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) = x - 1. For x between 2 and 3, x - 1 is between 1 and 2, so max(0, x - 1) = 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. But wait, that seems too straightforward. Let me check if there is any overlap between the cases. Wait, in Case 1, we have x ≤ 1 and y ≤ 1. In Cases 2 and 3, we have x ≥ 1 or y ≥ 1. So, there is no overlap between the cases. Therefore, adding them is okay.

But wait, let me think again. Suppose x and y are both greater than 1. Then, neither x nor y is the largest side? No, if both are greater than 1, then the largest side could be x or y. So, in that case, they would be covered in Case 2 or Case 3. However, if both x and y are greater than 1, then the largest side is either x or y, so we need to check whether 1 + the other side is greater than the largest side. So, for example, if x and y are both greater than 1, then if x is the largest, then 1 + y > x must hold, and similarly for y. So, the Cases 2 and 3 cover those scenarios. However, when both x and y are greater than 1, they are included in either Case 2 or Case 3, depending on which is larger. So, the total area is indeed Case1 + Case2 + Case3.

But let me check with an example. Suppose x=2 and y=2. Then, the sides are 1, 2, 2. The triangle inequalities: 1 + 2 > 2 (3 > 2, yes), 1 + 2 > 2 (same), and 2 + 2 > 1 (yes). So, that's valid. So, this point is included in Case 2 or 3? Since x and y are both greater than 1, and since they are equal, it's included in both? Wait, but in our Cases 2 and 3, we considered x being the largest or y being the largest. However, if x and y are equal, then both conditions would be satisfied? Wait, but in our Case 2, we considered x being the largest side, which would require x ≥ y. Similarly, Case 3 requires y ≥ x. So, if x = y, then it's included in both? But since we are integrating over x from 1 to 3 and y from x - 1 to x, and similarly for Case 3, there might be some overlap? Wait, but in reality, when x = y, it's only counted once? Wait, no. Because in Case 2, we have x ≥ y, so when x = y, it's included in Case 2. Similarly, in Case 3, y ≥ x, so when x = y, it's included in Case 3. Therefore, there is an overlap along the line x = y. However, since it's a line, which has area zero, it doesn't affect the total area. Therefore, adding Case 2 and Case 3 is okay.

But let me check if the total area is indeed 0.5 + 2 + 2 = 4.5. Then, probability is 4.5 / 9 = 0.5. But let me think again. Suppose I consider the entire square (0,3)x(0,3). The area where triangle inequalities hold is 4.5? Let me check with another approach.

Alternatively, maybe I can consider the three inequalities:

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 these two lines. However, we also need to be above x + y > 1. So, the intersection of these regions.

So, let me try to visualize. The area between the two lines y = x - 1 and y = x + 1 is a band around the line y = x. However, we also need to be above x + y = 1. So, the region is the band between the two lines, but above the line x + y = 1.

So, perhaps the area can be calculated by subtracting the regions that are below x + y = 1 from the band between the two lines.

But maybe it's easier to split the area into parts.

Alternatively, since we have already calculated using cases and got 4.5, which is 1/2, but let me check with another method.

Let me consider integrating over the square. The total area is 9. The area where triangle inequalities hold is the set of (x,y) such that:

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

So, let me try to find the area where all three hold.

First, let's consider the region where y < x + 1 and y > x - 1. This is the band between the two lines. Then, within that band, we need to also satisfy x + y > 1.

So, perhaps we can split the integration into regions where x + y > 1 is automatically satisfied or not.

Alternatively, since the band between y = x - 1 and y = x + 1 overlaps with the region x + y > 1.

But maybe it's better to find the area where all three inequalities hold by integrating.

Alternatively, since we have already considered the cases and found 4.5, but let me check with coordinates.

Let me think of the entire square. The area where triangle inequalities hold is the area where:

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

So, let me consider the overlap of these three.

First, the region between y = x - 1 and y = x + 1. Let me find where this band overlaps with x + y > 1.

For x between 0 and 1:

The line x + y = 1 intersects the band. Let me see. For x between 0 and 1, the lower bound of y is x - 1, which is negative, but since y is in (0,3), the lower bound is actually 0. The upper bound is y = x + 1. However, we also have the condition x + y > 1. So, for x between 0 and 1, the lower bound for y is max(0, 1 - x). Wait, because x + y > 1 implies y > 1 - x. So, for x between 0 and 1, 1 - x ranges from 1 to 0. Therefore, the lower bound for y is max(0, 1 - x) = 1 - x (since 1 - x is positive for x < 1). The upper bound is y < x + 1. However, since y is also less than 3, but for x between 0 and 1, x + 1 is between 1 and 2, so y is between 1 - x and x + 1. Therefore, the area for x between 0 and 1 is the integral from x=0 to x=1 of [ (x + 1) - (1 - x) ] dx = integral of 2x dx from 0 to 1? Wait, (x + 1) - (1 - x) = x + 1 - 1 + x = 2x. Therefore, integral from 0 to 1 of 2x dx = [x²] from 0 to 1 = 1 - 0 = 1.

