Circumcenter, Incenter, Orthocenter and the Euler Line

2. The CircumcenterLink to 2-the-circumcenter

The circumcenter is the point equidistant from all three vertices. It is the center of the circumcircle — the unique circle that passes through $A$, $B$, and $C$.

ConstructionLink to 2-the-circumcenter-construction

The circumcenter $O$ is found by intersecting the perpendicular bisectors of any two sides.

For side $\overline{AB}$, the perpendicular bisector is the line of points equidistant from $A$ and $B$:

$\{P \;|\; PA = PB\}$

The intersection of the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ gives $O$, which automatically lies on the third bisector as well.

Location by Triangle TypeLink to 2-the-circumcenter-location-by-triangle-type

Coordinate FormulaLink to 2-the-circumcenter-coordinate-formula

Given $A = (x_1, y_1)$, $B = (x_2, y_2)$, , the circumcenter coordinates can be found by solving:

$(x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2$

Expanding and subtracting gives two linear equations. Let $D = 2(x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2))$. Then:

$O_x = \frac{(x_1^2 + y_1^2)(y_2 - y_3) + (x_2^2 + y_2^2)(y_3 - y_1) + (x_3^2 + y_3^2)(y_1 - y_2)}{D}$

$O_y = \frac{(x_1^2 + y_1^2)(x_3 - x_2) + (x_2^2 + y_2^2)(x_1 - x_3) + (x_3^2 + y_3^2)(x_2 - x_1)}{D}$

3. The IncenterLink to 3-the-incenter

The incenter is the point equidistant from all three sides. It is the center of the incircle — the unique circle tangent to all three sides.

ConstructionLink to 3-the-incenter-construction

The incenter $I$ is found by intersecting the angle bisectors of any two vertices.

For $\angle A$, the angle bisector is the set of points equidistant from sides $\overline{AB}$ and $\overline{AC}$:

$\{P \;|\; d(P, AB) = d(P, AC)\}$

Key PropertyLink to 3-the-incenter-key-property

Since $I$ is equidistant from all three sides, the perpendicular distance from $I$ to each side equals the inradius $r$:

$r = d(I, AB) = d(I, BC) = d(I, CA)$

Coordinate FormulaLink to 3-the-incenter-coordinate-formula

The incenter is a weighted average of the vertices, weighted by the side lengths:

$I = \frac{aA + bB + cC}{a + b + c}$

where $a = BC$, $b = CA$, $c = AB$ (side lengths opposite the respective vertices).

In coordinates:

$I = \left(\frac{ax_1 + bx_2 + cx_3}{a + b + c},\; \frac{ay_1 + by_2 + cy_3}{a + b + c}\right)$

4. The OrthocenterLink to 4-the-orthocenter

The orthocenter is the point where all three altitudes of a triangle intersect. An altitude is a line through a vertex perpendicular to the opposite side.

ConstructionLink to 4-the-orthocenter-construction

From vertex $A$, draw the line through $A$ perpendicular to $\overline{BC}$. Repeat for vertex $B$ and $\overline{AC}$. The three altitudes are always concurrent at the orthocenter $H$.

Location by Triangle TypeLink to 4-the-orthocenter-location-by-triangle-type

Coordinate FormulaLink to 4-the-orthocenter-coordinate-formula

If $A = (x_1, y_1)$, $B = (x_2, y_2)$, , the orthocenter is:

$H_x = x_1 + x_2 + x_3 - 2O_x$

where $O$ is the circumcenter. More directly:

$H = A + B + C - 2O$

5. The Euler LineLink to 5-the-euler-line

One of the most beautiful results in triangle geometry is that three of the four centers we have studied lie on a single straight line.

The incenter $I$ lies on the Euler line only for isosceles triangles.

A Concrete ExampleLink to 5-the-euler-line-a-concrete-example

Let $A = (0, 0)$, $B = (6, 0)$, $C = (2, 8)$.

Centroid:

$G = \left(\frac{0+6+2}{3},\; \frac{0+0+8}{3}\right) = \left(\frac{8}{3},\; \frac{8}{3}\right)$

Circumcenter (solving the perpendicular bisectors):

The perpendicular bisector of $\overline{AB}$ is $x = 3$.

The perpendicular bisector of $\overline{AC}$: midpoint of $AC$ is $(1, 4)$, slope of $AC$ is $$, so perpendicular slope is . Equation:

$y - 4 = -\frac{1}{4}(x - 1)$

Intersecting with $x = 3$:

$y - 4 = -\frac{1}{4}(3 - 1) = -\frac{1}{2} \implies y = \frac{7}{2}$

$O = \left(3,\; \frac{7}{2}\right)$

Orthocenter (using the altitude intersection):

Altitude from $A$ is perpendicular to $BC$. Slope of $BC$: $\frac{8-0}{2-6} = \frac{8}{-4} = -2$, so altitude slope is . Through : .

Altitude from $B$ is perpendicular to $AC$. Slope of $AC$ is $4$, so altitude slope is $-\frac{1}{4}$. Through : .

Intersecting:

$\frac{1}{2}x = -\frac{1}{4}(x - 6)$

$H = (2, 1)$

Verification of the Euler line:

Are $O$, $G$, $H$ collinear? Check that $G$ divides $OH$ in the ratio $2:1$:

$O = \left(3, \frac{7}{2}\right), \quad H = (2, 1)$

$G_{\text{predicted}} = \frac{O + 2H}{3} = \frac{(3, 3.5) + (4, 2)}{3} = \left(\frac{7}{3},\; \frac{5.5}{3}\right) = \left(\frac{7}{3},\; \frac{11}{6}\right)$

But our computed $G$ is $\left(\frac{8}{3},\; \frac{8}{3}\right) = \left(\frac{16}{6},\; \frac{16}{6}\right)$. These differ — why?

6. Summary of Triangle CentersLink to 6-summary-of-triangle-centers

Euler Line PositionsLink to 6-summary-of-triangle-centers-euler-line-positions

7. The Euler Line and the Nine-Point CircleLink to 7-the-euler-line-and-the-nine-point-circle

The Euler line leads to another remarkable discovery — the nine-point circle:

8. Special Case: The Equilateral TriangleLink to 8-special-case-the-equilateral-triangle

In an equilateral triangle, all four centers coincide into a single point:

$O = G = I = H$

There is no unique Euler line — the line is undefined because the points are not distinct.

9. Relationship SummaryLink to 9-relationship-summary

10. Putting It All TogetherLink to 10-putting-it-all-together

The four triangle centers — centroid, circumcenter, incenter, and orthocenter — each capture a different aspect of the triangle's geometry: balance, circumscribed circle, inscribed circle, and altitude concurrency. Together they form an interconnected web of relationships summarized by the Euler line and the nine-point circle, revealing that even the simplest geometric figure contains extraordinary depth.