Explanation:
Okay, so I need to figure out the maximum number of points where at least two of four distinct circles intersect. Hmm, let me think. I remember that when dealing with circles and their intersections, each pair of circles can intersect in at most two points. So maybe I can use that? Let me start by recalling some basics.

First, if I have two circles, they can intersect in 0, 1, or 2 points. The maximum number of intersection points between two circles is 2. So, if I want the maximum total number of intersection points for four circles, I need to arrange them so that every pair intersects in two points. But wait, is that possible? Let me check.

Let me consider the case with two circles first. If two circles intersect, they can have two points. Now, with three circles, each pair of circles can intersect in two points. So, for three circles, how many intersection points would that be? Well, there are three pairs: circle 1 & 2, circle 1 & 3, and circle 2 & 3. Each pair contributes two points, so that's 3*2=6 points. But wait, is there any overlap? If I have three circles, each pair intersecting in two points, but are those points unique? Unless the circles are arranged in some special way, maybe they don't share common intersection points. So if all intersections are distinct, then yes, three circles can have up to 6 points. Similarly, for four circles, the number of pairs is C(4,2)=6 pairs. Each contributing 2 points, so 6*2=12 points. So is the maximum number of intersection points 12? But wait, the question says "at least two of the circles intersect." Wait, but if each pair intersects in two points, then all those points are counted. So 12?

But maybe there's some overlap? Wait, but if we can arrange the four circles so that no two circles intersect in more than two points and that no three circles intersect at the same point. Then, each intersection is counted once. So, for four circles, the maximum number would be 2*C(4,2)=12. But let me check if that's possible.

Wait, but maybe when adding more circles, some constraints come in. Let me think about the case with three circles. If I have three circles, each pair intersecting in two points, then each circle would have two intersection points with each of the other two circles. But is that possible? For example, take three circles where each pair intersects in two points. Let me visualize. Let me imagine the first two circles intersecting at two points. Now, if I add a third circle, it needs to intersect both the first and second circles in two points each. So the third circle would have to pass through two points on the first circle and two points on the second circle? Wait, but if the first two circles intersect at two points, then those two points are already on both circles. If I want the third circle to intersect the first circle in two points, those can be the same two points? But then the third circle would pass through the two intersection points of the first two circles. However, if it also intersects the second circle in two points, then those would also have to be the same two points? But then the third circle is passing through the same two points as the first two circles. That would make three circles intersecting at two common points. But if three circles all intersect at two common points, then those two points are intersections of all three circles. But then, when counting the total number of intersection points, those two points would be counted multiple times. So if I have three circles intersecting at two common points, then each pair would have those two points as intersections, so instead of having 3*2=6 points, we have only 2 points, but each pair contributes those two points. But that's worse for maximum count. So we have to make sure that no three circles intersect at the same point. Therefore, for maximum number of intersection points, each pair of circles intersects in two distinct points, and no three circles share a common intersection point. Therefore, the total number of points would be 2*C(4,2)=12.

But is this possible? Let me think if arranging four circles so that each pair intersects in two points, and no three circles share a common intersection point. So, for example, if I can have four circles where each pair intersects in two unique points. How can I do that?

Let me think of circles in general position. If I can arrange them such that no two circles are tangent (so they intersect at two points), and no three circles meet at a single point. Then, the total number of intersections would be 2*C(4,2)=12. So is this possible?

Yes, for example, take four circles where each one is arranged such that they intersect every other circle in two distinct points, and no three circles meet at a single point. For example, imagine four circles, each pair intersecting in two unique points. For example, take two circles intersecting at two points, then add a third circle that intersects each of the first two circles at two new points, not shared with any other circle. Then, the fourth circle would similarly intersect each of the previous three circles at two new points, not overlapping with previous intersections. So, if I can do that, then the total number of intersection points would be 12.

Wait, but how can we ensure that? Let me try to think of an example.

Start with circle A and circle B, intersecting at points P1 and P2. Now, add circle C. To make sure that circle C intersects circle A at two new points, say Q1 and Q2, and circle B at two new points R1 and R2. So circle C doesn't pass through P1 or P2. Then add circle D. Circle D needs to intersect circle A at two new points, say S1 and S2, circle B at two new points T1 and T2, and circle C at two new points U1 and U2. If we can arrange circle D so that none of these intersection points are overlapping with any previous points, then we have 2*6=12 points.

But is there any reason why this wouldn't be possible? Maybe not. In the plane, you can always adjust the circles to avoid passing through existing points. So as long as you don't have overlapping intersections, this should be possible. Hence, the maximum number of intersection points is 12.

But wait, let me check with smaller numbers. For two circles, it's 2. For three circles, each pair intersects at two points, so 3*2=6, but if they are arranged so that no three circles meet at a single point. Which is possible. For example, three circles where each pair intersects at two unique points. So, yes, that's 6 points. Then for four circles, 6 pairs, so 12 points. That seems to be the case.

Wait, but I remember that in some cases, with circles, there might be some limitations. For example, with lines, the maximum number of intersections is C(n,2), but with circles, it's different because two circles can intersect at two points. However, for lines, each pair intersects at one point, so maximum is C(n,2). For circles, with n circles, the maximum number of intersections is 2*C(n,2). So for n=4, 2*C(4,2)=12. Therefore, the answer should be 12. But maybe I need to verify with a reference or think again.

Alternatively, maybe there is a mistake here. Wait, sometimes when you add more circles, there might be some dependencies. For example, if you have four circles, each pair intersects at two points. However, perhaps there's a constraint that requires some intersection points to overlap? Let me think.

Suppose I have four circles: A, B, C, D. Each pair intersects in two points. If I have circle A and circle B intersect at points P1 and P2, circle A and circle C intersect at Q1 and Q2, circle A and circle D intersect at R1 and R2, circle B and circle C intersect at S1 and S2, circle B and circle D intersect at T1 and T2, circle C and circle D intersect at U1 and U2. If all these points are distinct, then total is 12 points. But is there a way that when you add the fourth circle, you can't arrange it to intersect each existing circle at two new points?

Alternatively, maybe the problem is that when you have multiple circles, the number of intersection points is limited by some other factors? Hmm.

Let me think of an example. Let me take four circles in general position. For example, take four circles, each pair intersecting in two points, and no three circles sharing an intersection point. Is this possible? Let me try to construct such a configuration.

First, take circle A and circle B. Let them intersect at points P and Q. Then, take circle C such that it intersects circle A at two new points, R and S, and circle B at two new points, T and U. Now, circle C is crossing both A and B at new points. Then, take circle D. To intersect circle A at two new points, say V and W; circle B at two new points, X and Y; circle C at two new points, Z and... Wait, circle C already has points R, S, T, U. So circle D needs to intersect circle C at two points that aren't R, S, T, U. Similarly, circle D needs to intersect circle A at V, W not overlapping with existing points on A, and same with circle B. So as long as we can adjust the circle D to pass through these new points without overlapping with previous intersections, which seems possible.

But how do we ensure that? Because circles are defined by their centers and radii. Maybe by careful placement, we can adjust the circles so that each new circle intersects all previous ones at two unique points. Since there are infinitely many ways to place circles, I think this is possible.

Alternatively, maybe there is a mathematical formula or known result. I recall that for n circles, the maximum number of intersections is 2*C(n,2), provided that no two circles are tangent and no three circles meet at a single point. Therefore, for four circles, 2*(4*3)/2=12. Therefore, the answer is 12.

But let me check an authoritative source or recall previous knowledge. Wait, in general, for the maximum number of intersection points with n circles, it's 2*C(n,2). Similarly, for n lines, it's C(n,2); for n conic sections, it's 2*C(n,2) as well? Wait, no, conic sections can intersect in up to 4 points, but that's different.

But for circles, since they are a specific type of conic, their intersection is limited to 2 points. So, for circles, if you have n circles, the maximum number of intersections is 2*C(n,2). So, for n=4, 2*6=12. Therefore, the answer is 12.

But let me think if there is any mistake here. For example, maybe when adding the fourth circle, there is some restriction? Wait, for example, if you have three circles already, each pair intersecting in two points, making 6 points. Then adding a fourth circle. How many intersections can it have with the existing three circles? Each existing circle can intersect the new circle in up to two points, so 3*2=6. So total intersections added by the fourth circle is 6. So total intersections would be previous 6 + 6 = 12. Which is same as 2*C(4,2)=12. So that works. Therefore, if each time you add a circle, you can have it intersect all previous circles in two new points, and since there are no overlapping points, the total is 12. Therefore, 12 is achievable.

Alternatively, maybe I need to think about the problem differently. Suppose we have four circles. Each pair contributes two points, so 6 pairs * 2 points = 12. So the maximum number is 12. So the answer is 12. Wait, but why is this problem considered? Maybe I need to check with an example. Let me think of specific circles.

Take four circles:

Circle 1: x² + y² = 1 (unit circle centered at origin)

Circle 2: (x - 1)² + y² = 1 (unit circle centered at (1,0))

They intersect at two points: (0.5, sqrt(3)/2) and (0.5, -sqrt(3)/2). So two points.

Now add Circle 3: Let me choose a circle that intersects Circle 1 and Circle 2 at two new points each. Let me choose a circle that's centered somewhere else. Let's say Circle 3: (x - 0.5)^2 + (y - 1)^2 = 0.5. Let me check intersections with Circle 1 and Circle 2.

First, with Circle 1: x² + y² = 1 and (x - 0.5)^2 + (y - 1)^2 = 0.5.

Expanding the second equation: x² - x + 0.25 + y² - 2y + 1 = 0.5 => x² + y² - x - 2y + 1.25 = 0.5 => x² + y² - x - 2y + 0.75 = 0.

