Okay, so I need to find the probability that a triangle with sides 1, x, and y exists when x and y are chosen randomly from the interval (0,3). Hmm, let's start by recalling the triangle inequality theorem. For three lengths to form a triangle, the sum of any two sides must be greater than the third side. 

So, in this case, the three sides are 1, x, and y. That means three conditions must be satisfied:
1. 1 + x > y
2. 1 + y > x
3. x + y > 1

Since x and y are both selected from (0,3), each of them can be anywhere between 0 and 3. The problem is asking for the probability that all three inequalities hold true. 

First, let me visualize the sample space. Since x and y are both between 0 and 3, the sample space is a square in the coordinate plane with vertices at (0,0), (3,0), (3,3), and (0,3). The total area of this square is 3*3 = 9, so the probability will be the area of the region where all three inequalities hold divided by 9.

Now, let's break down each inequality:

1. 1 + x > y → y < x + 1
2. 1 + y > x → x < y + 1
3. x + y > 1 → y > 1 - x

So, the region where all three inequalities are satisfied is the intersection of the regions defined by y < x + 1, x < y + 1, and y > 1 - x, within the square [0,3] x [0,3].

Let me sketch this mentally. The square is from (0,0) to (3,3). The lines y = x + 1 and x = y + 1 are diagonals with slopes 1 and -1, respectively, intersecting at (0.5, 1.5). The line y = 1 - x is another diagonal with slope -1, but shifted up by 1 unit. It intersects the square at (0,1) and (1,0).

So, the region of interest is where y is between 1 - x and x + 1, and also x is between 1 - y and y + 1. But since x and y are both between 0 and 3, I need to find the overlapping area where all three inequalities are satisfied.

Wait, maybe it's better to draw a diagram. But since I can't draw, I'll try to imagine it. Let's consider the square from (0,0) to (3,3). The line y = x + 1 starts at (0,1) and goes up to (2,3). Similarly, the line x = y + 1 is the same as y = x - 1, which starts at (1,0) and goes up to (3,2). The line y = 1 - x starts at (0,1) and goes down to (1,0).

So, the region y > 1 - x is above the line from (0,1) to (1,0). The region y < x + 1 is below the line from (0,1) to (2,3). The region x < y + 1 is below the line from (1,0) to (3,2). So, the intersection of these regions is a polygon bounded by these lines.

But since x and y can go up to 3, I need to check if these lines intersect the square beyond the points mentioned. For example, y = x + 1 when x = 3 would be y = 4, which is outside the square. Similarly, x = y + 1 when y = 3 would be x = 4, which is also outside. So the lines only intersect the square at the points I mentioned.

Now, let's find the vertices of the feasible region. Let's see where the lines intersect each other and the boundaries.

First, the lines y = x + 1 and x = y + 1 intersect where y = x + 1 and x = y + 1. Substituting the second equation into the first: y = (y + 1) + 1 → y = y + 2 → 0 = 2. That's impossible, so they don't intersect. Wait, that can't be right. Wait, if we set y = x + 1 and x = y + 1, substituting gives y = (y + 1) + 1 → y = y + 2 → 0 = 2. Which is a contradiction, so they are parallel? Wait, no. Wait, y = x + 1 has a slope of 1, and x = y + 1 can be rewritten as y = x - 1, which also has a slope of 1. So they are parallel lines. Therefore, they never intersect. So the region between them is a strip between these two parallel lines.

Similarly, the line y = 1 - x is a diagonal with slope -1, and the other lines have slope 1. So the feasible region is bounded by the lines y = x + 1, y = 1 - x, and x = y + 1. Wait, but x = y + 1 is the same as y = x - 1. So the feasible region is where y is between x - 1 and x + 1, and also y is greater than 1 - x.

But I need to find the intersection of all these regions. Let me think. The feasible region is the intersection of three half-planes: y < x + 1, y > x - 1, and y > 1 - x.

So, combining these, the region is where y is greater than both x - 1 and 1 - x, and less than x + 1. But since x and y are between 0 and 3, I need to consider how these lines interact within that square.

