Centroid
A triangle cut from stiff card balances on a pencil point at exactly one spot. Where is that spot, and can it be found with a ruler or with arithmetic instead of by trial?
Where you are 4 in Triangle centers
Why this exists
Cut a triangle out of stiff card and try to balance it flat on the point of a pencil. Almost everywhere it tips over. There is exactly one spot where it stays level, and finding it by trial takes patience and gives only a rough answer.
Geometry finds it exactly, in two quite different ways. With a ruler: draw the medians, the segments from each corner to the midpoint of the opposite side, and the spot is where they cross. With arithmetic: if the corners have coordinates, average them, and the average is the spot.
Neither of those ought to work as well as it does. Two lines that are not parallel cross at one point, but a third line drawn through the picture almost always misses that point. That the third median always passes through the crossing of the other two is a theorem, not a lucky drawing. That an average of three corners lands on all three medians at once is a second surprise. This lesson proves both, and shows that they are one fact seen two ways.
The point is the centroid. It is one of the classic centers of a triangle, beside the Circumcenter and the Incenter, and it is the one that says where the triangle’s weight sits.
Definition
A median of a triangle joins a vertex to the midpoint of the opposite side, and every triangle has three. The centroid of a triangle is the point where its three medians meet. It is usually named G.
In Figure 1, D, E and F are the midpoints of \overline{BC}, \overline{CA} and \overline{AB} (matching tick marks show the equal halves), so \overline{AD}, \overline{BE} and \overline{CF} are the medians, and they meet at G.
The definition takes for granted that the three medians meet at a single point. Until that is proved, “the point where they meet” might name nothing at all. The first result below is that it always names exactly one point.
Intuition
Why a plate balances along a median. Think of the card triangle as a stack of very thin strips, all parallel to \overline{BC}, as in Figure 2. Each strip is a thin rod of even thickness, and such a rod balances at its own midpoint.
The midpoint of every strip lies on the median \overline{AD}. Each strip is a copy of \overline{BC} shrunk toward A, which is a dilation centered at A, and a dilation centered at A moves every point straight along its line to A. So the midpoint D of \overline{BC} moves along \overline{AD}, and it stays the midpoint of the shrunken copy.
Now lay the triangle on a straight knife edge along \overline{AD}. Every strip is held at its midpoint, so every strip balances, and so the whole triangle balances. The same is true of a knife edge along \overline{BE}, using strips parallel to \overline{CA}, and along \overline{CF}. A single point the triangle balances on has to lie on all three balance lines, and the only point on all three medians is the centroid.
Why three corner weights give two thirds. Put three equal weights at A, B and C on a board that weighs nothing. The weights at B and C on their own balance at their midpoint D, so together they act like one double weight at D. That leaves a weight of 1 at A and a weight of 2 at D. A lever balances when weight times distance is the same on both sides, so the balance point is twice as far from A as from D: two thirds of the way from A to D. That is where the centroid sits. It is the mean as a balance point, done in the plane instead of on a number line.
Results
The medians meet. The three medians of every triangle pass through one point, the centroid G.
The two-to-one split. On each median the centroid lies two thirds of the way from the vertex to the midpoint. With D the midpoint of \overline{BC},
In the same way BG = 2\,GE and CG = 2\,GF. The long piece is always the one at the vertex. Figure 3 marks the two pieces as 2k and k.
The centroid in coordinates. If the vertices are (x_1, y_1), (x_2, y_2) and (x_3, y_3), then
Each coordinate of the centroid is the Mean of the three vertices’ coordinates.
Equal areas. The segments \overline{GA}, \overline{GB} and \overline{GC} cut the triangle into three triangles of equal area, each a third of the whole. The three medians cut it into six small triangles of equal area, each a sixth of the whole.
Distance to a side. The distance from G to any side is one third of the altitude to that side.
Balance. A flat plate of even thickness in the shape of the triangle balances at G, and so do three equal weights placed at its vertices. (The weights are proved below; the plate is argued with strips, and making that argument exact takes calculus.)
Always inside. The centroid lies inside the triangle whatever its shape. The circumcenter and the Orthocenter and the Euler line both leave the triangle when it is obtuse; the centroid never does.
