Stick's dilemma

First of all, in a triangle, sum of 2 sides is greater than the third side

To solve this, draw an equilateral triangle of altitude = 3 cm ( stick's length )

Now, connect the midpoints of the 3 sides with straight lines to divide the triangle into 4 smaller equilateral triangles

Select a point in bigger triangle and drop perpendicular to all 3 sides.

Sum of these altitudes is always 3 cm inspite of where point is ( Why? )

Claim : whenever point is in shaded triangle, the lengths of altitudes on sides will form a triangle

Proof : assume the point to be on an edge of the shaded triangle

Then one of the perpendiculars from it on sides will be 1.5 cm ( midpoint theorem )

So, length of other 2 perpendicular will be equal to third and there will be no triangle formed by these perpendiculars

The same is the case if selected point is outside shaded triangle

So, point must be in shaded triangle so that perpendiculars from it make a triangle

The probability of point being in shaded triangle is $\boxed{\frac{1}{4}}$ as probability of point being in any 4 triangles is equal

The Triangle Inequality Theorem states that a triangle can be made from 3 sides if the sum of the two shorter sides is greater than the third side.

Let's split the stick into two halves.

If the 2 points lie on the same half of the stick, a triangle cannot be formed because the longest side is longer than half of the total length of the stick, so the other two sides have to be less than the longest side.

If the 2 points lie on different halves of the stick, it's possible for the 2 shorter sides to be longer than the third side. We can find the probability that a triangle can be made by trying different spots for the first point and seeing how different positions affect the probability.

First of all, if a triangle cannot be formed, then the middle piece has to be the longest piece because if the longest piece is a side piece, it would cover less than half of the stick since the 2 points are on different halves of the stick. Therefore, the sum of the lengths of the other two pieces would be greater than the length of the side piece that was originally designated to be the longest, meaning that a triangle can be formed.

Knowing that, all we need to figure out is if the sum of the lengths of the two side pieces is longer than the length of the middle piece.

Let's experiment by trying different spots for the point on the left half of the stick and seeing how that affects the size of the region that the point on the right side can be in in order for a triangle to be able to be formed. Notice that if a triangle cannot be formed, the length of the middle piece needs to be greater than or equal to 1.5 cm, or half of the length of the stick. This means that the distance between the 2 points needs to be greater than or equal to 1.5 cm if a triangle cannot be formed.

Let's suppose the 2 points are exactly 1.5 cm apart. In that case, the Triangle Inequality Theorem is not satisfied; however, we don't need to worry about this case because there are infinitely many possibilities for the distance between the two points, and the probability that the distance between them is exactly 1.5 cm is infinitely small. But notice that if the point on the right is moved right, the distance between the two points grows greater than 1.5 cm, while if the point is moved left, the distance becomes less than 1.5 cm, thus satisfying the inequality. Therefore, we can place a marker 1.5 cm to the right of the point on the left half, and if the second point is to the left of it and still on the right half of the stick, a triangle can be formed. This is our region. Any position on the right half of the stick for the marker is equally likely, which means that the average size of the region occurs when the marker is in the center of the right half of the stick, in which case the region is half of the length of the right half of the stick.

Therefore, the probability of the second point being in the region formed with the marker provided that it is on the right half of the stick is $\frac{1}{2}$ .

The probability of the 2 points being in different halves of the stick is also $\frac{1}{2}$ .

The final probability is $\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$ . $1 \times 4=\boxed4$ .

$\int _0^1\int _0^1\text{Boole}[\{a,b\}=\text{MinMax}[\{x,y\}];c=1-b;b\text{-=}a;a+b\geq c\land a+c\geq b\land b+c\geq a]dxdy \Longrightarrow \frac14$ .

Let's make the equations easier by saying the stick has length $1$ .

It's broken into three sections of length $x$ , $y$ and $z$ ; we can think of these as coordinates in 3D space. They satisfy the following:

$x+y+z=1$ (length of the stick)

$x,y,z>0$ (no negative lengths)

Additionally, sets that form a triangle also satisfy

$y+z>x$ , $z+x>y$ , $x+y>z$ (triangle inequality)

The first two conditions define a triangle with vertices $(1,0,0)$ , $(0,1,0)$ and $(0,0,1)$ . This is clearly equilateral. We'll call this triangle $S$ .

The third condition defines a subset of $S$ . Let's consider just one of the triangle inequalities, and to find the boundary of the subset we're interested in, make it an equality. We want to know which points of $S$ satisfy $y+z=x$ .

All points of $S$ satisfy $x+y+z=1$ ; combining this with $y+z=x$ and rearranging gives $x=\frac{1}{2}$ , which is a plane. The intersection with $S$ is a straight line segment joining the points $(\frac{1}{2},0,\frac{1}{2})$ and $(\frac{1}{2},\frac{1}{2},0)$

The subset in question is the triangular region $T$ with vertices $(0,\frac{1}{2},\frac{1}{2})$ , $(\frac{1}{2},0,\frac{1}{2})$ and $(\frac{1}{2},\frac{1}{2},0)$

Finally, the area of $T$ is $\frac{1}{4}$ the area of $S$ , so the answer is $\boxed4$ .