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A point P is chosen randomly in a square. Join P with the four vertices of the square so as to divide the square into four triangles. Find, correct to 2 decimal places, the probability that all interior angles of the four triangles do not exceed $120°$.

I find this question in a high school Maths competition and I do not know get the answer : $0.21$.

My idea:

  • The key point is to consider the four angles at P, since the angles at vertices should never exceed $120°$.
  • Let $a$, $b$, $c$ and $d$ be the four angles at P. Then we have $$\left\{\begin{array}{c} a+b+c+d = 360°\\0<a,b,c,d \leq 120°\end{array}\right.$$
  • If $a$, $b$, $c$ and $d$ are integers, then we can use $H_r^n$ to find the answer. But it is not the case!

I welcome any comments and ideas!

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    $\begingroup$ Hint: what is the figure, in which point $P$ would make $\angle APB$ (A and B are the square's vertices) greater than $120^\circ$...? $\endgroup$
    – CiaPan
    Commented Oct 2, 2015 at 12:06
  • $\begingroup$ @CiaPan : a part of a circle? $\endgroup$
    – Nighty
    Commented Oct 2, 2015 at 12:07
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    $\begingroup$ Right. So what is the figure, in which $\measuredangle APB < 120^\circ$...? And then, what is the figure, in which FOUR respective angles are ALL less than 120°...? $\endgroup$
    – CiaPan
    Commented Oct 2, 2015 at 12:10

1 Answer 1

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enter image description here

You just have to find the area of the curvilinear quadrilateral having its vertices in the four red points.

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