Explanation:
Okay, so I need to find the sum of the positive divisors of 500, which is called A, and then find the sum of the distinct prime divisors of A. Let me start by recalling how to calculate the sum of divisors of a number.

First, I remember that if a number is factored into primes as $ n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k} $, then the sum of its positive divisors is given by the product $ (1 + p_1 + p_1^2 + \dots + p_1^{a_1}) \times (1 + p_2 + p_2^2 + \dots + p_2^{a_2}) \times \dots \times (1 + p_k + p_k^2 + \dots + p_k^{a_k}) $.

So, my first step is to factorize 500 into its prime factors. Let me do that.

Starting with the smallest prime, 2. 500 divided by 2 is 250. Divided by 2 again is 125. Now, 125 is not divisible by 2. Next prime is 5. 125 divided by 5 is 25, divided by 5 again is 5, and once more is 1. So, the prime factorization of 500 is $ 2^2 \times 5^3 $. Let me check that: $ 2^2 = 4 $, $ 5^3 = 125 $, and 4 * 125 is indeed 500. Good.

Now, applying the formula for the sum of divisors. For each prime factor, we take the sum of powers from 0 up to the exponent.

For the prime 2 with exponent 2: the sum is $ 1 + 2 + 2^2 = 1 + 2 + 4 = 7 $.

For the prime 5 with exponent 3: the sum is $ 1 + 5 + 5^2 + 5^3 $. Let me calculate that. 1 + 5 is 6, plus 25 is 31, plus 125 is 156. Wait, 1 + 5 + 25 + 125. Let me check again: 1 + 5 = 6, 6 + 25 = 31, 31 + 125 = 156. Yes, that's correct.

Therefore, the sum of the divisors A is 7 multiplied by 156. Let me compute that. 7 * 156. Hmm, 7 * 150 is 1050, and 7 * 6 is 42, so total is 1050 + 42 = 1092. Therefore, A is 1092.

Wait, let me verify that again. Maybe I made a mistake in multiplication. Let me do 156 * 7. Breaking it down: 100*7=700, 50*7=350, 6*7=42. So 700 + 350 is 1050, plus 42 is 1092. Yes, that's correct. So A = 1092.

Now, the next part is to find the sum of the distinct prime divisors of A. So first, I need to factorize 1092 into its prime factors, then identify the distinct primes, and add them up.

Let me start by factoring 1092. Let me check divisibility by small primes.

First, check if it's even. 1092 ends with a 2, so yes, it's divisible by 2. Dividing by 2: 1092 / 2 = 546. Again, 546 is even, divide by 2 again: 546 / 2 = 273. Now, 273 is not even. Let me check divisibility by 3. The sum of digits of 273 is 2 + 7 + 3 = 12, which is divisible by 3. So divide by 3: 273 / 3 = 91.

Now, 91. Let me check if it's divisible by 7. 7 * 13 is 91. Yes, because 7*10=70, 7*13=91. So 91 factors into 7 and 13.

So putting it all together, the prime factors of 1092 are 2, 2, 3, 7, 13. But since we need distinct primes, they are 2, 3, 7, 13.

Therefore, the distinct prime divisors of A are 2, 3, 7, and 13. Now, sum them up: 2 + 3 + 7 + 13. Let me compute that. 2 + 3 is 5, 5 + 7 is 12, 12 + 13 is 25.

Wait, so the sum of the distinct prime divisors of A is 25? Let me check my steps again to make sure I didn't make a mistake.

First, verifying the sum of divisors of 500. 500 factors into 2² * 5³. Sum of divisors formula: (1 + 2 + 4) * (1 + 5 + 25 + 125) = 7 * 156 = 1092. That seems correct.

Then factoring 1092. Divided by 2 twice gives 273. Divided by 3 gives 91. Divided by 7 gives 13. So prime factors are 2² * 3 * 7 * 13. Therefore, distinct primes are 2, 3, 7, 13. Sum is 2 + 3 + 7 + 13 = 25.

Wait, but let me check if 1092 is indeed equal to 2² * 3 * 7 * 13. Let me compute 2² * 3 * 7 * 13. 2² is 4. 4 * 3 is 12. 12 * 7 is 84. 84 * 13. Let me compute 84 * 10 = 840, 84 * 3 = 252, so total is 840 + 252 = 1092. Yes, that's correct.

Therefore, the distinct primes are indeed 2, 3, 7, 13. Their sum is 25.

But wait, let me check if there's another way to factor 1092. Maybe I missed a prime factor? Let me try dividing 1092 by other primes.

We had 1092 divided by 2 is 546, divided by 2 is 273. Then divided by 3 is 91. Then divided by 7 is 13. 13 is prime. So that's all.

Alternatively, maybe I can check if 1092 is divisible by 5? 1092 ends with 2, so no. Divisible by 11? Let me check. The rule for 11 is alternating sum of digits. For 1092: 1 - 0 + 9 - 2 = 1 - 0 is 1, +9 is 10, -2 is 8. 8 is not divisible by 11, so no. Divisible by 17? Let me try 17 * 64 is 1088, 1092 - 1088 = 4, so no. 19? 19*57 is 1083, 1092 - 1083 = 9, not divisible. 23? 23*47 is 1081, 1092 - 1081 = 11, not divisible. So no other primes. So yes, the prime factors are 2, 3, 7, 13.

Therefore, the sum is 2 + 3 + 7 + 13 = 25.

Wait, but let me check once more if I didn't miscalculate the sum. 2 + 3 is 5, 5 + 7 is 12, 12 + 13 is 25. Yes.

But just to be thorough, let me check if there's a mistake in the initial calculation of the sum of divisors.

Original number: 500. Prime factors 2² * 5³. Sum of divisors formula: (2^0 + 2^1 + 2^2) * (5^0 + 5^1 + 5^2 + 5^3).

Calculating each part:

For 2: 1 + 2 + 4 = 7. Correct.

For 5: 1 + 5 + 25 + 125. Let me add them again: 1 + 5 = 6, 6 + 25 = 31, 31 + 125 = 156. Correct.

Multiply 7 * 156. Let me do 156 * 7 again. 150*7=1050, 6*7=42, total 1092. Correct.

So A is indeed 1092. Then factorizing 1092 gives 2² * 3 * 7 * 13. Therefore, distinct primes are 2, 3, 7, 13. Sum is 25.

Therefore, the answer should be 25.

But just to make sure, let me check if there's an alternative way to compute the sum of divisors or if I made a mistake in the factorization.

Alternatively, maybe I can list all the divisors of 500 and sum them up? Although that might take longer, but maybe for verification.

Divisors of 500. Since 500 = 2² * 5³, the divisors are of the form 2^a * 5^b where a = 0,1,2 and b = 0,1,2,3.

So list them all:

For a=0:

- 5^0 = 1

- 5^1 = 5

- 5^2 = 25

- 5^3 = 125

For a=1:

- 2*1 = 2

- 2*5 = 10

- 2*25 = 50

- 2*125 = 250

For a=2:

- 4*1 = 4

- 4*5 = 20

- 4*25 = 100

- 4*125 = 500

So all divisors are: 1, 5, 25, 125, 2, 10, 50, 250, 4, 20, 100, 500.

Now let me sum them up:

Start adding step by step:

Start with 1 + 5 = 6

6 + 25 = 31

31 + 125 = 156

156 + 2 = 158

158 + 10 = 168

168 + 50 = 218

218 + 250 = 468

468 + 4 = 472

472 + 20 = 492

492 + 100 = 592

592 + 500 = 1092.

Yes! The total sum is indeed 1092. So that confirms A is 1092.

Therefore, my previous steps are correct.

So, moving on, the prime factors of 1092 are 2, 3, 7, 13. Their sum is 25.

But just to make sure, let me check if 1092 has any other prime factors. For example, is 13 a prime? Yes. 7 is prime, 3 is prime, 2 is prime. So no other primes.

Alternatively, maybe I can check if 1092 is divisible by any other primes. Let me check 17 again. 17*64 is 1088, 1092-1088=4, so no. 19*57=1083, 1092-1083=9, not divisible. 23*47=1081, 1092-1081=11, not divisible. 29? 29*37=1073, 29*38=1102, which is higher. So no. So no other primes.

Therefore, I think my answer is correct.

But just to recap:

Sum of divisors of 500 is 1092. Prime factors of 1092 are 2, 3, 7, 13. Sum of distinct primes is 2 + 3 + 7 + 13 = 25.

Therefore, the answer is 25.

**Final Answer**
\boxed{25}

To solve the problem, we are asked to find the **sum of the distinct prime divisors of $ A $**, where $ A $ is the **sum of the positive divisors of 500**.

---

### Step 1: Prime Factorization of 500

We begin by factoring 500 into its prime components:

$$
500 = 2^2 \times 5^3
$$

---

### Step 2: Use the Formula for the Sum of Divisors

The sum of the positive divisors of a number $ n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k} $ is given by:

$$
\sigma(n) = (1 + p_1 + p_1^2 + \dots + p_1^{a_1}) \times (1 + p_2 + p_2^2 + \dots + p_2^{a_2}) \times \dots
$$

Applying this to $ 500 = 2^2 \times 5^3 $:

- For $ 2^2 $: $ 1 + 2 + 4 = 7 $
- For $ 5^3 $: $ 1 + 5 + 25 + 125 = 156 $

So, the sum of the divisors of 500 is:

$$
A = 7 \times 156 = 1092
$$

---

### Step 3: Prime Factorization of $ A = 1092 $

Now, we factor 1092 into its prime components:

- $ 1092 \div 2 = 546 $
- $ 546 \div 2 = 273 $
- $ 273 \div 3 = 91 $
- $ 91 \div 7 = 13 $

So, the prime factorization of 1092 is:

$$
1092 = 2^2 \times 3 \times 7 \times 13
$$

The **distinct prime divisors** of 1092 are: $ 2, 3, 7, 13 $

---

### Step 4: Sum of the Distinct Prime Divisors

Now, we sum these distinct primes:

$$
2 + 3 + 7 + 13 = 25
$$

---

### Final Answer

$$
\boxed{25}
$$

[Assistant]
Explanation:
Okay, so I need to find the sum of the positive divisors of 500, which is called A, and then find the sum of the distinct prime divisors of A. Let me start by recalling how to calculate the sum of divisors of a number.