Then, for x between 1 and 2:

The lower bound for y is max(0, 1 - x). But 1 - x is negative here, so lower bound is 0. However, we also have the band between y = x - 1 and y = x + 1. So, for x between 1 and 2, the lower bound is y > x - 1 (since x - 1 is between 0 and 1), and upper bound is y < x + 1. However, we also have the condition x + y > 1. But since x ≥ 1 and y ≥ 0, x + y ≥ 1 + 0 = 1. Therefore, x + y > 1 is automatically satisfied for x ≥ 1 and y ≥ 0. Wait, but if x is 1 and y is 0, then x + y = 1, which is not greater than 1. So, actually, for x between 1 and 2, the condition x + y > 1 is equivalent to y > 1 - x. But since x ≥ 1, 1 - x ≤ 0. Since y ≥ 0, then y > 1 - x is always true. Therefore, for x between 1 and 2, the condition x + y > 1 is automatically satisfied. Therefore, the area for x between 1 and 2 is the integral from x=1 to x=2 of [ (x + 1) - (x - 1) ] dx = integral of 2 dx from 1 to 2 = 2*(2 - 1) = 2.

Then, for x between 2 and 3:

The lower bound for y is still y > x - 1 (since x - 1 is between 1 and 2). The upper bound is y < x + 1. However, since y is at most 3, we need to check if x + 1 exceeds 3. For x between 2 and 3, x + 1 is between 3 and 4. Therefore, the upper bound for y is 3. Therefore, the area for x between 2 and 3 is integral from x=2 to x=3 of [3 - (x - 1)] dx = integral of (4 - x) dx from 2 to 3.

Calculating that integral:

Integral of 4 - x dx = 4x - (x²)/2 evaluated from 2 to 3.

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 is 7.5 - 6 = 1.5

Therefore, the total area is the sum of the three parts:

For x between 0 and 1: 1

For x between 1 and 2: 2

For x between 2 and 3: 1.5

Total area = 1 + 2 + 1.5 = 4.5

Which matches the previous result. Therefore, the area where the triangle inequalities hold is 4.5, so the probability is 4.5 / 9 = 0.5, which is 1/2. Therefore, the probability is 1/2.

Wait, but let me check if there are any mistakes here. For example, when x is between 2 and 3, the upper bound for y is 3, not x + 1. So, the calculation for that part was correct. And the integral from 2 to 3 of (4 - x) dx is indeed 1.5. So, adding up gives 4.5. Therefore, probability is 4.5 / 9 = 0.5. So, 1/2.

But let me think again. Suppose I pick x and y such that they are both greater than 1. For example, x=2, y=2. Then, the sides are 1, 2, 2. Which is a valid triangle. Similarly, if x=3, y=3, then sides are 1, 3, 3. Which is valid because 1 + 3 > 3 (4 > 3). So, that's okay. However, if x=3 and y=1, then sides are 1, 3, 1. Then, 1 + 1 = 2 < 3, which is invalid. But according to our calculation, when x=3, y ranges from 2 to 3. So, y=1 is not included. Therefore, our calculation seems to exclude such cases. So, seems okay.

Wait, but in the case where x=3 and y=1, which is in the square, but since y=1 is less than x - 1 = 3 - 1 = 2, so y=1 is below the lower bound of the band. Therefore, it's excluded. Therefore, our calculation seems correct.

Another check: when x=0.5, y=0.5. Then, sides are 1, 0.5, 0.5. Then, 0.5 + 0.5 = 1, which is not greater than 1. Therefore, invalid. However, according to our calculation, for x between 0 and 1, the lower bound for y is 1 - x. For x=0.5, lower bound is 1 - 0.5 = 0.5. Therefore, y must be greater than 0.5. Therefore, y=0.5 is excluded. Therefore, the area where x + y > 1 is considered. Therefore, the point (0.5, 0.5) is excluded, which is correct. Therefore, our calculation seems correct.

Therefore, the probability is 4.5 / 9 = 1/2. Therefore, the answer is 1/2. But wait, let me check with another approach.

Alternatively, since the problem is symmetric in x and y, maybe we can compute the area where 1 + x > y, 1 + y > x, and x + y > 1. Which is equivalent to the area where |x - y| < 1 and x + y > 1. Wait, because 1 + x > y and 1 + y > x can be rewritten as y < x + 1 and x < y + 1, which is equivalent to |x - y| < 1. Therefore, the region is |x - y| < 1 and x + y > 1. Therefore, the area is the area between the lines y = x + 1 and y = x - 1, and above the line x + y = 1.

Which is exactly what we calculated earlier. Therefore, the area is 4.5, so probability is 1/2.

But let me check with coordinates. For example, if I consider the entire square, the area where |x - y| < 1 is the area between the lines y = x + 1 and y = x - 1. Which is the area of the square minus the areas of the two triangles where |x - y| ≥ 1. Each of those triangles has base and height of 2 (since from x=0 to 2, y=0 to 2, etc.), but wait, actually, for the square (0,3)x(0,3), the area where |x - y| ≥ 1 is two triangles. Each triangle has base and height of 3 - 1 = 2? Wait, no.

Wait, the area where |x - y| ≥ 1 is the area above y = x + 1 and below y = x - 1. For the square (0,3)x(0,3), the area above y = x + 1 is a triangle with vertices at (0,1), (0,3), and (2,3). Similarly, the area below y = x - 1 is a triangle with vertices at (1,0), (3,0), and (3,2). Each of these triangles has base 2 and height 2, so area is (2*2)/2 = 2. Therefore, total area where |x - y| ≥ 1 is 2 + 2 = 4. Therefore, the area where |x - y| < 1 is 9 - 4 = 5. However, we also need to consider the condition x + y > 1. Therefore, the area where |x - y| < 1 and x + y > 1 is equal to the area where |x - y| < 1 minus the area where |x - y| < 1 and x + y ≤ 1.

