The incircle configuration: five facts worth memorising

Incircle problems are rarely hard because of the geometry. They are hard because the problem is stated in sides and angles, and the incircle speaks a different language: tangent lengths, the semiperimeter, one special angle and one special circle. A student who knows five facts translates in the first minute. A student who does not spends twenty minutes rediscovering one of them, usually halfway.

This is the short list. Each fact comes with the phrase in a problem that should make you reach for it.

Fix the notation once. Triangle $ABC$ has sides $a = BC$, $b = CA$, $c = AB$ and semiperimeter $s = \tfrac{a+b+c}{2}$. The incircle has centre $I$ and radius $r$, and touches $BC$, $CA$, $AB$ at $D$, $E$, $F$.

Fact 1: the tangent lengths are s − a, s − b, s − c

The two tangents from a point to a circle have equal length. So $AE = AF$, $BD = BF$ and $CD = CE$, and each side of the triangle is cut into two of these three lengths. Solving the three equations gives

• $AE = AF = s - a$
• $BD = BF = s - b$
• $CD = CE = s - c$

The way to remember which is which: the tangent length from a vertex is $s$ minus the side opposite that vertex, the one side the vertex does not touch.

When it fires: the words "the incircle touches $BC$ at $D$". Write $BD = s - b$ and $CD = s - c$ on the figure before you do anything else. Nearly every incircle problem wants a length along a side, and this is the only way to get one.

A check you can do in your head: the three tangent lengths add up to $3s - (a+b+c) = s$. That same sum is the whole engine of the largest triangle for a given perimeter. With the perimeter fixed, $s-a$, $s-b$, $s-c$ are three positive numbers with a fixed total, and Heron's formula turns the area into their product.

Fact 2: the area is r times s

Join $I$ to the three vertices. That cuts the triangle into three pieces with bases $a$, $b$, $c$ and the same height $r$, so the area is $K = \tfrac12 r(a+b+c) = rs$.

So $r = K/s$, and with Heron's formula, $r^2 = \dfrac{(s-a)(s-b)(s-c)}{s}$.

The 13–14–15 triangle is the one to carry around. Here $s = 21$, the tangent lengths are $8$, $7$, $6$, Heron gives $K = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$, and so $r = 4$. For a right triangle with legs $a$, $b$ and hypotenuse $c$, the same formula collapses to $r = \tfrac{a+b-c}{2} = s - c$: the 3–4–5 triangle has $r = 1$.

When it fires: any problem that mentions the inradius, or asks for an area and gives you a perimeter. It is also the move behind Viviani's theorem: split the triangle from an interior point into three pieces, add the areas, and the heights have to sum to a constant. The incircle version is the same split, from the one point whose three heights are equal.

Fact 3: the angle BIC is 90° + A/2

$I$ lies on the bisectors from $B$ and $C$, so in triangle $BIC$ the angles at $B$ and $C$ are $\tfrac{B}{2}$ and $\tfrac{C}{2}$. The third angle is $180^\circ - \tfrac{B+C}{2} = 180^\circ - \tfrac{180^\circ - A}{2}$, which is $90^\circ + \tfrac{A}{2}$.

Two things follow that are worth more than the formula. It is always obtuse. And it depends only on $A$: move $B$ and $C$ around however you like with $A$ fixed, and $\angle BIC$ does not change.

When it fires: the angle $BIC$ appears in the statement, or you need to show that $B$, $I$, $C$ and some fourth point are concyclic. A fixed angle over a fixed segment is a circle, which is exactly what Fact 4 names.

Fact 4: the arc midpoint is equidistant from B, I and C

Let the bisector from $A$ meet the circumcircle again at $M$, the midpoint of the arc $BC$ that does not contain $A$. Then

$$MB = MI = MC.$$

The chords $MB$ and $MC$ are equal because they stand on equal arcs. For $MI$, chase angles in triangle $MBI$: the angle at $B$ is $\tfrac{B}{2} + \tfrac{A}{2}$, since $\angle MBC = \angle MAC = \tfrac{A}{2}$, and the exterior angle at $I$ of triangle $ABI$ is also $\tfrac{A}{2} + \tfrac{B}{2}$. Equal angles, so $MB = MI$.

The circle centred at $M$ through $B$ and $C$ also passes through $I$, and through the excentre opposite $A$ as well, which sits on the same bisector at the far end of a diameter. People call it the incentre–excentre lemma, or the trillium, and it is the single most useful fact in the configuration, because it turns a statement about $I$ into a statement about the circumcircle.

When it fires: the incentre and the circumcircle in the same problem, or the bisector from $A$ extended past $BC$. That extension is the angle bisector you already know, continued until it hits the circle, and the point it hits is almost always the point the problem is really about.

Fact 5: the excircle swaps s − b and s − c

The excircle opposite $A$ touches $BC$ and the extensions of $AB$ and $AC$. The same equal-tangent bookkeeping, done with that circle, gives two results:

• The tangent length from $A$ to the excircle is $s$ itself, the full semiperimeter.
• It touches $BC$ at a point $X$ with $BX = s - c$ and $CX = s - b$.

Compare Fact 1, where $BD = s - b$. The incircle and the excircle touch $BC$ at points that are mirror images in the midpoint of $BC$. And the excircle's radius is $r_a = K/(s-a)$, the same formula as Fact 2 with $s - a$ in place of $s$. In the 13–14–15 triangle the three exradii are $84/8 = 10.5$, $84/7 = 12$ and $84/6 = 14$.

When it fires: a length that looks like $s - b$ on the wrong side of the triangle, a problem asking you to prove two segments along $BC$ are equal, or anything with "excircle" in the statement. The symmetry about the midpoint of $BC$ is the answer more often than it looks.

How to use the list

The five facts are not five separate topics. They are one translation table. A problem arrives in sides and angles; the first minute of your solution rewrites it in $s$, $s-a$, $s-b$, $s-c$, $r$ and the arc midpoint $M$, and most of the time the rewritten problem is short.

To make it stick, take ten incircle problems and do only the translation: label the tangent lengths, mark $M$, and write which of the five facts the problem is secretly about. Do not solve them. The translation is the part that runs out of time in an exam.

The incircle configuration page has the same kit in its shortest form and where it sits in the ladder, and the problem archive has the statements above in full. For a daily cold start, the daily problem posts a new one every day, no account needed.