First, I remember that if a number is factored into primes as $ n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k} $, then the sum of its positive divisors is given by the product $ (1 + p_1 + p_1^2 + \dots + p_1^{a_1}) \times (1 + p_2 + p_2^2 + \dots + p_2^{a_2}) \times \dots \times (1 + p_k + p_k^2 + \dots + p_k^{a_k}) $.

So, my first step is to factorize 500 into its prime factors. Let me do that.

Starting with the smallest prime, 2. 500 divided by 2 is 250. Divided by 2 again is 125. Now, 125 is not divisible by 2. Next prime is 5. 125 divided by 5 is 25, divided by 5 again is 5, and once more is 1. So, the prime factorization of 500 is $ 2^2 \times 5^3 $. Let me check that: $ 2^2 = 4 $, $ 5^3 = 125 $, and 4 * 125 is indeed 500. Good.

Now, applying the formula for the sum of divisors. For each prime factor, we take the sum of powers from 0 up to the exponent.

For the prime 2 with exponent 2: the sum is $ 1 + 2 + 2^2 = 1 + 2 + 4 = 7 $.

For the prime 5 with exponent 3: the sum is $ 1 + 5 + 5^2 + 5^3 $. Let me calculate that. 1 + 5 is 6, plus 25 is 31, plus 125 is 156. Wait, 1 + 5 + 25 + 125. Let me check again: 1 + 5 = 6, 6 + 25 = 31, 31 + 125 = 156. Yes, that's correct.

Therefore, the sum of the divisors A is 7 multiplied by 156. Let me compute that. 7 * 156. Hmm, 7 * 150 is 1050, and 7 * 6 is 42, so total is 1050 + 42 = 1092. Therefore, A is 1092.

Wait, let me verify that again. Maybe I made a mistake in multiplication. Let me do 156 * 7. Breaking it down: 100*7=700, 50*7=350, 6*7=42. So 700 + 350 is 1050, plus 42 is 1092. Yes, that's correct. So A = 1092.

Now, the next part is to find the sum of the distinct prime divisors of A. So first, I need to factorize 1092 into its prime factors, then identify the distinct primes, and add them up.

Let me start by factoring 1092. Let me check divisibility by small primes.

First, check if it's even. 1092 ends with a 2, so yes, it's divisible by 2. Dividing by 2: 1092 / 2 = 546. Again, 546 is even, divide by 2 again: 546 / 2 = 273. Now, 273 is not even. Let me check divisibility by 3. The sum of digits of 273 is 2 + 7 + 3 = 12, which is divisible by 3. So divide by 3: 273 / 3 = 91.

Now, 91. Let me check if it's divisible by 7. 7 * 13 is 91. Yes, because 7*10=70, 7*13=91. So 91 factors into 7 and 13.

So putting it all together, the prime factors of 1092 are 2, 2, 3, 7, 13. But since we need distinct primes, they are 2, 3, 7, 13.

Therefore, the distinct prime divisors of A are 2, 3, 7, and 13. Now, sum them up: 2 + 3 + 7 + 13. Let me compute that. 2 + 3 is 5, 5 + 7 is 12, 12 + 13 is 25.

Wait, so the sum of the distinct prime divisors of A is 25? Let me check my steps again to make sure I didn't make a mistake.

First, verifying the sum of divisors of 500. 500 factors into 2² * 5³. Sum of divisors formula: (1 + 2 + 4) * (1 + 5 + 25 + 125) = 7 * 156 = 1092. That seems correct.

Then factoring 1092. Divided by 2 twice gives 273. Divided by 3 gives 91. Divided by 7 gives 13. So prime factors are 2² * 3 * 7 * 13. Therefore, distinct primes are 2, 3, 7, 13. Sum is 2 + 3 + 7 + 13 = 25.

Wait, but let me check if 1092 is indeed equal to 2² * 3 * 7 * 13. Let me compute 2² * 3 * 7 * 13. 2² is 4. 4 * 3 is 12. 12 * 7 is 84. 84 * 13. Let me compute 84 * 10 = 840, 84 * 3 = 252, so total is 840 + 252 = 1092. Yes, that's correct.

Therefore, the distinct primes are indeed 2, 3, 7, 13. Their sum is 25.

But wait, let me check if there's another way to factor 1092. Maybe I missed a prime factor? Let me try dividing 1092 by other primes.

We had 1092 divided by 2 is 546, divided by 2 is 273. Then divided by 3 is 91. Then divided by 7 is 13. 13 is prime. So that's all.

Alternatively, maybe I can check if 1092 is divisible by 5? 1092 ends with 2, so no. Divisible by 11? Let me check. The rule for 11 is alternating sum of digits. For 1092: 1 - 0 + 9 - 2 = 1 - 0 is 1, +9 is 10, -2 is 8. 8 is not divisible by 11, so no. Divisible by 17? Let me try 17 * 64 is 1088, 1092 - 1088 = 4, so no. 19? 19*57 is 1083, 1092 - 1083 = 9, not divisible. 23? 23*47 is 1081, 1092 - 1081 = 11, not divisible. So no other primes. So yes, the prime factors are 2, 3, 7, 13.

Therefore, the sum is 2 + 3 + 7 + 13 = 25.

Wait, but let me check once more if I didn't miscalculate the sum. 2 + 3 is 5, 5 + 7 is 12, 12 + 13 is 25. Yes.

But just to be thorough, let me check if there's a mistake in the initial calculation of the sum of divisors.

Original number: 500. Prime factors 2² * 5³. Sum of divisors formula: (2^0 + 2^1 + 2^2) * (5^0 + 5^1 + 5^2 + 5^3).

Calculating each part:

For 2: 1 + 2 + 4 = 7. Correct.

For 5: 1 + 5 + 25 + 125. Let me add them again: 1 + 5 = 6, 6 + 25 = 31, 31 + 125 = 156. Correct.

Multiply 7 * 156. Let me do 156 * 7 again. 150*7=1050, 6*7=42, total 1092. Correct.

So A is indeed 1092. Then factorizing 1092 gives 2² * 3 * 7 * 13. Therefore, distinct primes are 2, 3, 7, 13. Sum is 25.

Therefore, the answer should be 25.

But just to make sure, let me check if there's an alternative way to compute the sum of divisors or if I made a mistake in the factorization.

Alternatively, maybe I can list all the divisors of 500 and sum them up? Although that might take longer, but maybe for verification.

Divisors of 500. Since 500 = 2² * 5³, the divisors are of the form 2^a * 5^b where a = 0,1,2 and b = 0,1,2,3.

So list them all:

For a=0:

- 5^0 = 1

- 5^1 = 5

- 5^2 = 25

- 5^3 = 125

For a=1:

- 2*1 = 2

- 2*5 = 10

- 2*25 = 50

- 2*125 = 250

For a=2:

- 4*1 = 4

- 4*5 = 20

- 4*25 = 100

- 4*125 = 500

So all divisors are: 1, 5, 25, 125, 2, 10, 50, 250, 4, 20, 100, 500.

Now let me sum them up:

Start adding step by step:

Start with 1 + 5 = 6

6 + 25 = 31

31 + 125 = 156

156 + 2 = 158

158 + 10 = 168

168 + 50 = 218

218 + 250 = 468

468 + 4 = 472

472 + 20 = 492

492 + 100 = 592

592 + 500 = 1092.

Yes! The total sum is indeed 1092. So that confirms A is 1092.

Therefore, my previous steps are correct.

So, moving on, the prime factors of 1092 are 2, 3, 7, 13. Their sum is 25.

But just to make sure, let me check if 1092 has any other prime factors. For example, is 13 a prime? Yes. 7 is prime, 3 is prime, 2 is prime. So no other primes.

Alternatively, maybe I can check if 1092 is divisible by any other primes. Let me check 17 again. 17*64 is 1088, 1092-1088=4, so no. 19*57=1083, 1092-1083=9, not divisible. 23*47=1081, 1092-1081=11, not divisible. 29? 29*37=1073, 29*38=1102, which is higher. So no. So no other primes.

Therefore, I think my answer is correct.

But just to recap:

Sum of divisors of 500 is 1092. Prime factors of 1092 are 2, 3, 7, 13. Sum of distinct primes is 2 + 3 + 7 + 13 = 25.

Therefore, the answer is 25.

**Final Answer**
\boxed{25}

To solve the problem, we are asked to find the **sum of the distinct prime divisors of $ A $**, where $ A $ is the **sum of the positive divisors of 500**.

---

### Step 1: Prime Factorization of 500

We begin by factoring 500 into its prime components:

$$
500 = 2^2 \times 5^3
$$

---

### Step 2: Use the Formula for the Sum of Divisors

The sum of the positive divisors of a number $ n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k} $ is given by:

$$
\sigma(n) = (1 + p_1 + p_1^2 + \dots + p_1^{a_1}) \times (1 + p_2 + p_2^2 + \dots + p_2^{a_2}) \times \dots
$$

Applying this to $ 500 = 2^2 \times 5^3 $:

- For $ 2^2 $: $ 1 + 2 + 4 = 7 $
- For $ 5^3 $: $ 1 + 5 + 25 + 125 = 156 $

So, the sum of the divisors of 500 is:

$$
A = 7 \times 156 = 1092
$$

---

### Step 3: Prime Factorization of $ A = 1092 $

Now, we factor 1092 into its prime components:

- $ 1092 \div 2 = 546 $
- $ 546 \div 2 = 273 $
- $ 273 \div 3 = 91 $
- $ 91 \div 7 = 13 $

So, the prime factorization of 1092 is:

$$
1092 = 2^2 \times 3 \times 7 \times 13
$$

The **distinct prime divisors** of 1092 are: $ 2, 3, 7, 13 $

---

### Step 4: Sum of the Distinct Prime Divisors

Now, we sum these distinct primes:

$$
2 + 3 + 7 + 13 = 25
$$

---

### Final Answer

$$
\boxed{25}
$$

Solution:
First, we find $A$. The prime factorization of $500$ is $2^2 \cdot 5^3$. Therefore,  $$A=(1+2+2^2)(1+5+5^2+5^3)=(7)(156).$$To see why $(1+2+2^2)(1+5+5^2+5^3)$ equals the sum of the divisors of 500, note that if you distribute (without simplifying), you get 12 terms, with each divisor of $2^2\cdot 5^3$ appearing exactly once.

