#incenter

The incenter of a triangle is another simple construction. The angle bisectors, as you might guess, are the lines that divide the angles in half. They meet at a point I, the incenter.

The proof that the angle bisectors all meet is very simple, following book 4, proposition 4 of Euclid's Elements.

Theorem: the angle bisectors of a triangle coincide.

Draw angle bisectors from A and B. Label their point of intersection as I. From I, draw perpendicular lines to the sides at D, E, and F.

The two blue triangles AFI and AEI have the same angles at A, same (right) angles at E and F, and a common side AI, so by angle-side-angle they are congruent. Therefore IE = IF.

By the same argument, the green triangles are congruent, and ID = IF, and all three perpendiculars from I are equal.

Draw line CI (dashed). The two white triangles CEI and CDI have equal (right) angles at E and D, equal sides DI and EI, and shared side CI. By side-side-angle they are congruent, so their angles at C are equal, and CI is the angle bisector from C.

Of course while this is essentially Euclid's argument, his goal was not to prove something about this point I, it was to draw a circle inscribed in an arbitrary triangle. Proving that the three perpendiculars are equal proves that they are radii of a common circle, which we call the incircle. And immediately after demonstrating this construction, his next proposition was construction of the circumcircle, which we looked at before. Is it weird that I'm more focused on the point at the center than the circle? Well, maybe. We'll get back to that later. For now, let's look at the two circles.

Here they are together. What's the relationship between them? The circumcircle is always bigger, and they don't in general have the same center.

Unless we want to start fiddling around with magnitudes of the sides or angles, we don't have a lot of variables here. Just the radii of the two circles and the distance between the centers. And here's a good reason to consider only those three numbers as important in the incircle-circumcircle relationship:

For any arrangement of incircle and circumcircle, there are infinitely many triangles that could be drawn to fit them, with varying side lengths and angles. So we should expect that those values don't matter, and the relationship between the circles should depend only on the circumcircle radius R, the incircle radius r, and the distance d between their centers.

To figure out how those values are related, we will first prove a handy little theorem about intersecting chords.

Theorem: the products of the two segments of intersecting chords are equal.

That is, we are saying AP * PB = CP * PD. Proof is by vertical angles, meaning that the red angles are equal, and the inscribed angle theorem, meaning the blue angles are equal and so are the greens. Therefore the two triangles APD and CPB are similar, so

AP / PD = CP / PB,

AP * PB = CP * PD.

Okay, let's use our new theorem.

Extend the angle bisector CI until it meets the circumcircle at K (red). Draw OI and extend it to meet the circumcircle at U and V (blue). Then by intersecting chords,

1) CI * IK = UI * IV = (R - d) (R + d) = R^2 - d^2

Next, a little angling. Draw segments KA and AI. Let α be the angle at vertex A and γ be the angle at vertex C. Segment CI is an angle bisector, so the red angles ACK and KCB are γ/2, and by inscribed angles, so is the red angle BAK. Segment AI is also an angle bisector, so the blue angles are α/2.

The green angle completes triangle AIC, so

∠AIC = 180° - (α + γ)/2

The purple angle is supplemental to it, so

∠AIK = 180° - (180° - (α + γ)/2) = (α + γ)/2

Therefore triangle AIK has two angles equal to (α + γ)/2, so it is isoceles, so

2) AK = IK

Finally, draw a perpendicular from I to side AB to intersect at Z. Draw KO and extend to the circumcircle at L. Now the two red triangles, CZI and LAK, have one angle equal to γ/2 (inscribe angle theorem again), and they are both right angles, so they are similar. So,

CI / IZ = LK / AK

3) CI * AK = LK * IZ = 2 R r

Putting 1), 2), and 3) together, we get this:

Theorem: given a triangle with circumradius R, inradius r, and distance between circumcenter and incenter d, 2 R r = R^2 - d^2.

This is known as Euler's theorem in geometry, first published by William Chapple in 1746 and then by Leonhard Euler nineteen years later. Chapple also first proved the existence of the orthocenter.

So does the converse hold? That is, if we have a pair of circles that fits that formula, can we always draw a triangle inside the larger one that is tangent to the smaller one?

Yes! In this image, the circles are drawn according to the formula and then the triangle drawn to fit them, and it works! (And although I am satisfied with this image because it turned out the way I wanted, somehow looking at it makes me anxious. This triangle looks like it's not having a good time. Don't stare at it too long if it bugs you.)

This is an example of Poncelet's porism, a theorem that says this works for polygons of any number of sides. If you have a pair of circles that will fit as incircle and circumcircle for some n-gon, then there are infinitely many n-gons that fit those circles. Chapple proved it for triangles and then Jean-Victor Poncelet proved the general case in 1822.

What else can we say about the incenter? Last week, we saw that the altitudes of a triangle are also the angle bisectors of the orthic triangle, which means that the orthocenter of the reference triangle is the incenter of the orthic triangle. Triangle geometry is full of cases like this, where the X-center of the Y-triangle is also the Z-center of the W-triangle, or whatever.

The incenter always lies in the interior of the circle with the line segment form the orthocenter to the centroid as a diameter. And last week, we learned that the centroid, circumcenter, and orthocenter all lie on one line, the Euler line. The incenter does not, unless the triangle is isosceles, in which case all triangle centers lie on the Euler line.

