Olympiad MethodsDifficulty 60-10026 tagged problems

Barycentric Coordinates

Barycentric Coordinates is the habit of representing points by weighted triangle coordinates to handle cevians and ratios. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$P = (x : y : z), \quad x+y+z \ne 0, \qquad \vec{P} = \frac{x\,\vec{A} + y\,\vec{B} + z\,\vec{C}}{x+y+z}.$$

Relative to a reference triangle $ABC$, every point is written as a weighted average of the vertices; the weights (defined up to a common scale) make cevians, ratios, and collinearity purely algebraic.

Why It Works

1
Fix triangle $ABC$. Any point $P$ in the plane is a unique affine combination $\vec{P} = \alpha\vec{A} + \beta\vec{B} + \gamma\vec{C}$ with $\alpha+\beta+\gamma = 1$; these normalized weights are the barycentric coordinates.
2
Coordinates are homogeneous: $(x:y:z)$ and $(kx:ky:kz)$ denote the same point for $k\ne 0$, so we recover the point by dividing by $x+y+z$.
3
The vertices are $A=(1:0:0)$, $B=(0:1:0)$, $C=(0:0:1)$; a point on side $BC$ has first coordinate $0$.
4
Three points are collinear iff the determinant of their coordinate rows is $0$, and a cevian's foot inherits the opposite-side ratios directly from the weights — this is what makes ratio problems mechanical.

Worked Examples

Example 1

Prove the medians of triangle $ABC$ concur, and find the concurrency point in barycentric coordinates.

1
The midpoint of $BC$ is $M_A = (0:1:1)$ (equal weights on $B$ and $C$, none on $A$); similarly $M_B=(1:0:1)$ and $M_C=(1:1:0)$.
2
The median from $A$ passes through $A=(1:0:0)$ and $M_A=(0:1:1)$; points on it have the form $(t:1:1)$.
3
Setting all three medians to share a point forces equal weights, giving the candidate $G=(1:1:1)$.
4
Check $G=(1:1:1)$ lies on median $A M_A$: it has the form $(t:1:1)$ with $t=1$. ✓ By symmetry it lies on all three, so they concur. Normalizing, $\vec{G} = \frac{\vec{A}+\vec{B}+\vec{C}}{3}$.

Answer:

The centroid $G = (1:1:1)$, i.e. $\vec{G} = \frac{\vec{A}+\vec{B}+\vec{C}}{3}$.

Contest level

In triangle $ABC$, point $D$ lies on $BC$ with $BD:DC = 2:3$, and $E$ lies on $AD$ with $AE:ED = 4:1$. Find the barycentric coordinates of $E$.

1
$D$ divides $BC$ with $BD:DC = 2:3$, so $D$ is the weighted average putting weight on $B$ and $C$ inversely to the opposite segments: $D = (0 : DC : BD) = (0 : 3 : 2)$ (a point on $BC$ has zero $A$-weight; weights split as $DC$ on $B$, $BD$ on $C$). Normalize: $D = (0 : 3 : 2)$ with weight sum $5$.
2
$E$ divides $AD$ with $AE:ED = 4:1$, so $E$ puts weight $ED=1$ on $A$ and weight $AE=4$ on $D$ (again inverse to the segment lengths): $\vec{E} = \tfrac{1\cdot \vec{A} + 4\cdot \vec{D}}{5}$.
3
Use the normalized $\vec{D} = \tfrac{3\vec{B} + 2\vec{C}}{5}$, so $\vec{E} = \tfrac{1}{5}\vec{A} + \tfrac{4}{5}\cdot\tfrac{3\vec{B}+2\vec{C}}{5} = \tfrac{1}{5}\vec{A} + \tfrac{12}{25}\vec{B} + \tfrac{8}{25}\vec{C}$.
4
Clear denominators (multiply by $25$): weights $(5 : 12 : 8)$, which sum to $25$ — consistent.

Answer:

$E = (5 : 12 : 8)$.

Olympiad / Challenge

Prove that the cevians $AA'$, $BB'$, $CC'$ concur iff $\dfrac{BA'}{A'C}\cdot\dfrac{CB'}{B'A}\cdot\dfrac{AC'}{C'B} = 1$ (Ceva), using barycentrics.