But from Circle 1, x² + y² = 1. Substitute into the above equation: 1 - x - 2y + 0.75 = 0 => -x - 2y + 1.75 = 0 => x + 2y = 1.75. So the line of intersection between Circle 1 and Circle 3 is x + 2y = 1.75. Let me find the points of intersection.

Substitute x = 1.75 - 2y into Circle 1: (1.75 - 2y)^2 + y² = 1.

Calculating:

(3.0625 - 7y + 4y²) + y² = 1

3.0625 -7y +5y² = 1

5y² -7y + 2.0625 = 0

Multiply by 16 to eliminate decimals? Maybe not. Let me calculate discriminant.

Discriminant D = 49 - 4*5*(2.0625) = 49 - 20*2.0625 = 49 - 41.25 = 7.75

Which is positive, so two real solutions. So two intersection points between Circle 1 and Circle 3. Similarly, check intersections with Circle 2.

Circle 2: (x - 1)^2 + y² = 1.

Circle 3: (x - 0.5)^2 + (y - 1)^2 = 0.5.

Expanding Circle 3: x² - x + 0.25 + y² - 2y + 1 = 0.5 => x² + y² - x - 2y + 1.25 = 0.5 => same as before, x² + y² - x - 2y + 0.75 = 0.

From Circle 2: (x -1)^2 + y² = 1 => x² - 2x +1 + y² = 1 => x² + y² - 2x = 0. Substitute into Circle 3 equation:

(x² + y²) - x - 2y + 0.75 = 0 => (2x) - x - 2y + 0.75 = 0 => x - 2y + 0.75 = 0 => x = 2y - 0.75.

Substitute x into Circle 2: (2y - 0.75 - 1)^2 + y² = 1 => (2y - 1.75)^2 + y² = 1.

Expanding: 4y² - 7y + 3.0625 + y² = 1 => 5y² -7y + 2.0625 =0. Same equation as before. So the solutions are same? Wait, but this would imply that the intersections of Circle 2 and Circle 3 lie on the same line as Circle 1 and Circle 3? Wait, but that seems odd. Wait, but actually, since both intersections are derived from the same equation, perhaps the intersection points are the same as between Circle 1 and Circle 3? Wait, but that would mean that the three circles (1, 2, 3) intersect at two common points? Wait, but that would mean that the two circles 1 and 3 intersect at two points, and circles 2 and 3 also intersect at the same two points. Therefore, circles 1, 2, 3 all intersect at two common points. Which would mean that the two points are common to all three circles. Therefore, if that's the case, then those two points would be counted multiple times in the total intersection points. Hence, reducing the total number of unique points.

Wait, but in my previous calculation, when solving for intersections between Circle 1 and Circle 3, and between Circle 2 and Circle 3, I ended up with the same quadratic equation. Which suggests that the line of intersection between Circle 1 and 3 and the line of intersection between Circle 2 and 3 are different? Wait, no, the line x + 2y = 1.75 and x = 2y - 0.75. Let me check if they are the same.

From the first, x = 1.75 - 2y.

From the second, x = 2y - 0.75. Let me set them equal:

1.75 - 2y = 2y - 0.75 => 1.75 + 0.75 = 4y => 2.5 = 4y => y = 2.5 / 4 = 0.625. Then x = 2*(0.625) - 0.75 = 1.25 - 0.75 = 0.5. So the two lines intersect at (0.5, 0.625). Therefore, they are different lines. Therefore, the equations for intersections between 1 & 3 and between 2 & 3 are different lines. Therefore, their intersections are different points. Hence, perhaps the two circles 1 & 3 intersect at two points, and 2 & 3 also intersect at two different points. Wait, but why does solving for the intersection between 1 & 3 and between 2 & 3 give the same quadratic equation?

Wait, because when we substituted, we ended up with the same quadratic equation? Let me check again.

For Circle 1 and 3, we substituted x = 1.75 - 2y into Circle 1 equation and ended up with 5y² -7y + 2.0625 = 0.

Similarly, for Circle 2 and 3, substituting x = 2y - 0.75 into Circle 2 equation, we ended up with 5y² -7y + 2.0625 =0. So the same quadratic equation. Therefore, the solutions for y are the same? Wait, but if that's the case, then the two intersections for 1 & 3 are the same as for 2 & 3? That can't be, unless the circles 1 & 3 and 2 & 3 have the same intersection points. But how is that possible? Because we derived that the intersection points lie on both lines x + 2y = 1.75 and x = 2y - 0.75? Wait, no. Each intersection is between two circles, so for Circle 1 and 3, the intersection is on the line x + 2y = 1.75. For Circle 2 and 3, the intersection is on the line x = 2y - 0.75. So the points of intersection between 1 & 3 are on the first line, and those between 2 & 3 are on the second line. However, when we solved for the intersection of 1 & 3, we substituted into Circle 1, which gave us a quadratic with solutions. Similarly, for 2 & 3, substituting into Circle 2 gave us the same quadratic. Therefore, the solutions for y are same, but what about x?

Wait, if we have for the intersections between 1 & 3, the solutions for y come from 5y² -7y + 2.0625 = 0. Let me compute the solutions.

Discriminant D = 49 - 4*5*2.0625 = 49 - 41.25 = 7.75.

sqrt(7.75) ≈ 2.78

So y = [7 ± 2.78]/10 ≈ [7 ± 2.78]/10.

So approximately, y ≈ (7 + 2.78)/10 ≈ 9.78/10 ≈ 0.978 or y ≈ (7 - 2.78)/10 ≈ 4.22/10 ≈ 0.422.

Then x for 1 & 3 is 1.75 - 2y.

For y ≈ 0.978: x ≈ 1.75 - 2*(0.978) ≈ 1.75 - 1.956 ≈ -0.206

For y ≈ 0.422: x ≈ 1.75 - 2*(0.422) ≈ 1.75 - 0.844 ≈ 0.906

So two intersection points between 1 & 3: (-0.206, 0.978) and (0.906, 0.422)

Similarly, for intersections between 2 & 3, we have x = 2y - 0.75.

So for the same y ≈ 0.978 and 0.422?

Wait, no, because when we solved for Circle 2 and 3, we substituted into Circle 2. So if the quadratic equation is same, then these y's are same. But then x for 2 & 3 is 2y - 0.75.

So for y ≈ 0.978: x ≈ 2*(0.978) - 0.75 ≈ 1.956 - 0.75 ≈ 1.206

For y ≈ 0.422: x ≈ 2*(0.422) - 0.75 ≈ 0.844 - 0.75 ≈ 0.094

Therefore, points between Circle 2 & 3 are (1.206, 0.978) and (0.094, 0.422). These are two different points from the ones between 1 & 3. Therefore, there is no overlap. Therefore, circles 1, 2, 3 have 2 + 2 + 2 = 6 intersection points, with no three circles intersecting at the same point. Therefore, that works.

So, that seems okay. Then adding a fourth circle, say, Circle D, which intersects each of the previous three circles at two new points. So, Circle D needs to intersect Circle A at two new points, Circle B at two new points, and Circle C at two new points. If we can arrange it so that all these intersections are unique, then we add 6 new points, giving total of 6 + 6 = 12.

Therefore, the maximum number is 12. Therefore, the answer should be \boxed{12}.

Wait, but I need to make sure that with four circles, you can indeed have all those intersections. Maybe there's some geometric constraint? For example, maybe if you have too many circles, some of their intersection points have to coincide? But I think if you can place the circles in general position, such that no two circles are tangent and no three circles meet at a point, then it's possible. Since each circle can be adjusted independently to intersect others at two unique points.

Alternatively, maybe there's a different way to count. For example, the number of intersections for n circles can't exceed 2*C(n,2), since each pair can contribute at most 2. Therefore, 2*C(4,2)=12 is the upper bound, and since it's achievable, that's the maximum. So yeah, I think that's solid.

Alternatively, if I search my memory, I think that for the maximum number of regions created by circles, it's different, but here it's just intersection points. For regions, the formula is similar to the one for lines, but with different coefficients. But here, it's just about intersection points. So I think 12 is correct.

Therefore, my conclusion is that the maximum number of points where at least two of the circles intersect is 12. Therefore, the answer is \boxed{12}.

**Final Answer**
\boxed{12}

To determine the **maximum number of points where at least two of the four distinct circles intersect**, we must consider the fundamental geometric properties of circle intersections.

---

### Step 1: Understanding the Maximum Intersections Between Two Circles

Two distinct circles in a plane can intersect in **at most two points**. These can be visualized as two distinct points that both lie on the circumferences of the two circles. While they may intersect at **0** (no intersection), **1** (tangent), or (typically) **2** points.

Therefore, in the case of two circles, the **maximum** number of distinct intersection points is **2**.

---

### Step 2: Counting the Number of Pairs of Circles

We are given **four distinct circles**. The number of unordered pairs of circles we can choose from these four is:

$$
\binom{4}{2} = 6 \text{ pairs}
$$

Each of these six pairs of circles can, in theory, intersect at **two distinct points**, **provided** we avoid any overlap of these points (i.e., no three circles share a single intersection point).

---

### Step 3: Maximizing the Intersection Points

To achieve the **maximum** number of **distinct** intersection points, we must ensure:

- Each pair of circles intersects at **two distinct** points.
- No three circles intersect at the **same point**.

If we can arrange the circles in such a way, the total number of **distinct** intersection points is:

$$
\text{Number of pairs} \times \text{points per pair} = 6 \times 2 = 12
$$

This assumes **general position** for the circles — meaning that no two circles are tangent, and no three circles intersect at a single point.

---

### Step 4: Justification of the Result

This arrangement is **feasible**. By carefully choosing the radii and positions of the circles, it is possible to ensure that:

- Every pair of circles intersects at two unique points.
- No three circles intersect at the same point.
- All 12 intersection points are distinct.

