and that the quadrilateral OXCY is a square we have $r=OY=YC=\frac{1}{2} (a+b-c)$. Analogously, for the right triangles ACD and BCD we have $r_1=\frac{1}{2} (AD+CD-AC)$ and $r_2=\frac{1}{2} (BD+CD-BC)$. Now it is clear that r+r₁+r₂=CD, because AD+BD=c.

Problem 7. A circle is circumscribed about a triangle ABC. Knowing the radius R of that circle, find the radius of a circle passing through the centers I_a, I_b and I_c of the three excircles of the triangle ABC.

Solution. In order to find the radius x of the circumscribed circle of the triangle I_a I_b I_c we shall begin by determining the angles of this triangle. Since

$$\angle I_a BC = 90^{\circ}-\frac{1}{2}\angle B, \angle I_a CB = 90^{\circ}-\frac{1}{2}\angle C,$$

we have

$$\angle I_a=180^{\circ}-(90^{\circ}-\frac{1}{2}\angle B)-(90^{\circ}-\frac{1}{2}\angle C)=90^{\circ}-\frac{1}{2}\angle A.$$

Similarly $\angle I_b=90^{\circ}-\frac{1}{2}\angle B$, $\angle I_c=90^{\circ}-\frac{1}{2}\angle C$. So all the triangles I_a I_b I_c, I_a BC, I_b AC and I_c AB are similar since they have angles with the same magnitudes. The ratio of similitude of I_a BC and I_a I_b I_c is:

$$\frac{I_a B}{I_a I_b}=\cos \angle I_a=\sin \frac{1}{2} \angle A.$$

So we have that $\frac{R_1}{x}=\sin \frac{1}{2} \angle A$, where R₁ is radius of the circumscribed circle of the triangle I_a BC. Using the Sine rule for the triangles I_aBC and ABC we obtain that

$$R_1=\frac{a}{2\sin \angle I_a}=\frac{a}{2\cos \angle A}=\frac{2R\sin \angle A}{2\cos \angle A}=2R\sin \frac{1}{2} \angle A.$$

Substituting this value in the preceding expression for x, we find x=2R.