So, let me compute that. The area where |x - y| < 1 and x + y ≤ 1. This is the region that is between the lines y = x - 1 and y = x + 1, and below the line x + y = 1. But since x and y are positive, and x + y ≤ 1, this region is a small area near the origin.

Let me visualize. The line x + y = 1 intersects the band |x - y| < 1. The intersection of these two regions is the area where x + y ≤ 1 and |x - y| < 1. Which is the region bounded by x + y ≤ 1, y < x + 1, and y > x - 1. But since x + y ≤ 1 and y > x - 1, let me see.

For x between 0 and 1, y between max(0, x - 1) and min(1 - x, x + 1). But since x - 1 is negative, max(0, x - 1) = 0. And 1 - x is between 0 and 1, and x + 1 is between 1 and 2. Therefore, the upper bound is 1 - x. Therefore, the area is the integral from x=0 to x=1 of (1 - x - 0) dx = integral of (1 - x) dx from 0 to 1 = [x - (x²)/2] from 0 to 1 = (1 - 0.5) - 0 = 0.5. Therefore, the area where |x - y| < 1 and x + y ≤ 1 is 0.5. Therefore, the area where |x - y| < 1 and x + y > 1 is 5 - 0.5 = 4.5, which matches our previous result. Therefore, the probability is 4.5 / 9 = 0.5. Therefore, the probability is 1/2.

But wait, this seems to contradict my initial intuition. Let me check with another example. Suppose I pick x and y uniformly from (0,3). What is the probability that they can form a triangle with 1? If the answer is 1/2, that seems plausible? Let me think of another way.

Alternatively, maybe I can use geometric probability. The total area is 9. The area where triangle inequalities hold is 4.5. Therefore, probability is 1/2. But let me check with specific numbers.

For example, if x and y are both in (0,1), then the triangle inequalities require x + y > 1. The area of that is 0.5 as calculated. Then, for x and y in (1,3), the triangle inequalities are automatically satisfied? No, not necessarily. For example, if x=2 and y=2, it's okay. If x=2 and y=1, then sides are 1, 2, 1. Which is invalid. But according to our previous calculation, when x=2, y must be between 1 and 3. Wait, for x=2, the lower bound for y is x - 1 = 1, so y must be between 1 and 3. Therefore, y=1 is excluded. Therefore, in the region x between 1 and 3, y is between max(0, x - 1) and min(3, x + 1). But since x is between 1 and 3, max(0, x - 1) is x - 1, and min(3, x + 1) is 3 when x ≥ 2, and x + 1 when x < 2. Therefore, the area for x between 1 and 2 is integral of 2 dx from 1 to 2 = 2, and for x between 2 and 3, integral of (4 - x) dx from 2 to 3 = 1.5. So, total area is 2 + 1.5 = 3.5. Then, adding the area for x and y in (0,1) which is 0.5, total area is 3.5 + 0.5 = 4.0? Wait, no. Wait, earlier we had 4.5. Wait, maybe I miscalculated.

Wait, no. Wait, when we split into cases, Case 1 was 0.5, Case 2 was 2, Case 3 was 2, total 4.5. But when I considered the area where |x - y| < 1 and x + y > 1, it was 4.5. So, where is the discrepancy?

Wait, when I considered x between 0 and 1, the area was 1. Then, x between 1 and 2, area 2, and x between 2 and 3, area 1.5. Total 1 + 2 + 1.5 = 4.5. But when I considered splitting into regions where x and y are both in (0,1), which is 0.5, and then the rest, which is 4.5 - 0.5 = 4.0? Wait, no. Wait, the area where |x - y| < 1 and x + y > 1 is 4.5. Which includes all regions where x and y are in (0,3) with |x - y| < 1 and x + y > 1. So, that includes the area where both x and y are in (0,1) (but with x + y > 1), which is 0.5, and the rest. So, the total is 4.5. Therefore, the previous calculation where I thought of splitting into cases was correct.

Therefore, the answer is 1/2. However, I need to make sure that there isn't a mistake in the integration steps. Let me check again.

For x between 0 and 1:

The area is integral from 0 to 1 of [ (x + 1) - (1 - x) ] dx = integral of 2x dx from 0 to 1 = [x²] from 0 to 1 = 1. Correct.

For x between 1 and 2:

Integral from 1 to 2 of [ (x + 1) - (x - 1) ] dx = integral of 2 dx from 1 to 2 = 2*(2 - 1) = 2. Correct.

For x between 2 and 3:

Integral from 2 to 3 of [3 - (x - 1)] dx = integral of (4 - x) dx from 2 to 3 = [4x - (x²)/2] from 2 to 3 = (12 - 4.5) - (8 - 2) = 7.5 - 6 = 1.5. Correct.

Total area: 1 + 2 + 1.5 = 4.5. Correct.

Therefore, probability is 4.5 / 9 = 0.5. Therefore, the probability is 1/2.

But let me check with another approach. Let me consider the probability that 1, x, y can form a triangle. Which is equivalent to the probability that x + y > 1, |x - y| < 1. Which is the same as the area we calculated. So, since the answer is 1/2, maybe that's correct.