This configuration can be constructed in Euclidean geometry, and thus the **maximum** number of intersection points is indeed:

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the maximum number of points where at least two of four distinct circles intersect. Hmm, let me think. I remember that when dealing with circles and their intersections, each pair of circles can intersect in at most two points. So maybe I can use that? Let me start by recalling some basics.

First, if I have two circles, they can intersect in 0, 1, or 2 points. The maximum number of intersection points between two circles is 2. So, if I want the maximum total number of intersection points for four circles, I need to arrange them so that every pair intersects in two points. But wait, is that possible? Let me check.

Let me consider the case with two circles first. If two circles intersect, they can have two points. Now, with three circles, each pair of circles can intersect in two points. So, for three circles, how many intersection points would that be? Well, there are three pairs: circle 1 & 2, circle 1 & 3, and circle 2 & 3. Each pair contributes two points, so that's 3*2=6 points. But wait, is there any overlap? If I have three circles, each pair intersecting in two points, but are those points unique? Unless the circles are arranged in some special way, maybe they don't share common intersection points. So if all intersections are distinct, then yes, three circles can have up to 6 points. Similarly, for four circles, the number of pairs is C(4,2)=6 pairs. Each contributing 2 points, so 6*2=12 points. So is the maximum number of intersection points 12? But wait, the question says "at least two of the circles intersect." Wait, but if each pair intersects in two points, then all those points are counted. So 12?

But maybe there's some overlap? Wait, but if we can arrange the four circles so that no two circles intersect in more than two points and that no three circles intersect at the same point. Then, each intersection is counted once. So, for four circles, the maximum number would be 2*C(4,2)=12. But let me check if that's possible.

Wait, but maybe when adding more circles, some constraints come in. Let me think about the case with three circles. If I have three circles, each pair intersecting in two points, then each circle would have two intersection points with each of the other two circles. But is that possible? For example, take three circles where each pair intersects in two points. Let me visualize. Let me imagine the first two circles intersecting at two points. Now, if I add a third circle, it needs to intersect both the first and second circles in two points each. So the third circle would have to pass through two points on the first circle and two points on the second circle? Wait, but if the first two circles intersect at two points, then those two points are already on both circles. If I want the third circle to intersect the first circle in two points, those can be the same two points? But then the third circle would pass through the two intersection points of the first two circles. However, if it also intersects the second circle in two points, then those would also have to be the same two points? But then the third circle is passing through the same two points as the first two circles. That would make three circles intersecting at two common points. But if three circles all intersect at two common points, then those two points are intersections of all three circles. But then, when counting the total number of intersection points, those two points would be counted multiple times. So if I have three circles intersecting at two common points, then each pair would have those two points as intersections, so instead of having 3*2=6 points, we have only 2 points, but each pair contributes those two points. But that's worse for maximum count. So we have to make sure that no three circles intersect at the same point. Therefore, for maximum number of intersection points, each pair of circles intersects in two distinct points, and no three circles share a common intersection point. Therefore, the total number of points would be 2*C(4,2)=12.

But is this possible? Let me think if arranging four circles so that each pair intersects in two points, and no three circles share a common intersection point. So, for example, if I can have four circles where each pair intersects in two unique points. How can I do that?

Let me think of circles in general position. If I can arrange them such that no two circles are tangent (so they intersect at two points), and no three circles meet at a single point. Then, the total number of intersections would be 2*C(4,2)=12. So is this possible?

Yes, for example, take four circles where each one is arranged such that they intersect every other circle in two distinct points, and no three circles meet at a single point. For example, imagine four circles, each pair intersecting in two unique points. For example, take two circles intersecting at two points, then add a third circle that intersects each of the first two circles at two new points, not shared with any other circle. Then, the fourth circle would similarly intersect each of the previous three circles at two new points, not overlapping with previous intersections. So, if I can do that, then the total number of intersection points would be 12.

Wait, but how can we ensure that? Let me try to think of an example.

Start with circle A and circle B, intersecting at points P1 and P2. Now, add circle C. To make sure that circle C intersects circle A at two new points, say Q1 and Q2, and circle B at two new points R1 and R2. So circle C doesn't pass through P1 or P2. Then add circle D. Circle D needs to intersect circle A at two new points, say S1 and S2, circle B at two new points T1 and T2, and circle C at two new points U1 and U2. If we can arrange circle D so that none of these intersection points are overlapping with any previous points, then we have 2*6=12 points.

But is there any reason why this wouldn't be possible? Maybe not. In the plane, you can always adjust the circles to avoid passing through existing points. So as long as you don't have overlapping intersections, this should be possible. Hence, the maximum number of intersection points is 12.

But wait, let me check with smaller numbers. For two circles, it's 2. For three circles, each pair intersects at two points, so 3*2=6, but if they are arranged so that no three circles meet at a single point. Which is possible. For example, three circles where each pair intersects at two unique points. So, yes, that's 6 points. Then for four circles, 6 pairs, so 12 points. That seems to be the case.

Wait, but I remember that in some cases, with circles, there might be some limitations. For example, with lines, the maximum number of intersections is C(n,2), but with circles, it's different because two circles can intersect at two points. However, for lines, each pair intersects at one point, so maximum is C(n,2). For circles, with n circles, the maximum number of intersections is 2*C(n,2). So for n=4, 2*C(4,2)=12. Therefore, the answer should be 12. But maybe I need to verify with a reference or think again.

Alternatively, maybe there is a mistake here. Wait, sometimes when you add more circles, there might be some dependencies. For example, if you have four circles, each pair intersects at two points. However, perhaps there's a constraint that requires some intersection points to overlap? Let me think.

Suppose I have four circles: A, B, C, D. Each pair intersects in two points. If I have circle A and circle B intersect at points P1 and P2, circle A and circle C intersect at Q1 and Q2, circle A and circle D intersect at R1 and R2, circle B and circle C intersect at S1 and S2, circle B and circle D intersect at T1 and T2, circle C and circle D intersect at U1 and U2. If all these points are distinct, then total is 12 points. But is there a way that when you add the fourth circle, you can't arrange it to intersect each existing circle at two new points?

Alternatively, maybe the problem is that when you have multiple circles, the number of intersection points is limited by some other factors? Hmm.

Let me think of an example. Let me take four circles in general position. For example, take four circles, each pair intersecting in two points, and no three circles sharing an intersection point. Is this possible? Let me try to construct such a configuration.

First, take circle A and circle B. Let them intersect at points P and Q. Then, take circle C such that it intersects circle A at two new points, R and S, and circle B at two new points, T and U. Now, circle C is crossing both A and B at new points. Then, take circle D. To intersect circle A at two new points, say V and W; circle B at two new points, X and Y; circle C at two new points, Z and... Wait, circle C already has points R, S, T, U. So circle D needs to intersect circle C at two points that aren't R, S, T, U. Similarly, circle D needs to intersect circle A at V, W not overlapping with existing points on A, and same with circle B. So as long as we can adjust the circle D to pass through these new points without overlapping with previous intersections, which seems possible.

But how do we ensure that? Because circles are defined by their centers and radii. Maybe by careful placement, we can adjust the circles so that each new circle intersects all previous ones at two unique points. Since there are infinitely many ways to place circles, I think this is possible.

Alternatively, maybe there is a mathematical formula or known result. I recall that for n circles, the maximum number of intersections is 2*C(n,2), provided that no two circles are tangent and no three circles meet at a single point. Therefore, for four circles, 2*(4*3)/2=12. Therefore, the answer is 12.

But let me check an authoritative source or recall previous knowledge. Wait, in general, for the maximum number of intersection points with n circles, it's 2*C(n,2). Similarly, for n lines, it's C(n,2); for n conic sections, it's 2*C(n,2) as well? Wait, no, conic sections can intersect in up to 4 points, but that's different.

But for circles, since they are a specific type of conic, their intersection is limited to 2 points. So, for circles, if you have n circles, the maximum number of intersections is 2*C(n,2). So, for n=4, 2*6=12. Therefore, the answer is 12.

But let me think if there is any mistake here. For example, maybe when adding the fourth circle, there is some restriction? Wait, for example, if you have three circles already, each pair intersecting in two points, making 6 points. Then adding a fourth circle. How many intersections can it have with the existing three circles? Each existing circle can intersect the new circle in up to two points, so 3*2=6. So total intersections added by the fourth circle is 6. So total intersections would be previous 6 + 6 = 12. Which is same as 2*C(4,2)=12. So that works. Therefore, if each time you add a circle, you can have it intersect all previous circles in two new points, and since there are no overlapping points, the total is 12. Therefore, 12 is achievable.

Alternatively, maybe I need to think about the problem differently. Suppose we have four circles. Each pair contributes two points, so 6 pairs * 2 points = 12. So the maximum number is 12. So the answer is 12. Wait, but why is this problem considered? Maybe I need to check with an example. Let me think of specific circles.

Take four circles:

Circle 1: x² + y² = 1 (unit circle centered at origin)

Circle 2: (x - 1)² + y² = 1 (unit circle centered at (1,0))

They intersect at two points: (0.5, sqrt(3)/2) and (0.5, -sqrt(3)/2). So two points.

Now add Circle 3: Let me choose a circle that intersects Circle 1 and Circle 2 at two new points each. Let me choose a circle that's centered somewhere else. Let's say Circle 3: (x - 0.5)^2 + (y - 1)^2 = 0.5. Let me check intersections with Circle 1 and Circle 2.

First, with Circle 1: x² + y² = 1 and (x - 0.5)^2 + (y - 1)^2 = 0.5.

Expanding the second equation: x² - x + 0.25 + y² - 2y + 1 = 0.5 => x² + y² - x - 2y + 1.25 = 0.5 => x² + y² - x - 2y + 0.75 = 0.

But from Circle 1, x² + y² = 1. Substitute into the above equation: 1 - x - 2y + 0.75 = 0 => -x - 2y + 1.75 = 0 => x + 2y = 1.75. So the line of intersection between Circle 1 and Circle 3 is x + 2y = 1.75. Let me find the points of intersection.