Now we prime factorize $7 \cdot 156 = 7 \cdot 2^2 \cdot 3 \cdot 13$. The sum of the prime divisors of $A$ is $2+3+7+13=\boxed{25}$.


Explanation:
Okay, so I need to find the probability that a triangle with sides of length 1, x, and y exists, where x and y are selected at random from the interval (0,3). Hmm, let me think about how to approach this.

First, I remember that for three lengths to form a triangle, they must satisfy the triangle inequality. That is, the sum of any two sides must be greater than the third side. So, for sides 1, x, and y, the following three inequalities must hold:

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

But since x and y are both selected from (0,3), they can be anywhere between 0 and 3. However, since they are lengths, they must be positive, which they are because the interval is (0,3). So, I need to find the region in the coordinate plane where x and y satisfy all three inequalities, and then divide that area by the total area of the possible region (which is a square with side length 3, so area 9) to get the probability.

Wait, but actually, since x and y are selected from (0,3), the total area is indeed 3*3=9. So, the probability will be the area of the region where the triangle inequalities hold divided by 9.

But let me make sure I have the triangle inequalities right. Let me write them again:

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

So, these three inequalities. Let me think about how to visualize this. Since x and y are both between 0 and 3, we can consider the coordinate system with x on the horizontal axis and y on the vertical axis. The entire sample space is the square with vertices at (0,0), (3,0), (3,3), and (0,3).

Now, I need to find the area within this square where all three inequalities hold. Let me tackle each inequality one by one.

Starting with the third inequality: x + y > 1. This is the region above the line x + y = 1. Since x and y are both positive, the line x + y = 1 intersects the square at (1,0) and (0,1). So, the area where x + y > 1 is the entire square except the triangle below the line x + y = 1. The area of that triangle is (1*1)/2 = 0.5, so the area where x + y > 1 is 9 - 0.5 = 8.5. But we also have the other two inequalities to consider.

Now, the first inequality: 1 + x > y. This can be rewritten as y < x + 1. Similarly, the second inequality: 1 + y > x, which is x < y + 1 or y > x - 1.

So, combining these, we have y < x + 1 and y > x - 1. So, between the lines y = x - 1 and y = x + 1. So, the region between these two lines. However, since x and y are between 0 and 3, we need to consider how these lines intersect the square.

Let me sketch mentally. The line y = x + 1: when x=0, y=1; when x=3, y=4. But since y can only go up to 3, the line y = x + 1 intersects the square at (0,1) and (2,3). Similarly, the line y = x - 1: when x=1, y=0; when x=3, y=2. So, this line intersects the square at (1,0) and (3,2).

Therefore, the region between y = x - 1 and y = x + 1 within the square is a band around the line y = x. However, we also have the previous condition x + y > 1. So, the intersection of all three inequalities is the region where:

- y > x - 1
- y < x + 1
- x + y > 1

But since we already have the square from (0,3) x (0,3), maybe I need to find the overlap of these regions.

Alternatively, maybe I can think of the three inequalities together. Let me try to visualize.

First, the region where x + y > 1 is everything except the triangle near the origin. Then, the region between y = x - 1 and y = x + 1 is a band around the line y = x. So, the intersection of these two regions would be the area where both conditions hold.

But perhaps it's better to consider all three inequalities together. Let me try to find the area where all three inequalities hold.

Alternatively, maybe I can think of the triangle inequalities. Since the sides are 1, x, y, the triangle inequalities are:

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

But since x and y are positive, the third inequality x + y > 1 is automatically satisfied if both x and y are greater than 0? Wait, no. For example, if x = 0.2 and y = 0.2, then x + y = 0.4 < 1. So, that's not automatically satisfied. So, we need to include that.

But since x and y are selected from (0,3), we need to consider that. So, the three inequalities are necessary.

Alternatively, maybe I can think of the triangle inequalities as:

The sum of the two smaller sides must be greater than the largest side. So, depending on which side is the largest, the inequalities change.

But since one of the sides is fixed at 1, and the other two are variables x and y. So, depending on whether 1 is the largest side, or x is the largest, or y is the largest.

So, maybe breaking it into cases.

Case 1: 1 is the largest side. Then, we need x + y > 1. But since 1 is the largest, we also need x ≤ 1 and y ≤ 1. Wait, no. If 1 is the largest side, then both x and y must be less than or equal to 1. But since x and y are in (0,3), but if they are less than or equal to 1, then the triangle inequalities would be satisfied if x + y > 1. However, if 1 is the largest side, then the other two sides x and y must satisfy x + y > 1. But if x and y are both less than or equal to 1, then x + y can be as small as approaching 0 (if both x and y approach 0) up to 2 (if both are 1). So, in this case, the condition is x + y > 1.

Case 2: x is the largest side. Then, we need 1 + y > x. Also, since x is the largest, we need x ≥ 1 and x ≥ y. Similarly, Case 3: y is the largest side. Then, 1 + x > y, and y ≥ 1 and y ≥ x.

So, maybe breaking into these three cases would help. Let me try that.

First, let me note that the total area is 9. So, the probability is the area where the triangle inequalities hold divided by 9.

So, let me compute the area for each case.

Case 1: 1 is the largest side. So, x ≤ 1, y ≤ 1, and x + y > 1.

So, in this case, x and y are in [0,1]x[0,1], and x + y > 1. The area of this region is the area of the square [0,1]x[0,1] minus the area where x + y ≤ 1. The area where x + y ≤ 1 is a triangle with area 0.5, so the area for Case 1 is 1*1 - 0.5 = 0.5.

Case 2: x is the largest side. So, x ≥ 1, x ≥ y, and 1 + y > x. Also, since x is in (0,3), but here x ≥ 1 and y ≤ x. Also, since 1 + y > x, which can be rewritten as y > x - 1. So, in this case, x ranges from 1 to 3, and for each x, y ranges from max(0, x - 1) to x. However, since y must also be less than or equal to x (because x is the largest side). But also, since y is in (0,3), but since x is between 1 and 3, and y is between max(0, x - 1) and x. However, we need to check if x - 1 is less than 0. For x between 1 and 2, x - 1 is between 0 and 1, so max(0, x - 1) is x - 1. For x between 2 and 3, x - 1 is between 1 and 2, so max(0, x - 1) is x - 1. Therefore, for all x in [1,3], y ranges from x - 1 to x. However, we also need to ensure that y is less than 3, but since x is at most 3, and y is at most x, which is at most 3, so that's okay.

Therefore, the area for Case 2 is the integral from x=1 to x=3 of [x - (x - 1)] dx = integral from 1 to 3 of 1 dx = 3 - 1 = 2. Wait, that seems too simple. Wait, the length of y for each x is x - (x - 1) = 1. So, integrating 1 over x from 1 to 3 gives 2. So, area is 2.

But wait, let me check. For example, when x is between 1 and 2, y ranges from x - 1 to x. So, for x=1, y ranges from 0 to 1. For x=2, y ranges from 1 to 2. For x=3, y ranges from 2 to 3. So, the region is a parallelogram? Or a trapezoid? Wait, integrating 1 over x from 1 to 3 gives 2, which is the area. But let me visualize. If you plot x from 1 to 3, and for each x, y from x - 1 to x, then the region is a band between the lines y = x - 1 and y = x, from x=1 to x=3. The area between these two lines over that interval is indeed a parallelogram with base 2 (from x=1 to x=3) and height 1 (the vertical distance between the lines y = x and y = x - 1). Wait, but the area of a parallelogram is base times height. However, here the base is along the x-axis from 1 to 3, which is length 2, and the height is 1, so area is 2*1=2. So, that checks out. So, Case 2 area is 2.

Case 3: y is the largest side. Similarly, this should be symmetric to Case 2. So, the area would also be 2. Therefore, total area where triangle inequalities hold is Case1 + Case2 + Case3 = 0.5 + 2 + 2 = 4.5. Therefore, the probability is 4.5 / 9 = 0.5. So, 1/2. Wait, but that seems too straightforward. Let me check if there is any overlap between the cases. Because when we do Case1, Case2, Case3, are they mutually exclusive?

Case1: 1 is the largest side, so x ≤ 1 and y ≤ 1.

Case2: x is the largest side, so x ≥ 1 and x ≥ y.

Case3: y is the largest side, so y ≥ 1 and y ≥ x.

So, these cases are mutually exclusive. Because if 1 is the largest, then x and y are both ≤1. If x is the largest, then x ≥1 and x ≥ y, so y ≤x. Similarly for y. So, there is no overlap between the cases. Therefore, adding them up is okay.

But wait, let me check if there are any other cases. For example, if both x and y are greater than 1, but neither is the largest? Wait, no. If both x and y are greater than 1, then the largest side is either x or y. So, those are covered in Case2 or Case3. So, yes, the three cases cover all possibilities.

Therefore, total area is 0.5 + 2 + 2 = 4.5. Therefore, probability is 4.5 / 9 = 0.5. So, 1/2. But wait, that seems high. Let me think again.

Wait, but let me check with an example. Suppose x and y are both 2. Then, sides are 1, 2, 2. That's a valid triangle. If x and y are both 1.5, then 1, 1.5, 1.5 is valid. If x and y are both 0.5, then 1, 0.5, 0.5. Wait, that's not a valid triangle because 0.5 + 0.5 = 1, which is not greater than 1. So, that's on the boundary. So, in Case1, we have x and y between 0 and 1, but with x + y > 1. So, the area there is 0.5. Then, in Cases 2 and 3, we have areas of 2 each. So, total area 4.5. So, 4.5 / 9 = 0.5. But let me think if there's a mistake here.

