Geometry Lecture: Integral Triangles and Rules

Instructor: Ravi Prakash

Key Topics Covered

1. Integral Triangles
2. Perimeter of Triangles
3. Triangle Formation Conditions
4. Cosine Rule
5. Sine Rule

Integral Triangles with Perimeter 39

All Sides Odd

• Given: Perimeter = 39

• Let sides be A, B, C

• Transform sides to odd: A = 2a - 1, B = 2b - 1, C = 2c - 1

• Equation: 2a - 1 + 2b - 1 + 2c - 1 = 39

  • Simplifies to: 2a + 2b + 2c = 42 ⟹ a + b + c = 21

• Total number of triangles for a + b + c with odd perimeter,

  [\frac{(a+b+c+3)^2}{48}]

  • For a + b + c = 21, solution is:
    • [24^2 / 48 = 576 / 48 = 12]
    • So, 12 triangles have all odd sides.

Two Even, One Odd

• Given: Perimeter = 39
• Two cases for odd perimeter:
  1. All sides odd.
  2. Two sides even, one side odd.
• Total number of triangles calculation with odd perimeter for the second case:
  • Polynomial formula: [\frac{42^2}{48} = 1764 / 48 ⟹ 36.75 \ and approximately 37]
• Total triangles = 37
• Triangles for case 1 (all sides odd) = 12
• Triangles for case 2 = 37 - 12 = 25
• Answer: 25 triangles have two even sides and one odd side.

Perimeter 60: Maximum Area Calculation

• Problem: Which of the following cannot be the area: 151, 159, 169, 176?
• Concept: Maximum area for fixed perimeter when triangle is equilateral.
  • Equilateral triangle with side 20 (perimeter = 60)
  • Area: [rac{\sqrt{3}}{4} imes 20^2 = 100\sqrt{3} ≈ 173]
• Answer: Maximum possible area = 173
  • Thus, 176 cannot be the area.

Sine Rule and Cosine Rule

Sine Rule

• For any triangle ABC:

  [rac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}]

Cosine Rule

• For any triangle ABC, cosine of angle can be derived using sides:

  [\cos A = \frac{b^2 + c^2 - a^2}{2bc}]

  [\cos B = \frac{a^2 + c^2 - b^2}{2ac}]

  [\cos C = \frac{a^2 + b^2 - c^2}{2ab}]

Application of Cosine Rule and Example Problems

Example Problem 1: Triangle PQR

• Given:
  • PM is an angle bisector.
  • ( \angle Q + \angle R = 120°)
  • PQ = 9, PR = 12
  • Questions:
    1. Find QR
    2. Find PM

Solution

• Total angles: ( \angle Q + \angle R + \angle P = 180° )

• With ( \angle Q + \angle R = 120° ), ( \angle P = 60° )

• Use cosine rule to find QR:

  [\cos 60° = \frac{9^2 + 12^2 - QR^2}{2 \times 9 \times 12}] ⟹ (QR = 3\sqrt{13})

• Use area and angle bisector theorem for PM calculation:

  [\text{Area} = \frac{1}{2}ab\sin \theta]

  (PM = \frac{36\sqrt{3}}{7})

Summary

• Important to review conditions for forming triangles with given constraints.
• Application of rules (sine and cosine) for solving triangle-related problems.
• Key techniques such as transformation for maintaining odd sides, and maximum area consideration for equilateral triangles with fixed perimeter.