Why it is true
The medians meet two thirds along: a proof by midsegments. Take two of the medians, \overline{BE} and \overline{CF}. They cross: \overline{BE} cuts the triangle into two pieces, with C in one and F in the other, so \overline{CF} has to cross it. Call the crossing G. Let P be the midpoint of \overline{BG} and Q the midpoint of \overline{CG}, as in Figure 4.
| Statements | Reasons |
|---|---|
| FE \parallel BC and FE = \tfrac12 BC | F and E are the midpoints of two sides of \triangle ABC (midsegment theorem) |
| PQ \parallel BC and PQ = \tfrac12 BC | P and Q are the midpoints of two sides of \triangle GBC (midsegment theorem) |
| FE \parallel PQ and FE = PQ | both are parallel to \overline{BC}, and both equal \tfrac12 BC |
| FEQP is a parallelogram | one pair of opposite sides is both parallel and equal (proving a parallelogram) |
| GE = GP and GF = GQ | the diagonals \overline{EP} and \overline{FQ} of a parallelogram bisect each other, and they cross at G, since they lie along \overline{BE} and \overline{CF} |
| BP = PG = GE | P is the midpoint of \overline{BG}, and the step above |
| BG = 2\,GE and CG = 2\,GF | BG = BP + PG, and in the same way CG = CQ + QG |
So any two medians cross at the point two thirds of the way along each, measured from the vertex. Now bring in the third median, \overline{AD}. The same argument, applied to the pair \overline{AD} and \overline{BE}, says that they cross at the point two thirds of the way along \overline{BE} from B. There is only one such point, and it is G. So \overline{AD} passes through G as well: all three medians meet there, and each is split 2 : 1.
The coordinates: a proof by averaging. Let the vertices be A(x_1, y_1), B(x_2, y_2) and C(x_3, y_3). By the midpoint formula, the midpoint of \overline{BC} is D\left(\tfrac{x_2 + x_3}{2}, \tfrac{y_2 + y_3}{2}\right). The point two thirds of the way from A to D is found by partitioning the segment: start at A and add two thirds of the change from A to D. Its x-coordinate is
and the y-coordinate works out the same way. The answer does not change if the names of the vertices are swapped, so the point two thirds of the way along the median from B, and the one from C, are this same point. That is a second proof that the medians meet, and it gives the formula.
The equal areas. A median halves the area of its triangle, because the two halves have equal bases and the same height (medians). Use that on the small triangles first. In \triangle GBC, \overline{GD} joins a vertex to the midpoint of the opposite side, so \triangle GBD and \triangle GDC have equal areas; call each p. In the same way \triangle GCE and \triangle GEA each have area q, and \triangle GAF and \triangle GFB each have area r. Now use the big medians. \overline{AD} halves \triangle ABC, and \triangle ABD is made of the pieces r, r and p while \triangle ADC is made of q, q and p:
\overline{BE} halves \triangle ABC too, and \triangle ABE is made of r, r and q while \triangle CBE is made of p, p and q:
So all six pieces have the same area, a sixth of the whole, and \triangle GBC, made of two of them, has a third.
The distance to a side. Put \overline{BC} on the x-axis, with B(b, 0), C(c, 0) and A(a, h) above it, so that the altitude from A is h. The centroid’s y-coordinate is \tfrac{h + 0 + 0}{3} = \tfrac{h}{3}, and the distance from a point above the x-axis to the axis is its y-coordinate. Any side can be put on the x-axis, so the result holds for every side.
Inside, and balanced. G lies on \overline{AD} strictly between A and D, and every point of a median other than its two ends lies inside the triangle. The three-weights argument in the Intuition section is a complete reason for the corner weights. The strip argument is the reason for the plate; making it exact means adding up infinitely many thin strips, which is the work of the integral in calculus.
Worked examples
Lengths along two medians
G is the centroid of \triangle PQR. The median from P is \overline{PM}, with PM = 18. The median from Q is \overline{QN}, with GN = 5. Find PG, GM, QG and QN.
SolutionThe centroid is two thirds of the way along each median from the vertex. On \overline{PM}:
As a check, PG = 2\,GM, since 12 = 2 \cdot 6.