Substitute x = 1.75 - 2y into Circle 1: (1.75 - 2y)^2 + y² = 1.

Calculating:

(3.0625 - 7y + 4y²) + y² = 1

3.0625 -7y +5y² = 1

5y² -7y + 2.0625 = 0

Multiply by 16 to eliminate decimals? Maybe not. Let me calculate discriminant.

Discriminant D = 49 - 4*5*(2.0625) = 49 - 20*2.0625 = 49 - 41.25 = 7.75

Which is positive, so two real solutions. So two intersection points between Circle 1 and Circle 3. Similarly, check intersections with Circle 2.

Circle 2: (x - 1)^2 + y² = 1.

Circle 3: (x - 0.5)^2 + (y - 1)^2 = 0.5.

Expanding Circle 3: x² - x + 0.25 + y² - 2y + 1 = 0.5 => x² + y² - x - 2y + 1.25 = 0.5 => same as before, x² + y² - x - 2y + 0.75 = 0.

From Circle 2: (x -1)^2 + y² = 1 => x² - 2x +1 + y² = 1 => x² + y² - 2x = 0. Substitute into Circle 3 equation:

(x² + y²) - x - 2y + 0.75 = 0 => (2x) - x - 2y + 0.75 = 0 => x - 2y + 0.75 = 0 => x = 2y - 0.75.

Substitute x into Circle 2: (2y - 0.75 - 1)^2 + y² = 1 => (2y - 1.75)^2 + y² = 1.

Expanding: 4y² - 7y + 3.0625 + y² = 1 => 5y² -7y + 2.0625 =0. Same equation as before. So the solutions are same? Wait, but this would imply that the intersections of Circle 2 and Circle 3 lie on the same line as Circle 1 and Circle 3? Wait, but that seems odd. Wait, but actually, since both intersections are derived from the same equation, perhaps the intersection points are the same as between Circle 1 and Circle 3? Wait, but that would mean that the three circles (1, 2, 3) intersect at two common points? Wait, but that would mean that the two circles 1 and 3 intersect at two points, and circles 2 and 3 also intersect at the same two points. Therefore, circles 1, 2, 3 all intersect at two common points. Which would mean that the two points are common to all three circles. Therefore, if that's the case, then those two points would be counted multiple times in the total intersection points. Hence, reducing the total number of unique points.

Wait, but in my previous calculation, when solving for intersections between Circle 1 and Circle 3, and between Circle 2 and Circle 3, I ended up with the same quadratic equation. Which suggests that the line of intersection between Circle 1 and 3 and the line of intersection between Circle 2 and 3 are different? Wait, no, the line x + 2y = 1.75 and x = 2y - 0.75. Let me check if they are the same.

From the first, x = 1.75 - 2y.

From the second, x = 2y - 0.75. Let me set them equal:

1.75 - 2y = 2y - 0.75 => 1.75 + 0.75 = 4y => 2.5 = 4y => y = 2.5 / 4 = 0.625. Then x = 2*(0.625) - 0.75 = 1.25 - 0.75 = 0.5. So the two lines intersect at (0.5, 0.625). Therefore, they are different lines. Therefore, the equations for intersections between 1 & 3 and between 2 & 3 are different lines. Therefore, their intersections are different points. Hence, perhaps the two circles 1 & 3 intersect at two points, and 2 & 3 also intersect at two different points. Wait, but why does solving for the intersection between 1 & 3 and between 2 & 3 give the same quadratic equation?

Wait, because when we substituted, we ended up with the same quadratic equation? Let me check again.

For Circle 1 and 3, we substituted x = 1.75 - 2y into Circle 1 equation and ended up with 5y² -7y + 2.0625 = 0.

Similarly, for Circle 2 and 3, substituting x = 2y - 0.75 into Circle 2 equation, we ended up with 5y² -7y + 2.0625 =0. So the same quadratic equation. Therefore, the solutions for y are the same? Wait, but if that's the case, then the two intersections for 1 & 3 are the same as for 2 & 3? That can't be, unless the circles 1 & 3 and 2 & 3 have the same intersection points. But how is that possible? Because we derived that the intersection points lie on both lines x + 2y = 1.75 and x = 2y - 0.75? Wait, no. Each intersection is between two circles, so for Circle 1 and 3, the intersection is on the line x + 2y = 1.75. For Circle 2 and 3, the intersection is on the line x = 2y - 0.75. So the points of intersection between 1 & 3 are on the first line, and those between 2 & 3 are on the second line. However, when we solved for the intersection of 1 & 3, we substituted into Circle 1, which gave us a quadratic with solutions. Similarly, for 2 & 3, substituting into Circle 2 gave us the same quadratic. Therefore, the solutions for y are same, but what about x?

Wait, if we have for the intersections between 1 & 3, the solutions for y come from 5y² -7y + 2.0625 = 0. Let me compute the solutions.

Discriminant D = 49 - 4*5*2.0625 = 49 - 41.25 = 7.75.

sqrt(7.75) ≈ 2.78

So y = [7 ± 2.78]/10 ≈ [7 ± 2.78]/10.

So approximately, y ≈ (7 + 2.78)/10 ≈ 9.78/10 ≈ 0.978 or y ≈ (7 - 2.78)/10 ≈ 4.22/10 ≈ 0.422.

Then x for 1 & 3 is 1.75 - 2y.

For y ≈ 0.978: x ≈ 1.75 - 2*(0.978) ≈ 1.75 - 1.956 ≈ -0.206

For y ≈ 0.422: x ≈ 1.75 - 2*(0.422) ≈ 1.75 - 0.844 ≈ 0.906

So two intersection points between 1 & 3: (-0.206, 0.978) and (0.906, 0.422)

Similarly, for intersections between 2 & 3, we have x = 2y - 0.75.

So for the same y ≈ 0.978 and 0.422?

Wait, no, because when we solved for Circle 2 and 3, we substituted into Circle 2. So if the quadratic equation is same, then these y's are same. But then x for 2 & 3 is 2y - 0.75.

So for y ≈ 0.978: x ≈ 2*(0.978) - 0.75 ≈ 1.956 - 0.75 ≈ 1.206

For y ≈ 0.422: x ≈ 2*(0.422) - 0.75 ≈ 0.844 - 0.75 ≈ 0.094

Therefore, points between Circle 2 & 3 are (1.206, 0.978) and (0.094, 0.422). These are two different points from the ones between 1 & 3. Therefore, there is no overlap. Therefore, circles 1, 2, 3 have 2 + 2 + 2 = 6 intersection points, with no three circles intersecting at the same point. Therefore, that works.

So, that seems okay. Then adding a fourth circle, say, Circle D, which intersects each of the previous three circles at two new points. So, Circle D needs to intersect Circle A at two new points, Circle B at two new points, and Circle C at two new points. If we can arrange it so that all these intersections are unique, then we add 6 new points, giving total of 6 + 6 = 12.

Therefore, the maximum number is 12. Therefore, the answer should be \boxed{12}.

Wait, but I need to make sure that with four circles, you can indeed have all those intersections. Maybe there's some geometric constraint? For example, maybe if you have too many circles, some of their intersection points have to coincide? But I think if you can place the circles in general position, such that no two circles are tangent and no three circles meet at a point, then it's possible. Since each circle can be adjusted independently to intersect others at two unique points.

Alternatively, maybe there's a different way to count. For example, the number of intersections for n circles can't exceed 2*C(n,2), since each pair can contribute at most 2. Therefore, 2*C(4,2)=12 is the upper bound, and since it's achievable, that's the maximum. So yeah, I think that's solid.

Alternatively, if I search my memory, I think that for the maximum number of regions created by circles, it's different, but here it's just intersection points. For regions, the formula is similar to the one for lines, but with different coefficients. But here, it's just about intersection points. So I think 12 is correct.

Therefore, my conclusion is that the maximum number of points where at least two of the circles intersect is 12. Therefore, the answer is \boxed{12}.

**Final Answer**
\boxed{12}

To determine the **maximum number of points where at least two of the four distinct circles intersect**, we must consider the fundamental geometric properties of circle intersections.

---

### Step 1: Understanding the Maximum Intersections Between Two Circles

Two distinct circles in a plane can intersect in **at most two points**. These can be visualized as two distinct points that both lie on the circumferences of the two circles. While they may intersect at **0** (no intersection), **1** (tangent), or (typically) **2** points.

Therefore, in the case of two circles, the **maximum** number of distinct intersection points is **2**.

---

### Step 2: Counting the Number of Pairs of Circles

We are given **four distinct circles**. The number of unordered pairs of circles we can choose from these four is:

$$
\binom{4}{2} = 6 \text{ pairs}
$$

Each of these six pairs of circles can, in theory, intersect at **two distinct points**, **provided** we avoid any overlap of these points (i.e., no three circles share a single intersection point).

---

### Step 3: Maximizing the Intersection Points

To achieve the **maximum** number of **distinct** intersection points, we must ensure:

- Each pair of circles intersects at **two distinct** points.
- No three circles intersect at the **same point**.

If we can arrange the circles in such a way, the total number of **distinct** intersection points is:

$$
\text{Number of pairs} \times \text{points per pair} = 6 \times 2 = 12
$$

This assumes **general position** for the circles — meaning that no two circles are tangent, and no three circles intersect at a single point.

---

### Step 4: Justification of the Result

This arrangement is **feasible**. By carefully choosing the radii and positions of the circles, it is possible to ensure that:

- Every pair of circles intersects at two unique points.
- No three circles intersect at the same point.
- All 12 intersection points are distinct.

