MHB Prove $x=-y$: A Math Challenge

AI Thread Summary
The discussion centers on proving that if \( (\sqrt{y^{2} - x} - x)(\sqrt{x^{2} + y} - y) = y \), then \( x = -y \). Participants suggest substituting \( -y \) for \( x \) in the equation to simplify and verify the equality. The identity \( (a+b)(a-b) = a^2 - b^2 \) is highlighted as a useful tool for the proof. Some participants clarify that demonstrating \( x = -y \) is not the same as proving the original statement. The conversation emphasizes the need for careful interpretation of the mathematical challenge.
ipaper
Messages
4
Reaction score
0
Let $x, y$ be real numbers such that
$$(\sqrt{y^{2} - x\,\,}\, - x)(\sqrt{x^{2} + y\,\,}\, - y)=y.$$
Prove $x=-y$.

Any suggestion would be appreciated.
 
Mathematics news on Phys.org
Another way of stating the question is:

Show that x=-y is a solution to:

\left[\sqrt{y^2-x}-x \right]\left[\sqrt{x^2+y}-y\right]=y

Writing it is this way it is more obvious that all you have to do is substitute (-y) for each of the x's in the left hand side and then simplify to show that the left hand side is equal to y.

To do this it will be helpful to remember the identity:
(a+b)(a-b)=(a^2-b^2)
 
Kiwi said:
Another way of stating the question is:

Show that x=-y is a solution to:

\left[\sqrt{y^2-x}-x \right]\left[\sqrt{x^2+y}-y\right]=y
No, this is the converse of what the original question is asking.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Back
Top