Wait, but when I considered Case2, I integrated from x=1 to 3, and for each x, y from x - 1 to x. However, when x is between 1 and 2, x - 1 is between 0 and 1, so y ranges from x - 1 to x. However, since y must be in (0,3), but since x is between 1 and 2, x - 1 is between 0 and 1, so y ranges from x - 1 to x, which is within (0,3). Similarly, when x is between 2 and 3, y ranges from x - 1 (which is between 1 and 2) to x (which is between 2 and 3). So, that's okay. So, integrating 1 over x from 1 to 3 gives 2. Similarly for Case3. So, that seems okay.

But let me check another way. Let me think of the entire region where the triangle inequalities hold. The three inequalities are:

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

So, combining these, the region is between the lines y = x - 1 and y = x + 1, and above the line x + y = 1.

So, maybe I can graph this region and compute the area.

First, the square is from (0,0) to (3,3). The lines y = x - 1 and y = x + 1. The line y = x - 1 crosses the square at (1,0) and (3,2). The line y = x + 1 crosses the square at (0,1) and (2,3). The line x + y = 1 crosses the square at (0,1) and (1,0).

So, the region where y > x - 1 and y < x + 1 is between those two lines. However, we also need to be above x + y > 1. So, the intersection of these regions.

So, let me try to find the area where y is between x - 1 and x + 1, and x + y > 1.

But maybe it's easier to split the square into regions.

Alternatively, since we have already considered the three cases and found the total area as 4.5, which is 1/2, but let me check with another approach.

Alternatively, the area where the triangle inequalities hold is the area of the square minus the areas where the triangle inequalities are not satisfied.

The triangle inequalities are not satisfied if any of the three inequalities fail. So, the complement is:

Either 1 + x ≤ y, or 1 + y ≤ x, or x + y ≤ 1.

So, the area where the triangle does not exist is the union of three regions:

A: y ≥ x + 1

B: x ≥ y + 1

C: x + y ≤ 1

So, the total area where triangle does not exist is Area(A) + Area(B) + Area(C) - Area(A∩B) - Area(A∩C) - Area(B∩C) + Area(A∩B∩C)

But since A and B are disjoint? Let me check.

A is y ≥ x + 1, which is above the line y = x + 1.

B is x ≥ y + 1, which is to the right of the line x = y + 1, which is the same as y ≤ x - 1.

So, A and B are disjoint because one is above y = x + 1 and the other is below y = x - 1. So, their intersection is empty. Similarly, A and C: A is y ≥ x + 1, and C is x + y ≤ 1. Let me see if they can overlap. Suppose y ≥ x + 1 and x + y ≤ 1. Then substituting y ≥ x + 1 into x + y ≤ 1 gives x + (x + 1) ≤ 1 => 2x + 1 ≤ 1 => 2x ≤ 0 => x ≤ 0. But x is in (0,3), so x cannot be ≤ 0. Therefore, A and C are disjoint. Similarly, B and C: x ≥ y + 1 and x + y ≤ 1. Substitute x ≥ y + 1 into x + y ≤ 1: (y + 1) + y ≤ 1 => 2y + 1 ≤ 1 => 2y ≤ 0 => y ≤ 0. But y is in (0,3), so again, no overlap. Therefore, A, B, C are pairwise disjoint. Also, their intersection is empty. Therefore, the total area where triangle does not exist is Area(A) + Area(B) + Area(C).

So, compute each:

Area(C): x + y ≤ 1. As before, this is a triangle with area 0.5.

Area(A): y ≥ x + 1. This is the area above the line y = x + 1 within the square. The line y = x + 1 intersects the square at (0,1) and (2,3). So, the region is a triangle with vertices at (0,1), (0,3), and (2,3). Wait, let me check. When x=0, y ranges from 1 to 3. When x=2, y=3. So, the area is a right triangle with base along the y-axis from (0,1) to (0,3), which is length 2, and base along the line x=2 from (2,3) to (2, y). Wait, actually, the region is a triangle with vertices at (0,1), (0,3), and (2,3). The base is from (0,1) to (0,3), which is 2 units, and the height is from x=0 to x=2, which is 2 units. Wait, but the area of a triangle is ½ * base * height. If we consider the base as 2 (from y=1 to y=3 at x=0) and the height as 2 (from x=0 to x=2), then area is ½ * 2 * 2 = 2. Alternatively, integrating over x from 0 to 2, y from x + 1 to 3. So, integral from x=0 to 2 of (3 - (x + 1)) dx = integral from 0 to 2 of (2 - x) dx = [2x - (x²)/2] from 0 to 2 = (4 - 2) - 0 = 2. So, Area(A) is 2.

Similarly, Area(B): x ≥ y + 1. This is symmetric to Area(A). The line x = y + 1 intersects the square at (1,0) and (3,2). The region is a triangle with vertices at (1,0), (3,0), and (3,2). The area is also 2. So, integrating over y from 0 to 2, x from y + 1 to 3. Integral from y=0 to 2 of (3 - (y + 1)) dy = integral from 0 to 2 of (2 - y) dy = [2y - (y²)/2] from 0 to 2 = (4 - 2) - 0 = 2. So, Area(B) is 2.

Therefore, total area where triangle does not exist is 0.5 + 2 + 2 = 4.5. Therefore, the area where triangle exists is total area 9 - 4.5 = 4.5. Therefore, probability is 4.5 / 9 = 0.5. So, 1/2. So, same result as before. Therefore, the probability is 1/2.

But wait, this seems counterintuitive. Let me check with some examples. For example, if x and y are both 1.5, then sides are 1, 1.5, 1.5. That's a valid triangle. If x and y are both 2, sides are 1, 2, 2. Valid. If x and y are both 0.5, then sides are 1, 0.5, 0.5. Not valid. But according to our calculation, the area where triangle exists is 4.5, which is half of the square. So, half the time, the triangle exists. Is that correct?

Alternatively, maybe I can think of the problem in terms of coordinates. The region where triangle exists is between the lines y = x - 1 and y = x + 1, and above y = 1 - x. So, maybe integrating over the square.

Alternatively, let me think of the entire area where the triangle inequalities hold. The three inequalities:

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

So, combining these, the region is the intersection of the three. So, to compute this area, maybe split the square into regions based on the lines.

First, the line x + y = 1 divides the square into two regions. The area above it is where x + y > 1, which we already considered. Then, within that area, we have the band between y = x - 1 and y = x + 1.

So, perhaps the area can be calculated by integrating over the square, but maybe splitting into regions.

Alternatively, since we already did two methods and both gave 4.5, which is 1/2, maybe that is correct. However, let me check with another approach.

Suppose we consider the three inequalities:

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

Let me try to find the area where all three hold.

First, the region where y > x - 1 and y < x + 1 is between the two lines. Then, within that region, we also need x + y > 1.

So, perhaps the area is the area between the two lines, minus the area where x + y ≤ 1 within that band.

But since the band between y = x - 1 and y = x + 1 overlaps with the region x + y ≤ 1?

Wait, earlier we saw that the intersection of A (y ≥ x + 1) and C (x + y ≤ 1) is empty. Similarly for B and C. But what about the band between y = x - 1 and y = x + 1? Does it intersect with x + y ≤ 1?

Let me check. Suppose we have y between x - 1 and x + 1. Then, x + y is between x + (x - 1) = 2x - 1 and x + (x + 1) = 2x + 1. So, for x + y to be ≤ 1, we need 2x - 1 ≤ 1 => 2x ≤ 2 => x ≤ 1. Also, since x + y ≤ 1, and y ≥ x - 1, then x + y ≥ x + x - 1 = 2x - 1. So, if 2x - 1 ≤ 1 and 2x + 1 ≥ something? Wait, maybe this is getting too convoluted.

Alternatively, let me consider the overlap between the band and the region x + y ≤ 1. The region x + y ≤ 1 is the triangle with area 0.5. The band between y = x - 1 and y = x + 1. The overlap between them would be the part of the triangle x + y ≤ 1 that is within the band. But since the band is between y = x - 1 and y = x + 1, and the triangle x + y ≤ 1 is below the line x + y = 1. Let me see if the band intersects this triangle.

For example, take the line y = x - 1. Where does it intersect x + y = 1? Substitute y = x - 1 into x + y = 1: x + x - 1 = 1 => 2x = 2 => x = 1, y = 0. Similarly, the line y = x + 1 intersects x + y = 1 at x + x + 1 = 1 => 2x = 0 => x = 0, y = 1. Therefore, the band between y = x - 1 and y = x + 1 intersects the line x + y = 1 at (1,0) and (0,1). Therefore, the region where x + y ≤ 1 and between the band is exactly the triangle with vertices at (0,1), (1,0), and the overlap? Wait, actually, the region where x + y ≤ 1 and between the band is the triangle itself, because the band includes the lines y = x - 1 and y = x + 1, which intersect x + y = 1 at (1,0) and (0,1). Therefore, the area where x + y ≤ 1 and between the band is exactly the triangle with area 0.5. Therefore, the area where all three inequalities hold is the area between the band (which is 9 - 2 - 2 = 5? Wait, no. Wait, the area between the band is the area between y = x - 1 and y = x + 1. Wait, the entire area between the band is the area of the square minus the areas above y = x + 1 and below y = x - 1. Which is 9 - 2 - 2 = 5. But then, within that area, we have to subtract the area where x + y ≤ 1, which is 0.5. Therefore, the area where all three inequalities hold is 5 - 0.5 = 4.5. Which matches our previous result. Therefore, the area is indeed 4.5, so probability is 4.5 / 9 = 0.5. Therefore, 1/2.

But let me check with coordinates. Suppose I pick x and y uniformly in (0,3). What is the probability that 1, x, y can form a triangle? According to this, it's 1/2. But intuitively, since the triangle inequalities are symmetric, and the area where they hold is half the square, maybe that's correct. However, I want to make sure there isn't a mistake in considering the cases.