This configuration can be constructed in Euclidean geometry, and thus the **maximum** number of intersection points is indeed:

$$
\boxed{12}
$$

Solution:
Each pair of circles has at most two intersection points. There are $\binom{4}{2} = 6$ pairs of circles, so there are at most $6\times 2 = 12$ points of intersection. The following configuration shows that $\boxed{12}$ points of intersection are indeed possible:

[asy]
draw(Circle((0,0),2));
draw(Circle((1,1.3),2));
draw(Circle((-1,1.3),2));
draw(Circle((0,0.7),2));
[/asy]


Explanation:
Okay, so I need to figure out how many digits repeat in the decimal expansion of 31 divided by 11111. Let me start by recalling that when a fraction is converted to a decimal, if the denominator has prime factors other than 2 and 5, the decimal will be repeating. Since 11111 is the denominator here, I should check its prime factors first.

Wait, actually, maybe I should just divide 31 by 11111 directly? But that might be tedious. Let me think. Alternatively, maybe there's a pattern in the decimal expansion based on the denominator. Let me recall that 1/11111 might have a repeating decimal with some period. Let me try to see.

First, let me check if 11111 is a prime number or if it can be factored. Let me try dividing it by some small primes. Let's see, 11111 divided by 3: 3 times 3703 is 11109, which leaves a remainder of 2. Not divisible by 3. Divided by 7? Let me do 7*1587=11109, which again leaves remainder 2. Not divisible by 7. How about 11? Let's apply the divisibility rule for 11: subtract and add digits alternately. For 11111: 1 - 1 + 1 - 1 + 1 = 1. Since 1 isn't divisible by 11, 11111 isn't divisible by 11.

How about 101? Let me try dividing 11111 by 101. 101*110 is 11110. So 11111 - 11110 = 1. So remainder 1. Not divisible by 101. How about 11111 divided by 41? Let me try that. 41*271 is 11111? Let me check: 41*270 = 11070, plus 41 is 11111. Yes! So 41*271 = 11111. So the denominator factors into 41 and 271. So 11111 = 41 * 271. Both of these are primes? Let me check 41 is prime, yes. 271: Let me check divisibility. Let's see, square root of 271 is about 16.4, so I need to check primes up to 17. Divided by 2: no, it's odd. Divided by 3: 2+7+1=10, not divisible by 3. Divided by 5: ends with 1, no. Divided by 7: 7*38=266, 271-266=5, not divisible. 11: 2 -7 +1 = -4, not divisible. 13: 13*20=260, 271-260=11, not divisible. 17: 17*15=255, 271-255=16, not divisible. So 271 is prime. Therefore, denominator factors into 41 and 271, both primes other than 2 and 5, so the decimal will repeat.

Now, the question is how long is the repeating part? For fractions with denominators that are co-prime to 10, the period of the decimal is the multiplicative order of 10 modulo the denominator. However, since our denominator is 11111, which factors into 41 and 271, the multiplicative order of 10 modulo 11111 would be the least common multiple of the multiplicative orders modulo 41 and 271. So, I need to compute the order of 10 modulo 41 and modulo 271, then take LCM of those two numbers.

Let me first compute the multiplicative order of 10 modulo 41. The multiplicative order is the smallest positive integer k such that 10^k ≡ 1 mod 41.

First, φ(41) = 40. So the order must divide 40. The divisors of 40 are 1, 2, 4, 5, 8, 10, 20, 40.

Let me check 10^k mod 41:

10^1 = 10 mod 41 ≠ 1

10^2 = 100 mod 41. 41*2=82, 100-82=18, so 18 ≠1

10^4: Let's compute (10^2)^2 = 18^2 = 324. 324 divided by 41: 41*7=287, 324-287=37. 37 mod 41 is 37, not 1.

10^5: 10^4 * 10 = 37 * 10 = 370. 370 divided by 41: 41*9=369. So 370 - 369 = 1. So 10^5 ≡ 1 mod 41. Therefore, the order of 10 modulo 41 is 5.

Now, compute the order of 10 modulo 271. φ(271) = 270. So the order divides 270. Let me factor 270: 270 = 2 * 3^3 * 5. Therefore, the possible orders are divisors of 270. Let me find the smallest k such that 10^k ≡ 1 mod 271.

This might take some time. Let me try to find it step by step.

First, check if 10^k ≡ 1 mod 271. Let me check the divisors of 270. Let me start with the prime factors. Let me note that 270 = 2 * 3^3 * 5. So the exponents are 2, 3^3, 5. The possible orders are the divisors of 270. Let me start with the factors. Let me check if 10^something is congruent to 1 mod 271.

First, check if 10^5 ≡ 1 mod 271? Let me compute 10^5 mod 271.

Compute 10^1 = 10

10^2 = 100

10^3 = 1000. 1000 divided by 271: 271*3=813, 1000-813=187. So 10^3 ≡ 187 mod 271.

10^4 = 10^3 * 10 ≡ 187*10 = 1870 mod 271. Let me divide 1870 by 271. 271*6=1626. 1870 - 1626 = 244. So 10^4 ≡ 244 mod 271.

10^5 = 244 * 10 = 2440 mod 271. Let me divide 2440 by 271. Let's see, 271*9 = 2439. So 2440 - 2439 = 1. Therefore, 10^5 ≡ 1 mod 271. Wait, so does that mean that the order of 10 modulo 271 is 5? But 5 divides 270? Yes, 270 divided by 5 is 54. But is there a smaller exponent?

Wait, but let me check if 5 is the minimal exponent. Let me check 10^1 mod 271 = 10 ≠1. 10^2=100≠1. 10^3=187≠1. 10^4=244≠1. 10^5=1. Therefore, the order is indeed 5. Wait, but that seems surprising. Is that correct?

Wait, 10^5 ≡ 1 mod 271? Let me verify again.

Starting with 10^1 = 10.

10^2 = 100.

10^3 = 10*100 = 1000.

1000 divided by 271:

271 * 3 = 813, as above. 1000 - 813 = 187. Correct.

10^4 = 10 * 187 = 1870. 271*6 = 1626, 1870 - 1626 = 244. Correct.

10^5 = 10 * 244 = 2440. 271*9 = 2439, so 2440 - 2439 = 1. Yes, that's correct. So 10^5 ≡ 1 mod 271. Therefore, the multiplicative order of 10 modulo 271 is 5.

Therefore, since the multiplicative orders modulo 41 and 271 are 5 and 5, the multiplicative order modulo 11111 is LCM(5,5) = 5. Therefore, the decimal expansion of 1/11111 has a repeating cycle of length 5? But wait, let me check with an example.

Wait, if 1/11111 has period 5, then 1/11111 = 0.\overline{abcde} where abcde is a 5-digit repeating block. However, let me check with actual division. Let me compute 1 divided by 11111.

But that might be tedious. Alternatively, since we know that 1/11111 = 1/(41*271). But since the multiplicative order of 10 modulo 11111 is 5, then the decimal repeats every 5 digits. Therefore, the decimal expansion of 31/11111 would also have the same period, since multiplying by 31 doesn't change the period, only the digits. So the number of repeating digits is 5? But let me verify with actual division.

Alternatively, maybe I made a mistake in assuming that the order is LCM(5,5) = 5. Let me think again. For modulus m = p*q, where p and q are coprime, then the multiplicative order of a modulo m is LCM of multiplicative orders modulo p and modulo q. So since 41 and 271 are coprime, then yes, the order modulo 11111 is LCM of 5 and 5, which is 5. Therefore, the period should be 5. Therefore, the decimal expansion of 31/11111 would have a repeating cycle of length 5. Therefore, the answer should be 5? Wait, but let me check an example.

Let me check 1/11111. If the period is 5, then 1/11111 = 0.000090000900009..., repeating every 5 digits? Wait, that seems odd. Wait, maybe my calculation is wrong. Let me check with another approach.

Alternatively, note that 1/11111. Let me note that 1/11111 is equal to 1/(10^5 - 1)/9. Wait, 10^5 -1 is 99999, so 99999/9 = 11111. Therefore, 1/11111 = 9/99999 = 0.\overline{00009}. Wait, but 0.\overline{00009} would have a repeating cycle of 5 digits: 00009. However, that seems like the repeating part is 5 digits. So then, if 1/11111 is 0.\overline{00009}, then multiplying numerator and denominator by 31 would give 31/11111 = 31 * 9 / 99999. Let me compute 31*9 = 279. Therefore, 279/99999 = 0.\overline{00279}. Wait, because 279 divided by 99999... Let me see, 0.002790027900279..., so the repeating part is 00279, which is 5 digits. Therefore, the number of repeating digits is 5. Therefore, the answer is 5?

But let me confirm by doing the division manually. Let me divide 31 by 11111.

So 31 divided by 11111 is 0.00... something. Let me perform the division step by step.

First, 31 divided by 11111. Since 31 is smaller than 11111, the decimal starts as 0. Then, we add decimal places.

Multiply numerator by 10: 310. Still less than 11111.

Multiply by 10 again: 3100. Still less.

Multiply by 10 again: 31000. Still less than 11111? No, 11111 is 11,111. 31,000 is bigger. Wait, 31,000 divided by 11,111 is approximately 2.8. So the first digit after decimal is 0, then 0, then 0, then 2? Wait, let me do it properly.

Let me write this out:

Dividing 31 by 11111:

31.000000...

11111 goes into 31 zero times. 0.

11111 goes into 310 zero times. 0.

11111 goes into 3100 zero times. 0.

11111 goes into 31000 two times (2*11111=22222). Subtract 22222 from 31000: 31000 - 22222 = 8778. Bring down next 0: 87780.

Now, 11111 goes into 87780 how many times? Let me calculate 11111*7 = 77777. Subtract that from 87780: 87780 - 77777 = 10003. Bring down 0: 100030.

11111 goes into 100030 nine times (11111*9 = 99999). Subtract 99999 from 100030: 100030 - 99999 = 31. Bring down 0: 310.