1
Put $A' = (0 : m : n)$ on $BC$; then $BA':A'C = n:m$ (the weight on $B$ is proportional to the FAR segment $A'C$, the weight on $C$ to $BA'$), i.e. $\tfrac{BA'}{A'C} = \tfrac{n}{m}$. Similarly $B' = (p : 0 : q)$ with $\tfrac{CB'}{B'A} = \tfrac{p}{q}$ and $C' = (r : s : 0)$ with $\tfrac{AC'}{C'B} = \tfrac{s}{r}$.
2
Cevian $AA'$ is the line through $A=(1:0:0)$ and $A'=(0:m:n)$, i.e. the locus $\{(t : m : n)\}$, which in coordinates is the line $n\,y = m\,z$ (eliminating $t$).
3
Likewise $BB'$ is $q\,x = p\,z$ and $CC'$ is $s\,x = r\,y$. A common point exists iff these three linear relations are consistent, i.e. iff the product around the cycle returns: from the three, $\tfrac{x}{z}=\tfrac{p}{q}$, $\tfrac{y}{z}=\tfrac{m}{n}$, $\tfrac{x}{y}=\tfrac{r}{s}$, and consistency requires $\tfrac{p}{q}\cdot\tfrac{n}{m}\cdot\tfrac{s}{r}=1$.
4
Translate back: $\tfrac{n}{m}\cdot\tfrac{p}{q}\cdot\tfrac{s}{r} = \dfrac{BA'}{A'C}\cdot\dfrac{CB'}{B'A}\cdot\dfrac{AC'}{C'B} = 1$, which is exactly Ceva's condition. ✓

Answer:

Concurrency $\iff \dfrac{BA'}{A'C}\cdot\dfrac{CB'}{B'A}\cdot\dfrac{AC'}{C'B} = 1$ (Ceva, recovered from barycentric line equations).

Going Deeper

Barycentrics generalize to the incenter $I = (a:b:c)$, circumcenter $O = (a^2(b^2+c^2-a^2):\dots)$, and symmedian point $K=(a^2:b^2:c^2)$ — using side lengths as weights handles 'triangle-center' problems that pure coordinates make ugly.
It is the engine behind many olympiad geometry bashes (USAMO/IMO): collinearity is a $3\times 3$ determinant, concurrency is its dual, and conics (circumcircle, etc.) have clean barycentric equations.
Pitfall: barycentric coordinates are AFFINE, not Euclidean — distance, perpendicularity, and circles require the displacement vector $(x,y,z)$ with $x+y+z=0$ and the metric $-a^2 yz - b^2 zx - c^2 xy$. Treating raw coordinates as Cartesian gives wrong lengths and angles.

Spot the Signal

Look for problems where the key step is representing points by weighted triangle coordinates to handle cevians and ratios.
You can describe the hard part as representing points by weighted triangle coordinates to handle cevians and ratios, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the olympiad structure that calls for barycentric coordinates, then rewrite the givens around it.
Name the relation that makes Barycentric Coordinates legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Barycentric Coordinates just because the surface looks familiar; verify the required condition first.
Applying Barycentric Coordinates because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let barycentric coordinates reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is representing points by weighted triangle coordinates to handle cevians and ratios.

Back to techniques

Olympiad MethodsDifficulty 60-10026 tagged problems

Barycentric Coordinates

Practice this technique Match my level

The Key Result

$$P = (x : y : z), \quad x+y+z \ne 0, \qquad \vec{P} = \frac{x\,\vec{A} + y\,\vec{B} + z\,\vec{C}}{x+y+z}.$$

Why It Works

1
Fix triangle $ABC$. Any point $P$ in the plane is a unique affine combination $\vec{P} = \alpha\vec{A} + \beta\vec{B} + \gamma\vec{C}$ with $\alpha+\beta+\gamma = 1$; these normalized weights are the barycentric coordinates.
2
Coordinates are homogeneous: $(x:y:z)$ and $(kx:ky:kz)$ denote the same point for $k\ne 0$, so we recover the point by dividing by $x+y+z$.
3
The vertices are $A=(1:0:0)$, $B=(0:1:0)$, $C=(0:0:1)$; a point on side $BC$ has first coordinate $0$.
4
Three points are collinear iff the determinant of their coordinate rows is $0$, and a cevian's foot inherits the opposite-side ratios directly from the weights — this is what makes ratio problems mechanical.

Worked Examples

Example 1

Prove the medians of triangle $ABC$ concur, and find the concurrency point in barycentric coordinates.

1
The midpoint of $BC$ is $M_A = (0:1:1)$ (equal weights on $B$ and $C$, none on $A$); similarly $M_B=(1:0:1)$ and $M_C=(1:1:0)$.
2
The median from $A$ passes through $A=(1:0:0)$ and $M_A=(0:1:1)$; points on it have the form $(t:1:1)$.
3
Setting all three medians to share a point forces equal weights, giving the candidate $G=(1:1:1)$.
4
Check $G=(1:1:1)$ lies on median $A M_A$: it has the form $(t:1:1)$ with $t=1$. ✓ By symmetry it lies on all three, so they concur. Normalizing, $\vec{G} = \frac{\vec{A}+\vec{B}+\vec{C}}{3}$.

Answer:

The centroid $G = (1:1:1)$, i.e. $\vec{G} = \frac{\vec{A}+\vec{B}+\vec{C}}{3}$.

Contest level

In triangle $ABC$, point $D$ lies on $BC$ with $BD:DC = 2:3$, and $E$ lies on $AD$ with $AE:ED = 4:1$. Find the barycentric coordinates of $E$.

