GeometryDifficulty 35-750 tagged problems

Mass Points

Mass Points is the habit of assigning weights to vertices so cevian ratios become balance equations. 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.

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The Key Result

$$\text{Place mass } m_B \text{ at } B,\ m_C \text{ at } C \text{ with } m_B\cdot BD = m_C\cdot DC \;\Longrightarrow\; \frac{BD}{DC} = \frac{m_C}{m_B}$$

You hang weights on the vertices so each cevian balances at its foot, turning ratio problems into simple lever arithmetic.

Why It Works

1
Model the triangle as a physical lever: a mass at $B$ and a mass at $C$ balance at the point $D$ on $BC$ where the torques cancel, $m_B \cdot BD = m_C \cdot DC$.
2
So the foot of a cevian divides the opposite side in the inverse ratio of the endpoint masses: $\dfrac{BD}{DC} = \dfrac{m_C}{m_B}$.
3
At a cevian's foot, the mass placed is the sum $m_B + m_C$; the cevian then balances the opposite vertex against this combined mass.
4
Two cevians meet where all masses balance, so the intersection ratio is read directly from vertex-mass sums — no algebra-heavy ratio chase needed.

Worked Examples

Example 1

In $\triangle ABC$, cevians $AD$ and $BE$ meet at $P$, where $D$ is on $BC$ with $BD:DC = 2:1$ and $E$ is the midpoint of $AC$. Find $AP:PD$.

1
Balance $BC$ at $D$ with $BD:DC = 2:1$: put mass $1$ at $B$ and mass $2$ at $C$ (so $1\cdot BD = 2\cdot DC$ requires $BD:DC = 2:1$). The foot $D$ carries mass $1 + 2 = 3$.
2
$E$ is the midpoint of $AC$, so masses at $A$ and $C$ must be equal; since $C$ has mass $2$, set $A$'s mass to $2$. Check $BE$ later if needed.
3
Now cevian $AD$ balances vertex $A$ (mass $2$) against foot $D$ (mass $3$), so $\dfrac{AP}{PD} = \dfrac{\text{mass at }D}{\text{mass at }A} = \dfrac{3}{2}$.

Answer:

$AP:PD = 3:2$

Warm-up

In $\triangle ABC$, the medians $AD$ and $BE$ meet at the centroid $G$. Use mass points to find $AG:GD$.

1
A median has equal-length halves, so for $D$ the midpoint of $BC$ put equal masses $1$ at $B$ and $1$ at $C$; then $D$ carries mass $1+1 = 2$.
2
For median $BE$ ($E$ the midpoint of $AC$) we need equal masses at $A$ and $C$; $C$ already has mass $1$, so set $A$'s mass to $1$.
3
Cevian $AD$ balances vertex $A$ (mass $1$) against foot $D$ (mass $2$), so $\dfrac{AG}{GD} = \dfrac{\text{mass at }D}{\text{mass at }A} = \dfrac{2}{1}$.

Answer:

$AG:GD = 2:1$ (the centroid divides each median $2:1$).

Contest level

In $\triangle ABC$, $D$ on $BC$ with $BD:DC = 1:2$ and $E$ on $AB$ with $AE:EB = 3:1$. Cevians $CE$ and $AD$ meet at $P$. Find $CP:PE$.

1
Balance $BC$ at $D$: $BD:DC = 1:2$ needs $m_B:m_C = 2:1$, so put $m_B = 2$, $m_C = 1$; then $m_D = 3$.
2
Balance $AB$ at $E$: $AE:EB = 3:1$ needs $m_A:m_B = 1:3$. Since $m_B = 2$, scale so masses stay consistent: $m_A = \tfrac{1}{3}m_B = \tfrac{2}{3}$. To clear fractions multiply all masses by $3$: $m_A = 2$, $m_B = 6$, $m_C = 3$, and recompute $m_D = m_B + m_C = 9$.
3
Now $E$ carries $m_E = m_A + m_B = 2 + 6 = 8$.
4
Cevian $CE$ balances vertex $C$ (mass $3$) against foot $E$ (mass $8$), so $\dfrac{CP}{PE} = \dfrac{m_E}{m_C} = \dfrac{8}{3}$.

Answer:

$CP:PE = 8:3$

Olympiad / Challenge

In $\triangle ABC$, $D$ on $BC$ with $BD:DC = 2:3$ and $E$ on $CA$ with $CE:EA = 5:2$. Cevians $AD$ and $BE$ meet at $P$. Find $\dfrac{AP}{PD}$ and $\dfrac{BP}{PE}$.

1
Balance $BC$ at $D$ ($BD:DC = 2:3$): $m_B:m_C = 3:2$. Balance $CA$ at $E$ ($CE:EA = 5:2$): $m_C:m_A = 2:5$.
2
Make $m_C$ consistent. From the first, $m_B = 3,\, m_C = 2$. From the second $m_C:m_A = 2:5$ gives $m_A = 5$ (with $m_C = 2$). All consistent: $m_A = 5,\, m_B = 3,\, m_C = 2$.
3
Foot $D$ carries $m_B + m_C = 5$; foot $E$ carries $m_C + m_A = 7$.
4
Cevian $AD$: $\dfrac{AP}{PD} = \dfrac{m_D}{m_A} = \dfrac{5}{5} = 1$. Cevian $BE$: $\dfrac{BP}{PE} = \dfrac{m_E}{m_B} = \dfrac{7}{3}$.

Answer:

$\dfrac{AP}{PD} = 1$ and $\dfrac{BP}{PE} = \dfrac{7}{3}$.

Going Deeper

Generalization: mass points are projective barycentric coordinates in disguise — the mass at a vertex is its barycentric weight. When a cevian foot divides a side in a ratio involving the triangle's own sides (e.g. an angle bisector with $BD:DC = c:b$), assign masses $b$ and $c$ and the incenter/Gergonne-type points fall out for free. For NON-cevian intersections (a side split by a non-through-vertex line) you need 'split masses' or switch to Menelaus.
Where it appears: mass points are the fastest route through AMC/AIME cevian-ratio problems — they replace a two-equation Menelaus computation with bookkeeping. The centroid $2:1$ result and any 'two cevians meet at $P$, find a ratio' problem are immediate.
Pitfall: masses must be CONSISTENT across the whole figure — a single vertex cannot carry two different masses, so scale (multiply through) to reconcile the two balancing conditions before reading any ratio. Forgetting to rescale (as in the Contest example) silently corrupts every downstream ratio.

Spot the Signal

Look for problems where the key step is assigning weights to vertices so cevian ratios become balance equations.
You can describe the hard part as assigning weights to vertices so cevian ratios become balance equations, 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 diagram relation that calls for mass points, then rewrite the givens around it.
Name the relation that makes Mass Points 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 Mass Points just because the surface looks familiar; verify the required condition first.
Applying Mass Points 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 mass points reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is assigning weights to vertices so cevian ratios become balance equations.