Prove that the probability that two positive numbers, x and y both less than 1, written down at random with unity, yield a triplet(x,y,1) whish are the sides of an obtuse angled triangle is $\displaystyle \frac{\pi - 2}{4}$

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Malay

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- Jun 19th 2006, 09:16 PMmalaygoelProbability of a triangle
Prove that the probability that two positive numbers, x and y both less than 1, written down at random with unity, yield a triplet(x,y,1) whish are the sides of an obtuse angled triangle is $\displaystyle \frac{\pi - 2}{4}$

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Malay - Jun 20th 2006, 08:32 AMThePerfectHacker
I do not understand what you mean by a triplet?

- Jun 20th 2006, 09:41 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

A set of three numbers that are the lengths of the sides of some triangle,

(or one of my chidren :confused: :eek: )

RonL - Jun 20th 2006, 09:43 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

1) do not make posts of cardinality greater than 1

2) do not discuss illegal operations on the forum

3) do not use improper integrals

4) do not abuse notation

5) Axiom of Choice is allowed.

:p

RonL - Jun 20th 2006, 11:49 AMJakeDQuote:

Originally Posted by**malaygoel**

$\displaystyle \setlength{\unitlength}{2.5cm}

\begin{picture}(1,1)

\qbezier(0,0)(0,0)(1,0)

\qbezier(0,0)(0,0)(0,1)

\qbezier(1,0)(1,0)(1,1)

\qbezier(0,1)(0,1)(1,1)

\linethickness{1.3pt}

\qbezier(1,0)(1,0)(0,1)

\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71)

\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0)

\end{picture}

$ - Jun 20th 2006, 12:04 PMMathGuruwow! +repQuote:

Originally Posted by**JakeD**

- Jun 20th 2006, 06:42 PMmalaygoelQuote:

Originally Posted by**JakeD**

- Jun 20th 2006, 06:53 PMmalaygoelQuote:

Originally Posted by**malaygoel**

I drew the graphs of

$\displaystyle x + y = 1$

$\displaystyle x^2 + y^2 = 1$

and got the same picture as yours.

I got the answer also from the picture.

Did you also arrive at the picture using Analytic Geometry?If no, what method you adopted.

Thanks for your help,JakeD

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Malay - Jun 20th 2006, 08:09 PMJakeDQuote:

Originally Posted by**malaygoel**

- Jun 21st 2006, 02:32 AMmalaygoelQuote:

Originally Posted by**JakeD**

Could the answer be obtained without using Analytic Geometry?

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Malay - Jun 21st 2006, 05:46 AMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

Maybe once we have rules on the forum link, then I will take it off. - Jun 21st 2006, 07:05 AMJakeDQuote:

Originally Posted by**malaygoel**

$\displaystyle \begin{aligned}

x + y &\ge 1 \\

x^2 + y^2 &\le 1 \\

0 \le x &\le 1 \\

0 \le y &\le 1. \\

\end{aligned}$ - Jun 21st 2006, 08:42 AMmalaygoelQuote:

Originally Posted by**JakeD**

We have used analytic geometry to obtain the picture!

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Malay - Jun 21st 2006, 09:13 AMJakeDQuote:

Originally Posted by**malaygoel**

- Jun 21st 2006, 06:03 PMmalaygoelQuote:

Originally Posted by**JakeD**

Quote:

Then the leap to the probability comes from seeing that the picture is of a unit circle and a triangle. That analytical geometry and trigonometry is not necessary. As I said you can use calculus there. But then the trig comes back in solving the integral for the circle. The answer contains $\displaystyle \pi$ remember.

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Malay