⚖️ Mass Points & Ceva/Menelaus

Mass point geometry and Ceva/Menelaus theorems are powerful tools for AMC 12 problems involving concurrency and collinearity.

🎯 Key Concepts

Mass Point Geometry

Basic Idea: Assign masses to points in a triangle to find ratios using the principle of moments.

Key Principles:

  • Masses are assigned to vertices
  • Masses are proportional to lengths of opposite sides
  • Center of mass is at the intersection of cevians

Common Applications:

  • Finding ratios in triangles
  • Proving concurrency
  • Solving cevian problems

Ceva’s Theorem

Purpose: Determine if three cevians are concurrent Formula: $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1$

When to Use: Given three cevians, check if they meet at one point AMC Appearance: Common in problems about concurrency

Menelaus’s Theorem

Purpose: Determine if three points are collinear Formula: $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1$

When to Use: Given three points, check if they lie on one line AMC Appearance: Common in problems about collinearity

🔍 Micro-Examples

Mass Points Example

In triangle $ABC$ with $AB = 3$ and $AC = 4$, if $D$ is on $BC$ such that $BD:DC = 3:4$, assign masses:

  • Mass at $B$: 4 (opposite to $AC$)
  • Mass at $C$: 3 (opposite to $AB$)
  • Mass at $A$: 7 (sum of masses at $B$ and $C$)

Ceva’s Theorem Example

In triangle $ABC$, if $AD$, $BE$, $CF$ are concurrent and $\frac{AF}{FB} = 2$, $\frac{BD}{DC} = 3$, then $\frac{CE}{EA} = \frac{1}{6}$.

Menelaus’s Theorem Example

In triangle $ABC$, if points $D$, $E$, $F$ are collinear and $\frac{AF}{FB} = 2$, $\frac{BD}{DC} = 3$, then $\frac{CE}{EA} = \frac{1}{6}$.

⚠️ Common Traps

Pitfall: Wrong mass assignment

  • Fix: Masses are proportional to lengths of opposite sides

Pitfall: Wrong Ceva/Menelaus setup

  • Fix: Remember the order: $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}$

Pitfall: Confusing Ceva and Menelaus

  • Fix: Ceva for concurrency, Menelaus for collinearity

Pitfall: Forgetting about directed segments

  • Fix: Use directed segments for proper sign

🎯 AMC-Style Worked Example

Problem: In triangle $ABC$, points $D$, $E$, $F$ are on sides $BC$, $CA$, $AB$ respectively. If $AD$, $BE$, $CF$ are concurrent and $\frac{AF}{FB} = 2$, $\frac{BD}{DC} = 3$, find $\frac{CE}{EA}$.

Solution: Using Ceva’s theorem for concurrent cevians: $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1$

Substituting the given values: $2 \cdot 3 \cdot \frac{CE}{EA} = 1$ $6 \cdot \frac{CE}{EA} = 1$ $\frac{CE}{EA} = \frac{1}{6}$

Answer: $\frac{CE}{EA} = \frac{1}{6}$

🔗 Related Topics

💡 Quick Reference

Mass Point Rules

  • Mass assignment: Proportional to opposite side lengths
  • Center of mass: At intersection of cevians
  • Ratio finding: Use mass ratios

Ceva’s Theorem

  • Purpose: Check concurrency of three cevians
  • Formula: $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1$
  • Order: Follow the triangle around

Menelaus’s Theorem

  • Purpose: Check collinearity of three points
  • Formula: $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1$
  • Order: Follow the triangle around

Common Applications

  • Concurrency: Use Ceva’s theorem
  • Collinearity: Use Menelaus’s theorem
  • Ratios: Use mass point geometry
  • Cevians: Use mass points or Ceva’s theorem

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