Alternatively, maybe I can think of the problem as follows: For two numbers x and y selected uniformly from (0,3), what is the probability that they satisfy |x - y| < 1 and x + y > 1. Which is the same as the area we calculated. Therefore, 4.5 / 9 = 1/2.

Alternatively, maybe I can think of the problem in terms of the unit square, but scaled. However, since the interval is (0,3), it's a bit different.

Alternatively, maybe I can use geometric probability with coordinates. But I think the previous calculations are consistent.

Therefore, I think the answer is 1/2. However, I recall that sometimes when dealing with triangle inequalities, there might be some edge cases or miscalculations. Let me check with specific values.

For example, take x=1.5 and y=1.5. Then, sides are 1, 1.5, 1.5. The triangle inequalities: 1 + 1.5 > 1.5 (yes), 1 + 1.5 > 1.5 (yes), 1.5 + 1.5 > 1 (yes). Valid.

Take x=1.5 and y=0.5. Then, sides are 1, 1.5, 0.5. Check triangle inequalities: 1 + 0.5 = 1.5, which is not greater than 1.5. Therefore, invalid. However, according to our calculation, for x=1.5, y must be between 0.5 and 2.5. Wait, y=0.5 is the lower bound? Wait, for x=1.5, the lower bound is x - 1 = 0.5. Therefore, y must be greater than 0.5. Therefore, y=0.5 is excluded. Therefore, the point (1.5, 0.5) is excluded. Therefore, the calculation is correct.

Another example: x=0.6, y=0.6. Then, sides are 1, 0.6, 0.6. Check triangle inequality: 0.6 + 0.6 = 1.2 > 1. So, valid. However, according to our calculation, for x=0.6, y must be between 1 - 0.6 = 0.4 and 0.6 + 1 = 1.6. Therefore, y=0.6 is within that range. Therefore, included. Therefore, valid. So, correct.

Another example: x=0.4, y=0.4. Then, sides are 1, 0.4, 0.4. 0.4 + 0.4 = 0.8 < 1. Invalid. According to our calculation, for x=0.4, the lower bound for y is 1 - 0.4 = 0.6. Therefore, y must be greater than 0.6. Therefore, y=0.4 is excluded. Correct.

Therefore, all these examples check out. Therefore, I think the answer is indeed 1/2.

But wait, I just thought of something. The original problem says "two numbers, x and y are selected at random from the interval (0,3)". So, is it continuous uniform distribution? Yes, I assume so. Therefore, the area calculation is correct.

Therefore, the probability is 1/2. Therefore, the 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 any three sides $ a, b, c $ to form a triangle, the following must hold:

$$
a + b > c, \quad a + c > b, \quad b + c > a
$$

In our case, the sides are $ 1, x, y $, so the triangle inequalities become:

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

We are selecting $ x $ and $ y $ uniformly at random from the interval $ (0, 3) $, so the sample space is the square $ (0, 3) \times (0, 3) $, with area $ 9 $.

---

### Step 1: Geometric Interpretation

We interpret the conditions geometrically in the $ xy $-plane:

- The inequality $ x + y > 1 $ defines the region **above** the line $ x + y = 1 $.
- The inequalities $ 1 + x > y $ and $ 1 + y > x $ can be rewritten as:
  - $ y < x + 1 $
  - $ y > x - 1 $

These define the region **between** the lines $ y = x + 1 $ and $ y = x - 1 $, i.e., the **band** around the line $ y = x $.

So, the region where all three triangle inequalities hold is the **intersection** of:
- The band between $ y = x - 1 $ and $ y = x + 1 $
- The region above the line $ x + y = 1 $

This region lies within the square $ (0, 3) \times (0, 3) $, and we are to compute its area.

---

### Step 2: Compute the Area of the Valid Region

We divide the square into three parts based on the value of $ x $:

#### **Case 1: $ x \in [0, 1] $**

- The lower bound for $ y $ is $ \max(0, 1 - x) = 1 - x $
- The upper bound for $ y $ is $ \min(3, x + 1) = x + 1 $
- So, the vertical slice at each $ x $ has length $ (x + 1) - (1 - x) = 2x $

$$
\text{Area} = \int_0^1 2x \, dx = \left[ x^2 \right]_0^1 = 1
$$

#### **Case 2: $ x \in [1, 2] $**

- The lower bound for $ y $ is $ x - 1 $
- The upper bound for $ y $ is $ x + 1 $
- So, the vertical slice at each $ x $ has length $ (x + 1) - (x - 1) = 2 $

$$
\text{Area} = \int_1^2 2 \, dx = 2(2 - 1) = 2
$$

#### **Case 3: $ x \in [2, 3] $**

- The lower bound for $ y $ is $ x - 1 $
- The upper bound for $ y $ is $ 3 $ (since $ x + 1 > 3 $)
- So, the vertical slice at each $ x $ has length $ 3 - (x - 1) = 4 - x $

$$
\text{Area} = \int_2^3 (4 - x) \, dx = \left[ 4x - \frac{x^2}{2} \right]_2^3 = (12 - 4.5) - (8 - 2) = 7.5 - 6 = 1.5
$$

---

### Step 3: Total Valid Area

$$
\text{Total Area} = 1 + 2 + 1.5 = 4.5
$$

---

### Step 4: Compute the Probability

The total area of the square is $ 9 $, so the probability that a triangle with sides $ 1, x, y $ exists is:

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

---

### 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) = x - 1. For x between 2 and 3, x - 1 is between 1 and 2, so max(0, x - 1) = 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. But wait, that seems too straightforward. Let me check if there is any overlap between the cases. Wait, in Case 1, we have x ≤ 1 and y ≤ 1. In Cases 2 and 3, we have x ≥ 1 or y ≥ 1. So, there is no overlap between the cases. Therefore, adding them is okay.

