技术求证：以√(a²+(b-c)²)等为边的三角形为锐角三角形

Let's break down how to prove this step by step — it's all about leveraging the Law of Cosines and basic triangle properties.

First, let's define the sides of our new triangle to make notation easier:

• ( x = \sqrt{a^2 + (b - c)^2} )
• ( y = \sqrt{b^2 + (a - c)^2} )
• ( z = \sqrt{c^2 + (a - b)^2} )

To confirm a triangle is acute, we just need to show that for every pair of sides, the sum of their squares is greater than the square of the third side. This comes directly from the Law of Cosines: for any angle ( \theta ) opposite side ( z ), ( \cos\theta = \frac{x^2 + y^2 - z^2}{2xy} ). If ( x^2 + y^2 > z^2 ), then ( \cos\theta > 0 ), so ( \theta ) is an acute angle. We need to verify this holds for all three combinations.

Step 1: Calculate ( x^2 + y^2 - z^2 ) #

First, expand each squared term (we can ignore the square roots for now since we're working with squared side lengths):

• ( x^2 = a^2 + (b - c)^2 = a^2 + b^2 - 2bc + c^2 )
• ( y^2 = b^2 + (a - c)^2 = b^2 + a^2 - 2ac + c^2 )
• ( z^2 = c^2 + (a - b)^2 = c^2 + a^2 - 2ab + b^2 )

Substitute these into ( x^2 + y^2 - z^2 ):

x² + y² - z² = [a² + b² - 2bc + c²] + [b² + a² - 2ac + c²] - [c² + a² - 2ab + b²]

Now simplify term by term:

• Combine ( a^2 ) terms: ( a² + a² - a² = a² )
• Combine ( b^2 ) terms: ( b² + b² - b² = b² )
• Combine ( c^2 ) terms: ( c² + c² - c² = c² )
• Combine linear terms: ( -2bc - 2ac + 2ab )

Putting it all together, we get:

x² + y² - z² = a² + b² + c² - 2bc - 2ac + 2ab

We can rearrange and factor this expression neatly:

= a² + 2ab + b² - 2ac - 2bc + c²
= (a + b)² - 2c(a + b) + c²
= (a + b - c)²

Step 2: Verify the result is positive #

Since ( \triangle ABC ) is a valid acute triangle, it's non-degenerate — meaning the sum of any two sides is greater than the third side. So ( a + b > c ), which means ( a + b - c > 0 ). Squaring this positive value gives ( (a + b - c)^2 > 0 ).

This means ( x^2 + y^2 - z^2 > 0 ), so ( x^2 + y^2 > z^2 ). The angle opposite side ( z ) is acute.

Step 3: Use symmetry for the other sides #

By the same logic, we can compute the other two combinations:

• ( y^2 + z^2 - x^2 = (b + c - a)^2 > 0 ) (since ( b + c > a ))
• ( z^2 + x^2 - y^2 = (a + c - b)^2 > 0 ) (since ( a + c > b ))

All three angles of the new triangle have positive cosine values, so every angle is acute.

内容的提问来源于stack exchange，提问作者user513964