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Math Help - Probability of a triangle

  1. #1
    Super Member malaygoel's Avatar
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    Question Probability 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 \frac{\pi - 2}{4}

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    Malay
    Last edited by malaygoel; June 21st 2006 at 06:38 AM.
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  2. #2
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    I do not understand what you mean by a triplet?
    Last edited by CaptainBlack; June 20th 2006 at 09:40 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    I do not understand what you mean by a triplet?

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

    (or one of my chidren )

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    I do not understand what you mean by a triplet?
    Have you considered revising you signature to :

    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.



    RonL
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  5. #5
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    Quote Originally Posted by malaygoel
    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 \frac{\pi - 2}{4}

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    Malay
    Here's the picture. The probability is the area between the thicker lines.

    \setlength{\unitlength}{2.5cm}<br />
\begin{picture}(1,1)<br />
\qbezier(0,0)(0,0)(1,0)<br />
\qbezier(0,0)(0,0)(0,1)<br />
\qbezier(1,0)(1,0)(1,1)<br />
\qbezier(0,1)(0,1)(1,1)<br />
\linethickness{1.3pt}<br />
\qbezier(1,0)(1,0)(0,1)<br />
\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71)<br />
\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0)<br /> <br />
\end{picture}<br />
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  6. #6
    Site Founder MathGuru's Avatar
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    wow! +rep

    Quote Originally Posted by JakeD
    Here's the picture. The probability is the area between the thicker lines.

    \setlength{\unitlength}{2.5cm}<br />
\begin{picture}(1,1)<br />
\qbezier(0,0)(0,0)(1,0)<br />
\qbezier(0,0)(0,0)(0,1)<br />
\qbezier(1,0)(1,0)(1,1)<br />
\qbezier(0,1)(0,1)(1,1)<br />
\linethickness{1.3pt}<br />
\qbezier(1,0)(1,0)(0,1)<br />
\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71)<br />
\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0)<br /> <br />
\end{picture}<br />
    wow! +rep
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  7. #7
    Super Member malaygoel's Avatar
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    Quote Originally Posted by JakeD
    Here's the picture. The probability is the area between the thicker lines.

    \setlength{\unitlength}{2.5cm}<br />
\begin{picture}(1,1)<br />
\qbezier(0,0)(0,0)(1,0)<br />
\qbezier(0,0)(0,0)(0,1)<br />
\qbezier(1,0)(1,0)(1,1)<br />
\qbezier(0,1)(0,1)(1,1)<br />
\linethickness{1.3pt}<br />
\qbezier(1,0)(1,0)(0,1)<br />
\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71)<br />
\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0)<br /> <br />
\end{picture}<br />
    Will you please explain how did you get that picture(how it is linked to the question)?
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  8. #8
    Super Member malaygoel's Avatar
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    Quote Originally Posted by malaygoel
    Will you please explain how did you get that picture(how it is linked to the question)?
    Ok, I understand it.
    I drew the graphs of
    x + y = 1
    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

    KeepSmiling
    Malay
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  9. #9
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    Quote Originally Posted by malaygoel
    Ok, I understand it.
    I drew the graphs of
    x + y = 1
    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

    KeepSmiling
    Malay
    I used the same thinking. Glad to help. --JakeD
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  10. #10
    Super Member malaygoel's Avatar
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    Quote Originally Posted by JakeD
    I used the same thinking. Glad to help. --JakeD
    But the problem is that we are discouraged to use Analytic Geometry.
    Could the answer be obtained without using Analytic Geometry?

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    Malay
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  11. #11
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    Quote Originally Posted by CaptainBlack
    Have you considered revising you signature to :

    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.



    RonL
    I like that
    Maybe once we have rules on the forum link, then I will take it off.
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  12. #12
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    Quote Originally Posted by malaygoel
    But the problem is that we are discouraged to use Analytic Geometry.
    Could the answer be obtained without using Analytic Geometry?

    Keep Smiling
    Malay
    Yes. Use integral calculus to find the area of the region

    \begin{aligned}<br />
x + y &\ge 1 \\<br />
x^2 + y^2 &\le 1 \\<br />
0 \le x &\le 1 \\<br />
0 \le y &\le 1. \\<br />
\end{aligned}
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  13. #13
    Super Member malaygoel's Avatar
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    Quote Originally Posted by JakeD
    Yes. Use integral calculus to find the area of the region

    \begin{aligned}<br />
x + y &\ge 1 \\<br />
x^2 + y^2 &\le 1 \\<br />
0 \le x &\le 1 \\<br />
0 \le y &\le 1. \\<br />
\end{aligned}
    Thanks, but what I wanted to say that can the answer be obtained without using analytic geometry at any stage?
    We have used analytic geometry to obtain the picture!

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    Malay
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  14. #14
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    Quote Originally Posted by malaygoel
    Thanks, but what I wanted to say that can the answer be obtained without using analytic geometry at any stage?
    We have used analytic geometry to obtain the picture!

    Keep Smiling

    Malay
    The inequalities involving x and y come from considering what it means to be an obtuse triangle. That comes from geometry. But the picture comes from graphing those inequalities. 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 \pi remember.
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  15. #15
    Super Member malaygoel's Avatar
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    Quote Originally Posted by JakeD
    The inequalities involving x and y come from considering what it means to be an obtuse triangle. That comes from geometry. But the picture comes from graphing those inequalities.
    We have to draw arough sketch, we can't use graphs.
    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 \pi remember.
    Trigonometry can be used as extensively as you want. \pi can arise from trigonometry.

    Keep Smiling

    Malay
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