But wait, let me think again. Suppose x and y are both greater than 1. Then, neither x nor y is the largest side? No, if both are greater than 1, then the largest side could be x or y. So, in that case, they would be covered in Case 2 or Case 3. However, if both x and y are greater than 1, then the largest side is either x or y, so we need to check whether 1 + the other side is greater than the largest side. So, for example, if x and y are both greater than 1, then if x is the largest, then 1 + y > x must hold, and similarly for y. So, the Cases 2 and 3 cover those scenarios. However, when both x and y are greater than 1, they are included in either Case 2 or Case 3, depending on which is larger. So, the total area is indeed Case1 + Case2 + Case3.

But let me check with an example. Suppose x=2 and y=2. Then, the sides are 1, 2, 2. The triangle inequalities: 1 + 2 > 2 (3 > 2, yes), 1 + 2 > 2 (same), and 2 + 2 > 1 (yes). So, that's valid. So, this point is included in Case 2 or 3? Since x and y are both greater than 1, and since they are equal, it's included in both? Wait, but in our Cases 2 and 3, we considered x being the largest or y being the largest. However, if x and y are equal, then both conditions would be satisfied? Wait, but in our Case 2, we considered x being the largest side, which would require x ≥ y. Similarly, Case 3 requires y ≥ x. So, if x = y, then it's included in both? But since we are integrating over x from 1 to 3 and y from x - 1 to x, and similarly for Case 3, there might be some overlap? Wait, but in reality, when x = y, it's only counted once? Wait, no. Because in Case 2, we have x ≥ y, so when x = y, it's included in Case 2. Similarly, in Case 3, y ≥ x, so when x = y, it's included in Case 3. Therefore, there is an overlap along the line x = y. However, since it's a line, which has area zero, it doesn't affect the total area. Therefore, adding Case 2 and Case 3 is okay.

But let me check if the total area is indeed 0.5 + 2 + 2 = 4.5. Then, probability is 4.5 / 9 = 0.5. But let me think again. Suppose I consider the entire square (0,3)x(0,3). The area where triangle inequalities hold is 4.5? Let me check with another approach.

Alternatively, maybe I can consider the three inequalities:

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 these two lines. However, we also need to be above x + y > 1. So, the intersection of these regions.

So, let me try to visualize. The area between the two lines y = x - 1 and y = x + 1 is a band around the line y = x. However, we also need to be above x + y = 1. So, the region is the band between the two lines, but above the line x + y = 1.

So, perhaps the area can be calculated by subtracting the regions that are below x + y = 1 from the band between the two lines.

But maybe it's easier to split the area into parts.

Alternatively, since we have already calculated using cases and got 4.5, which is 1/2, but let me check with another method.

Let me consider integrating over the square. The total area is 9. The area where triangle inequalities hold is the set of (x,y) such that:

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

So, let me try to find the area where all three hold.

First, let's consider the region where y < x + 1 and y > x - 1. This is the band between the two lines. Then, within that band, we need to also satisfy x + y > 1.

So, perhaps we can split the integration into regions where x + y > 1 is automatically satisfied or not.

Alternatively, since the band between y = x - 1 and y = x + 1 overlaps with the region x + y > 1.

But maybe it's better to find the area where all three inequalities hold by integrating.

Alternatively, since we have already considered the cases and found 4.5, but let me check with coordinates.

Let me think of the entire square. The area where triangle inequalities hold is the area where:

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

So, let me consider the overlap of these three.

First, the region between y = x - 1 and y = x + 1. Let me find where this band overlaps with x + y > 1.

For x between 0 and 1:

The line x + y = 1 intersects the band. Let me see. For x between 0 and 1, the lower bound of y is x - 1, which is negative, but since y is in (0,3), the lower bound is actually 0. The upper bound is y = x + 1. However, we also have the condition x + y > 1. So, for x between 0 and 1, the lower bound for y is max(0, 1 - x). Wait, because x + y > 1 implies y > 1 - x. So, for x between 0 and 1, 1 - x ranges from 1 to 0. Therefore, the lower bound for y is max(0, 1 - x) = 1 - x (since 1 - x is positive for x < 1). The upper bound is y < x + 1. However, since y is also less than 3, but for x between 0 and 1, x + 1 is between 1 and 2, so y is between 1 - x and x + 1. Therefore, the area for x between 0 and 1 is the integral from x=0 to x=1 of [ (x + 1) - (1 - x) ] dx = integral of 2x dx from 0 to 1? Wait, (x + 1) - (1 - x) = x + 1 - 1 + x = 2x. Therefore, integral from 0 to 1 of 2x dx = [x²] from 0 to 1 = 1 - 0 = 1.

