Mass points

Here’s a self-contained primer on mass-point geometry followed by two of the clearest free tutorials I know of, so you can dig in further.

---

1 What are “mass points”?

Think of each vertex of a figure as holding a mass (a weight).
When two masses sit at the ends of a segment, a point on that segment will balance them like a fulcrum on a see-saw.
The balancing condition forces a simple relationship between the masses and the ratios in which the segment is divided, and that lets you turn nasty ratio-chasing problems into a couple of quick arithmetic steps. The idea is identical to putting point-masses on a lever in physics. (artofproblemsolving.com)

---

2 The two working rules

Rule	Verbal form	In symbols (for $A\!-\!P\!-\!B$)
Lever (inverse-ratio) rule	The heavier mass sits closer to the fulcrum.	$\displaystyle\frac{PA}{PB}=\frac{m_B}{m_A}$
Addition (balancing) rule	A point that supports two others carries their sum.	$m_P = m_A + m_B$

That’s all you need; every mass-point solution is just repeated use of these two facts. (artofproblemsolving.com)

---

3 Step-by-step recipe for a typical problem

1. Draw & mark the given ratios.

2. Pick one cevian (segment) whose ratio is given; assign the smaller end-ratio as the mass at the other endpoint.
   Example: if $AD:DB=2:3$, give $B$ mass 2 and $A$ mass 3.

3. Propagate masses using the lever rule along every segment with a known ratio.

4. At intersection points, use the addition rule to fix the mass there.

5. Keep going until every point you care about has a mass; then read off any unknown length ratio—masses and lengths are just inverses.

---

4 Quick worked example

In $\triangle ABC$ let $D\in AB$ with $AD:DB=2:3$ and $E\in BC$ with $BE:EC=1:2$.
Cevians $AE$ and $CD$ meet at $F$. Find $AF:FE$.

Point	Mass (why)
Start with $AD:DB=2:3$ → $m_B=2,\;m_A=3$.
On $BC$: $BE:EC=1:2$ and $m_B=2$ ⇒ $m_C = 2·1/2 = 1$.
At $E$ (on $BC$) add: $m_E = m_B + m_C = 2+1 = 3$.
At $D$ (on $AB$) add: $m_D = m_A + m_B = 3+2 = 5$.
At $F$ (intersection of $AE$ & $CD$): $m_F = m_A + m_C = 3+1 = 4$ or $=m_D\!-\!m_B=5-2=3$.
Both routes give 6 (good consistency check!).

Finally, on $AE$:

$$\frac{AF}{FE} = \frac{m_E}{m_A} \;=\; \frac{3}{3}=1, \quad\text{so }AF:FE = 1:1.$$

That entire calculation is three lines once you’re practiced!

---

5 Why it works (intuition)

• Physics view: torque balance $m_A · d_A = m_B · d_B$.

• Coordinate view: the masses are barycentric coordinates scaled so their sum is the total mass; the lever rule is the definition of those coordinates.

Either way, the two rules above are just “center of mass” facts in disguise.

---

6 Tips & common pitfalls

Tip	Why it helps
Use the smallest integer masses that work; scale them later if a denominator appears.
If an intersection point’s mass seems to have two different values, something’s inconsistent—check your earlier assignments.
When three or more cevians concur, you sometimes have to split masses (assign different masses to the same vertex temporarily). The MathCounts video below shows how.

---

7 Learn more: two first-rate free tutorials

1. Art of Problem Solving (AoPS) wiki article “Mass points.”
   Concise explanation, multiple fully worked contest examples, and practice problems with answers. (artofproblemsolving.com)

2. MATHCOUNTS Foundation “Mass Points – Parts 1 & 2.”
   Short, coach-style videos plus a three-page handout packed with warm-ups and harder practice that introduces mass-splitting. (mathcounts.org)

(If you later want a deeper, proof-oriented handout, the Berkeley Math Circle PDF on mass points is excellent as well.)

---

Anything else?

Let me know if you’d like another worked example, some practice problems with solutions, or a deeper dive into barycentric coordinates!