Wait, another way: Let me think of the problem as selecting x and y from (0,3). The triangle inequalities are:

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

Which can be rewritten as:

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

So, combining these, the region is defined by:

x + y > 1,

and |x - y| < 1.

So, the area where |x - y| < 1 and x + y > 1.

So, maybe I can compute this area.

The area where |x - y| < 1 is the area between the lines y = x + 1 and y = x - 1, which we already calculated as 5. Then, within that area, we have to intersect with x + y > 1, which subtracts the triangle of area 0.5, giving 4.5. So, same result.

Alternatively, maybe I can compute the area where |x - y| < 1 and x + y > 1 directly.

Let me try integrating. The region |x - y| < 1 is between the two lines. So, for x from 0 to 3, y is between max(0, x - 1) and min(3, x + 1). But we also have x + y > 1. So, for each x, the lower bound of y is max( max(0, x - 1), 1 - x ). Wait, because x + y > 1 implies y > 1 - x. So, combining with the previous bounds.

Therefore, for each x, the lower bound of y is max( max(0, x - 1), 1 - x ), and upper bound is min(3, x + 1). Therefore, the area is the integral over x from 0 to 3 of [ min(3, x + 1) - max( max(0, x - 1), 1 - x ) ] dx.

This seems complicated, but maybe we can split the integral into regions where the expressions change.

First, let's find the ranges of x where max(0, x - 1) and 1 - x change.

The term max(0, x - 1) is 0 when x < 1, and x - 1 when x ≥ 1.

Similarly, 1 - x is positive when x < 1, and negative when x > 1.

Therefore, for x < 1:

max( max(0, x - 1), 1 - x ) = max(0, 1 - x) = 1 - x, since 1 - x > 0.

For x ≥ 1:

max( max(0, x - 1), 1 - x ) = max(x - 1, 1 - x). Since x ≥ 1, 1 - x ≤ 0, so max(x - 1, 1 - x) = x - 1.

Therefore, the lower bound of y is:

For x < 1: 1 - x

For x ≥ 1: x - 1

The upper bound of y is min(3, x + 1). Let's see when x + 1 ≤ 3. That is when x ≤ 2. For x > 2, x + 1 > 3, so upper bound is 3.

Therefore, we can split the integral into three parts:

1. x from 0 to 1:

Lower bound: 1 - x

Upper bound: x + 1 (since x ≤ 1 < 2, so x + 1 ≤ 2 < 3)

So, the integrand is (x + 1) - (1 - x) = 2x

Integral from 0 to 1 of 2x dx = [x²] from 0 to 1 = 1 - 0 = 1

2. x from 1 to 2:

Lower bound: x - 1

Upper bound: x + 1

Integrand: (x + 1) - (x - 1) = 2

Integral from 1 to 2 of 2 dx = 2*(2 - 1) = 2

3. x from 2 to 3:

Lower bound: x - 1

Upper bound: 3

Integrand: 3 - (x - 1) = 4 - x

Integral from 2 to 3 of (4 - x) dx = [4x - (x²)/2] from 2 to 3

At x=3: 12 - 9/2 = 12 - 4.5 = 7.5

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

So, difference is 7.5 - 6 = 1.5

Therefore, total area is 1 + 2 + 1.5 = 4.5. Same result! Therefore, the area is indeed 4.5, so probability is 4.5 / 9 = 0.5. Therefore, the probability is 1/2.

But let me check once more with a different approach. Let me consider the probability that |x - y| < 1 and x + y > 1. Since the problem is symmetric in x and y, maybe we can compute the area where |x - y| < 1 and then subtract the area where |x - y| < 1 and x + y ≤ 1.

Earlier, we found that the area where |x - y| < 1 is 5, and the area where |x - y| < 1 and x + y ≤ 1 is 0.5, so 5 - 0.5 = 4.5. Which again gives the same result.

Alternatively, since the problem is symmetric, maybe we can compute the probability that |x - y| < 1 and x + y > 1 by considering the entire square and subtracting the regions where |x - y| ≥ 1 or x + y ≤ 1. But we already did that.

Therefore, after multiple approaches, I keep getting 4.5 as the area where the triangle exists, leading to probability 1/2. Therefore, I think this is correct.

But just to be thorough, let me check with specific values. For example, if x and y are both in (0,1), then the triangle inequality x + y > 1 must hold. The area of that is 0.5 as calculated. Then, for x and y where one is greater than 1 and the other is less than 1, but such that |x - y| < 1. For example, if x is between 1 and 2, and y is between x - 1 and x. But since x is between 1 and 2, x - 1 is between 0 and 1. So, y ranges from x - 1 to x. For each x, the length is 1, so integrating from 1 to 2 gives 1. Similarly for y between 1 and 2 and x between y - 1 and y, which is another 1. Then, for x between 2 and 3, y ranges from x - 1 to 3. The length is 3 - (x - 1) = 4 - x, which integrates to 1.5. Similarly for y between 2 and 3. Wait, but in our previous calculation, the total area was 1 + 2 + 1.5 = 4.5. Wait, but in this breakdown, the area where |x - y| < 1 is 1 (from 0 to 1) + 2 (from 1 to 2) + 1.5 (from 2 to 3) = 4.5? Wait, no. Wait, earlier when we split the integral into three parts, we had:

For x from 0 to 1: integral was 1

For x from 1 to 2: integral was 2

For x from 2 to 3: integral was 1.5

Total 1 + 2 + 1.5 = 4.5. But that was the area where |x - y| < 1 and x + y > 1. Wait, no, actually, in this case, the integral was already considering x + y > 1. Wait, no, actually, in this approach, we were computing the area where |x - y| < 1 and x + y > 1. Wait, no, actually, in this case, the integral was set up to account for both |x - y| < 1 and x + y > 1. Wait, no, actually, the integral was set up as:

For each x, the lower bound of y is max( max(0, x - 1), 1 - x ), which ensures x + y > 1, and upper bound is min(3, x + 1), which ensures |x - y| < 1. Therefore, the integral is indeed the area where both |x - y| < 1 and x + y > 1. Therefore, the total area is 4.5, which is half of 9. Therefore, probability is 1/2.

Therefore, after multiple methods and checks, I think the probability is indeed 1/2. Therefore, the answer is 1/2, so boxed answer is \boxed{\dfrac{1}{2}}.

**Final Answer**
\boxed{\dfrac{1}{2}}

To determine the probability that a triangle with sides of length 1, $ x $, and $ y $ exists, where $ x $ and $ y $ are selected at random from the interval $ (0, 3) $, we must apply the **triangle inequality**. For three lengths to form a triangle, the sum of any two sides must be **greater than** the third side. This gives us the following three conditions:

1. $ 1 + x > y $
2. $ 1 + y > x $
3. $ x + y > 1 $

These inequalities must all be satisfied simultaneously for the triangle to exist.

---

### Step 1: Define the Sample Space

Since $ x $ and $ y $ are selected from the interval $ (0, 3) $, the total sample space is the square in the $ xy $-plane with vertices at $ (0,0) $, $ (3,0) $, $ (3,3) $, and $ (0,3) $. The area of this square is:

$$
\text{Total Area} = 3 \times 3 = 9
$$

---

### Step 2: Determine the Region Where Triangle Inequalities Hold

We now find the region within this square where all three triangle inequalities are satisfied:

- $ x + y > 1 $: This is the region **above** the line $ x + y = 1 $, which cuts the square at $ (0,1) $ and $ (1,0) $.
- $ y < x + 1 $: This is the region **below** the line $ y = x + 1 $, which intersects the square at $ (0,1) $ and $ (2,3) $.
- $ y > x - 1 $: This is the region **above** the line $ y = x - 1 $, which intersects the square at $ (1,0) $ and $ (3,2) $.

So, the region where all three inequalities hold is the **intersection** of the following:

- Above the line $ x + y = 1 $
- Between the lines $ y = x - 1 $ and $ y = x + 1 $

This region is a **band** around the line $ y = x $, **above** the line $ x + y = 1 $, and within the square $ (0,3) \times (0,3) $.

---

### Step 3: Compute the Area of the Valid Region

We can compute the area of the region where the triangle inequalities hold by subtracting the areas of the regions where the triangle inequalities **fail** from the total area of the square.

#### Regions Where Triangle Inequalities Fail:

1. **Region A**: $ y \geq x + 1 $ — above the line $ y = x + 1 $
2. **Region B**: $ x \geq y + 1 $ — to the right of the line $ x = y + 1 $
3. **Region C**: $ x + y \leq 1 $ — below the line $ x + y = 1 $

These regions are **disjoint**, so we can compute their areas separately and sum them.

- **Area of Region C** (where $ x + y \leq 1 $): This is a triangle with vertices at $ (0,0) $, $ (0,1) $, and $ (1,0) $, so its area is:

$$
\text{Area of C} = \frac{1}{2} \times 1 \times 1 = 0.5
$$

- **Area of Region A** (where $ y \geq x + 1 $): This is a triangle with vertices at $ (0,1) $, $ (0,3) $, and $ (2,3) $, so its area is:

$$
\text{Area of A} = \frac{1}{2} \times 2 \times 2 = 2
$$

- **Area of Region B** (where $ x \geq y + 1 $): This is a triangle with vertices at $ (1,0) $, $ (3,0) $, and $ (3,2) $, so its area is:

$$
\text{Area of B} = \frac{1}{2} \times 2 \times 2 = 2
$$

So, the total area where the triangle inequalities **fail** is:

$$
\text{Area of failure} = 0.5 + 2 + 2 = 4.5
$$

Therefore, the area where the triangle **does exist** is:

$$
\text{Area of success} = 9 - 4.5 = 4.5
$$

---

### Step 4: Compute the Probability

The probability that a triangle with sides 1, $ x $, and $ y $ exists is the ratio of the area where the triangle exists to the total area of the sample space:

$$
\text{Probability} = \frac{4.5}{9} = \frac{1}{2}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that a triangle with sides of length 1, x, and y exists, where x and y are selected at random from the interval (0,3). Hmm, let me think about how to approach this.