Now we see that we have a remainder of 31, which is where we started. Therefore, the decimal expansion is 0.002790027900279..., repeating every 5 digits: 00279. Wait, but the remainders start repeating from the beginning after 5 digits. Wait, but let me check:

First steps:

After 0.002 (three decimal places?), wait let me track the steps:

After first division steps:

After 0.002, then the remainder was 8778. Then we brought down a 0, making 87780. Then divided by 11111, got 7, remainder 10003. Then brought down 0, making 100030. Divided by 11111, got 9, remainder 31. Then brought down 0, making 310. Then we had 0.00279... and then remainder 31, which was our original dividend. So the cycle repeats. So the decimal expansion is 0.002790027900279..., so the repeating part is "00279", which is 5 digits. Therefore, the number of repeating digits is 5.

Therefore, the answer should be 5. But let me check if there is any possibility that the repeating part could be shorter. For example, if there was a shorter cycle. But since the period of 1/11111 is 5, and multiplying by 31 doesn't change the period length, only the digits, so the period should still be 5.

Alternatively, if there's a mistake in my order calculations. Wait, earlier, I found that 10^5 ≡ 1 mod 41 and mod 271, so LCM(5,5)=5, so the order modulo 11111 is 5. Therefore, the decimal expansion has period 5. Therefore, the answer is 5.

Alternatively, maybe the question is a trick question? Let me check 31/11111.

Alternatively, note that 11111 is 10^5 - 1 over 9, but maybe there's another way. Wait, but since we've done the division and saw that the decimal repeats every 5 digits, and confirmed with the multiplicative order, I think 5 is correct.

Wait, but let me check an example with 1/11111. If I do 1 divided by 11111, and if it's 0.\overline{00009}, then that's 5 digits. So 00009 repeating. But multiplying 00009 by 31 gives 279, so 0.\overline{00279}. That is, shifting the decimal? Wait, but 1/11111 is 0.0000900009..., so multiplying by 31 gives 0.0027900279..., which is 0.\overline{00279}. So the repeating part is 00279, which is 5 digits. Therefore, the answer is 5. Therefore, the number of digits repeating is 5. So I think the answer is 5. But let me check once again if maybe there's a shorter cycle.

Wait, suppose that the decimal repeats with a shorter period. For example, if 00279 repeats, but perhaps there is a smaller period? For example, could it be that the decimal is 0.0027900279..., but maybe the 00279 is 5 digits, but perhaps the first few 0s are non-repeating? But no, because when you have a fraction with denominator that's co-prime to 10, the decimal is purely repeating. Therefore, there are no non-repeating digits. So the entire decimal is repeating from the first decimal place. Therefore, the repeating cycle is 5 digits. Therefore, the answer is 5.

Therefore, after all this calculation and verification, I think the answer is 5.

**Final Answer**
\boxed{5}

To determine the number of repeating digits in the decimal expansion of the fraction $\frac{31}{11111}$, we begin by analyzing the nature of its decimal expansion.

---

### 1. Understanding the Structure of the Denominator

The given denominator is $11111$. Since it is not divisible by $2$ or $5$, the decimal expansion of $\frac{31}{11111}$ will **not terminate**, but instead be a **repeating** decimal.

To find the length of the **repeating period**, we use a key concept in number theory:
> The **length of the repeating block** in the decimal expansion of a fraction $\frac{a}{b}$ (where $b$ is coprime to 10) is equal to the **multiplicative order of 10 modulo $b$**.

So, we need to compute the **multiplicative order** of $10$ modulo $11111$.

---

### 2. Factoring the Denominator

Instead of directly computing the multiplicative order of 10 modulo 11111, we factor the denominator:
$$
11111 = 41 \times 271
$$
Both 41 and 271 are **prime numbers**.

Since the multiplicative order of 10 modulo a composite number $n = p \cdot q$, where $p$ and $q$ are coprime, is given by:
$$
\text{ord}_{10}(n) = \text{lcm}(\text{ord}_{10}(p), \text{ord}_{10}(q))
$$
we compute the multiplicative orders modulo 41 and 271.

---

### 3. Computing the Multiplicative Orders

- **Modulo 41**:

  The order of 10 mod 41 is the smallest $k$ such that $10^k \equiv 1 \mod 41$. Through trials, we find:
  $$
  10^5 \equiv 1 \mod 41
  $$
  So, $\text{ord}_{10}(41) = 5$

- **Modulo 271**:

  Similarly, we compute:
  $$
  10^5 \equiv 1 \mod 271
  $$
  So, $\text{ord}_{10}(271) = 5$

---

### 4. Applying the LCM Rule

$$
\text{ord}_{10}(11111) = \text{lcm}(\text{ord}_{10}(41), \text{ord}_{10}(271)) = \text{lcm}(5, 5) = 5
$$

This means that the decimal expansion of $\frac{1}{11111}$ (and any multiple of it) will have a **repeating block of length 5**.

---

### 5. Confirming with Direct Division (Optional)

If we perform the division $\frac{31}{11111}$ manually, or using a calculator, we find the decimal starts as:
$$
0.002790027900279\ldots
$$
The repeating block is **00279**, which contains **5 digits**.

---

### ✅ Final Conclusion

Since the decimal expansion of $\frac{31}{11111}$ has a repeating block of exactly 5 digits, the number of digits that repeat in the decimal is:

$$
\boxed{5}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many digits repeat in the decimal expansion of 31 divided by 11111. Let me start by recalling that when a fraction is converted to a decimal, if the denominator has prime factors other than 2 and 5, the decimal will be repeating. Since 11111 is the denominator here, I should check its prime factors first.

Wait, actually, maybe I should just divide 31 by 11111 directly? But that might be tedious. Let me think. Alternatively, maybe there's a pattern in the decimal expansion based on the denominator. Let me recall that 1/11111 might have a repeating decimal with some period. Let me try to see.

First, let me check if 11111 is a prime number or if it can be factored. Let me try dividing it by some small primes. Let's see, 11111 divided by 3: 3 times 3703 is 11109, which leaves a remainder of 2. Not divisible by 3. Divided by 7? Let me do 7*1587=11109, which again leaves remainder 2. Not divisible by 7. How about 11? Let's apply the divisibility rule for 11: subtract and add digits alternately. For 11111: 1 - 1 + 1 - 1 + 1 = 1. Since 1 isn't divisible by 11, 11111 isn't divisible by 11.

How about 101? Let me try dividing 11111 by 101. 101*110 is 11110. So 11111 - 11110 = 1. So remainder 1. Not divisible by 101. How about 11111 divided by 41? Let me try that. 41*271 is 11111? Let me check: 41*270 = 11070, plus 41 is 11111. Yes! So 41*271 = 11111. So the denominator factors into 41 and 271. So 11111 = 41 * 271. Both of these are primes? Let me check 41 is prime, yes. 271: Let me check divisibility. Let's see, square root of 271 is about 16.4, so I need to check primes up to 17. Divided by 2: no, it's odd. Divided by 3: 2+7+1=10, not divisible by 3. Divided by 5: ends with 1, no. Divided by 7: 7*38=266, 271-266=5, not divisible. 11: 2 -7 +1 = -4, not divisible. 13: 13*20=260, 271-260=11, not divisible. 17: 17*15=255, 271-255=16, not divisible. So 271 is prime. Therefore, denominator factors into 41 and 271, both primes other than 2 and 5, so the decimal will repeat.

Now, the question is how long is the repeating part? For fractions with denominators that are co-prime to 10, the period of the decimal is the multiplicative order of 10 modulo the denominator. However, since our denominator is 11111, which factors into 41 and 271, the multiplicative order of 10 modulo 11111 would be the least common multiple of the multiplicative orders modulo 41 and 271. So, I need to compute the order of 10 modulo 41 and modulo 271, then take LCM of those two numbers.

Let me first compute the multiplicative order of 10 modulo 41. The multiplicative order is the smallest positive integer k such that 10^k ≡ 1 mod 41.

First, φ(41) = 40. So the order must divide 40. The divisors of 40 are 1, 2, 4, 5, 8, 10, 20, 40.

Let me check 10^k mod 41:

10^1 = 10 mod 41 ≠ 1

10^2 = 100 mod 41. 41*2=82, 100-82=18, so 18 ≠1

10^4: Let's compute (10^2)^2 = 18^2 = 324. 324 divided by 41: 41*7=287, 324-287=37. 37 mod 41 is 37, not 1.

10^5: 10^4 * 10 = 37 * 10 = 370. 370 divided by 41: 41*9=369. So 370 - 369 = 1. So 10^5 ≡ 1 mod 41. Therefore, the order of 10 modulo 41 is 5.

Now, compute the order of 10 modulo 271. φ(271) = 270. So the order divides 270. Let me factor 270: 270 = 2 * 3^3 * 5. Therefore, the possible orders are divisors of 270. Let me find the smallest k such that 10^k ≡ 1 mod 271.

This might take some time. Let me try to find it step by step.

First, check if 10^k ≡ 1 mod 271. Let me check the divisors of 270. Let me start with the prime factors. Let me note that 270 = 2 * 3^3 * 5. So the exponents are 2, 3^3, 5. The possible orders are the divisors of 270. Let me start with the factors. Let me check if 10^something is congruent to 1 mod 271.

First, check if 10^5 ≡ 1 mod 271? Let me compute 10^5 mod 271.

Compute 10^1 = 10

10^2 = 100

10^3 = 1000. 1000 divided by 271: 271*3=813, 1000-813=187. So 10^3 ≡ 187 mod 271.

10^4 = 10^3 * 10 ≡ 187*10 = 1870 mod 271. Let me divide 1870 by 271. 271*6=1626. 1870 - 1626 = 244. So 10^4 ≡ 244 mod 271.

