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A right angled triangle has sides of length X, Y and Z (all lengths in cm.). It is known that Z is the length of the longest side.

The lengths of the other two sides satisfy the inequality $$\sqrt{x^2-4\sqrt{2}x + 12} + \sqrt{y^2-6\sqrt{3}y + 31} \leq 4$$

What is the length (in cm.) of the hypotenuse of this triangle?

How can we get the exact value of $x$ and $y$? Since during solving equation get in the power of 8 and very complicated.

Thanks in advance.

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2 Answers 2

up vote 4 down vote accepted

Note that $x^2-4\sqrt{2}+12 = (x-2\sqrt{2})^2 + 4$ and $y^2-6\sqrt{3}y+31 = (y-3\sqrt{3})^2 + 4$. That means that each square root is at least equal to $2$, and strictly larger than $2$ if the first square is nonzero.

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we start by rewriting the first radicand \begin{align*} x^2 - 4\sqrt 2 x + 12 &= (x - 2\sqrt 2)^2 - (2\sqrt 2)^2 + 12\\ &= (x - 2\sqrt 2)^2 + 4 \end{align*} for the second one we get \begin{align*} y^2 - 6\sqrt 3y + 31 &= (y - 3\sqrt 3)^2 - (3\sqrt 3)^2 + 31\\ &= (y - 3\sqrt 3)^2 + 4 \end{align*}Now, as squares are positive, we obtain $$ \sqrt{x^2 - 4\sqrt 2 x + 12} = \sqrt{(x - 2\sqrt 2)^2 + 4} \ge \sqrt 4 = 2 $$ and similarly $\sqrt{y^2 - 6\sqrt 3y + 31} \ge 2$. Now since each term is at least 2, there sum is only less than 4 if they are both exactly equal 2. This happens if $x - 2\sqrt 2 = y - 3\sqrt 3 = 0$.

So we have $x = 2\sqrt 2$, $y = 3\sqrt 3$ and $z = \sqrt{x^2 + y^2} = \sqrt{8 + 27} = \sqrt{35}$.

AB,

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+1 , as you gave the whole solution. –  vikiiii Mar 24 '12 at 3:30

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