If a wire of length 5 is cut into 3 segments of any length, the probability of forming a triangle is

The above answer is obviously wrong, this requires that the sum of the two sides of the triangle is greater than the third side, and the difference between the two sides is less than the third side constraint, the algorithm will be very complicated.

I haven't studied this, but since it's three sections, it should be able to form a triangle of any length.

100% ah, as long as it's not an equilateral triangle, you can put it together into a triangle.

Let one of the segments be x, the second segment be y, then the third segment is 5 x y, then:

x+y-5+x+y>0

5-x-y+x-y>0

5-x-y+y-x>0

Solution. x+y>5/2

x<5/2

y<5/2

There is also x, y is greater than zero.

Make use of the coordinates handle.

x+y=5/2

The image of x=5 2y=5 2 is drawn, and the triangle region that satisfies the inequality occupies 1 2 of the square region, which is the answer.

Let the line segment (0,a) arbitrarily fold into three segments of length x, y, a-x-y, obviously there are x>0 , y>0 , a-x-y>0 , and the feasible domain of (x,y) in the plane Cartesian coordinate system satisfies these three constraints is a right triangle with an area of: (1 2) a 2

The condition for a triangle to be formed is that the sum of any two sides is greater than the third side, i.e.,

x+y>a-x-y,a-x-y+x>y,a-x-y+y>x at the same time becomes immediate x+y>a 2,ya 2,y, so the probability that the three segments can form a triangle is:

p=[(1/8)a^2]/[(1/2)a^2]=1/4=

If it's long, there's no answer.

Let the truncation be three segments of a, b, and c.

As long as a+b c,|a-b|＜c

Because the Great Hunger is arbitrarily truncated in three sections, so |a-b|The probability of return to c is 1 2, and the probability of a+b c in this case is 1 2

So satisfying a+b imitation scrambling c,|a-b|The probability of c is 1 2*1 2 = 1 4, then the probability that these three segments can form a triangle is 1 4

Let the line segment (0,a) be arbitrarily folded into three sections, respectively.

x, y, a-x-y, obviously x>0 , y>0 , a-x-y>0 , and the feasible domain of (x,y) satisfying these three constraints in the plane Cartesian coordinate system is a right-angled triangle with an area of: (1 2)a 2

The condition for the formation of a triangle is that the sum of any two sides is known to be on the third side, that is:

x+y>a-x-y,a-x-y+x>y,a-x-y+y>x at the same time as immediate x+y>a 2,y

Let the wire be arbitrarily divided into three sections: x, y, 7-x-y, 1There are 0 x 7, 0 y 7, 0 l-x-y 7, which can be represented as regions on a planar domain: x=7, y=7, y=7-x, and the area is:

s1=1/2·7·7=49/2（1）

2.When x,y,7-x-y can form a triangle, it should also satisfy:

x+y 7-x-y, get x+y 7 2, 7-x-y+y x, get x 7 Qi Zhaosen 2, high acres.

7-x-y+x y, we get y 7 2, which can be expressed as: x=7 2, y=7 2, y=7 2-x, and the area is: s2=1 2·7 2·7 2=49 8,(2).

From classical generalizations: Guess Kai p = (49 8) (49 2) = 1 4

The answer is 1 4

Combine the coordinate graph to find the probability.

Let the truncation be divided into three sections, which are x, y, 5-x-y, then 000<5-x-y y<-x+5

To form a triangle, the sum of the two sides is greater than the third side.

x+y>5-x-y y<-x+5/2

x+5-x-y>y gets 0y+5-x-y>x 0, which is represented by the coordinates as follows (since I don't know how to draw on a computer, I took a picture).

As you can see from the figure, the area of the red shaded area divided by the area of the entire shaded area is 1 4

Solution: Let the lengths of the three sections be a, b, and c respectively, then a, b, and c need to meet the following conditions (the first part constraint):

a+b+c = 5;

0c；b+c>a;

c+a>b；

a+b+c =5;

000 on the spatial graph:

The first part of the constraint represents the plane a+b+c=5 at 0s1 = 1 2 * 5 2 *(5 2 * 3 2) = 25 3 2

The first part of the constraint represents the plane a+b+c=5 at 0s2 =1 2 * 3 2) = 25 3 8

When the points represented by the values of a, b, and c fall within the range of s2, a triangle can be formed, so the probability of forming a triangle is.

p = s2/s1 = 1/4

5 is divided into 3 segments. It can only be divided into ).

and ) cannot form a triangle ) can.

So it's 1 out of 2

The wire is truncated into 3 pieces of any length of an integer: the sum of the two sides of the triangle must be greater than the third side.

The truncated three sides are a, b c

When a+b c b+c a c+a b form a triangle.

When a+b c b+c a c+a b, the group does not form a triangle.

Probability of the above result: 1 2

The probability of forming a triangle is 1 in 2.

Let the length of the wire with length 1 cut into 3 sections is x, y and z=1-(x+y), x +y 1

Three segments can form a triangle, then.

x+y z, i.e. x+y>(1-x-y), x +y>1 2y+z x, i.e. y+(1-x-y)>x, x 1 2z+x y, i.e. (1-x-y)+x>y, y 1 2 The probability is equal to x+y=1 2, x=1 2, y=1 2 The area of the graph surrounded by the three straight lines divided by the area of the graph surrounded by the straight line (x+y)=1 and the x-axis and y-axis (the figure cannot be inserted).

Therefore, the length of 1 wire is cut into 3 sections, and the probability that these three sections can form a triangle is (1 2*1 2*1 2) (1*1*1 2)=1 8 1 2=1 4

According to the nature between the sides of the triangle.

The sum of the two sides is greater than the third side.

The difference between the two sides is less than the third side.

There are countless ways to intercept the three sections, and if you want to find the probability, let the master solve it.

1 4 Let the truncation be three segments of a, b, and c.

As long as a+b c,|a-b|＜c

Because three sections are arbitrarily truncated, |a-b|The probability of c is 1 2, in this case the probability of a+b c is 1 2, so it satisfies a+b c, |a-b|The probability of c is 1 2*1 2 = 1 4, then the probability that these three segments can form a triangle is 1 4

Let the first section be long and x, and the second section be y, then the third length is a-x-y, so the basic conditions that x,y satisfies are x>0,y>0,x+ya-x-y,x-yy,a-y>x (that is, the sum of the two sides is greater than the third side, and the difference between the two sides is less than the third side), and the solution is x+y>a 2,0

Let the line segment (0,a) be arbitrarily folded into three sections, respectively.

x, y, a-x-y, obviously there are x>0 , y>0 , a-x-y>0 , and the feasible domain of (x,y) in a planar Cartesian coordinate system satisfies these three constraints is a right-angled triangle with an area of: (1 2)a 2

The condition for a triangle to be formed is that the sum of any two sides is greater than the third side, i.e.,

x+y>a-x-y,a-x-y+x>y,a-x-y+y>x at the same time.

i.e. x+y>a 2,ya 2,y, so the probability that the three segments can form a triangle is:

p=[(1 8)a 2] shouting [(1 2)a 2]=1 4=