10^5 = 244 * 10 = 2440 mod 271. Let me divide 2440 by 271. Let's see, 271*9 = 2439. So 2440 - 2439 = 1. Therefore, 10^5 ≡ 1 mod 271. Wait, so does that mean that the order of 10 modulo 271 is 5? But 5 divides 270? Yes, 270 divided by 5 is 54. But is there a smaller exponent?

Wait, but let me check if 5 is the minimal exponent. Let me check 10^1 mod 271 = 10 ≠1. 10^2=100≠1. 10^3=187≠1. 10^4=244≠1. 10^5=1. Therefore, the order is indeed 5. Wait, but that seems surprising. Is that correct?

Wait, 10^5 ≡ 1 mod 271? Let me verify again.

Starting with 10^1 = 10.

10^2 = 100.

10^3 = 10*100 = 1000.

1000 divided by 271:

271 * 3 = 813, as above. 1000 - 813 = 187. Correct.

10^4 = 10 * 187 = 1870. 271*6 = 1626, 1870 - 1626 = 244. Correct.

10^5 = 10 * 244 = 2440. 271*9 = 2439, so 2440 - 2439 = 1. Yes, that's correct. So 10^5 ≡ 1 mod 271. Therefore, the multiplicative order of 10 modulo 271 is 5.

Therefore, since the multiplicative orders modulo 41 and 271 are 5 and 5, the multiplicative order modulo 11111 is LCM(5,5) = 5. Therefore, the decimal expansion of 1/11111 has a repeating cycle of length 5? But wait, let me check with an example.

Wait, if 1/11111 has period 5, then 1/11111 = 0.\overline{abcde} where abcde is a 5-digit repeating block. However, let me check with actual division. Let me compute 1 divided by 11111.

But that might be tedious. Alternatively, since we know that 1/11111 = 1/(41*271). But since the multiplicative order of 10 modulo 11111 is 5, then the decimal repeats every 5 digits. Therefore, the decimal expansion of 31/11111 would also have the same period, since multiplying by 31 doesn't change the period, only the digits. So the number of repeating digits is 5? But let me verify with actual division.

Alternatively, maybe I made a mistake in assuming that the order is LCM(5,5) = 5. Let me think again. For modulus m = p*q, where p and q are coprime, then the multiplicative order of a modulo m is LCM of multiplicative orders modulo p and modulo q. So since 41 and 271 are coprime, then yes, the order modulo 11111 is LCM of 5 and 5, which is 5. Therefore, the period should be 5. Therefore, the decimal expansion of 31/11111 would have a repeating cycle of length 5. Therefore, the answer should be 5? Wait, but let me check an example.

Let me check 1/11111. If the period is 5, then 1/11111 = 0.000090000900009..., repeating every 5 digits? Wait, that seems odd. Wait, maybe my calculation is wrong. Let me check with another approach.

Alternatively, note that 1/11111. Let me note that 1/11111 is equal to 1/(10^5 - 1)/9. Wait, 10^5 -1 is 99999, so 99999/9 = 11111. Therefore, 1/11111 = 9/99999 = 0.\overline{00009}. Wait, but 0.\overline{00009} would have a repeating cycle of 5 digits: 00009. However, that seems like the repeating part is 5 digits. So then, if 1/11111 is 0.\overline{00009}, then multiplying numerator and denominator by 31 would give 31/11111 = 31 * 9 / 99999. Let me compute 31*9 = 279. Therefore, 279/99999 = 0.\overline{00279}. Wait, because 279 divided by 99999... Let me see, 0.002790027900279..., so the repeating part is 00279, which is 5 digits. Therefore, the number of repeating digits is 5. Therefore, the answer is 5?

But let me confirm by doing the division manually. Let me divide 31 by 11111.

So 31 divided by 11111 is 0.00... something. Let me perform the division step by step.

First, 31 divided by 11111. Since 31 is smaller than 11111, the decimal starts as 0. Then, we add decimal places.

Multiply numerator by 10: 310. Still less than 11111.

Multiply by 10 again: 3100. Still less.

Multiply by 10 again: 31000. Still less than 11111? No, 11111 is 11,111. 31,000 is bigger. Wait, 31,000 divided by 11,111 is approximately 2.8. So the first digit after decimal is 0, then 0, then 0, then 2? Wait, let me do it properly.

Let me write this out:

Dividing 31 by 11111:

31.000000...

11111 goes into 31 zero times. 0.

11111 goes into 310 zero times. 0.

11111 goes into 3100 zero times. 0.

11111 goes into 31000 two times (2*11111=22222). Subtract 22222 from 31000: 31000 - 22222 = 8778. Bring down next 0: 87780.

Now, 11111 goes into 87780 how many times? Let me calculate 11111*7 = 77777. Subtract that from 87780: 87780 - 77777 = 10003. Bring down 0: 100030.

11111 goes into 100030 nine times (11111*9 = 99999). Subtract 99999 from 100030: 100030 - 99999 = 31. Bring down 0: 310.

Now we see that we have a remainder of 31, which is where we started. Therefore, the decimal expansion is 0.002790027900279..., repeating every 5 digits: 00279. Wait, but the remainders start repeating from the beginning after 5 digits. Wait, but let me check:

First steps:

After 0.002 (three decimal places?), wait let me track the steps:

After first division steps:

After 0.002, then the remainder was 8778. Then we brought down a 0, making 87780. Then divided by 11111, got 7, remainder 10003. Then brought down 0, making 100030. Divided by 11111, got 9, remainder 31. Then brought down 0, making 310. Then we had 0.00279... and then remainder 31, which was our original dividend. So the cycle repeats. So the decimal expansion is 0.002790027900279..., so the repeating part is "00279", which is 5 digits. Therefore, the number of repeating digits is 5.

Therefore, the answer should be 5. But let me check if there is any possibility that the repeating part could be shorter. For example, if there was a shorter cycle. But since the period of 1/11111 is 5, and multiplying by 31 doesn't change the period length, only the digits, so the period should still be 5.

Alternatively, if there's a mistake in my order calculations. Wait, earlier, I found that 10^5 ≡ 1 mod 41 and mod 271, so LCM(5,5)=5, so the order modulo 11111 is 5. Therefore, the decimal expansion has period 5. Therefore, the answer is 5.

Alternatively, maybe the question is a trick question? Let me check 31/11111.

Alternatively, note that 11111 is 10^5 - 1 over 9, but maybe there's another way. Wait, but since we've done the division and saw that the decimal repeats every 5 digits, and confirmed with the multiplicative order, I think 5 is correct.

Wait, but let me check an example with 1/11111. If I do 1 divided by 11111, and if it's 0.\overline{00009}, then that's 5 digits. So 00009 repeating. But multiplying 00009 by 31 gives 279, so 0.\overline{00279}. That is, shifting the decimal? Wait, but 1/11111 is 0.0000900009..., so multiplying by 31 gives 0.0027900279..., which is 0.\overline{00279}. So the repeating part is 00279, which is 5 digits. Therefore, the answer is 5. Therefore, the number of digits repeating is 5. So I think the answer is 5. But let me check once again if maybe there's a shorter cycle.

Wait, suppose that the decimal repeats with a shorter period. For example, if 00279 repeats, but perhaps there is a smaller period? For example, could it be that the decimal is 0.0027900279..., but maybe the 00279 is 5 digits, but perhaps the first few 0s are non-repeating? But no, because when you have a fraction with denominator that's co-prime to 10, the decimal is purely repeating. Therefore, there are no non-repeating digits. So the entire decimal is repeating from the first decimal place. Therefore, the repeating cycle is 5 digits. Therefore, the answer is 5.

Therefore, after all this calculation and verification, I think the answer is 5.

**Final Answer**
\boxed{5}

To determine the number of repeating digits in the decimal expansion of the fraction $\frac{31}{11111}$, we begin by analyzing the nature of its decimal expansion.

---

### 1. Understanding the Structure of the Denominator

The given denominator is $11111$. Since it is not divisible by $2$ or $5$, the decimal expansion of $\frac{31}{11111}$ will **not terminate**, but instead be a **repeating** decimal.

To find the length of the **repeating period**, we use a key concept in number theory:
> The **length of the repeating block** in the decimal expansion of a fraction $\frac{a}{b}$ (where $b$ is coprime to 10) is equal to the **multiplicative order of 10 modulo $b$**.

So, we need to compute the **multiplicative order** of $10$ modulo $11111$.

---

### 2. Factoring the Denominator

Instead of directly computing the multiplicative order of 10 modulo 11111, we factor the denominator:
$$
11111 = 41 \times 271
$$
Both 41 and 271 are **prime numbers**.

Since the multiplicative order of 10 modulo a composite number $n = p \cdot q$, where $p$ and $q$ are coprime, is given by:
$$
\text{ord}_{10}(n) = \text{lcm}(\text{ord}_{10}(p), \text{ord}_{10}(q))
$$
we compute the multiplicative orders modulo 41 and 271.

---

### 3. Computing the Multiplicative Orders

- **Modulo 41**:

  The order of 10 mod 41 is the smallest $k$ such that $10^k \equiv 1 \mod 41$. Through trials, we find:
  $$
  10^5 \equiv 1 \mod 41
  $$
  So, $\text{ord}_{10}(41) = 5$

- **Modulo 271**:

  Similarly, we compute:
  $$
  10^5 \equiv 1 \mod 271
  $$
  So, $\text{ord}_{10}(271) = 5$

---

### 4. Applying the LCM Rule

$$
\text{ord}_{10}(11111) = \text{lcm}(\text{ord}_{10}(41), \text{ord}_{10}(271)) = \text{lcm}(5, 5) = 5
$$

This means that the decimal expansion of $\frac{1}{11111}$ (and any multiple of it) will have a **repeating block of length 5**.

