Circumcenter and Incenter of a Triangle

Try this

Drag the triangle into an obtuse triangle. Where is the circumcenter $O$ ? What about a right triangle?
However you drag, does the incenter $I$ ever leave the triangle?
Look at the readout $∠ B I C$ and the small text under it. When you drag $A$ , do they always agree?
Compare $R$ and $r$ . Can you drag the triangle so that $R$ is less than $2 r$ ?

The circumcenter $O$ is where the perpendicular bisectors of the three sides meet. It is equally far from the three vertices and is the center of the circumcircle. The incenter $I$ is where the three angle bisectors meet. It is equally far from the three sides and is the center of the incircle.

The angle at the incenter depends only on $∠ A$ :

∠ B I C = 18 0^{\circ} - \frac{∠ B + ∠ C}{2} = 9 0^{\circ} + \frac{1}{2} ∠ A

Predict first

In $△ A B C$ , $∠ A = 6 0^{\circ}$ and $I$ is the incenter. What is $∠ B I C$ , in degrees?

Choose an answer first; then you can see the result.

Break it

Counterexample lab: remove one condition

Conclusion: The circumcenter $O$ is inside the triangle

All three angles are acute: the circumcenter is inside.

The incenter–excenter lemma

Extend $A I$ to meet the circumcircle at $M$ . Let $J$ be the excenter opposite $A$ (where the bisector of $∠ A$ meets the bisectors of the exterior angles at $B$ and $C$ ). Then

M B = M C = M I = M J

The four equal segments fanning out from $M$ look like a chicken's foot, which is why Chinese students call this the “chicken-feet theorem”.

Proof: MB = MC = MI = MJ

Step 1 / 6

Get to know the figure
$M$ is where the extension of $A I$ meets the circumcircle, and $J$ is the excenter. The four readouts are equal right now; drag a vertex and they all change together, but stay equal.
MB = MC
Find ∠MBI
Find ∠MIB
MB = MI
The excenter J

Euler's formula

The distance $d = O I$ between the circumcenter and the incenter is completely determined by the two radii:

d^{2} = R (R - 2 r)

Drag the vertices and compare

O I^{2}

with

R (R - 2 r)

. Then click “Make it equilateral”.

Why? Use the power of a point for the incenter $I$ . $I$ is inside the circumcircle, so the chord $A M$ through $I$ gives

I A \cdot I M = R^{2} - d^{2} .

By the incenter–excenter lemma, $I M = M B$ . Now draw the diameter $M N$ , and let the incircle touch $A B$ at $F$ . In right triangles $A F I$ and $N B M$ , $∠ F A I = ∠ B N M$ (both stand on arc $B M$ ), so they are similar, and $\frac{I F}{I A} = \frac{M B}{M N}$ , that is, $I A \cdot M B = I F \cdot M N = r \cdot 2 R$ .

So $R^{2} - d^{2} = 2 R r$ , which is $d^{2} = R (R - 2 r)$ .

Since $d^{2} \geq 0$ , we immediately get $R \geq 2 r$ : the circumradius is at least twice the inradius, with equality only for an equilateral triangle.

Practice

Exercise 1Basic

A right triangle has sides $6$ , $8$ and $10$ . Find its inradius.

Exercise 2Basic

$I$ is the incenter of $△ A B C$ , and $∠ A = 7 0^{\circ}$ . Find $∠ B I C$ in degrees.

Exercise 3Advanced

A right triangle has sides $6$ , $8$ and $10$ , circumcenter $O$ and incenter $I$ . Find $O I^{2}$ .

Exercise 4Challenge

$△ A B C$ is inscribed in a circle of radius $6$ , $∠ A = 6 0^{\circ}$ , $I$ is the incenter, and the extension of $A I$ meets the circumcircle at $M$ . Find the length $M I$ .

Try this

Counterexample lab: remove one condition

The incenter–excenter lemma

Proof: MB = MC = MI = MJ

Euler's formula

Practice

Mistake Museum · Exhibits from this stop

Thinking the incenter is equally far from the three vertices

Taking ∠BIC to be 2∠A