Alternatively, maybe it's easier to split the square into regions where the inequalities hold. Let's try to find the boundaries.

First, y > 1 - x. This is the region above the line from (0,1) to (1,0). Then, y < x + 1 is the region below the line from (0,1) to (2,3). And y < x + 1 is also above x = y - 1, which is the same as y > x - 1. Wait, but y > x - 1 is the same as x < y + 1. So, the feasible region is the area where y > max(1 - x, x - 1) and y < x + 1.

But since x and y are positive, we need to consider different cases based on the value of x.

Let me consider different intervals for x:

Case 1: x is between 0 and 1.

In this case, 1 - x is between 0 and 1, and x - 1 is negative. So, since y must be greater than both 1 - x and x - 1, and x - 1 is negative, the lower bound for y is 1 - x. The upper bound is x + 1.

But since x is between 0 and 1, x + 1 is between 1 and 2. So, for x in [0,1], y must be between 1 - x and x + 1.

Case 2: x is between 1 and 2.

Here, 1 - x is negative (since x > 1), and x - 1 is between 0 and 1. So, the lower bound for y is x - 1, and the upper bound is x + 1. However, we also need to check if y < x + 1 is within the square, which it is as x + 1 is up to 3 when x is 2. But wait, when x is 2, x + 1 is 3, which is the upper limit. However, the upper bound is 3, but the line y = x + 1 intersects the square at (2,3). So, for x in [1,2], y ranges from x - 1 to x + 1.

Case 3: x is between 2 and 3.

Here, 1 - x is negative, and x - 1 is between 1 and 2. The upper bound is x + 1, which is between 3 and 4, but since y is limited to 3, the upper bound becomes 3. So, y must be between x - 1 and 3.

But we also need to check if the line y = x + 1 intersects the square at x=3, which would be y=4, which is outside. So, for x in [2,3], the upper bound is 3, and the lower bound is x - 1. However, we must also ensure that y > 1 - x, but since x - 1 is greater than 1 - x when x >1. Let's check:

For x in [2,3], 1 - x is negative, and x - 1 is positive. So, y must be greater than x - 1, which is already greater than 1 - x because x -1 > 1 - x when x >1. For x=2, x -1 =1, 1 -x = -1. For x=3, x -1=2, 1 -x= -2. So, in this case, the lower bound is x -1, and the upper bound is 3.

But we also need to check if y = x -1 intersects the square. When x is 2, y =1, which is within the square. When x is 3, y=2, which is also within the square. So, for x in [2,3], the region is y from x -1 to 3.

Wait, but we also need to make sure that the upper bound y < x +1 is less than 3. Since x +1 is up to 4, but y is limited to 3, so the upper bound is min(x +1, 3). However, for x in [2,3], x +1 is greater than 3 when x > 2. So for x in [2,3], the upper bound is 3.

Now, let's check if the lines y = x -1 and y = x +1 intersect within the square. Since they are parallel, they don't. So the feasible region is a polygon defined by the lines y = x -1, y = x +1, y =1 -x, and the boundaries of the square.

Wait, but to find the exact shape, maybe it's better to find the intersection points of these lines within the square.

First, the line y = x +1 intersects the square at (0,1) and (2,3). The line y = x -1 intersects the square at (1,0) and (3,2). The line y =1 -x intersects the square at (0,1) and (1,0). 

So, the intersection points of the lines are at (0,1), (1,0), (2,3), and (3,2). Wait, but these points are the corners of the polygon.

Wait, the lines y = x +1 and y = x -1 are parallel, so they don't intersect. The line y =1 -x intersects y = x +1 at (1,2). Because solving 1 -x = x +1 → 1 -x =x +1 → -2x =0 → x=0, but that's (0,1). Wait, maybe not. Wait, 1 -x = x +1 → 1 -x =x +1 → -2x =0 → x=0. So, the only intersection is at (0,1). Similarly, y =1 -x and y =x -1 intersect at (1,0). So, the lines y =1 -x and y =x -1 intersect at (1,0). The lines y =x +1 and y =x -1 are parallel and don't intersect. The line y =x +1 intersects the square at (0,1) and (2,3), and the line y =x -1 intersects the square at (1,0) and (3,2). The line y =1 -x intersects the square at (0,1) and (1,0). 

