đź“Ź Length & Area Classics

These classical formulas are essential tools for AMC geometry. They provide direct paths to solutions when other methods become complex.

🎯 Key Formulas

Heron’s Formula

Purpose: Find triangle area from side lengths Formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$

When to Use: Given three side lengths, need area AMC Appearance: Common in problems with specific side lengths

Brahmagupta’s Formula

Purpose: Find area of cyclic quadrilateral from side lengths Formula: $A = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s = \frac{a+b+c+d}{2}$

When to Use: Cyclic quadrilateral with given side lengths AMC Appearance: Less common but powerful when applicable

Stewart’s Theorem

Purpose: Find length of cevian from side lengths and cevian segments Formula: $b^2m + c^2n = a(d^2 + mn)$ where $m + n = a$

When to Use: Given cevian and its segments, need cevian length AMC Appearance: Common in problems with medians, angle bisectors

Apollonius’s Theorem

Purpose: Find median length from side lengths Formula: $m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}$

When to Use: Need median length from side lengths AMC Appearance: Common in problems involving centroids

🔍 Micro-Examples

Heron’s Formula

Triangle with sides 3, 4, 5: $s = \frac{3+4+5}{2} = 6$ $A = \sqrt{6(6-3)(6-4)(6-5)} = \sqrt{6 \cdot 3 \cdot 2 \cdot 1} = \sqrt{36} = 6$

Brahmagupta’s Formula

Cyclic quadrilateral with sides 3, 4, 5, 6: $s = \frac{3+4+5+6}{2} = 9$ $A = \sqrt{(9-3)(9-4)(9-5)(9-6)} = \sqrt{6 \cdot 5 \cdot 4 \cdot 3} = \sqrt{360} = 6\sqrt{10}$

Stewart’s Theorem

Triangle with sides 3, 4, 5, cevian length 3, segments 2 and 3: $4^2 \cdot 2 + 5^2 \cdot 3 = 3(3^2 + 2 \cdot 3)$ $32 + 75 = 3(9 + 6) = 3 \cdot 15 = 45$ $107 = 45$ (This example doesn’t work - need valid triangle)

Apollonius’s Theorem

Triangle with sides 3, 4, 5, median to side 5: $m^2 = \frac{2(3^2) + 2(4^2) - 5^2}{4} = \frac{18 + 32 - 25}{4} = \frac{25}{4}$ $m = \frac{5}{2}$

⚠️ Common Traps

Pitfall: Wrong semi-perimeter calculation

  • Fix: $s = \frac{a+b+c}{2}$ for triangles, $s = \frac{a+b+c+d}{2}$ for quadrilaterals

Pitfall: Forgetting to check if triangle exists

  • Fix: Use triangle inequality: $a + b > c$ for all sides

Pitfall: Wrong Stewart’s theorem setup

  • Fix: Remember $m + n = a$ and the formula structure

Pitfall: Confusing Heron and Brahmagupta

  • Fix: Heron for triangles, Brahmagupta for cyclic quadrilaterals

🎯 AMC-Style Worked Example

Problem: Triangle $ABC$ has sides $AB = 5$, $BC = 6$, and $CA = 7$. Find the area of the triangle.

Solution: Using Heron’s formula: $s = \frac{5+6+7}{2} = 9$

$A = \sqrt{s(s-a)(s-b)(s-c)}$ $A = \sqrt{9(9-5)(9-6)(9-7)}$ $A = \sqrt{9 \cdot 4 \cdot 3 \cdot 2}$ $A = \sqrt{216}$ $A = 6\sqrt{6}$

Answer: $A = 6\sqrt{6}$

đź”— Related Topics

đź’ˇ Quick Reference

Formula Summary

  • Heron: $A = \sqrt{s(s-a)(s-b)(s-c)}$ (triangles)
  • Brahmagupta: $A = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ (cyclic quadrilaterals)
  • Stewart: $b^2m + c^2n = a(d^2 + mn)$ (cevians)
  • Apollonius: $m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}$ (medians)

When to Use

  • Heron: Three side lengths given
  • Brahmagupta: Cyclic quadrilateral with side lengths
  • Stewart: Cevian with known segments
  • Apollonius: Median length from side lengths

Common Values

  • 3-4-5: Area = 6, semi-perimeter = 6
  • 5-12-13: Area = 30, semi-perimeter = 15
  • 8-15-17: Area = 60, semi-perimeter = 20

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