On \overline{QN} the given piece \overline{GN} is the one next to the midpoint N, so it is the short piece, and the piece at the vertex is twice as long:
The two medians have different lengths, 18 and 15, and that is normal. The 2 : 1 split holds on each median separately. It does not say that PG and QG are equal, and here they are not.
A split written in algebra
In \triangle ABC the centroid G lies on the median \overline{AD}, with AG = 3x + 2 and GD = 2x - 1. Find x, AG, GD and AD.
Solution\overline{AG} is the piece at the vertex, so it is the long one, twice the other: AG = 2\,GD.
Then AG = 3(4) + 2 = 14 and GD = 2(4) - 1 = 7, so AD = 14 + 7 = 21. Both pieces are positive, as lengths must be, and 14 = 2 \cdot 7.
Putting the factor 2 on the wrong side, GD = 2\,AG, gives 2x - 1 = 6x + 4, so x = -\tfrac54, and then AG = -\tfrac74. A negative length is the sign that the equation was set up backwards.
The centroid from coordinates
A triangle has vertices A(-3, 4), B(5, 6) and C(4, -4). Find its centroid G, and check that G lies two thirds of the way along the median from A.
Each coordinate of the centroid is the mean of the three vertices’ coordinates:
So G = (2, 2).
For the check, the midpoint of \overline{BC} is D = \left(\tfrac{5 + 4}{2}, \tfrac{6 + (-4)}{2}\right) = (\tfrac92, 1) by the midpoint formula. Going from A to D, the x-coordinate increases by \tfrac92 - (-3) = \tfrac{15}{2} and the y-coordinate changes by 1 - 4 = -3. Two thirds of those changes are 5 and -2, so the point two thirds of the way along is (-3 + 5,\ 4 - 2) = (2, 2). It is the same point, as Figure 5 shows.
A missing vertex
The centroid of \triangle ABC is G(1, 3), with A(-4, 2) and B(5, 8). Find C.
SolutionCall C = (x, y). Each coordinate of G is the mean of the vertices’ coordinates, so
and
So C = (2, -1).
Check. The vertices A, B and C average to
There is a shortcut once the idea is clear: the three vertices add up to three times the centroid, so C = 3G - A - B, which is (3 + 4 - 5,\ 9 - 2 - 8) = (2, -1) again.
Two routes to one distance
A right triangle has vertices A(0, 0), B(12, 0) and C(0, 9). Find the distance AG from the right-angled vertex to the centroid in two ways, and find the distance from G to side \overline{AB}.
SolutionThe centroid is
Route one, the distance formula.
Route two, the two-to-one split. The midpoint of \overline{BC} is D(6, \tfrac92), so the median from A has length
and AG = \tfrac23 \cdot \tfrac{15}{2} = 5. The two routes agree.
Side \overline{AB} lies along the x-axis, so the distance from G to it is G’s y-coordinate, 3. The altitude from C to \overline{AB} is the leg \overline{AC}, of length 9, and 3 = \tfrac13 \cdot 9, as the distance result says.
Areas cut out by the medians
The medians \overline{AD}, \overline{BE} and \overline{CF} of \triangle ABC meet at G, labeled as in Figure 1, and the area of \triangle ABC is 54 square units. Find the areas of \triangle GBD, \triangle GBC and the quadrilateral AFGE.
SolutionThe three medians cut the triangle into six small triangles of equal area, so each has area 54 \div 6 = 9 square units.
\triangle GBD is one of the six, so its area is 9 square units.
\triangle GBC is made of two of them, \triangle GBD and \triangle GDC, so its area is 9 + 9 = 18 square units. That is a third of 54, as the three-triangle result says.
The segment \overline{AG} cuts the quadrilateral AFGE into \triangle AFG and \triangle AGE, which are two more of the six. Its area is also 18 square units, although its shape is nothing like \triangle GBC.
A triangle rebuilt from its midpoints
The midpoints of the sides of \triangle ABC are D(4, 1) on \overline{BC}, E(0, 3) on \overline{CA} and F(2, -1) on \overline{AB}. Find the centroid of \triangle ABC without finding its vertices, and then find the vertices.