First, I remember that for three lengths to form a triangle, they must satisfy the triangle inequality. That is, the sum of any two sides must be greater than the third side. So, for sides 1, x, and y, the following three inequalities must hold:

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

But since x and y are both selected from (0,3), they can be anywhere between 0 and 3. However, since they are lengths, they must be positive, which they are because the interval is (0,3). So, I need to find the region in the coordinate plane where x and y satisfy all three inequalities, and then divide that area by the total area of the possible region (which is a square with side length 3, so area 9) to get the probability.

Wait, but actually, since x and y are selected from (0,3), the total area is indeed 3*3=9. So, the probability will be the area of the region where the triangle inequalities hold divided by 9.

But let me make sure I have the triangle inequalities right. Let me write them again:

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

So, these three inequalities. Let me think about how to visualize this. Since x and y are both between 0 and 3, we can consider the coordinate system with x on the horizontal axis and y on the vertical axis. The entire sample space is the square with vertices at (0,0), (3,0), (3,3), and (0,3).

Now, I need to find the area within this square where all three inequalities hold. Let me tackle each inequality one by one.

Starting with the third inequality: x + y > 1. This is the region above the line x + y = 1. Since x and y are both positive, the line x + y = 1 intersects the square at (1,0) and (0,1). So, the area where x + y > 1 is the entire square except the triangle below the line x + y = 1. The area of that triangle is (1*1)/2 = 0.5, so the area where x + y > 1 is 9 - 0.5 = 8.5. But we also have the other two inequalities to consider.

Now, the first inequality: 1 + x > y. This can be rewritten as y < x + 1. Similarly, the second inequality: 1 + y > x, which is x < y + 1 or y > x - 1.

So, combining these, we have y < x + 1 and y > x - 1. So, between the lines y = x - 1 and y = x + 1. So, the region between these two lines. However, since x and y are between 0 and 3, we need to consider how these lines intersect the square.

Let me sketch mentally. The line y = x + 1: when x=0, y=1; when x=3, y=4. But since y can only go up to 3, the line y = x + 1 intersects the square at (0,1) and (2,3). Similarly, the line y = x - 1: when x=1, y=0; when x=3, y=2. So, this line intersects the square at (1,0) and (3,2).

Therefore, the region between y = x - 1 and y = x + 1 within the square is a band around the line y = x. However, we also have the previous condition x + y > 1. So, the intersection of all three inequalities is the region where:

- y > x - 1
- y < x + 1
- x + y > 1

But since we already have the square from (0,3) x (0,3), maybe I need to find the overlap of these regions.

Alternatively, maybe I can think of the three inequalities together. Let me try to visualize.

First, the region where x + y > 1 is everything except the triangle near the origin. Then, the region between y = x - 1 and y = x + 1 is a band around the line y = x. So, the intersection of these two regions would be the area where both conditions hold.

But perhaps it's better to consider all three inequalities together. Let me try to find the area where all three inequalities hold.

Alternatively, maybe I can think of the triangle inequalities. Since the sides are 1, x, y, the triangle inequalities are:

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

But since x and y are positive, the third inequality x + y > 1 is automatically satisfied if both x and y are greater than 0? Wait, no. For example, if x = 0.2 and y = 0.2, then x + y = 0.4 < 1. So, that's not automatically satisfied. So, we need to include that.

But since x and y are selected from (0,3), we need to consider that. So, the three inequalities are necessary.

Alternatively, maybe I can think of the triangle inequalities as:

The sum of the two smaller sides must be greater than the largest side. So, depending on which side is the largest, the inequalities change.

But since one of the sides is fixed at 1, and the other two are variables x and y. So, depending on whether 1 is the largest side, or x is the largest, or y is the largest.

So, maybe breaking it into cases.

Case 1: 1 is the largest side. Then, we need x + y > 1. But since 1 is the largest, we also need x ≤ 1 and y ≤ 1. Wait, no. If 1 is the largest side, then both x and y must be less than or equal to 1. But since x and y are in (0,3), but if they are less than or equal to 1, then the triangle inequalities would be satisfied if x + y > 1. However, if 1 is the largest side, then the other two sides x and y must satisfy x + y > 1. But if x and y are both less than or equal to 1, then x + y can be as small as approaching 0 (if both x and y approach 0) up to 2 (if both are 1). So, in this case, the condition is x + y > 1.

Case 2: x is the largest side. Then, we need 1 + y > x. Also, since x is the largest, we need x ≥ 1 and x ≥ y. Similarly, Case 3: y is the largest side. Then, 1 + x > y, and y ≥ 1 and y ≥ x.

So, maybe breaking into these three cases would help. Let me try that.

First, let me note that the total area is 9. So, the probability is the area where the triangle inequalities hold divided by 9.

So, let me compute the area for each case.

Case 1: 1 is the largest side. So, x ≤ 1, y ≤ 1, and x + y > 1.

So, in this case, x and y are in [0,1]x[0,1], and x + y > 1. The area of this region is the area of the square [0,1]x[0,1] minus the area where x + y ≤ 1. The area where x + y ≤ 1 is a triangle with area 0.5, so the area for Case 1 is 1*1 - 0.5 = 0.5.

Case 2: x is the largest side. So, x ≥ 1, x ≥ y, and 1 + y > x. Also, since x is in (0,3), but here x ≥ 1 and y ≤ x. Also, since 1 + y > x, which can be rewritten as y > x - 1. So, in this case, x ranges from 1 to 3, and for each x, y ranges from max(0, x - 1) to x. However, since y must also be less than or equal to x (because x is the largest side). But also, since y is in (0,3), but since x is between 1 and 3, and y is between max(0, x - 1) and x. However, we need to check if x - 1 is less than 0. For x between 1 and 2, x - 1 is between 0 and 1, so max(0, x - 1) is x - 1. For x between 2 and 3, x - 1 is between 1 and 2, so max(0, x - 1) is x - 1. Therefore, for all x in [1,3], y ranges from x - 1 to x. However, we also need to ensure that y is less than 3, but since x is at most 3, and y is at most x, which is at most 3, so that's okay.

Therefore, the area for Case 2 is the integral from x=1 to x=3 of [x - (x - 1)] dx = integral from 1 to 3 of 1 dx = 3 - 1 = 2. Wait, that seems too simple. Wait, the length of y for each x is x - (x - 1) = 1. So, integrating 1 over x from 1 to 3 gives 2. So, area is 2.

But wait, let me check. For example, when x is between 1 and 2, y ranges from x - 1 to x. So, for x=1, y ranges from 0 to 1. For x=2, y ranges from 1 to 2. For x=3, y ranges from 2 to 3. So, the region is a parallelogram? Or a trapezoid? Wait, integrating 1 over x from 1 to 3 gives 2, which is the area. But let me visualize. If you plot x from 1 to 3, and for each x, y from x - 1 to x, then the region is a band between the lines y = x - 1 and y = x, from x=1 to x=3. The area between these two lines over that interval is indeed a parallelogram with base 2 (from x=1 to x=3) and height 1 (the vertical distance between the lines y = x and y = x - 1). Wait, but the area of a parallelogram is base times height. However, here the base is along the x-axis from 1 to 3, which is length 2, and the height is 1, so area is 2*1=2. So, that checks out. So, Case 2 area is 2.

Case 3: y is the largest side. Similarly, this should be symmetric to Case 2. So, the area would also be 2. Therefore, total area where triangle inequalities hold is Case1 + Case2 + Case3 = 0.5 + 2 + 2 = 4.5. Therefore, the probability is 4.5 / 9 = 0.5. So, 1/2. Wait, but that seems too straightforward. Let me check if there is any overlap between the cases. Because when we do Case1, Case2, Case3, are they mutually exclusive?

Case1: 1 is the largest side, so x ≤ 1 and y ≤ 1.

Case2: x is the largest side, so x ≥ 1 and x ≥ y.

Case3: y is the largest side, so y ≥ 1 and y ≥ x.

So, these cases are mutually exclusive. Because if 1 is the largest, then x and y are both ≤1. If x is the largest, then x ≥1 and x ≥ y, so y ≤x. Similarly for y. So, there is no overlap between the cases. Therefore, adding them up is okay.

But wait, let me check if there are any other cases. For example, if both x and y are greater than 1, but neither is the largest? Wait, no. If both x and y are greater than 1, then the largest side is either x or y. So, those are covered in Case2 or Case3. So, yes, the three cases cover all possibilities.

Therefore, total area is 0.5 + 2 + 2 = 4.5. Therefore, probability is 4.5 / 9 = 0.5. So, 1/2. But wait, that seems high. Let me think again.

Wait, but let me check with an example. Suppose x and y are both 2. Then, sides are 1, 2, 2. That's a valid triangle. If x and y are both 1.5, then 1, 1.5, 1.5 is valid. If x and y are both 0.5, then 1, 0.5, 0.5. Wait, that's not a valid triangle because 0.5 + 0.5 = 1, which is not greater than 1. So, that's on the boundary. So, in Case1, we have x and y between 0 and 1, but with x + y > 1. So, the area there is 0.5. Then, in Cases 2 and 3, we have areas of 2 each. So, total area 4.5. So, 4.5 / 9 = 0.5. But let me think if there's a mistake here.

Wait, but when I considered Case2, I integrated from x=1 to 3, and for each x, y from x - 1 to x. However, when x is between 1 and 2, x - 1 is between 0 and 1, so y ranges from x - 1 to x. However, since y must be in (0,3), but since x is between 1 and 2, x - 1 is between 0 and 1, so y ranges from x - 1 to x, which is within (0,3). Similarly, when x is between 2 and 3, y ranges from x - 1 (which is between 1 and 2) to x (which is between 2 and 3). So, that's okay. So, integrating 1 over x from 1 to 3 gives 2. Similarly for Case3. So, that seems okay.

But let me check another way. Let me think of the entire region where the triangle inequalities hold. The three inequalities are:

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

So, combining these, the region is between the lines y = x - 1 and y = x + 1, and above the line x + y = 1.

So, maybe I can graph this region and compute the area.

