The Timpson Lyceum · Geometry (2024)

Centroid, Incenter, Circumcenter, and Orthocenter – Lecture 16 (Geometry)

Overview

This lecture builds toward four remarkable concurrency properties of triangles. It opens by recapping set theory and introducing the idea of a locus — a set of points satisfying a condition — then uses the locus viewpoint to prove a powerful, reusable proof technique: to show two figures are the same set of points, show each is a subset of the other. With that tool in hand, the lecture proves that each of the four classic families of triangle lines is concurrent (all three meet at one point), giving the incenter, circumcenter, centroid, and orthocenter. These four points are foundational in geometry and recur constantly in later coursework.

Key Concepts

Set basics

• A set is an unordered, unique collection: $\{A, B\} = \{B, A\} = \{A, B, A\}$. Order and duplicates do not matter, so $\{A, B\}$ has exactly two elements.
• Membership $x \in A$ means $x$ is an element of the set $A$. Note $1 \in \{1, 2\}$ but $\{1\} \notin \{1, 2\}$, since the set $\{1\}$ is a different object from the number $1$.

Subsets and equality

• Subset: $A \subseteq B$ means $x \in A \implies x \in B$ (everything in $A$ is also in $B$). Every set is a subset of itself.
• Set equality (the key proof tool): $A = B$ if and only if $A \subseteq B$ and $B \subseteq A$.
• Proper subset: $A \subsetneq B$ means $A \subseteq B$ and $A \neq B$. (E.g. $\mathbb{N} \subsetneq \mathbb{Z}$.)

Set-builder notation and operations

• Set-builder: $\{x : P(x)\}$ is the set of all $x$ satisfying condition $P$. Example: the evens are $\{x : x \in \mathbb{Z} \text{ and } 2 \mid x\}$; the perfect squares are $\{y : y = x^2,\ x \in \mathbb{N}\}$.
• Intersection: $A \cap B = \{x : x \in A \text{ and } x \in B\}$.
• Union: $A \cup B = \{x : x \in A \text{ or } x \in B\}$.

Locus

• A locus (plural loci) is a set of points satisfying a condition — a fancy word for a set whose elements are points, signaling a geometric figure. Examples: points a distance $r$ from a point $O$ form a circle; points a distance $d$ from a line $L$ form two parallel lines; points equidistant from two parallel lines form a third parallel line midway between them.

Distance from a point to a line

• Defined as the length of the perpendicular segment from the point to the line. This definition is used throughout the proofs.

Supporting theorems

• Perpendicular-bisector locus: the locus of points equidistant from two points $A$ and $B$ is the perpendicular bisector of $\overline{AB}$.
• Angle-bisector locus: the locus of points interior to an angle and equidistant from its two sides is the angle's bisector (minus the vertex).
• Inscribed-angle theorem: an inscribed angle is half the central angle subtending the same arc.

The four triangle centers

• Incenter: the three angle bisectors are concurrent. The meeting point is equidistant from all three sides and is the center of the inscribed circle.
• Circumcenter: the three perpendicular bisectors of the sides are concurrent. The meeting point is equidistant from all three vertices and is the center of the circumscribed circle.
• Centroid: the three medians (vertex to opposite midpoint) are concurrent. The meeting point lies $\tfrac{2}{3}$ of the way from each vertex along its median — it cuts each median in a $1 : 2$ ratio.
• Orthocenter: the three altitudes (vertex perpendicular to the opposite side) are concurrent.

Worked Example

Proving the locus of points equidistant from $A$ and $B$ is the perpendicular bisector of $\overline{AB}$. The strategy is set equality: with $O$ the midpoint of $\overline{AB}$, let $S_1$ be the equidistant locus and $S_2$ the perpendicular bisector; show $S_1 \subseteq S_2$ and $S_2 \subseteq S_1$.

For $S_1 \subseteq S_2$, take any $P \in S_1$ (the case $P = O$ is immediate). Then $\overline{PA} \cong \overline{PB}$, so $\triangle PAB$ is isosceles and the base angles are congruent; together with $\overline{OA} \cong \overline{OB}$, the side-angle-side criterion gives $\triangle PAO \cong \triangle PBO$:

These two angles are congruent and supplementary, so each is a right angle. Hence $\overline{OP} \perp \overline{AB}$ through the midpoint $O$, so $P$ lies on the perpendicular bisector: $P \in S_2$. The reverse inclusion $S_2 \subseteq S_1$ runs symmetrically — any $P$ on the perpendicular bisector gives $\triangle PAO \cong \triangle PBO$ by SAS, forcing $\overline{PA} \cong \overline{PB}$, so $P \in S_1$. With both inclusions, $S_1 = S_2$. $\blacksquare$

Summary

• A locus is just a set of points meeting a stated condition; thinking of geometric figures as sets of points lets us prove two figures identical by a clean technique.
• Two sets are equal exactly when each is a subset of the other — prove it by taking an arbitrary element of one and showing it lies in the other, then repeating in reverse.
• The angle bisectors of a triangle meet at the incenter, equidistant from the sides (center of the inscribed circle).
• The perpendicular bisectors of the sides meet at the circumcenter, equidistant from the vertices (center of the circumscribed circle).
• The medians meet at the centroid, which divides each median in a $1 : 2$ ratio ($\tfrac{2}{3}$ of the way from each vertex).
• The altitudes meet at the orthocenter; this is proved by enclosing the triangle in a larger one whose perpendicular bisectors are the original altitudes, reusing the circumcenter result.