Orthocentric System — Definition, Formula & Examples

An orthocentric system is a set of four points in a plane such that each point is the orthocenter of the triangle formed by the remaining three points. This means all four triangles share the same collection of altitudes.

A set of four points $\{A, B, C, H\}$ constitutes an orthocentric system if and only if $H$ is the orthocenter of $\triangle ABC$ , $C$ is the orthocenter of $\triangle ABH$ , $B$ is the orthocenter of $\triangle ACH$ , and $A$ is the orthocenter of $\triangle BCH$ . Equivalently, every pair of line segments connecting these four points is perpendicular to the line segment connecting the other two points.

How It Works

Start with any triangle

\triangle ABC

and find its orthocenter

H

by constructing the three altitudes. Now consider the triangle formed by any three of the four points

A

B

C

H

. Remarkably, the fourth point is always the orthocenter of that triangle. This works because the altitude from

A

BC

is also the altitude from

C

AB

\triangle ABH

, and similarly for the other altitudes. The six line segments connecting pairs of these four points form three pairs, and each pair is mutually perpendicular to one other pair.

Worked Example

Problem: Given triangle ABC with vertices A(0, 6), B(0, 0), and C(8, 0), find the orthocenter H and verify that A is the orthocenter of triangle BCH.

Step 1: Find the orthocenter H of triangle ABC. The altitude from A to BC (the x-axis) is a vertical line through A, so x = 0. The altitude from C to AB (the y-axis) is a horizontal line through C, so y = 0. These meet at the origin, but that is B. Since angle B is 90°, the orthocenter is at B itself — so pick a non-right triangle instead. Let A(1, 5), B(0, 0), C(6, 0). The altitude from A perpendicular to BC (y = 0) is x = 1. The altitude from C perpendicular to AB: slope of AB = 5, so slope of altitude = -1/5, giving y - 0 = -1/5(x - 6), i.e., y = -(x-6)/5. Set x = 1: y = -(1-6)/5 = 1. So H = (1, 1).

H = (1,\, 1)

Step 2: Now verify that A(1, 5) is the orthocenter of triangle BCH with B(0, 0), C(6, 0), H(1, 1). The altitude from B perpendicular to CH: slope of CH = (1-0)/(1-6) = -1/5, so the perpendicular slope is 5. Line through B(0,0) with slope 5: y = 5x. The altitude from C perpendicular to BH: slope of BH = (1-0)/(1-0) = 1, so the perpendicular slope is -1. Line through C(6,0): y = -(x-6) = -x + 6.

5x = -x + 6 \implies 6x = 6 \implies x = 1,\; y = 5

Step 3: The altitudes of triangle BCH intersect at (1, 5), which is exactly point A. This confirms that the four points form an orthocentric system.

A = (1,\, 5) \checkmark

Answer: H = (1, 1), and A = (1, 5) is confirmed as the orthocenter of triangle BCH, verifying the orthocentric system {A, B, C, H}.

Why It Matters

Orthocentric systems appear in competition geometry and advanced Euclidean geometry courses. They reveal deep symmetry in how triangle centers behave: the four triangles in the system all share the same nine-point circle and the same set of altitude lines.

Common Mistakes

Mistake: Assuming the orthocenter of a triangle is always inside the triangle.

Correction: For obtuse triangles, the orthocenter lies outside the triangle. In an orthocentric system, some of the four triangles will be obtuse, so expect exterior orthocenters.