In a right angle triangle,sa and sb denote the medians that belong to the legs of triangle,the median belonging to hypotenuse is sc . Find the Maximum value of the expression sa+sb÷ sc .

To tackle the problem of finding the maximum value of the expression \( \frac{s_a + s_b}{s_c} \) in a right triangle, we first need to understand the properties of medians in a triangle, particularly in the context of a right-angled triangle. Let's break this down step by step.

Understanding Medians in a Right Triangle

In any triangle, a median is a line segment joining a vertex to the midpoint of the opposite side. In a right triangle, we have the following sides:

• Let \( a \) and \( b \) be the lengths of the legs.
• Let \( c \) be the length of the hypotenuse, which can be calculated using the Pythagorean theorem: \( c = \sqrt{a^2 + b^2} \).

Calculating the Medians

The formulas for the medians \( s_a \), \( s_b \), and \( s_c \) from the vertices opposite to sides \( a \), \( b \), and \( c \) respectively are given by:

• Median to side \( a \): s_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}
• Median to side \( b \): s_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}
• Median to side \( c \): s_c = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2} = \frac{1}{2} \sqrt{2a^2 + 2b^2 - (a^2 + b^2)} = \frac{1}{2} \sqrt{a^2 + b^2} = \frac{c}{2}

Substituting the Medians into the Expression

Now, substituting the expressions for \( s_a \) and \( s_b \) into our original expression:

Expression: \( \frac{s_a + s_b}{s_c} = \frac{\frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2} + \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}}{\frac{c}{2}} = \frac{\sqrt{2b^2 + 2c^2 - a^2} + \sqrt{2a^2 + 2c^2 - b^2}}{c} \end{strong>

Finding the Maximum Value

To find the maximum value of \( \frac{s_a + s_b}{s_c} \), we can analyze specific cases of right triangles. A notable case is when the triangle is isosceles, meaning \( a = b \). In this case, we can simplify our calculations:

• Let \( a = b = x \). Then, \( c = \sqrt{x^2 + x^2} = x\sqrt{2} \).
• Calculating the medians:
  • \( s_a = s_b = \frac{1}{2} \sqrt{2x^2 + 2(x\sqrt{2})^2 - x^2} = \frac{1}{2} \sqrt{2x^2 + 4x^2 - x^2} = \frac{1}{2} \sqrt{5x^2} = \frac{x\sqrt{5}}{2} \)
  • \( s_c = \frac{x\sqrt{2}}{2} \)

Substituting these values back into our expression gives:

\( \frac{s_a + s_b}{s_c} = \frac{\frac{x\sqrt{5}}{2} + \frac{x\sqrt{5}}{2}}{\frac{x\sqrt{2}}{2}} = \frac{x\sqrt{5}}{\frac{x\sqrt{2}}{2}} = \frac{2\sqrt{5}}{\sqrt{2}} = \sqrt{10} \)

Conclusion

Thus, the maximum value of the expression \( \frac{s_a + s_b}{s_c} \) in a right triangle occurs when the triangle is isosceles, yielding a maximum value of \( \sqrt{10} \). This result illustrates how the properties of medians and the specific geometry of the triangle can lead to interesting mathematical conclusions.