So, the feasible region is bounded by the lines y =x +1, y =x -1, and y =1 -x. But since y =x +1 and y =x -1 are parallel, the feasible region is a quadrilateral bounded by:

- From (0,1) to (1,0): along y =1 -x.
- From (1,0) to (3,2): along y =x -1.
- From (3,2) to (2,3): along y =x +1.
- From (2,3) back to (0,1): but wait, that's not possible. Wait, maybe I need to check the intersection points.

Wait, actually, the lines y =x +1 and y =x -1 are parallel, so they don't intersect. Therefore, the feasible region is bounded by:

- The line y =1 -x from (0,1) to (1,0).
- The line y =x -1 from (1,0) to (3,2).
- The line y =x +1 from (0,1) to (2,3), but beyond (2,3), y can't go higher than 3. So, the upper boundary is y =3.

Wait, this is getting a bit confusing. Maybe it's better to plot the lines and see where the feasible region is.

Alternatively, let's think about the inequalities again:

1. y < x +1
2. y > x -1
3. y >1 -x

So, combining these, the feasible region is where y is between max(x -1, 1 -x) and x +1.

But since x and y are positive, let's consider different ranges for x.

When x is between 0 and 1:

- max(x -1, 1 -x) is 1 -x because x -1 is negative.
- So y must be between 1 -x and x +1.

When x is between 1 and 2:

- max(x -1, 1 -x) is x -1 because x -1 is positive and greater than 1 -x (since x >1, x -1 >1 -x → 2x >2 →x>1).
- So y must be between x -1 and x +1.

When x is between 2 and 3:

- max(x -1, 1 -x) is x -1 because x -1 >1 -x (since x >2, x -1 >1 -x →2x >2 →x>1, which is true here).
- So y must be between x -1 and 3 (since y <x +1 but x +1 >3 when x>2, so upper bound is 3).

So, summarizing:

For x ∈ [0,1]:

y ∈ [1 -x, x +1]

For x ∈ [1,2]:

y ∈ [x -1, x +1]

For x ∈ [2,3]:

y ∈ [x -1, 3]

Now, we need to calculate the area of the region defined by these y ranges for each x interval.

Let's compute each part separately.

First, for x ∈ [0,1]:

The area is the integral from x=0 to x=1 of [upper(x) - lower(x)] dx, where upper(x) = x +1 and lower(x) =1 -x.

So, integral from 0 to1 of (x +1 - (1 -x)) dx = integral from 0 to1 of (2x) dx = [x²] from 0 to1 =1.

Second, for x ∈ [1,2]:

The area is integral from x=1 to x=2 of [upper(x) - lower(x)] dx, where upper(x) =x +1 and lower(x)=x -1.

So, integral from1 to2 of (x +1 - (x -1)) dx = integral of 2 dx from1 to2 =2*(2 -1)=2*1=2.

Third, for x ∈ [2,3]:

The area is integral from x=2 to x=3 of [upper(x) - lower(x)] dx, where upper(x)=3 and lower(x)=x -1.

So, integral from2 to3 of (3 - (x -1)) dx = integral from2 to3 of (4 -x) dx.

Compute the integral:

∫(4 -x) dx from2 to3 = [4x - (x²)/2] from2 to3.

At x=3: 4*3 - (9)/2 =12 -4.5=7.5

At x=2:4*2 -4/2=8 -2=6

So, the integral is7.5 -6=1.5

Therefore, the total area is1 +2 +1.5=4.5

But wait, the total area of the square is9, so the probability would be4.5/9=0.5. But that can't be right because the probability is 0.5, but intuitively, the triangle inequality might not be 50%. Wait, maybe I made a mistake.

Wait, let's check the integrals again.

First part, x from0 to1:

Integral of (x +1 - (1 -x))=2x dx from0 to1. The integral is [x²] from0 to1=1. Correct.

