# Thread: An unusual problem: probability meets geometry

1. ## An unusual problem: probability meets geometry

A good mixture of probability and plane geometry:
Creative Unusual Geometry Problems: Fox 198

2. Originally Posted by oswaldo
A good mixture of probability and plane geometry:
Creative Unusual Geometry Problems: Fox 198
Why is 0 not an option?

CB

3. It must be "any triangle" or "within the overall picture there should be one triangle".

Expecting C collide with O has a zero chance, but there should be other ways of getting "a" triangle -as part of the final figure.

4. Hello, oswaldo!

A random process starts from point $O,$
Point $A$ is obtained in a uniformly selected direction, 1 unit from $O.$
The same process is repeated at point $A$ to obtain point $B$,
and at point $B$ to obtain point $C.$

What is the probability that a triangle is obtained when point $C$ is generated?

. . $(A)\;\frac{1}{8} \qquad (B)\;\frac{1}{9} \qquad (C)\;\frac{1}{12} \qquad (D)\;\frac{1}{16} \qquad (E)\;\frac{1}{18}$

Code:
                B
o
*  *
*     *
*        *
O o * * * * * * * * o A
*
o
C
I got: . ${\color{blue}(E)\;\frac{1}{18}}$

Segment $OA$ can be any direction.

Code:
            B
o
\
\
\
\
60° \
O o * * * * * o A
60° /
/
/
/
/
o
B
Segment $AB$ must be within 60° on either side of $OA$.
Otherwise, $OB > 1$ and we will have no triangle.

$P(\angle OAB \leq 60^o) \:=\:\frac{120^o}{360^o} \:=\:\frac{1}{3}$

Code:
            B
o
/ *
/60°*
/     *
/       *
*
O o * * * * * o A
Segment $BC$ must be within 60° of $AB.$

$P(\angle ABC \leq 60^o) \:=\:\frac{60^o}{360^o} \:=\:\frac{1}{6}$

Therefore: . $P(OABC\text{ forms a triangle}) \;=\;\frac{1}{3}\cdot\frac{1}{6} \;=\;\frac{1}{18}$

5. Originally Posted by Soroban
Hello, oswaldo!

I got: . ${\color{blue}(E)\;\frac{1}{18}}$

Segment $OA$ can be any direction.

Code:
            B
o
\
\
\
\
60° \
O o * * * * * o A
60° /
/
/
/
/
o
B
Segment $AB$ must be within 60° on either side of $OA$.
Otherwise, $OB > 1$ and we will have no triangle.

$P(\angle OAB \leq 60^o) \:=\:\frac{120^o}{360^o} \:=\:\frac{1}{3}$

Code:
            B
o
/ *
/60°*
/     *
/       *
*
O o * * * * * o A
Segment $BC$ must be within 60° of $AB.$

$P(\angle ABC \leq 60^o) \:=\:\frac{60^o}{360^o} \:=\:\frac{1}{6}$

Therefore: . $P(OABC\text{ forms a triangle}) \;=\;\frac{1}{3}\cdot\frac{1}{6} \;=\;\frac{1}{18}$
Ahh .. different question from my reading which required O and C be coincident, not that segment BC intersect segment OA

CB

6. As Captain was objecting, assuming "that segment BC intersect segment OA" the answer is NOT 1/18. I think it is because angle OAB can be bigger than 60 degrees. Assume it is 70. A triangle is still possible. I think AOB can have values upto 90 degrees. That will change the solution with new combinations.

$P( 0 \leq \angle OAB \leq 90^o) \:=\: \ldots{}$

7. Actually this can be simulated in Excel. And you should have an answer.
Here is my approach:
a: first angle randomly generated
b: second angle randomly generated

a in [0,60]
Then 0 < b < (180-a)/2 forms a triangle.

a in [60,90]
Then 0 < b < (180-2a) forms a triangle.
Compute probabilities in both case.
Then multiple by 2 (why?)

8. Originally Posted by oswaldo
Actually this can be simulated in Excel. And you should have an answer.
Here is my approach:
a: first angle randomly generated
b: second angle randomly generated

a in [0,60]
Then 0 < b < (180-a)/2 forms a triangle.

a in [60,90]
Then 0 < b < (180-2a) forms a triangle.
Compute probabilities in both case.
Then multiple by 2 (why?)
That is correct. Working in radians rather than degrees, the probability is given by $\frac1\pi\left(\int_0^{\pi/3}\frac{\pi-\theta}{4\pi}d\theta + \int_{\pi/3}^{\pi/2}\frac{\pi-2\theta}{2\pi}d\theta\right)$, which works out as 1/12.