I suppose it's about time I explained what makes these points special and why we're looking at them. The four points I've been talking about belong to a class points called triangle centers. Though these four constructions were known to the ancient Greeks, they weren't identified as belonging to a special class until the 19th century, by which time many other triangle centers had been identified.

So what are they, exactly? Many of them are in fact the centers of notable circles, such as the circumcircle and incircle, or the centers of perspectivity of some pairs of triangles (I will have to explain perspectivity another day), and I suppose that's where the name comes from, but that's not what defines them.

Introducing a formal definition would be premature at the moment, but for now, we can observe one of the properties that applies to the four centers we've looked at so far. They are all constructed symmetrically with regard to sides and vertices. That is, if we use the midpoint of a side, we use the midpiont of all three sides. If we use the angle bisector of one angle, we use the bisectors of the other two as well. For a given triangle center, the roles of vertices A, B, and C are interchangeable.

Let's look at how things can work out if we don't construct things symmetrically. We constructed the incenter by bisecting the triangle's angles. But if we extend the sides into lines, we see that there are actually two bisectors at each intersection of the side lines, an interior bisector and an exterior bisector.

Here we have the exterior bisectors of vertices B and C (green) meeting the interior bisector of vertex A (blue) at a point Ja. This is called an excentral point, and it's the center of an exscribed circle for the triangle ABC.

But it's not a triangle center, because it wasn't constructed symmetrically. If we switched the roles of A and B in this construction, we'd get a different point. If we switch A and C, we get a third point.

Here are the three excircles and the incircle together. And just as with the incircle, there is a formula describing the relationship between the radius of the circumcircle, the radius of an excircle, and the distance between their centers. It's almost the same as the formula for the incircle:

2 R r_a = d_a^2 - R^2

where R is again the radius of the circumcircle, r_a is the radius of excircle A, and d_a is the distance from its center to the circumcenter. The equations for the other excircles are formed analogously.

And what about their positions? We can't just plonk them down anywhere and say those are the excircles of some triangle. Let's look at the angle bisectors again. The exterior bisectors, in green, form the excentral triangle. The interior bisectors, in blue, meet them perpendicularly at the vertices. That means that internal bisectors form altitudes of the excentral triangle, which demonstrates today's final result.

Theorem: The incenter of a triangle is the orthocenter of its excentral triangle, and the incenter and excenters form an orthocentric system.

It's all connected!

If you found this interesting, please try drawing some of this stuff for yourself! You can use a compass and straightedge, or software such as Geogebra, which I used to make all my drawings. You can try it on the web here or download apps to run on your own computer here.

Triangles are nice with the three sides and all, but what if we wanted to add more sides?

Let's add two more copies of side a attached to vertex A, extending the other two sides. Then we'll do the same with sides b and c. Now we have three times as many sides as your standard triangle, terminating in six points. Sweet bling!

But the amazing thing about this configuration is that all six endpoints lie on a common circle. And the other amazing thing is that this was discovered by John Conway. Yes, that John Conway. So this is called the Conway circle.

Proof: consider a circle with center I and a line segment PQ tangent to the circle at Z. Then we have two right triangles PZI and QZI with equal sides PZ and QZ and common side IZ. Then their hypotenuses are also equal, and are radii of a common larger circle, which makes PQ a chord of this circle.

In the Conway configuration, our tangent circle will be the incircle of triangle ABC. Now clearly the three extended line segments are equal, each being made of segments equal to a, b, and c in some order. It only remains, then, to prove that the incircle of triangle ABC touches the midpoints of these extended sides.

Let the incircle touch the sides at X, Y, and Z. Now consider the two segments AZ and AY. They are equal, since they are both tangents to the incircle from the same point A. Let their length be d. In the same way, the segments BX and BZ have equal length e and and segments CX and CY have equal length f. We can write the length of each side in terms of these shorter segments:

a = e + f

b = d + f

c = d + e

Now let's show one of the extended sides divided into those shorter segments.

The segment PQ is divided at Z, with PZ = f + e + d and QZ = e + d + f. Therefore Z is the midpoint of PQ, which is a chord of the larger circle.

The same argument applies to the other two extended sides, so X and Y are the midpoints of those segments as well, and all are chords of a common circle. QED.

That's all for this time. I wanted to do something quick and easy while I"m working on a longer piece that I hope to have done next week. Plus, I got to name-drop a celebrity, which is always fun.

If you found this interesting, please try drawing some of this stuff for yourself! You can use a compass and straightedge, or software such as Geogebra, which I used to make all my drawings. You can try it on the web here or download apps to run on your own computer here.

In this video, we will explain these fundamental concepts of geometry in an easy-to-understand manner.

We will begin by defining what the incenter, inradius, and incircle of a triangle are, and then move on to explore their properties and how they relate to each other. You will learn how the incenter is the center of the circle inscribed within the triangle, the inradius is the radius of that circle, and the incircle is the circle inscribed within the triangle itself.

We will also cover some practical applications of these concepts, such as how they are used in finding the area and perimeter of a triangle. By the end of the video, you will have a solid understanding of the incenter, inradius, and incircle of a triangle, and be able to apply this knowledge to solve a variety of geometry problems.