Question

If \( p_1, p_2, p_3 \) are the altitudes and \( a=4, b=5, c=6 \) are the sides of a triangle ABC, then \( \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \)

• A. \( \frac{77}{225} \)
• B. \( \frac{44}{225} \)
• C. \( \frac{308}{225} \)
• D. \( \frac{22}{75} \)

Hint:

Expressing geometrical quantities (like altitudes) in terms of Area (\( \Delta \)) and sides often simplifies algebraic manipulation significantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

The area of a triangle \( \Delta \) is related to altitude \( p_1 \) by \( \Delta = \frac{1}{2} a p_1 \). Therefore, \( \frac{1}{p_1} = \frac{a}{2\Delta} \). We substitute this into the required sum.
Step 2: Key Formula or Approach:

Required Sum \( S = \frac{a^2 + b^2 + c^2}{4\Delta^2} \). Heron's Formula: \( \Delta^2 = s(s-a)(s-b)(s-c) \).
Step 3: Detailed Explanation:

Given \( a=4, b=5, c=6 \). Semi-perimeter \( s = \frac{4+5+6}{2} = \frac{15}{2} \). Calculate \( \Delta^2 \): \[ \Delta^2 = \frac{15}{2} \left(\frac{15}{2}-4\right) \left(\frac{15}{2}-5\right) \left(\frac{15}{2}-6\right) \] \[ \Delta^2 = \frac{15}{2} \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} = \frac{1575}{16} \] Calculate sum of squares of sides: \[ a^2 + b^2 + c^2 = 16 + 25 + 36 = 77 \] Calculate the expression: \[ S = \frac{77}{4 \times \frac{1575}{16}} = \frac{77}{\frac{1575}{4}} = \frac{308}{1575} \] Simplify by dividing numerator and denominator by 7: \[ S = \frac{44}{225} \]
Step 4: Final Answer:

The value is \( \frac{44}{225} \).