First, the square is from (0,0) to (3,3). The lines y = x - 1 and y = x + 1. The line y = x - 1 crosses the square at (1,0) and (3,2). The line y = x + 1 crosses the square at (0,1) and (2,3). The line x + y = 1 crosses the square at (0,1) and (1,0).

So, the region where y > x - 1 and y < x + 1 is between those two lines. However, we also need to be above x + y > 1. So, the intersection of these regions.

So, let me try to find the area where y is between x - 1 and x + 1, and x + y > 1.

But maybe it's easier to split the square into regions.

Alternatively, since we have already considered the three cases and found the total area as 4.5, which is 1/2, but let me check with another approach.

Alternatively, the area where the triangle inequalities hold is the area of the square minus the areas where the triangle inequalities are not satisfied.

The triangle inequalities are not satisfied if any of the three inequalities fail. So, the complement is:

Either 1 + x ≤ y, or 1 + y ≤ x, or x + y ≤ 1.

So, the area where the triangle does not exist is the union of three regions:

A: y ≥ x + 1

B: x ≥ y + 1

C: x + y ≤ 1

So, the total area where triangle does not exist is Area(A) + Area(B) + Area(C) - Area(A∩B) - Area(A∩C) - Area(B∩C) + Area(A∩B∩C)

But since A and B are disjoint? Let me check.

A is y ≥ x + 1, which is above the line y = x + 1.

B is x ≥ y + 1, which is to the right of the line x = y + 1, which is the same as y ≤ x - 1.

So, A and B are disjoint because one is above y = x + 1 and the other is below y = x - 1. So, their intersection is empty. Similarly, A and C: A is y ≥ x + 1, and C is x + y ≤ 1. Let me see if they can overlap. Suppose y ≥ x + 1 and x + y ≤ 1. Then substituting y ≥ x + 1 into x + y ≤ 1 gives x + (x + 1) ≤ 1 => 2x + 1 ≤ 1 => 2x ≤ 0 => x ≤ 0. But x is in (0,3), so x cannot be ≤ 0. Therefore, A and C are disjoint. Similarly, B and C: x ≥ y + 1 and x + y ≤ 1. Substitute x ≥ y + 1 into x + y ≤ 1: (y + 1) + y ≤ 1 => 2y + 1 ≤ 1 => 2y ≤ 0 => y ≤ 0. But y is in (0,3), so again, no overlap. Therefore, A, B, C are pairwise disjoint. Also, their intersection is empty. Therefore, the total area where triangle does not exist is Area(A) + Area(B) + Area(C).

So, compute each:

Area(C): x + y ≤ 1. As before, this is a triangle with area 0.5.

Area(A): y ≥ x + 1. This is the area above the line y = x + 1 within the square. The line y = x + 1 intersects the square at (0,1) and (2,3). So, the region is a triangle with vertices at (0,1), (0,3), and (2,3). Wait, let me check. When x=0, y ranges from 1 to 3. When x=2, y=3. So, the area is a right triangle with base along the y-axis from (0,1) to (0,3), which is length 2, and base along the line x=2 from (2,3) to (2, y). Wait, actually, the region is a triangle with vertices at (0,1), (0,3), and (2,3). The base is from (0,1) to (0,3), which is 2 units, and the height is from x=0 to x=2, which is 2 units. Wait, but the area of a triangle is ½ * base * height. If we consider the base as 2 (from y=1 to y=3 at x=0) and the height as 2 (from x=0 to x=2), then area is ½ * 2 * 2 = 2. Alternatively, integrating over x from 0 to 2, y from x + 1 to 3. So, integral from x=0 to 2 of (3 - (x + 1)) dx = integral from 0 to 2 of (2 - x) dx = [2x - (x²)/2] from 0 to 2 = (4 - 2) - 0 = 2. So, Area(A) is 2.

Similarly, Area(B): x ≥ y + 1. This is symmetric to Area(A). The line x = y + 1 intersects the square at (1,0) and (3,2). The region is a triangle with vertices at (1,0), (3,0), and (3,2). The area is also 2. So, integrating over y from 0 to 2, x from y + 1 to 3. Integral from y=0 to 2 of (3 - (y + 1)) dy = integral from 0 to 2 of (2 - y) dy = [2y - (y²)/2] from 0 to 2 = (4 - 2) - 0 = 2. So, Area(B) is 2.

Therefore, total area where triangle does not exist is 0.5 + 2 + 2 = 4.5. Therefore, the area where triangle exists is total area 9 - 4.5 = 4.5. Therefore, probability is 4.5 / 9 = 0.5. So, 1/2. So, same result as before. Therefore, the probability is 1/2.

But wait, this seems counterintuitive. Let me check with some examples. For example, if x and y are both 1.5, then sides are 1, 1.5, 1.5. That's a valid triangle. If x and y are both 2, sides are 1, 2, 2. Valid. If x and y are both 0.5, then sides are 1, 0.5, 0.5. Not valid. But according to our calculation, the area where triangle exists is 4.5, which is half of the square. So, half the time, the triangle exists. Is that correct?

Alternatively, maybe I can think of the problem in terms of coordinates. The region where triangle exists is between the lines y = x - 1 and y = x + 1, and above y = 1 - x. So, maybe integrating over the square.

Alternatively, let me think of the entire area where the triangle inequalities hold. The three inequalities:

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

So, combining these, the region is the intersection of the three. So, to compute this area, maybe split the square into regions based on the lines.

First, the line x + y = 1 divides the square into two regions. The area above it is where x + y > 1, which we already considered. Then, within that area, we have the band between y = x - 1 and y = x + 1.

So, perhaps the area can be calculated by integrating over the square, but maybe splitting into regions.

Alternatively, since we already did two methods and both gave 4.5, which is 1/2, maybe that is correct. However, let me check with another approach.

Suppose we consider the three inequalities:

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

Let me try to find the area where all three hold.

First, the region where y > x - 1 and y < x + 1 is between the two lines. Then, within that region, we also need x + y > 1.

So, perhaps the area is the area between the two lines, minus the area where x + y ≤ 1 within that band.

But since the band between y = x - 1 and y = x + 1 overlaps with the region x + y ≤ 1?

Wait, earlier we saw that the intersection of A (y ≥ x + 1) and C (x + y ≤ 1) is empty. Similarly for B and C. But what about the band between y = x - 1 and y = x + 1? Does it intersect with x + y ≤ 1?

Let me check. Suppose we have y between x - 1 and x + 1. Then, x + y is between x + (x - 1) = 2x - 1 and x + (x + 1) = 2x + 1. So, for x + y to be ≤ 1, we need 2x - 1 ≤ 1 => 2x ≤ 2 => x ≤ 1. Also, since x + y ≤ 1, and y ≥ x - 1, then x + y ≥ x + x - 1 = 2x - 1. So, if 2x - 1 ≤ 1 and 2x + 1 ≥ something? Wait, maybe this is getting too convoluted.

Alternatively, let me consider the overlap between the band and the region x + y ≤ 1. The region x + y ≤ 1 is the triangle with area 0.5. The band between y = x - 1 and y = x + 1. The overlap between them would be the part of the triangle x + y ≤ 1 that is within the band. But since the band is between y = x - 1 and y = x + 1, and the triangle x + y ≤ 1 is below the line x + y = 1. Let me see if the band intersects this triangle.

For example, take the line y = x - 1. Where does it intersect x + y = 1? Substitute y = x - 1 into x + y = 1: x + x - 1 = 1 => 2x = 2 => x = 1, y = 0. Similarly, the line y = x + 1 intersects x + y = 1 at x + x + 1 = 1 => 2x = 0 => x = 0, y = 1. Therefore, the band between y = x - 1 and y = x + 1 intersects the line x + y = 1 at (1,0) and (0,1). Therefore, the region where x + y ≤ 1 and between the band is exactly the triangle with vertices at (0,1), (1,0), and the overlap? Wait, actually, the region where x + y ≤ 1 and between the band is the triangle itself, because the band includes the lines y = x - 1 and y = x + 1, which intersect x + y = 1 at (1,0) and (0,1). Therefore, the area where x + y ≤ 1 and between the band is exactly the triangle with area 0.5. Therefore, the area where all three inequalities hold is the area between the band (which is 9 - 2 - 2 = 5? Wait, no. Wait, the area between the band is the area between y = x - 1 and y = x + 1. Wait, the entire area between the band is the area of the square minus the areas above y = x + 1 and below y = x - 1. Which is 9 - 2 - 2 = 5. But then, within that area, we have to subtract the area where x + y ≤ 1, which is 0.5. Therefore, the area where all three inequalities hold is 5 - 0.5 = 4.5. Which matches our previous result. Therefore, the area is indeed 4.5, so probability is 4.5 / 9 = 0.5. Therefore, 1/2.

But let me check with coordinates. Suppose I pick x and y uniformly in (0,3). What is the probability that 1, x, y can form a triangle? According to this, it's 1/2. But intuitively, since the triangle inequalities are symmetric, and the area where they hold is half the square, maybe that's correct. However, I want to make sure there isn't a mistake in considering the cases.

Wait, another way: Let me think of the problem as selecting x and y from (0,3). The triangle inequalities are:

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

Which can be rewritten as:

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

So, combining these, the region is defined by:

x + y > 1,

and |x - y| < 1.

So, the area where |x - y| < 1 and x + y > 1.

So, maybe I can compute this area.

The area where |x - y| < 1 is the area between the lines y = x + 1 and y = x - 1, which we already calculated as 5. Then, within that area, we have to intersect with x + y > 1, which subtracts the triangle of area 0.5, giving 4.5. So, same result.

Alternatively, maybe I can compute the area where |x - y| < 1 and x + y > 1 directly.

Let me try integrating. The region |x - y| < 1 is between the two lines. So, for x from 0 to 3, y is between max(0, x - 1) and min(3, x + 1). But we also have x + y > 1. So, for each x, the lower bound of y is max( max(0, x - 1), 1 - x ). Wait, because x + y > 1 implies y > 1 - x. So, combining with the previous bounds.