Then, for x between 1 and 2:

The lower bound for y is max(0, 1 - x). But 1 - x is negative here, so lower bound is 0. However, we also have the band between y = x - 1 and y = x + 1. So, for x between 1 and 2, the lower bound is y > x - 1 (since x - 1 is between 0 and 1), and upper bound is y < x + 1. However, we also have the condition x + y > 1. But since x ≥ 1 and y ≥ 0, x + y ≥ 1 + 0 = 1. Therefore, x + y > 1 is automatically satisfied for x ≥ 1 and y ≥ 0. Wait, but if x is 1 and y is 0, then x + y = 1, which is not greater than 1. So, actually, for x between 1 and 2, the condition x + y > 1 is equivalent to y > 1 - x. But since x ≥ 1, 1 - x ≤ 0. Since y ≥ 0, then y > 1 - x is always true. Therefore, for x between 1 and 2, the condition x + y > 1 is automatically satisfied. Therefore, the area for x between 1 and 2 is the integral from x=1 to x=2 of [ (x + 1) - (x - 1) ] dx = integral of 2 dx from 1 to 2 = 2*(2 - 1) = 2.

Then, for x between 2 and 3:

The lower bound for y is still y > x - 1 (since x - 1 is between 1 and 2). The upper bound is y < x + 1. However, since y is at most 3, we need to check if x + 1 exceeds 3. For x between 2 and 3, x + 1 is between 3 and 4. Therefore, the upper bound for y is 3. Therefore, the area for x between 2 and 3 is integral from x=2 to x=3 of [3 - (x - 1)] dx = integral of (4 - x) dx from 2 to 3.

Calculating that integral:

Integral of 4 - x dx = 4x - (x²)/2 evaluated from 2 to 3.

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 is 7.5 - 6 = 1.5

Therefore, the total area is the sum of the three parts:

For x between 0 and 1: 1

For x between 1 and 2: 2

For x between 2 and 3: 1.5

Total area = 1 + 2 + 1.5 = 4.5

Which matches the previous result. Therefore, the area where the triangle inequalities hold is 4.5, so the probability is 4.5 / 9 = 0.5, which is 1/2. Therefore, the probability is 1/2.

Wait, but let me check if there are any mistakes here. For example, when x is between 2 and 3, the upper bound for y is 3, not x + 1. So, the calculation for that part was correct. And the integral from 2 to 3 of (4 - x) dx is indeed 1.5. So, adding up gives 4.5. Therefore, probability is 4.5 / 9 = 0.5. So, 1/2.

But let me think again. Suppose I pick x and y such that they are both greater than 1. For example, x=2, y=2. Then, the sides are 1, 2, 2. Which is a valid triangle. Similarly, if x=3, y=3, then sides are 1, 3, 3. Which is valid because 1 + 3 > 3 (4 > 3). So, that's okay. However, if x=3 and y=1, then sides are 1, 3, 1. Then, 1 + 1 = 2 < 3, which is invalid. But according to our calculation, when x=3, y ranges from 2 to 3. So, y=1 is not included. Therefore, our calculation seems to exclude such cases. So, seems okay.

Wait, but in the case where x=3 and y=1, which is in the square, but since y=1 is less than x - 1 = 3 - 1 = 2, so y=1 is below the lower bound of the band. Therefore, it's excluded. Therefore, our calculation seems correct.

Another check: when x=0.5, y=0.5. Then, sides are 1, 0.5, 0.5. Then, 0.5 + 0.5 = 1, which is not greater than 1. Therefore, invalid. However, according to our calculation, for x between 0 and 1, the lower bound for y is 1 - x. For x=0.5, lower bound is 1 - 0.5 = 0.5. Therefore, y must be greater than 0.5. Therefore, y=0.5 is excluded. Therefore, the area where x + y > 1 is considered. Therefore, the point (0.5, 0.5) is excluded, which is correct. Therefore, our calculation seems correct.

Therefore, the probability is 4.5 / 9 = 1/2. Therefore, the answer is 1/2. But wait, let me check with another approach.

Alternatively, since the problem is symmetric in x and y, maybe we can compute the area where 1 + x > y, 1 + y > x, and x + y > 1. Which is equivalent to the area where |x - y| < 1 and x + y > 1. Wait, because 1 + x > y and 1 + y > x can be rewritten as y < x + 1 and x < y + 1, which is equivalent to |x - y| < 1. Therefore, the region is |x - y| < 1 and x + y > 1. Therefore, the area is the area between the lines y = x + 1 and y = x - 1, and above the line x + y = 1.

Which is exactly what we calculated earlier. Therefore, the area is 4.5, so probability is 1/2.

But let me check with coordinates. For example, if I consider the entire square, the area where |x - y| < 1 is the area between the lines y = x + 1 and y = x - 1. Which is the area of the square minus the areas of the two triangles where |x - y| ≥ 1. Each of those triangles has base and height of 2 (since from x=0 to 2, y=0 to 2, etc.), but wait, actually, for the square (0,3)x(0,3), the area where |x - y| ≥ 1 is two triangles. Each triangle has base and height of 3 - 1 = 2? Wait, no.

Wait, the area where |x - y| ≥ 1 is the area above y = x + 1 and below y = x - 1. For the square (0,3)x(0,3), the area above y = x + 1 is a triangle with vertices at (0,1), (0,3), and (2,3). Similarly, the area below y = x - 1 is a triangle with vertices at (1,0), (3,0), and (3,2). Each of these triangles has base 2 and height 2, so area is (2*2)/2 = 2. Therefore, total area where |x - y| ≥ 1 is 2 + 2 = 4. Therefore, the area where |x - y| < 1 is 9 - 4 = 5. However, we also need to consider the condition x + y > 1. Therefore, the area where |x - y| < 1 and x + y > 1 is equal to the area where |x - y| < 1 minus the area where |x - y| < 1 and x + y ≤ 1.