SolutionBy the midpoint formula, B + C = 2D, C + A = 2E and A + B = 2F, adding coordinates point by point. Adding all three,
So the three midpoints add up to the same total as the three vertices, and their average is the same point. The centroid of \triangle ABC is the centroid of \triangle DEF:
For the vertices, take one pair away from the total. The total A + B + C is 3G = (6, 3), and B + C = 2D = (8, 2), so
In the same way B = 3G - 2E and C = 3G - 2F, so
Check. The midpoint of \overline{BC} is \left(\tfrac{6 + 2}{2}, \tfrac{-3 + 5}{2}\right) = (4, 1), which is D, and the other two midpoints come out as E and F in the same way.
The formula A = 3G - 2D is the two-to-one split in disguise: rearranged, it says G = \tfrac13 A + \tfrac23 D, the point two thirds of the way from A to D.
Boundaries
It needs a real triangle. If the three points lie on one line, the formula still gives a point, their average, but there is no triangle around it: no area, nothing to balance, and the three “medians” all lie along the one line.
A plate, a frame and three weights are different objects. Pieces balance together at the average of their own balance points, each weighted by how much it weighs. A plate of even thickness weighs in proportion to its area, and it balances at G. Three equal weights at the vertices balance at G too. A frame made of three rods of the same wire is a third object: each rod weighs in proportion to its length, so the longest side pulls the balance point toward itself, and in general the frame does not balance at G. Exercise 20 finds one that does not.
Averaging the corners works for a triangle, not for every shape. The average of the four vertices of a quadrilateral is where four equal corner weights balance, but in general it is not where a quadrilateral plate balances (Exercise 21). For a triangle the plate and the corner weights happen to balance at the same point, and that is special to triangles.
The centroid is not equally far from things. In general it is not the same distance from the three vertices (the circumcenter is) or from the three sides (the incenter is). The 2 : 1 split compares the two pieces of one median; it says nothing about AG, BG and CG being equal. In an equilateral triangle all these centers are the same point (Exercises 23 and 25).
The split has a direction. The long piece is at the vertex and the short piece at the midpoint. AG : GD = 2 : 1, never 1 : 2.
Equal area is not congruence. The six small triangles made by the medians all have the same area, but in general they have different shapes.
Where this goes next
The centroid lies on a famous line. The circumcenter, the centroid and the orthocenter of any triangle lie on one straight line, the Euler line, with the centroid between the other two, twice as far from the orthocenter as from the circumcenter. (In an equilateral triangle the three are a single point, and there is no line to speak of.)
The rule that pieces balance at a weighted average of their own balance points is the idea of a center of mass, and it reaches well beyond triangles. In calculus the center of mass of any flat shape is found by cutting it into infinitely many strips, the argument of Figure 2 made exact. In probability the same weighted average, with chances in place of weights, is the expected value.
The averaging proof also works in space. The four vertices of a tetrahedron average to a point that lies on each line from a vertex to the centroid of the opposite face, three quarters of the way along. And the step at the heart of it, finding a point a given fraction of the way along a segment, is partitioning a segment in a ratio, used for any ratio at all.
Exercises
Lengths on a median
In Exercises 1 to 7, G is the centroid of \triangle ABC, and \overline{AD}, \overline{BE} and \overline{CF} are its medians, with D, E and F the midpoints of \overline{BC}, \overline{CA} and \overline{AB}.
- AD = 27. Find AG and GD.
- BG = 14. Find GE and BE.
- GF = 4.5. Find CG and CF.
- AG = 3x - 4 and GD = x + 3. Find x and AD.
- BG = 5y and BE = 6y + 18. Find y, BG and GE.
- CG = x^2 and GF = 2x + 6. Find every value of x that works, and the two lengths each one gives.
- The medians have lengths AD = 9, BE = 12 and CF = 15. Find AG, BG and CG.
The centroid from coordinates
Find the centroid of the triangle with the given vertices.
- (0, 0), (6, 0), (0, 9)
- (-5, 2), (3, 8), (5, -4)
- (1, -2), (4, 7), (-8, 1)
- (\tfrac12, 3), (2, -1), (-1, 1)
- The centroid of \triangle ABC is G(2, -1), with A(7, 3) and B(-4, 0). Find C.
