PrevNext

Overview

In a triangle, a median is a segment from a vertex to the midpoint of the opposite side. Medians intersect at the centroid, which splits each median in a 2:12:1 ratio. This point is also the balance point of the triangle. They also give rise to triangles with equal areas.

Definitions and Key Ideas

  • The midpoint of a segment is the point that divides the segment into two equal lengths.
  • A median connects a vertex to the midpoint of the opposite side.
  • In a triangle, the three medians intersect at a single point called the centroid.
  • The centroid divides each median 2:12:1 from the vertex.
  • The three medians divide the triangle into 6 equal-area small triangles.
  • Each median splits the triangle into two equal-area triangles.
  • The centroid's coordinates are the average of the coordinates of the three vertices.

Area Properties

A B CL N MG

In △ABC,\triangle ABC,

[AMB]=[AMC],[AMB] = [AMC],

and similarly for the other three "half" triangles.

Also, the six triangles the median splits the larger triangle into have equal area. In other words,

[GLA]=[GBL]=[GBM]=[GCM]=[GNC]=[GAN][GLA] = [GBL] = [GBM] = [GCM] = [GNC] = [GAN]

Proof: If we drop an altitude from AA to BC,BC, triangles AMBAMB and AMCAMC will share this altitude. Their bases are also the same length since BM=MC.BM=MC. Then using Area=12bh,Area = \frac{1}{2} bh, these triangles have the same area. Also, if we drop an altitude from GG to BC,BC, we can show [GBM]=[GMC]=x.[GBM] = [GMC] = x. Repeat for the other three pairs of triangles, obtaining [GLA]=[GBL]=y[GLA] = [GBL] = y and [GAN]=[GNC]=z.[GAN] = [GNC] = z. Then from our previous result, 2y+x=2z+x.2y+x = 2z+x. Thus y=z.y=z. By a similar argument, x=y,x=y, so x=y=zx=y=z and we are done.

Ratio Properties

In △ABC,\triangle ABC, with MM as the midpoint of BCBC and GG the centroid, it holds that

$$AG : GM = 2 : 1.$

Proof: We follow the opposite reasoning as our area properties proof. We know that [BAG]=2[BGM].[BAG] = 2[BGM]. Therefore since these triangles share an altitude from BB to AM,AM, AG=2GM.AG = 2GM. Repeat for the other three medians symmetrically.

Connection to Coordinate Geometry

In a triangle with vertices (x1,y1),(x2,y2),(x3,y3),(x_1, y_1), (x_2,y_2), (x_3,y_3), the centroid has coordinates

(x1+x2+x33,y1+y2+y33).\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3} \right).

You may recognize this as the average of the corresponding xx and yy coordinates.

Proof: First, the midpoint of (x2,y2),(x3,y3)(x_2, y_2), (x_3,y_3) is

(x2+x32,y2+y32).\left(\dfrac{x_2+x_3}{2},\dfrac{y_2+y_3}{2} \right).

Next, the point 2/32/3 of the way to this point along the median from (x1,y1)(x_1,y_1) will be

13[2(x2+x32,y2+y32)+(x1,y1)]\dfrac{1}{3} \left [ 2\left(\dfrac{x_2+x_3}{2},\dfrac{y_2+y_3}{2} \right) + (x_1,y_1) \right]
(x1+x2+x33,y1+y2+y33).\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3} \right).

Common Pitfalls

  • Flipping the 2:12:1 ratio, or mistaking it for 1:1.1:1.
  • Assuming medians are perpendicular from their appearance on a diagram.
  • Confusing medians with angle bisectors or altitudes.

Worked Example

In triangle ABCABC, the centroid GG lies on median AMAM. If AG=10AG=10, find GMGM.

The centroid ratio is AG:GM=2:1AG:GM = 2:1, so GM=5GM = 5.

More Examples

Example 1: Full Median Length

If AG=8AG=8 on median AMAM, find AMAM.

AG=23AMAG = \frac{2}{3}AM, so AM=12AM = 12.

Example 2: Area Split

In triangle ABCABC, median AMAM splits the triangle into two triangles of equal area. If [ABC]=42[ABC]=42, find [ABM][ABM].

[ABM]=21[ABM]=21.

Example 3: Six Equal Areas

The medians divide a triangle into six equal areas. If one of the small triangles has area 55, what is the area of the whole triangle?

6⋅5=306\cdot 5 = 30.

Practice Problems

StatusSourceProblem NameDifficultyTags
AMC 10AMedium
Show TagsCentroid, Geometry, Isosceles Triangle, Triangle Medians
AMC 10BHard
Show TagsCentroid, Geometry, Locus, Triangle Medians
AMC 12BHard
Show TagsArea, Centroid, Geometry, Triangle Medians
Berkeley Math Circle Monthly Contest 3Hard
Show TagsAngle chasing, Cyclic quadrilaterals, Homothety, Tangents, Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle
Berkeley Math Circle: Monthly Contest 7Hard
Show TagsDistance chasing, Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle, Triangle trigonometry
Harvard-MIT Mathematics TournamentHard
Show TagsInduction / smoothing, Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle
Harvard-MIT Mathematics TournamentHard
Show TagsInversion, Tangents, Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle
HMMT November 2018Hard
Show TagsDistance chasing, Tangents, Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle, Triangle trigonometry
Berkeley Math CircleHard
Show TagsTriangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle, Triangle inequalities, Triangle trigonometry
Berkeley Math Circle Monthly Contest 4Hard
Show TagsDistance chasing, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
MathNetHard
Show TagsAngle chasing, Distance chasing, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle, Vectors
Berkeley Math Circle Monthly Contest 1Hard
Show TagsConstructions and loci, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math CircleHard
Show TagsDistance chasing, Quadrilaterals, Tangents, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math Circle Monthly Contest 5Hard
Show TagsConstructions and loci, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math CircleHard
Show TagsCartesian coordinates, Optimization in geometry, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math Circle: Monthly Contest 6Hard
Show TagsAngle chasing, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math CircleHard
Show TagsAngle chasing, Constructions and loci, Optimization in geometry, Rotation, Tangents, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math CircleHard
Show TagsAngle chasing, Cyclic quadrilaterals, Tangents, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math Circle Monthly Contest 2Hard
Show TagsAngle chasing, Tangents, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math Circle: Monthly Contest 8Hard
Show TagsAngle chasing, Tangents, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math Circle Monthly Contest 7Hard
Show TagsAngle chasing, Tangents, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math Circle Monthly Contest 3Hard
Show TagsAngle chasing, Distance chasing, Quadrilaterals with perpendicular diagonals, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math Circle Monthly Contest 7Hard
Show TagsAngle chasing, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math Circle Monthly Contest 1Hard
Show TagsAngle chasing, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle
Berkeley Math Circle Monthly Contest 8Hard
Show TagsAngle chasing, Homothety, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle

Module Progress:

Join the Discord Community!

Stuck on a problem, or don't understand a module? Join the Discord and get help with your doubts while making more math friends.

PrevNext