1
$D$ divides $BC$ with $BD:DC = 2:3$, so $D$ is the weighted average putting weight on $B$ and $C$ inversely to the opposite segments: $D = (0 : DC : BD) = (0 : 3 : 2)$ (a point on $BC$ has zero $A$-weight; weights split as $DC$ on $B$, $BD$ on $C$). Normalize: $D = (0 : 3 : 2)$ with weight sum $5$.
2
$E$ divides $AD$ with $AE:ED = 4:1$, so $E$ puts weight $ED=1$ on $A$ and weight $AE=4$ on $D$ (again inverse to the segment lengths): $\vec{E} = \tfrac{1\cdot \vec{A} + 4\cdot \vec{D}}{5}$.
3
Use the normalized $\vec{D} = \tfrac{3\vec{B} + 2\vec{C}}{5}$, so $\vec{E} = \tfrac{1}{5}\vec{A} + \tfrac{4}{5}\cdot\tfrac{3\vec{B}+2\vec{C}}{5} = \tfrac{1}{5}\vec{A} + \tfrac{12}{25}\vec{B} + \tfrac{8}{25}\vec{C}$.
4
Clear denominators (multiply by $25$): weights $(5 : 12 : 8)$, which sum to $25$ — consistent.

Answer:

$E = (5 : 12 : 8)$.

Olympiad / Challenge

Prove that the cevians $AA'$, $BB'$, $CC'$ concur iff $\dfrac{BA'}{A'C}\cdot\dfrac{CB'}{B'A}\cdot\dfrac{AC'}{C'B} = 1$ (Ceva), using barycentrics.

1
Put $A' = (0 : m : n)$ on $BC$; then $BA':A'C = n:m$ (the weight on $B$ is proportional to the FAR segment $A'C$, the weight on $C$ to $BA'$), i.e. $\tfrac{BA'}{A'C} = \tfrac{n}{m}$. Similarly $B' = (p : 0 : q)$ with $\tfrac{CB'}{B'A} = \tfrac{p}{q}$ and $C' = (r : s : 0)$ with $\tfrac{AC'}{C'B} = \tfrac{s}{r}$.
2
Cevian $AA'$ is the line through $A=(1:0:0)$ and $A'=(0:m:n)$, i.e. the locus $\{(t : m : n)\}$, which in coordinates is the line $n\,y = m\,z$ (eliminating $t$).
3
Likewise $BB'$ is $q\,x = p\,z$ and $CC'$ is $s\,x = r\,y$. A common point exists iff these three linear relations are consistent, i.e. iff the product around the cycle returns: from the three, $\tfrac{x}{z}=\tfrac{p}{q}$, $\tfrac{y}{z}=\tfrac{m}{n}$, $\tfrac{x}{y}=\tfrac{r}{s}$, and consistency requires $\tfrac{p}{q}\cdot\tfrac{n}{m}\cdot\tfrac{s}{r}=1$.
4
Translate back: $\tfrac{n}{m}\cdot\tfrac{p}{q}\cdot\tfrac{s}{r} = \dfrac{BA'}{A'C}\cdot\dfrac{CB'}{B'A}\cdot\dfrac{AC'}{C'B} = 1$, which is exactly Ceva's condition. ✓

Answer:

Concurrency $\iff \dfrac{BA'}{A'C}\cdot\dfrac{CB'}{B'A}\cdot\dfrac{AC'}{C'B} = 1$ (Ceva, recovered from barycentric line equations).

Going Deeper

Barycentrics generalize to the incenter $I = (a:b:c)$, circumcenter $O = (a^2(b^2+c^2-a^2):\dots)$, and symmedian point $K=(a^2:b^2:c^2)$ — using side lengths as weights handles 'triangle-center' problems that pure coordinates make ugly.
It is the engine behind many olympiad geometry bashes (USAMO/IMO): collinearity is a $3\times 3$ determinant, concurrency is its dual, and conics (circumcircle, etc.) have clean barycentric equations.
Pitfall: barycentric coordinates are AFFINE, not Euclidean — distance, perpendicularity, and circles require the displacement vector $(x,y,z)$ with $x+y+z=0$ and the metric $-a^2 yz - b^2 zx - c^2 xy$. Treating raw coordinates as Cartesian gives wrong lengths and angles.

Spot the Signal

Look for problems where the key step is representing points by weighted triangle coordinates to handle cevians and ratios.
You can describe the hard part as representing points by weighted triangle coordinates to handle cevians and ratios, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the olympiad structure that calls for barycentric coordinates, then rewrite the givens around it.
Name the relation that makes Barycentric Coordinates legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Barycentric Coordinates just because the surface looks familiar; verify the required condition first.
Applying Barycentric Coordinates because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let barycentric coordinates reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is representing points by weighted triangle coordinates to handle cevians and ratios.