Therefore, for each x, the lower bound of y is max( max(0, x - 1), 1 - x ), and upper bound is min(3, x + 1). Therefore, the area is the integral over x from 0 to 3 of [ min(3, x + 1) - max( max(0, x - 1), 1 - x ) ] dx.

This seems complicated, but maybe we can split the integral into regions where the expressions change.

First, let's find the ranges of x where max(0, x - 1) and 1 - x change.

The term max(0, x - 1) is 0 when x < 1, and x - 1 when x ≥ 1.

Similarly, 1 - x is positive when x < 1, and negative when x > 1.

Therefore, for x < 1:

max( max(0, x - 1), 1 - x ) = max(0, 1 - x) = 1 - x, since 1 - x > 0.

For x ≥ 1:

max( max(0, x - 1), 1 - x ) = max(x - 1, 1 - x). Since x ≥ 1, 1 - x ≤ 0, so max(x - 1, 1 - x) = x - 1.

Therefore, the lower bound of y is:

For x < 1: 1 - x

For x ≥ 1: x - 1

The upper bound of y is min(3, x + 1). Let's see when x + 1 ≤ 3. That is when x ≤ 2. For x > 2, x + 1 > 3, so upper bound is 3.

Therefore, we can split the integral into three parts:

1. x from 0 to 1:

Lower bound: 1 - x

Upper bound: x + 1 (since x ≤ 1 < 2, so x + 1 ≤ 2 < 3)

So, the integrand is (x + 1) - (1 - x) = 2x

Integral from 0 to 1 of 2x dx = [x²] from 0 to 1 = 1 - 0 = 1

2. x from 1 to 2:

Lower bound: x - 1

Upper bound: x + 1

Integrand: (x + 1) - (x - 1) = 2

Integral from 1 to 2 of 2 dx = 2*(2 - 1) = 2

3. x from 2 to 3:

Lower bound: x - 1

Upper bound: 3

Integrand: 3 - (x - 1) = 4 - x

Integral from 2 to 3 of (4 - x) dx = [4x - (x²)/2] from 2 to 3

At x=3: 12 - 9/2 = 12 - 4.5 = 7.5

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

So, difference is 7.5 - 6 = 1.5

Therefore, total area is 1 + 2 + 1.5 = 4.5. Same result! Therefore, the area is indeed 4.5, so probability is 4.5 / 9 = 0.5. Therefore, the probability is 1/2.

But let me check once more with a different approach. Let me consider the probability that |x - y| < 1 and x + y > 1. Since the problem is symmetric in x and y, maybe we can compute the area where |x - y| < 1 and then subtract the area where |x - y| < 1 and x + y ≤ 1.

Earlier, we found that the area where |x - y| < 1 is 5, and the area where |x - y| < 1 and x + y ≤ 1 is 0.5, so 5 - 0.5 = 4.5. Which again gives the same result.

Alternatively, since the problem is symmetric, maybe we can compute the probability that |x - y| < 1 and x + y > 1 by considering the entire square and subtracting the regions where |x - y| ≥ 1 or x + y ≤ 1. But we already did that.

Therefore, after multiple approaches, I keep getting 4.5 as the area where the triangle exists, leading to probability 1/2. Therefore, I think this is correct.

But just to be thorough, let me check with specific values. For example, if x and y are both in (0,1), then the triangle inequality x + y > 1 must hold. The area of that is 0.5 as calculated. Then, for x and y where one is greater than 1 and the other is less than 1, but such that |x - y| < 1. For example, if x is between 1 and 2, and y is between x - 1 and x. But since x is between 1 and 2, x - 1 is between 0 and 1. So, y ranges from x - 1 to x. For each x, the length is 1, so integrating from 1 to 2 gives 1. Similarly for y between 1 and 2 and x between y - 1 and y, which is another 1. Then, for x between 2 and 3, y ranges from x - 1 to 3. The length is 3 - (x - 1) = 4 - x, which integrates to 1.5. Similarly for y between 2 and 3. Wait, but in our previous calculation, the total area was 1 + 2 + 1.5 = 4.5. Wait, but in this breakdown, the area where |x - y| < 1 is 1 (from 0 to 1) + 2 (from 1 to 2) + 1.5 (from 2 to 3) = 4.5? Wait, no. Wait, earlier when we split the integral into three parts, we had:

For x from 0 to 1: integral was 1

For x from 1 to 2: integral was 2

For x from 2 to 3: integral was 1.5

Total 1 + 2 + 1.5 = 4.5. But that was the area where |x - y| < 1 and x + y > 1. Wait, no, actually, in this case, the integral was already considering x + y > 1. Wait, no, actually, in this approach, we were computing the area where |x - y| < 1 and x + y > 1. Wait, no, actually, in this case, the integral was set up to account for both |x - y| < 1 and x + y > 1. Wait, no, actually, the integral was set up as:

For each x, the lower bound of y is max( max(0, x - 1), 1 - x ), which ensures x + y > 1, and upper bound is min(3, x + 1), which ensures |x - y| < 1. Therefore, the integral is indeed the area where both |x - y| < 1 and x + y > 1. Therefore, the total area is 4.5, which is half of 9. Therefore, probability is 1/2.

Therefore, after multiple methods and checks, I think the probability is indeed 1/2. Therefore, the answer is 1/2, so boxed answer is \boxed{\dfrac{1}{2}}.

**Final Answer**
\boxed{\dfrac{1}{2}}

To determine the probability that a triangle with sides of length 1, $ x $, and $ y $ exists, where $ x $ and $ y $ are selected at random from the interval $ (0, 3) $, we must apply the **triangle inequality**. For three lengths to form a triangle, the sum of any two sides must be **greater than** the third side. This gives us the following three conditions:

1. $ 1 + x > y $
2. $ 1 + y > x $
3. $ x + y > 1 $

These inequalities must all be satisfied simultaneously for the triangle to exist.

---

### Step 1: Define the Sample Space

Since $ x $ and $ y $ are selected from the interval $ (0, 3) $, the total sample space is the square in the $ xy $-plane with vertices at $ (0,0) $, $ (3,0) $, $ (3,3) $, and $ (0,3) $. The area of this square is:

$$
\text{Total Area} = 3 \times 3 = 9
$$

---

### Step 2: Determine the Region Where Triangle Inequalities Hold

We now find the region within this square where all three triangle inequalities are satisfied:

- $ x + y > 1 $: This is the region **above** the line $ x + y = 1 $, which cuts the square at $ (0,1) $ and $ (1,0) $.
- $ y < x + 1 $: This is the region **below** the line $ y = x + 1 $, which intersects the square at $ (0,1) $ and $ (2,3) $.
- $ y > x - 1 $: This is the region **above** the line $ y = x - 1 $, which intersects the square at $ (1,0) $ and $ (3,2) $.

So, the region where all three inequalities hold is the **intersection** of the following:

- Above the line $ x + y = 1 $
- Between the lines $ y = x - 1 $ and $ y = x + 1 $

This region is a **band** around the line $ y = x $, **above** the line $ x + y = 1 $, and within the square $ (0,3) \times (0,3) $.

---

### Step 3: Compute the Area of the Valid Region

We can compute the area of the region where the triangle inequalities hold by subtracting the areas of the regions where the triangle inequalities **fail** from the total area of the square.

#### Regions Where Triangle Inequalities Fail:

1. **Region A**: $ y \geq x + 1 $ — above the line $ y = x + 1 $
2. **Region B**: $ x \geq y + 1 $ — to the right of the line $ x = y + 1 $
3. **Region C**: $ x + y \leq 1 $ — below the line $ x + y = 1 $

These regions are **disjoint**, so we can compute their areas separately and sum them.

- **Area of Region C** (where $ x + y \leq 1 $): This is a triangle with vertices at $ (0,0) $, $ (0,1) $, and $ (1,0) $, so its area is:

$$
\text{Area of C} = \frac{1}{2} \times 1 \times 1 = 0.5
$$

- **Area of Region A** (where $ y \geq x + 1 $): This is a triangle with vertices at $ (0,1) $, $ (0,3) $, and $ (2,3) $, so its area is:

$$
\text{Area of A} = \frac{1}{2} \times 2 \times 2 = 2
$$

- **Area of Region B** (where $ x \geq y + 1 $): This is a triangle with vertices at $ (1,0) $, $ (3,0) $, and $ (3,2) $, so its area is:

$$
\text{Area of B} = \frac{1}{2} \times 2 \times 2 = 2
$$

So, the total area where the triangle inequalities **fail** is:

$$
\text{Area of failure} = 0.5 + 2 + 2 = 4.5
$$

Therefore, the area where the triangle **does exist** is:

$$
\text{Area of success} = 9 - 4.5 = 4.5
$$

---

### Step 4: Compute the Probability

The probability that a triangle with sides 1, $ x $, and $ y $ exists is the ratio of the area where the triangle exists to the total area of the sample space:

$$
\text{Probability} = \frac{4.5}{9} = \frac{1}{2}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{2}}
$$

Solution:
If a triangle with sides of length 1, $x$, and $y$ exists, the triangle inequality must be satisfied, which states that $x+y>1$, $1+x>y$, and $1+y>x$.  We can draw a plane with $x$ and $y$ axes and shade in the area where all of these inequalities are satisfied.

[asy]
draw((0,0)--(3,0)--(3,3)--(0,3));
draw((0,0)--(0,3));
label("$x$",(3,0),S);
label("$y$",(0,3),W);
fill((1,0)--(3,2)--(3,3)--(2,3)--(0,1)--cycle,gray(.7));
draw((1,-.1)--(1,.1));
draw((2,-.1)--(2,.1));
draw((.1,1)--(-.1,1));
draw((.1,2)--(-.1,2));

draw((1,0)--(0,1));
draw((1,0)--(3,2));
draw((0,1)--(2,3));

[/asy]

The total area of the square is $3^2=9$.  The area of the unshaded region is $2^2+1/2=9/2$.  Thus, the shaded area is $9/2$ and the probability that such a triangle exists is $(9/2)/9=\boxed{\frac{1}{2}}$.