So, let me compute that. The area where |x - y| < 1 and x + y ≤ 1. This is the region that is between the lines y = x - 1 and y = x + 1, and below the line x + y = 1. But since x and y are positive, and x + y ≤ 1, this region is a small area near the origin.

Let me visualize. The line x + y = 1 intersects the band |x - y| < 1. The intersection of these two regions is the area where x + y ≤ 1 and |x - y| < 1. Which is the region bounded by x + y ≤ 1, y < x + 1, and y > x - 1. But since x + y ≤ 1 and y > x - 1, let me see.

For x between 0 and 1, y between max(0, x - 1) and min(1 - x, x + 1). But since x - 1 is negative, max(0, x - 1) = 0. And 1 - x is between 0 and 1, and x + 1 is between 1 and 2. Therefore, the upper bound is 1 - x. Therefore, the area is the integral from x=0 to x=1 of (1 - x - 0) dx = integral of (1 - x) dx from 0 to 1 = [x - (x²)/2] from 0 to 1 = (1 - 0.5) - 0 = 0.5. Therefore, the area where |x - y| < 1 and x + y ≤ 1 is 0.5. Therefore, the area where |x - y| < 1 and x + y > 1 is 5 - 0.5 = 4.5, which matches our previous result. Therefore, the probability is 4.5 / 9 = 0.5. Therefore, the probability is 1/2.

But wait, this seems to contradict my initial intuition. Let me check with another example. Suppose I pick x and y uniformly from (0,3). What is the probability that they can form a triangle with 1? If the answer is 1/2, that seems plausible? Let me think of another way.

Alternatively, maybe I can use geometric probability. The total area is 9. The area where triangle inequalities hold is 4.5. Therefore, probability is 1/2. But let me check with specific numbers.

For example, if x and y are both in (0,1), then the triangle inequalities require x + y > 1. The area of that is 0.5 as calculated. Then, for x and y in (1,3), the triangle inequalities are automatically satisfied? No, not necessarily. For example, if x=2 and y=2, it's okay. If x=2 and y=1, then sides are 1, 2, 1. Which is invalid. But according to our previous calculation, when x=2, y must be between 1 and 3. Wait, for x=2, the lower bound for y is x - 1 = 1, so y must be between 1 and 3. Therefore, y=1 is excluded. Therefore, in the region x between 1 and 3, y is between max(0, x - 1) and min(3, x + 1). But since x is between 1 and 3, max(0, x - 1) is x - 1, and min(3, x + 1) is 3 when x ≥ 2, and x + 1 when x < 2. Therefore, the area for x between 1 and 2 is integral of 2 dx from 1 to 2 = 2, and for x between 2 and 3, integral of (4 - x) dx from 2 to 3 = 1.5. So, total area is 2 + 1.5 = 3.5. Then, adding the area for x and y in (0,1) which is 0.5, total area is 3.5 + 0.5 = 4.0? Wait, no. Wait, earlier we had 4.5. Wait, maybe I miscalculated.

Wait, no. Wait, when we split into cases, Case 1 was 0.5, Case 2 was 2, Case 3 was 2, total 4.5. But when I considered the area where |x - y| < 1 and x + y > 1, it was 4.5. So, where is the discrepancy?

Wait, when I considered x between 0 and 1, the area was 1. Then, x between 1 and 2, area 2, and x between 2 and 3, area 1.5. Total 1 + 2 + 1.5 = 4.5. But when I considered splitting into regions where x and y are both in (0,1), which is 0.5, and then the rest, which is 4.5 - 0.5 = 4.0? Wait, no. Wait, the area where |x - y| < 1 and x + y > 1 is 4.5. Which includes all regions where x and y are in (0,3) with |x - y| < 1 and x + y > 1. So, that includes the area where both x and y are in (0,1) (but with x + y > 1), which is 0.5, and the rest. So, the total is 4.5. Therefore, the previous calculation where I thought of splitting into cases was correct.

Therefore, the answer is 1/2. However, I need to make sure that there isn't a mistake in the integration steps. Let me check again.

For x between 0 and 1:

The area is integral from 0 to 1 of [ (x + 1) - (1 - x) ] dx = integral of 2x dx from 0 to 1 = [x²] from 0 to 1 = 1. Correct.

For x between 1 and 2:

Integral from 1 to 2 of [ (x + 1) - (x - 1) ] dx = integral of 2 dx from 1 to 2 = 2*(2 - 1) = 2. Correct.

For x between 2 and 3:

Integral from 2 to 3 of [3 - (x - 1)] dx = integral of (4 - x) dx from 2 to 3 = [4x - (x²)/2] from 2 to 3 = (12 - 4.5) - (8 - 2) = 7.5 - 6 = 1.5. Correct.

Total area: 1 + 2 + 1.5 = 4.5. Correct.

Therefore, probability is 4.5 / 9 = 0.5. Therefore, the probability is 1/2.

But let me check with another approach. Let me consider the probability that 1, x, y can form a triangle. Which is equivalent to the probability that x + y > 1, |x - y| < 1. Which is the same as the area we calculated. So, since the answer is 1/2, maybe that's correct.

