The Four Centers of a Triangle

In the Chinese book: 三角形的四心

Four centers in one triangle. Which can leave it? Which stay in a line?

Try this

Keep only “Centroid G”: the three medians meet at one point. However you drag, $G$ stays inside the triangle.
Turn on “Circumcenter O”: the perpendicular bisectors of the three sides meet at one point, and the circle centered there passes through all three vertices. Drag the triangle into a right triangle: $O$ lands on the midpoint of the hypotenuse. Drag further to an obtuse triangle and $O$ goes outside.
Turn on “Orthocenter H”: the three altitudes meet at one point. In a right triangle $H$ is the right-angle vertex; in an obtuse triangle the altitudes only meet once they are extended, and $H$ is outside.
Turn on “Incenter I”: the three angle bisectors meet at one point, and the circle centered there touches all three sides. Like the centroid, it never leaves.
Turn on “Euler line”: however the triangle changes, $O$ , $G$ , $H$ stay on one straight line, and the $G H : O G$ reading is always $2$ . The incenter is usually not on this line.

Predict first

Drag the triangle into an obtuse triangle. How many of the four centers are outside it?

1

2

3

4

Choose an answer first; then you can see the result.

Why: one line through three centers

Where the Euler line comes from

Step 1 / 6

Centroid: splits each median 2 : 1
The three medians meet at the centroid $G$ . $G$ splits each median into two pieces in the ratio $2 : 1$ , with the longer piece next to the vertex.

$A G : G M_{a} = 2 : 1$
Circumcenter: where the perpendicular bisectors meet
Orthocenter: where the altitudes meet
Three points on one line
A right triangle
The nine-point circle

Why the centroid splits each median 2 : 1

Step 1 / 4

Let two medians meet at G
The medians $A M_{a}$ and $B M_{b}$ meet at $G$ . Join $C G$ . Write $S_{△ GB C} = x$ , $S_{△ GC A} = y$ , $S_{△ G A B} = z$ .
The common-side theorem, twice
The third median also passes through G
2 : 1

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: All four centers are inside the triangle

Drag the vertices, but keep all three angles acute. All four readings say “inside”.

Practice

Exercise 1Basic

A right triangle has legs $6$ and $8$ . Find the radius of its circumcircle.

Exercise 2Basic

The same right triangle as before (sides $6$ , $8$ , $10$ ). Find the radius of its incircle.

Exercise 3Advanced

$G$ is the centroid of $△ A B C$ , and the median to side $B C$ is $A D = 9$ . Find the length of $A G$ .

Exercise 4Challenge

In the acute triangle $△ A B C$ , $∠ A = 5 0^{\circ}$ , and $O$ , $H$ , $I$ are its circumcenter, orthocenter and incenter. Find $∠ B O C + ∠ B H C + ∠ B I C$ in degrees.

Mistake Museum · Exhibits from this stop

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Exhibit

Mixing up the definitions of incenter and circumcenter

The incenter is where the perpendicular bisectors of the sides meet, so it is the same distance from all three vertices.

What went wrong? Think first, then open

Diagnosis The point at equal distances from the three vertices lies on the perpendicular bisectors of the sides. It is the center of the circumcircle: the circumcenter. The incenter is where the three angle bisectors meet; it is at equal distances from the three sides, and it is the center of the incircle.

Circumcenter ↔ perpendicular bisectors ↔ equal distances to the vertices ↔ circumcircle. Incenter ↔ angle bisectors ↔ equal distances to the sides ↔ incircle.

“Circle through the vertices, circumcenter; circle hugging the sides, incenter”

Exhibit

Assuming all four centers are inside the triangle

O

is the circumcenter of

△ A B C

, so

O

is inside the triangle and

∠ B O C = 2∠ A

What went wrong? Think first, then open

Diagnosis In an obtuse triangle the circumcenter and the orthocenter move outside; in a right triangle they land on a side or a vertex. When $∠ A$ is obtuse, $∠ B O C = 36 0^{\circ} - 2∠ A$ .

In any problem with a circumcenter or an orthocenter, first check whether the triangle is acute, right or obtuse. If the problem does not say, split into cases. The centroid and the incenter are always inside, so they are never a worry.

“Circumcenter and orthocenter can go out; centroid and incenter never do”