probablity of random pick up three points inside a regular triangle which form a triangle and contain the center what is  the probablity of random pick up three points inside a regular triangle 
which form a triangle and contain the center of the regualr triangle

the three points are randomly picked within the regular triangle and then form a new triangle
and the new triangle have to contain the center of the original regular triangle
what is the probability
Let $R$ be a convex region of the plane with unit area, and choose $n\ge 3$ points at random from the region.  We will derive a general expression for the probability $P_{n}$ that the convex hull of these points contains a particular point $\tau \in R$ (which we will take as the origin of our coordinate system). We can then evaluate that expression for the particular question posed (in which $n=3$, $R$ is an equilateral triangle, and $\tau$ is its center).

Suppose the hull does not contain $\tau$, and assume that $\tau$ is not collinear with any two of the points (this is true with probability 1).  Then we can draw a ray from the center point through one of the $n$ points (the "leftmost" point) such that the remaining $n-1$ points are to the right of the ray.  (Note that this condition is both necessary and sufficient.  Note also that if the convex hull does contain $\tau$, then there is no "leftmost" point.)  We can calculate the desired probability by integrating over all such configurations.  In particular, take $f(\varphi)$ to be the area of the subregion that lies to the right of the radial ray at angle $\varphi$, and take $r(\varphi)$ to be the distance from $\tau$ to the boundary of $R$ along that ray.  Then the probability that a particular point will lie in the angular interval $[\varphi, \varphi+d\varphi]$ is $da = \frac{1}{2}r(\varphi)^2 d\varphi$, and the probability that each of the other $n-1$ points will lie to its right is $f(\varphi)$.  Since there are $n$ points to choose as the leftmost, the overall probability of not capturing $\tau$ in the convex hull is
$$
1-P_{n} = n \int f(\varphi)^{n-1} da = n \int_{-\pi}^{\pi} \frac{1}{2} f(\varphi)^{n-1} r(\varphi)^2 d\varphi,
$$
We can make a few observations at this point.  First, if the region $R$ is symmetric under reflection through $\tau$, then $f(\varphi)$ is identically $1/2$: each radial line splits the region in half.  The result then is
$$
P_{n} = 1 - \frac{n}{2^{n-1}},
$$
which gives the known result $P_{3} = 1/4$ for polygons with an even number of sides.  Second, if the region $R$ instead has bilateral symmetry across the $x$-axis, then $r(\varphi)$ is an even function, and $f(\varphi) = 1/2 + f^{-}(\varphi)$, where $f^{-}$ is an odd function.  Then terms in the integral with odd powers of $f^{-}$ must vanish: in particular, we have
$$
P_{3} = \frac{1}{4} - 3 \int{{f^{-}(\varphi)}^2 da} = \frac{1}{4} - 3\int_{-\pi}^{\pi}\frac{1}{2}{f^{-}(\varphi)}^2 r(\varphi)^2 d\phi \le \frac{1}{4},
$$
and interestingly the relation $P_{4} = 2P_{3}$ continues to hold.



It remains to evaluate $f^{-}(\varphi)$, $r(\varphi)$, and the resulting integral in the case of an equilateral triangle.  The figure above shows an equilateral triangle with vertices at $(-x,0)$ and $(x/2, \pm x\sqrt{3}/2)$.  For $\varphi \in [0,\pi/3]$, the function $f^{-}(\varphi)$ is the area of the blue triangle minus the area of the red triangle,
$$
f^{-}(\varphi) = \frac{x^2}{8}\tan\varphi - \frac{x^2}{2}\frac{1}{\cot\varphi + \cot\frac{\pi}{6}} = \frac{x^2}{8}\tan\varphi \left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right);
$$
and the radius $r(\varphi) = \frac{1}{2}x\sec\varphi$ over the same domain. By symmetry, the full integral is just six times its value over $[0,\pi/3]$:
$$
\begin{eqnarray}
P_3 &amp;=&amp; \frac{1}{4} - 18\int_{0}^{\pi/3}\frac{x^6}{512}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&amp;=&amp; \frac{1}{4} - \frac{1}{36\sqrt{3}}\int_{0}^{\pi/3}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&amp;=&amp; \frac{1}{4} - \frac{1}{12}\int_{0}^{1} u^2\left(1 - \frac{4}{1+3u}\right)^2 du,
\end{eqnarray}
$$
where we have used $x = 2/\sqrt{3\sqrt{3}}$ (in order for the triangle to have unit area) and introduced the transformed variable $u = \tan\varphi / \sqrt{3}$ (so $du = (\sec^2\varphi / \sqrt{3}) d\varphi$).  The final integral is a straightforward exercise for the reader; the result is
$$
P_3 = \frac{1}{4} - \frac{1}{324}\left(57 - 80\ln{2}\right) = \frac{2}{81}\left(3 + 10\ln{2}\right) = 0.2452215...,
$$
in agreement with the brute-force and numerical results already given.  This result can be generalized easily to the regular $m$-gon for any odd $m$ (changing only the values of some constants), and to the case where $n&gt;3$ (complicating the final integral).
If you ask for an beautiful solution. Then you should totally check this out.
Any point within the triangle can represent how the stick is broken into three pieces. For instance, if point P is selected, that represents that the stick is being broken into pieces with length a, b, and c. By the altitude theorem, it is known that a + b + c = x. Here x is the altitude
In this example it would not be possible to make a triangle, because a + b &lt; c.
To find the probability that the two smaller sides will sum to be larger than the third side it is helpful to divide the equilateral triangle into four smaller equilateral triangles, by connecting the midpoints of the sides.
As long as point P is selected in the middle triangle, it will be possible to create a triangle from the three pieces.
The probability that randomly selected P is in the middle triangle is 1/4 so the probability that a triangle can be formed.
This is a beautiful Geometric approach