- The centroid of \triangle ABC is the origin, with A(5, -2) and B(-1, 6). Find C.
Using the two-to-one split
- A median runs from A(1, 7) to D(7, 1), the midpoint of \overline{BC}. Find the centroid without finding B or C.
- A triangle has a vertex A(-6, -1) and centroid G(0, 2). Find the midpoint D of the side opposite A.
- A triangle has vertices (0, 0), (8, 0) and (2, 6). Find its centroid, the distance from the centroid to the side on the x-axis, and the altitude to that side.
- A triangle has vertices A(0, 0), B(9, 0) and C(0, 12). Find AG with the distance formula, then check it as two thirds of the median from A.
Areas
Use the labels of Figure 1.
- The area of \triangle ABC is 60. Find the areas of \triangle GBC, \triangle GBD and the quadrilateral BDGF.
- The area of \triangle ABC is 36. Find the area of \triangle GDE.
Weights and frames
Use the rule from the Boundaries section: pieces balance together at the average of their own balance points, each weighted by its weight.
- Three straight pieces of the same wire are joined into a frame in the shape of the triangle with vertices (0, 0), (3, 0) and (0, 4). Find where the frame balances, and compare it with the centroid of the triangle.
- A plate of even thickness has the shape of the quadrilateral with vertices (0, 0), (3, 0), (2, 2) and (0, 3), in that order. By cutting it along the diagonal from (0, 0) to (2, 2), find where the plate balances, and compare it with the average of the four vertices.
Reasoning
- Explain why the centroid always lies inside the triangle, even when the triangle is obtuse.
- Show that if the centroid and the circumcenter of a triangle are the same point, the triangle is equilateral.
- The line through G parallel to \overline{BC} meets \overline{AB} at X and \overline{AC} at Y. Show that XY = \tfrac23 BC, and that \triangle AXY has \tfrac49 of the area of \triangle ABC.
- Show that in an equilateral triangle the centroid is also the circumcenter and the incenter.
Answers
- AG = 18 and GD = 9. AG = \tfrac23 \cdot 27 and GD = \tfrac13 \cdot 27.
- GE = 7 and BE = 21. \overline{BG} is the long piece, so GE = 14 \div 2, and BE = 14 + 7.
- CG = 9 and CF = 13.5. CG = 2 \cdot 4.5, and CF = 9 + 4.5.
- x = 10 and AD = 39. AG = 2\,GD gives 3x - 4 = 2x + 6, so x = 10; then AG = 26 and GD = 13.
- y = 12, BG = 60 and GE = 30. BG = \tfrac23 BE gives 5y = 4y + 12. Then BE = 90 and GE = 90 - 60.
- x = 6, giving CG = 36 and GF = 18; or x = -2, giving CG = 4 and GF = 2. From CG = 2\,GF, x^2 = 4x + 12, so (x - 6)(x + 2) = 0. Both values give positive lengths, so both work: x may be negative, only the lengths must be positive.
- AG = 6, BG = 8 and CG = 10. Each is two thirds of its own median.
- (2, 3). \left(\tfrac{0 + 6 + 0}{3}, \tfrac{0 + 0 + 9}{3}\right).
- (1, 2). The x-coordinates add to 3 and the y-coordinates to 6.
- (-1, 2). The x-coordinates add to -3 and the y-coordinates to 6.
- (\tfrac12, 1). The x-coordinates add to \tfrac32, and \tfrac32 \div 3 = \tfrac12; the y-coordinates add to 3.
- C(3, -6). The vertices add to 3G, so C = 3G - A - B, which is (6 - 7 + 4,\ -3 - 3 - 0). Check: \tfrac{7 - 4 + 3}{3} = 2 and \tfrac{3 + 0 - 6}{3} = -1.
- C(-4, -4). The vertices must add to (0, 0), so C = (-5 + 1,\ 2 - 6).
- (5, 3). From A to D the coordinates change by 6 and -6. Two thirds of those are 4 and -4, so G = (1 + 4,\ 7 - 4).
- D(3, \tfrac72). G is two thirds of the way from A to D, so the change from A to D is \tfrac32 of the change from A to G. The change from A to G is (6, 3), so the change to D is (9, \tfrac92) and D = (-6 + 9,\ -1 + \tfrac92).