Second part, x from1 to2:

Integral of (x +1 - (x -1))=2 dx from1 to2. The integral is2*(2 -1)=2. Correct.

Third part, x from2 to3:

Integral of (3 - (x -1))=4 -x dx from2 to3. The integral is [4x - (x²)/2] from2 to3.

At x=3:12 -4.5=7.5

At x=2:8 -2=6

Difference:1.5. Correct.

Total area:1+2+1.5=4.5. So 4.5/9=0.5. So probability is 0.5. But let me check if this makes sense.

Wait, is there a symmetry here? Because the problem is symmetric in x and y. The region where x and y satisfy the triangle inequalities should be symmetric across the line x=y. But in our calculation, the total area is4.5, which is half of the square. So the probability is 0.5. But let me test with some specific values.

For example, if x=0.5, then y must be between1 -0.5=0.5 and0.5 +1=1.5. So from0.5 to1.5, which is a length of1. Similarly, if x=1.5, y must be between0.5 and2.5. But y can't exceed3, so it's from0.5 to2.5. The length is2. But in our integral, we accounted for this by using different limits for different x ranges. So maybe the calculation is correct.

Alternatively, maybe there's a mistake in the initial setup. Let me check the triangle inequalities again.

The three inequalities are:

1. 1 + x > y

2. 1 + y > x

3. x + y >1

These three inequalities must hold simultaneously.

Alternatively, we can combine them. The first two inequalities can be rewritten as:

y < x +1

and

y > x -1

The third inequality is y >1 -x.

So the feasible region is the intersection of y <x +1, y >x -1, and y >1 -x.

But this is exactly what we used in our calculation. So the area is4.5, so the probability is0.5.

But wait, actually, I recall that in similar problems, the probability is1/4. Wait, maybe not. Wait, maybe in this case, since the variables are selected from (0,3), the probability is different. Let me check.

Alternatively, maybe the answer is1/4. But according to my calculation, it's0.5. Hmm.

Wait, let's consider that when x and y are both in (0,3), the total area is9. The region where all three inequalities hold is4.5, which is half of the square. So probability is1/2. But is that correct?

Wait, let's take another example. Suppose x=3. Then the third inequality is y >1 -3= -2, which is always true. The first inequality is y <3 +1=4, which is also always true since y <3. The second inequality is y >3 -1=2. So for x=3, y must be between2 and3. The length is1. Similarly, for x approaching3, y must approach3.

But according to our integral, for x from2 to3, the area is1.5. Which aligns with this. So that part is correct.

Another test case: x=0. Let's see. For x=0, the first inequality is y <1, second inequality is y > -1 (always true), third inequality is y >1. So y must be between1 and1, which is impossible. Wait, but in our integral, when x=0, the lower bound is1 -0=1 and upper bound is0 +1=1. So y must be between1 and1, which is a line, area zero. So the integral at x=0 is zero, which is correct.

Similarly, for x approaching0, the lower bound is1 -x approaching1, and upper bound isx +1 approaching1. So the region is a line, area zero.

So the calculation seems correct. Therefore, the probability is4.5/9=0.5.

But wait, let's check with another approach. Let's consider that the triangle inequalities are:

1. x + y >1

2. x +1 > y

3. y +1 >x

These can be rewritten as:

y > max(1 -x, x -1)

and y <x +1

So the region is between y = max(1 -x, x -1) and y =x +1.

But since x and y are in (0,3), let's see where 1 -x and x -1 cross each other. They cross at x=1, where1 -1=0 and1 -1=0. So for x <1, max(1 -x, x -1)=1 -x, and for x >=1, max(1 -x, x -1)=x -1.

Therefore, the area can be split into two parts: x from0 to1 andx from1 to3.

Wait, but earlier we split x into0-1,1-2,2-3. Maybe to simplify, split into x from0-1 andx from1-3.

For x from0 to1:

y ∈ [1 -x, x +1]

For x from1 to3:

y ∈ [x -1, x +1]

But wait, for x >2, y <x +1 but y can't exceed3. So for x from2 to3, y ∈ [x -1,3].

