1. 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 $\displaystyle \frac{\pi - 2}{4}$

Keep Smiling
Malay

2. I do not understand what you mean by a triplet?

3. 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

4. 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

5. 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 $\displaystyle \frac{\pi - 2}{4}$

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

$\displaystyle \setlength{\unitlength}{2.5cm} \begin{picture}(1,1) \qbezier(0,0)(0,0)(1,0) \qbezier(0,0)(0,0)(0,1) \qbezier(1,0)(1,0)(1,1) \qbezier(0,1)(0,1)(1,1) \linethickness{1.3pt} \qbezier(1,0)(1,0)(0,1) \qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71) \qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0) \end{picture}$

6. wow! +rep

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

$\displaystyle \setlength{\unitlength}{2.5cm} \begin{picture}(1,1) \qbezier(0,0)(0,0)(1,0) \qbezier(0,0)(0,0)(0,1) \qbezier(1,0)(1,0)(1,1) \qbezier(0,1)(0,1)(1,1) \linethickness{1.3pt} \qbezier(1,0)(1,0)(0,1) \qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71) \qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0) \end{picture}$
wow! +rep

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

$\displaystyle \setlength{\unitlength}{2.5cm} \begin{picture}(1,1) \qbezier(0,0)(0,0)(1,0) \qbezier(0,0)(0,0)(0,1) \qbezier(1,0)(1,0)(1,1) \qbezier(0,1)(0,1)(1,1) \linethickness{1.3pt} \qbezier(1,0)(1,0)(0,1) \qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71) \qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0) \end{picture}$
Will you please explain how did you get that picture(how it is linked to the question)?

8. 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
$\displaystyle x + y = 1$
$\displaystyle x^2 + y^2 = 1$
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.

KeepSmiling
Malay

9. Originally Posted by malaygoel
Ok, I understand it.
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.

KeepSmiling
Malay
I used the same thinking. Glad to help. --JakeD

10. 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?

Keep Smiling
Malay

11. 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.

12. 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

\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}

13. Originally Posted by JakeD
Yes. Use integral calculus to find the area of the region

\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}
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

14. 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 $\displaystyle x$ and $\displaystyle y$ 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 $\displaystyle \pi$ remember.

15. Originally Posted by JakeD
The inequalities involving $\displaystyle x$ and $\displaystyle y$ 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 $\displaystyle \pi$ remember.
Trigonometry can be used as extensively as you want.$\displaystyle \pi$ can arise from trigonometry.

Keep Smiling

Malay

Page 1 of 2 12 Last