Math Is Fun Forum

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(multiplying both sides by

and reversing the inequality)

So the answer is

. The book is correct.

Why doesn't your diagram have OA as diameter? The equation of the circle is

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which can be rearranged as

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which is a circle centred on

= midpoint of OA with radius

.

Does not that make OA the diameter?

Let A be the point (p,q). The circle is

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Clearly the circle passes through the origin O and the point A and has radius OA/2. Thus OA is the diameter the circle.

Because of symmetry we can assume WLOG that p, q > 0 (i.e. A is in the first quadrant). Let a chord drawn from A intersect the x-axis at (t,0). One end of the chord is A; if the axis bisects it, the other end would be (P,Q) where:

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Susbstiting (x,y) = (P,Q) into the equation of the circle and simplifying gives the following quadratic equation in t:

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This equation has two real and distinct solutions in t; therefore its discriminant is strictly positive:

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hellacious

unfashioned

Yes, that works fine. It's along the lines of what zetafunc suggested in post #2.

Two integers

are relatively prime if and only if there exist integers

such that

.

Now,

. So try and find

such that

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I think it's like. this. Let

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And

. The problem is to show that

is independent of the choice of x, right?

Why are you considering

? Doesn't the question specifically say

?

Grrr, silly me. For #185 I used sin instead of tan. I am really angry with myself.

The domain of a function is important: it's the set of variables for which the function makes sense. In the case of the real-valued function

the domain is

. Always check the domain of a given function, or you might get confused.