Lesson · Incentre
The centres of the triangle
Every triangle carries three meeting points around with it: the internal bisectors meet, the perpendicular bisectors of the sides meet, the altitudes meet. Each meeting point is a character with fixed habits, and a problem about one of them is usually won by naming the centre and quoting the right habit - the statement itself will mention neither.
The incentre
Theorem
The three internal angle bisectors of a triangle meet at one point \(I\), the centre of the inscribed circle. (A point lies on a bisector exactly when it is equidistant from the two sides, so the three bisectors have no choice but to agree.) The incircle touches each side once, and the two tangent segments drawn from any vertex are equal.
The equal tangent segments are the incentre's length habit: they cut an isosceles triangle off every vertex. Its angle habit is one formula:
Theorem
If \(I\) is the incentre of a triangle \(ABC\) with angles \(\alpha\), \(\beta\), \(\gamma\), then
Proof: in the triangle \(BIC\) the angles at \(B\) and \(C\) are \(\tfrac{\beta}{2}\) and \(\tfrac{\gamma}{2}\), so the angle at \(I\) is \(180^\circ - \tfrac{\beta + \gamma}{2} = 180^\circ - \tfrac{180^\circ - \alpha}{2}\).
Read backwards, the formula is a locus: over the chord \(BC\) the incentre sits on the arc of points that see \(BC\) under \(90^\circ + \tfrac{\alpha}{2}\). Whenever a problem puts \(I\) on a circle, this is the number that meets the inscribed angle theorem.
Principle
In an isosceles triangle the bisector of the apex angle is also the perpendicular bisector of the base. Every isosceles triangle in a figure is therefore a converter: it turns angle bisectors, the incentre's material, into perpendicular bisectors, the circumcentre's material - and back.
The circumcentre
Theorem
The perpendicular bisectors of the three sides meet at one point \(O\), equidistant from the vertices (on a perpendicular bisector exactly when equidistant from the two endpoints - the same argument as before). So \(O\) is the centre of the circumscribed circle, of radius \(R = OA = OB = OC\).
Two habits follow at once:
- Direction. The triangle \(OAB\) is isosceles, and its apex angle is the central angle over \(AB\): for \(\gamma < 90^\circ\), \(\angle AOB = 2\gamma\) and hence \(\angle OAB = 90^\circ - \gamma\). The direction of a circumradius can be read off the angles of the triangle. (For \(\gamma > 90^\circ\) the centre crosses to the other side of \(AB\) and \(\angle OAB = \gamma - 90^\circ\).)
- Diameters. A chord is a diameter exactly when \(O\) is its midpoint. So "prove that \(XY\) is a diameter" means "prove that \(O\) is the midpoint of \(XY\)": then \(OX = OY = R\) puts both endpoints on the circle and finishes in one stroke.
Where the centre lies is decided by the largest angle:
| Triangle | Position of \(O\) |
|---|---|
| acute | inside the triangle |
| right | the midpoint of the hypotenuse |
| obtuse | outside, beyond the longest side |
The orthocentre
The three altitudes also meet at one point, \(H\) - the vector footnote below proves it in one line. It sits inside an acute triangle, at the right-angled vertex of a right triangle, outside an obtuse one. Statements almost never name it; it is an auxiliary point you summon. What makes it worth summoning is that its mirror images are already on the circumcircle:
Theorem
Let \(H\) be the orthocentre of \(ABC\). The reflection of \(H\) in the line \(BC\) lies on the circumcircle; and the reflection of \(H\) in the midpoint of \(BC\) lies on the circumcircle as well, at the point diametrically opposite \(A\). The same holds for the other two sides.
For the first reflection, let the altitude from \(A\) meet the circumcircle again at \(H_a\). Inscribed angles over the arc \(H_aC\) give \(\angle H_aBC = \angle H_aAC = 90^\circ - \gamma\), and the altitude from \(B\) gives \(\angle HBC = 90^\circ - \gamma\) too; so \(BC\) bisects the angle at \(B\) in the isosceles triangle \(BHH_a\) and is the perpendicular bisector of \(HH_a\). For the second, let \(A'\) be the antipode of \(A\): by Thales \(A'B \perp AB\) and \(A'C \perp AC\), so \(A'B \parallel CH\) and \(A'C \parallel BH\), and \(BA'CH\) is a parallelogram - its diagonals \(BC\) and \(HA'\) bisect each other.
Vector footnote. From the earlier course, the centroid satisfies \(\overrightarrow{XT} = \tfrac{1}{3}\bigl(\overrightarrow{XA} + \overrightarrow{XB} + \overrightarrow{XC}\bigr)\) for every origin \(X\). Measured from the circumcentre, the same three vectors, unaveraged, give the orthocentre: \(\overrightarrow{OH} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}\). The check is one dot product per altitude: with \(\vec{a}, \vec{b}, \vec{c}\) the position vectors from \(O\), the point \(\vec{a} + \vec{b} + \vec{c}\) differs from \(A\) by \(\vec{b} + \vec{c}\), and \((\vec{b} + \vec{c}) \cdot (\vec{c} - \vec{b}) = R^2 - R^2 = 0\). In passing this proves that the altitudes are concurrent, and comparing with the centroid it gives \(\overrightarrow{OH} = 3\,\overrightarrow{OT}\): the points \(O\), \(T\), \(H\) lie on one line, the Euler line.
Warning
Every formula above was read off a picture. The angle \(\angle OAB = 90^\circ - \gamma\) changes to \(\gamma - 90^\circ\) when \(\gamma\) passes \(90^\circ\), the orthocentre leaves the triangle when it turns obtuse, and an angle chase that adds two angles in one configuration subtracts them in another. Say which case you are in; at the regional round the configuration analysis is part of the score.
Warm-up
Example
Let \(I\) be the incentre of a triangle \(ABC\), and let the line \(AI\) meet the circumcircle of \(ABC\) again at \(M\). Prove that \(MB = MI = MC\) - that is, \(M\) is the circumcentre of the triangle \(BIC\).
Which point of the circle is \(M\)?
\(AI\) bisects the angle at \(A\), so the inscribed angles \(\angle BAM\) and \(\angle MAC\) are equal, and equal inscribed angles cut equal arcs: \(M\) is the midpoint of the arc \(BC\) not containing \(A\). Equal arcs are cut off by equal chords, so \(MB = MC\) at once.
Compute the angle \(\angle MBI\).
\(\angle MBC = \angle MAC = \tfrac{\alpha}{2}\), inscribed on the same arc \(MC\), and \(\angle CBI = \tfrac{\beta}{2}\) because \(BI\) is the bisector at \(B\). So \(\angle MBI = \tfrac{\alpha}{2} + \tfrac{\beta}{2}\).
Now the angle \(\angle MIB\), with no circle at all.
\(\angle MIB\) is the exterior angle of the triangle \(ABI\) at \(I\), so it is the sum of the two remote interior angles \(\tfrac{\alpha}{2}\) and \(\tfrac{\beta}{2}\). The triangle \(MBI\) has equal base angles, hence \(MI = MB\), and \(M\) is equidistant from \(B\), \(I\) and \(C\). Remember the configuration: the circle through \(B\), \(I\), \(C\) is centred at the arc midpoint - an incentre habit and a circumcentre habit meeting in one point.