- G(\tfrac{10}{3}, 2), distance 2, altitude 6. The side on the x-axis runs from (0, 0) to (8, 0), so the distance from G to it is G’s y-coordinate, 2, and the altitude from (2, 6) is 6. The distance is one third of the altitude.
- AG = 5. G = (3, 4) and \displaystyle AG = \sqrt{3^2 + 4^2} = 5. The midpoint of \overline{BC} is D(\tfrac92, 6), so \displaystyle AD = \sqrt{\tfrac{81}{4} + 36} = \sqrt{\tfrac{225}{4}} = \tfrac{15}{2} and \tfrac23 \cdot \tfrac{15}{2} = 5.
- 20, 10 and 20. Each of the six small triangles has area 60 \div 6 = 10. \triangle GBC is two of them, \triangle GBD is one, and BDGF is made of \triangle GBD and \triangle GBF, two more.
- 3. \overline{AD} is a median, so \triangle ADC has half the area, 18. In \triangle ADC, E is the midpoint of \overline{AC}, so \overline{DE} is a median of it and \triangle ADE has area 9. \triangle GDE and \triangle ADE have the same apex E and bases \overline{GD} and \overline{AD} on one line, so they have the same height, and GD = \tfrac13 AD gives \triangle GDE a third of the area: 3.
- The frame balances at (1, \tfrac32), and the centroid is (1, \tfrac43). The sides have lengths 3, 4 and 5 and balance at their midpoints (\tfrac32, 0), (0, 2) and (\tfrac32, 2). Weighting each midpoint by its length, out of a total of 12:
The centroid is \left(\tfrac{0 + 3 + 0}{3}, \tfrac{0 + 0 + 4}{3}\right). The frame’s point is higher, pulled up by the long hypotenuse. 21. The plate balances at (\tfrac76, \tfrac76), and the vertices average to (\tfrac54, \tfrac54), a different point. The diagonal makes the triangles (0, 0), (3, 0), (2, 2) and (0, 0), (2, 2), (0, 3). Each has area 3: the first has base 3 on the x-axis and height 2, the second base 3 on the y-axis and height 2. Their centroids are (\tfrac53, \tfrac23) and (\tfrac23, \tfrac53), and two equal weights balance halfway between, at (\tfrac76, \tfrac76). The four vertices add to (5, 5), which averages to (\tfrac54, \tfrac54). 22. G lies on the median \overline{AD} with AG = \tfrac23 AD, so it is strictly between A and D. The midpoint D always lies on side \overline{BC}, whatever the angles, so the median runs from a vertex to a point of the opposite side and every point of it other than its ends is inside the triangle. The circumcenter has no such guarantee: it is defined by perpendicular bisectors, which need not pass through any vertex. 23. Suppose G is the circumcenter. G is inside the triangle, so it is not on \overline{BC}, and G \ne D. The circumcenter lies on the perpendicular bisector of \overline{BC}, which passes through D, so the line through G and D is that perpendicular bisector. But the line through G and D is also the median line AD. So A lies on the perpendicular bisector of \overline{BC}, which makes AB = AC (perpendicular bisectors). The same argument from vertex B gives BA = BC. All three sides are equal. 24. The dilation centered at A that sends B to X sends line BC to the parallel line through X, which is line XY, so it sends C to Y. It sends D to the point where line AD meets line XY, and that point is G. So its scale factor is \tfrac{AG}{AD} = \tfrac23. A dilation multiplies every length by its scale factor, so XY = \tfrac23 BC, and \triangle AXY is similar to \triangle ABC (AA similarity says the same) with every length scaled by \tfrac23. Base and height both scale by \tfrac23, so the area scales by \left(\tfrac23\right)^2 = \tfrac49. 25. In an equilateral triangle each vertex has two equal sides meeting at it, so the median from each vertex is also the altitude from it and bisects the angle there, the isosceles fact from medians and altitudes. A median that meets its side at a right angle, at the side’s midpoint, lies along the perpendicular bisector of that side. So the three median lines are the three perpendicular bisectors and also the three angle bisectors, and their one meeting point G is both the circumcenter and the incenter.