Therefore, the area is:

Integral from0 to1 of (x +1 - (1 -x)) dx + Integral from1 to2 of (x +1 - (x -1)) dx + Integral from2 to3 of (3 - (x -1)) dx

Which is:

Integral0-1 (2x)dx + Integral1-2 (2)dx + Integral2-3 (4 -x)dx

Calculating each:

Integral0-1 (2x)dx = [x²]0-1 =1

Integral1-2 (2)dx =2*(2-1)=2

Integral2-3 (4 -x)dx = [4x - (x²)/2] from2 to3

At3:12 -4.5=7.5

At2:8 -2=6

Difference:7.5 -6=1.5

Total area=1 +2 +1.5=4.5

Probability=4.5/9=0.5=1/2

Therefore, the probability is1/2.

But wait, this seems high. Let me check with a simple case. Suppose both x and y are in (0,1). Then, the triangle inequalities:

x + y >1

|x - y| <1

But since x and y are in (0,1), x + y >1 is possible only if x + y >1. The area where x + y >1 in the unit square is1/2. But our problem here is with x and y in (0,3). So the calculation is different.

Alternatively, maybe the answer is indeed1/2. But I'm a bit unsure because I thought sometimes such problems have different answers. Let me check with another approach.

Alternatively, consider that the triangle inequalities can be rewritten as:

y <x +1

y >x -1

y >1 -x

These constraints define a region in the (x,y) plane. The area of this region divided by9 is the probability.

But since the inequalities are symmetric in x and y, the region is symmetric across the line x=y. So the area should be the same as the region where x and y satisfy the inequalities, which would be symmetric.

But according to our integral, it's4.5, which is half of the square. So the probability is0.5. Therefore, the answer is1/2.

But let me confirm with another test case. Let's pick x=1.5, y=2. Then check the triangle inequalities:

1 +1.5=2.5 >2: yes

1 +2=3 >1.5: yes

1.5 +2=3.5 >1: yes

So triangle is possible.

If x=2, y=2.5:1 +2=3 >2.5: yes, 1 +2.5=3.5 >2: yes, 2 +2.5=4.5 >1: yes. So triangle possible.

If x=2.5, y=3:1 +2.5=3.5 >3: yes, 1 +3=4 >2.5: yes, 2.5 +3=5.5 >1: yes. So triangle possible.

If x=0.5, y=0.5:1 +0.5=1.5 >0.5: yes, 1 +0.5=1.5 >0.5: yes, 0.5 +0.5=1 >1: no. So triangle not possible. So the region where y <x +1 and y >x -1 and y >1 -x excludes the case where y=0.5 andx=0.5.

So according to our integral, for x=0.5, y ranges from1 -0.5=0.5 toy=0.5 +1=1.5. But the third inequality y >1 -x=0.5. So y must be between0.5 and1.5. However, the third inequality is y >1 -x=0.5, which is already satisfied by y >=0.5. So the area is1.5, but according to the integral, it's2x=1. So maybe my earlier approach is wrong.

Wait, no. When x=0.5, the upper bound isx +1=1.5 and the lower bound is1 -x=0.5. So the length is1.5 -0.5=1, which matches the integral (2x=1). So that's correct.

But when x=0.5, the third inequality is y >1 -x=0.5. So y must be greater than0.5. But the lower bound is1 -x=0.5. So y must be between0.5 and1.5. So the length is1. Which is correct.

But in this case, the triangle inequality is not automatically satisfied. For example, when x=0.5 and y=0.5, the sum x + y=1, which is not greater than1. So the triangle inequality fails. Therefore, the region y > max(1 -x, x -1) and y <x +1 is correct, because it excludes the cases where y is too small.

Therefore, the integral correctly accounts for all the cases. Therefore, the probability is1/2. So the answer is\boxed{\dfrac{1}{2}}.

**Final Answer**
The probability that a triangle with sides of length 1, \(x\), and \(y\) exists is \boxed{\dfrac{1}{2}}.
Therefore, the probability that a triangle with sides of length 1, $x$, and $y$ exists is $\boxed{\frac{1}{2}}$.
</think>