Alternatively, maybe I can think of the problem as follows: For two numbers x and y selected uniformly from (0,3), what is the probability that they satisfy |x - y| < 1 and x + y > 1. Which is the same as the area we calculated. Therefore, 4.5 / 9 = 1/2.

Alternatively, maybe I can think of the problem in terms of the unit square, but scaled. However, since the interval is (0,3), it's a bit different.

Alternatively, maybe I can use geometric probability with coordinates. But I think the previous calculations are consistent.

Therefore, I think the answer is 1/2. However, I recall that sometimes when dealing with triangle inequalities, there might be some edge cases or miscalculations. Let me check with specific values.

For example, take x=1.5 and y=1.5. Then, sides are 1, 1.5, 1.5. The triangle inequalities: 1 + 1.5 > 1.5 (yes), 1 + 1.5 > 1.5 (yes), 1.5 + 1.5 > 1 (yes). Valid.

Take x=1.5 and y=0.5. Then, sides are 1, 1.5, 0.5. Check triangle inequalities: 1 + 0.5 = 1.5, which is not greater than 1.5. Therefore, invalid. However, according to our calculation, for x=1.5, y must be between 0.5 and 2.5. Wait, y=0.5 is the lower bound? Wait, for x=1.5, the lower bound is x - 1 = 0.5. Therefore, y must be greater than 0.5. Therefore, y=0.5 is excluded. Therefore, the point (1.5, 0.5) is excluded. Therefore, the calculation is correct.

Another example: x=0.6, y=0.6. Then, sides are 1, 0.6, 0.6. Check triangle inequality: 0.6 + 0.6 = 1.2 > 1. So, valid. However, according to our calculation, for x=0.6, y must be between 1 - 0.6 = 0.4 and 0.6 + 1 = 1.6. Therefore, y=0.6 is within that range. Therefore, included. Therefore, valid. So, correct.

Another example: x=0.4, y=0.4. Then, sides are 1, 0.4, 0.4. 0.4 + 0.4 = 0.8 < 1. Invalid. According to our calculation, for x=0.4, the lower bound for y is 1 - 0.4 = 0.6. Therefore, y must be greater than 0.6. Therefore, y=0.4 is excluded. Correct.

Therefore, all these examples check out. Therefore, I think the answer is indeed 1/2.

But wait, I just thought of something. The original problem says "two numbers, x and y are selected at random from the interval (0,3)". So, is it continuous uniform distribution? Yes, I assume so. Therefore, the area calculation is correct.

Therefore, the probability is 1/2. Therefore, the 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 any three sides $ a, b, c $ to form a triangle, the following must hold:

$$
a + b > c, \quad a + c > b, \quad b + c > a
$$

In our case, the sides are $ 1, x, y $, so the triangle inequalities become:

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

We are selecting $ x $ and $ y $ uniformly at random from the interval $ (0, 3) $, so the sample space is the square $ (0, 3) \times (0, 3) $, with area $ 9 $.

---

### Step 1: Geometric Interpretation

We interpret the conditions geometrically in the $ xy $-plane:

- The inequality $ x + y > 1 $ defines the region **above** the line $ x + y = 1 $.
- The inequalities $ 1 + x > y $ and $ 1 + y > x $ can be rewritten as:
  - $ y < x + 1 $
  - $ y > x - 1 $

These define the region **between** the lines $ y = x + 1 $ and $ y = x - 1 $, i.e., the **band** around the line $ y = x $.

So, the region where all three triangle inequalities hold is the **intersection** of:
- The band between $ y = x - 1 $ and $ y = x + 1 $
- The region above the line $ x + y = 1 $

This region lies within the square $ (0, 3) \times (0, 3) $, and we are to compute its area.

---

### Step 2: Compute the Area of the Valid Region

We divide the square into three parts based on the value of $ x $:

#### **Case 1: $ x \in [0, 1] $**

- The lower bound for $ y $ is $ \max(0, 1 - x) = 1 - x $
- The upper bound for $ y $ is $ \min(3, x + 1) = x + 1 $
- So, the vertical slice at each $ x $ has length $ (x + 1) - (1 - x) = 2x $

$$
\text{Area} = \int_0^1 2x \, dx = \left[ x^2 \right]_0^1 = 1
$$

#### **Case 2: $ x \in [1, 2] $**

- The lower bound for $ y $ is $ x - 1 $
- The upper bound for $ y $ is $ x + 1 $
- So, the vertical slice at each $ x $ has length $ (x + 1) - (x - 1) = 2 $

$$
\text{Area} = \int_1^2 2 \, dx = 2(2 - 1) = 2
$$

#### **Case 3: $ x \in [2, 3] $**

- The lower bound for $ y $ is $ x - 1 $
- The upper bound for $ y $ is $ 3 $ (since $ x + 1 > 3 $)
- So, the vertical slice at each $ x $ has length $ 3 - (x - 1) = 4 - x $

$$
\text{Area} = \int_2^3 (4 - x) \, dx = \left[ 4x - \frac{x^2}{2} \right]_2^3 = (12 - 4.5) - (8 - 2) = 7.5 - 6 = 1.5
$$

---

### Step 3: Total Valid Area

$$
\text{Total Area} = 1 + 2 + 1.5 = 4.5
$$

---

### Step 4: Compute the Probability

The total area of the square is $ 9 $, so the probability that a triangle with sides $ 1, x, y $ exists is:

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

---

### 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 $(9/2)/9=\boxed{\frac{1}{2}}$.