Sagot (Sa Ibang Tanong)

^Travis has a very interesting post over here. It's a fun geometric problem that calls for a two-part  proving - but let me just tackle the first part. It states that if given three sides of a right triangle with the expressions n²+ 1, n² – 1 and 2n (where n > 1), what expression represents the hypotenuse of the right triangle?

Before we even start, let's have quick brush up about the parts of the right triangle. The hypotenuse is the side directly opposite the right-angle of that triangle. Since the right-angle happens to be the largest angle in the triangle, it also means that the hypotenuse is the longest side of the triangle. With this in mind, we need to find out which among the three expressions is the largest.

If we look at it intuitively,  n² + 1 should be the longest side because 1 added to any square of any number (n) is always greater than 1 subtracted from the same square of any number (n). Also, the addition of 1 to the square of any number (n) would always be greater than the same number multiplied by 2.

But of course we can't just say, "It's intuitive!" on paper, right? So let's check the first statement.

Suppose:

n² + 1  >  n² – 1

Therefore:

n²  – n²  >  -1 – 1

Hence:

0  >  -2

But since 0 > -2 is always true, therefore n² + 1 > n² – 1 is always true.

Same logic applies to the second statement.

Suppose:

n² + 1 > 2n

Therefore:

n² + 1 - 2n > 0


By factoring the left side of the inequality, we get:

(n-1)² > 0

This statement is true because the square of any number, in this case (n-1), is always positive (read: it's greater than zero). Hence, n²+ 1 > 2n is always true, too.

So now, we finished proving that n² + 1 is bigger than n² – 1 or 2n.  We can then conclude that n² + 1, by definition, should the hypotenuse of the right triangle.
 

Oo na. Ako na. Ako na si Shaira Luna! Tseeeeeeeeh >:P hihihihihihi