---

### 5. Confirming with Direct Division (Optional)

If we perform the division $\frac{31}{11111}$ manually, or using a calculator, we find the decimal starts as:
$$
0.002790027900279\ldots
$$
The repeating block is **00279**, which contains **5 digits**.

---

### ✅ Final Conclusion

Since the decimal expansion of $\frac{31}{11111}$ has a repeating block of exactly 5 digits, the number of digits that repeat in the decimal is:

$$
\boxed{5}
$$

Solution:
We begin by noting that    $$\frac{31}{11111} = \frac{31 \times 9}{11111 \times 9} = \frac{279}{99999}.$$We will show that   $$\frac{279}{99999} = 0.\overline{00279},$$so our final answer is $\boxed{5}.$

Proof that $279/99999 = 0.\overline{00279}$:

Let $s = 0.\overline{00279}$. Then multiplying both sides $10^5$ gives   $$10^5 s = 279.\overline{00279}.$$Subtracting the left side by $s$ and the right side by $0.\overline{00279}$ gives   $$99999s = 279,$$so $s = 279/99999$. It follows that $0.\overline{00279} = 279 / 99999,$ as desired.


Explanation:
Okay, so I need to solve the equation √(2x)/√(3x - 1) = 3/2. Hmm, let me think about how to approach this. First, maybe I should get rid of the square roots to make it easier to handle. I remember that if I have square roots in the numerator and denominator, I can square both sides of the equation to eliminate them. Let me try that.

So, starting with the original equation:

√(2x) / √(3x - 1) = 3/2

If I square both sides, that should help. Let me write that out:

[√(2x) / √(3x - 1)]² = (3/2)²

Simplifying both sides. On the left side, squaring the fraction would mean squaring the numerator and denominator separately. So:

( (√(2x))² ) / ( (√(3x - 1))² ) = 9/4

Simplifying the square roots squared, they just become the expressions inside the roots:

(2x) / (3x - 1) = 9/4

Okay, that looks better. Now I have a rational equation: 2x divided by (3x - 1) equals 9/4. Let me write that:

2x / (3x - 1) = 9/4

Now, to solve for x, I can cross-multiply. So:

2x * 4 = 9 * (3x - 1)

Calculating both sides:

Left side: 8x

Right side: 27x - 9

So now the equation is:

8x = 27x - 9

Hmm, let me subtract 8x from both sides to get all the x terms on one side:

8x - 8x = 27x - 9 - 8x

Which simplifies to:

0 = 19x - 9

Then, adding 9 to both sides:

9 = 19x

So dividing both sides by 19:

x = 9/19

Wait, but I need to check if this solution is valid in the original equation. Sometimes when you square both sides of an equation, you can introduce extraneous solutions. Let me verify.

First, let me check the domain of the original equation. The expressions under the square roots must be non-negative. So:

For √(2x), the expression inside is 2x ≥ 0 ⇒ x ≥ 0.

For √(3x - 1), the expression inside is 3x - 1 ≥ 0 ⇒ 3x ≥ 1 ⇒ x ≥ 1/3.

Therefore, the domain is x ≥ 1/3. So our solution x = 9/19 must be checked if it is greater than or equal to 1/3.

Let me compute 9/19. Well, 1/3 is approximately 0.333..., and 9 divided by 19 is approximately 0.473... So 0.473 is greater than 0.333, so it is in the domain. Good.

Now, let me plug x = 9/19 back into the original equation to check.

Original equation: √(2x)/√(3x - 1) = 3/2

Compute numerator: √(2*(9/19)) = √(18/19)

Denominator: √(3*(9/19) - 1) = √(27/19 - 1) = √(27/19 - 19/19) = √(8/19)

So the left side is √(18/19) / √(8/19) = √[(18/19)/(8/19)] = √(18/8) = √(9/4) = 3/2.

Which matches the right side. Therefore, x = 9/19 is indeed the correct solution.

But just to make sure I didn't make any mistakes in my algebra steps. Let me go through them again:

Original equation: √(2x)/√(3x - 1) = 3/2.

Square both sides: 2x/(3x - 1) = 9/4.

Cross-multiplying: 8x = 27x - 9.

Subtracting 8x: 0 = 19x - 9 ⇒ 19x = 9 ⇒ x = 9/19.

Yes, that all looks correct. And checking the solution in the original equation works. So I think that's the answer.

Wait, hold on. Let me check my calculation when I plugged in x = 9/19 into the denominator. Let me recalculate that step again to be thorough.

Denominator: 3x - 1 where x = 9/19.

So 3*(9/19) is 27/19. Then subtract 1: 27/19 - 1 = 27/19 - 19/19 = 8/19. Correct. Then square root of 8/19 is √(8)/√(19). Similarly, the numerator is √(2x) = √(18/19) = √18 / √19. So dividing numerator by denominator:

(√18 / √19) / (√8 / √19) = √18 / √8 = √(18/8) = √(9/4) = 3/2. Perfect. So that's correct.

Therefore, I think x = 9/19 is the correct solution. I don't see any mistakes in my steps. So I think that's the answer.

**Final Answer**
The solution is \boxed{\dfrac{9}{19}}.

To solve the equation

$$
\frac{\sqrt{2x}}{\sqrt{3x - 1}} = \frac{3}{2},
$$

we begin by eliminating the square roots. A standard technique is to square both sides of the equation to remove the radicals:

$$
\left(\frac{\sqrt{2x}}{\sqrt{3x - 1}}\right)^2 = \left(\frac{3}{2}\right)^2.
$$

Now, simplify both sides:

$$
\frac{2x}{3x - 1} = \frac{9}{4}.
$$

Next, solve the resulting rational equation by cross-multiplying:

$$
4 \cdot 2x = 9 \cdot (3x - 1).
$$

Simplify both sides:

$$
8x = 27x - 9.
$$

Now, isolate $ x $ by subtracting $ 27x $ from both sides:

$$
8x - 27x = -9 \quad \Rightarrow \quad -19x = -9.
$$

Solving for $ x $:

$$
x = \frac{9}{19}.
$$

---

### Step 1: Check the Domain

Before finalizing the solution, we confirm that the expression is valid under the original square roots.

- $ 2x \geq 0 $ ⇒ $ x \geq 0 $
- $ 3x - 1 \geq 0 $ ⇒ $ x \geq \frac{1}{3} $

The more restrictive condition is $ x \geq \frac{1}{3} $. Since $ \frac{9}{19} \approx 0.4737 > \frac{1}{3} $, this solution is valid.

---

### Step 2: Verify the Solution

Now verify that $ x = \frac{9}{19} $ satisfies the original equation:

- Numerator: $ \sqrt{2x} = \sqrt{2 \cdot \frac{9}{19}} = \sqrt{\frac{18}{19}} $
- Denominator: $ \sqrt{3x - 1} = \sqrt{3 \cdot \frac{9}{19} - 1} = \sqrt{\frac{27}{19} - \frac{19}{19}} = \sqrt{\frac{8}{19}} $

So the left-hand side becomes:

$$
\frac{\sqrt{\frac{18}{19}}}{\sqrt{\frac{8}{19}}} = \sqrt{\frac{18}{8}} = \sqrt{\frac{9}{4}} = \frac{3}{2},
$$

which matches the right-hand side of the original equation.

---

### Final Answer

$$
\boxed{\dfrac{9}{19}}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation √(2x)/√(3x - 1) = 3/2. Hmm, let me think about how to approach this. First, maybe I should get rid of the square roots to make it easier to handle. I remember that if I have square roots in the numerator and denominator, I can square both sides of the equation to eliminate them. Let me try that.

So, starting with the original equation:

√(2x) / √(3x - 1) = 3/2

If I square both sides, that should help. Let me write that out:

[√(2x) / √(3x - 1)]² = (3/2)²

Simplifying both sides. On the left side, squaring the fraction would mean squaring the numerator and denominator separately. So:

( (√(2x))² ) / ( (√(3x - 1))² ) = 9/4

Simplifying the square roots squared, they just become the expressions inside the roots:

(2x) / (3x - 1) = 9/4

Okay, that looks better. Now I have a rational equation: 2x divided by (3x - 1) equals 9/4. Let me write that:

2x / (3x - 1) = 9/4

Now, to solve for x, I can cross-multiply. So:

2x * 4 = 9 * (3x - 1)

Calculating both sides:

Left side: 8x

Right side: 27x - 9

So now the equation is:

8x = 27x - 9

Hmm, let me subtract 8x from both sides to get all the x terms on one side:

8x - 8x = 27x - 9 - 8x

Which simplifies to:

0 = 19x - 9

Then, adding 9 to both sides:

9 = 19x

So dividing both sides by 19:

x = 9/19

Wait, but I need to check if this solution is valid in the original equation. Sometimes when you square both sides of an equation, you can introduce extraneous solutions. Let me verify.

First, let me check the domain of the original equation. The expressions under the square roots must be non-negative. So:

For √(2x), the expression inside is 2x ≥ 0 ⇒ x ≥ 0.

For √(3x - 1), the expression inside is 3x - 1 ≥ 0 ⇒ 3x ≥ 1 ⇒ x ≥ 1/3.

Therefore, the domain is x ≥ 1/3. So our solution x = 9/19 must be checked if it is greater than or equal to 1/3.

Let me compute 9/19. Well, 1/3 is approximately 0.333..., and 9 divided by 19 is approximately 0.473... So 0.473 is greater than 0.333, so it is in the domain. Good.

Now, let me plug x = 9/19 back into the original equation to check.

Original equation: √(2x)/√(3x - 1) = 3/2

Compute numerator: √(2*(9/19)) = √(18/19)

Denominator: √(3*(9/19) - 1) = √(27/19 - 1) = √(27/19 - 19/19) = √(8/19)

So the left side is √(18/19) / √(8/19) = √[(18/19)/(8/19)] = √(18/8) = √(9/4) = 3/2.

Which matches the right side. Therefore, x = 9/19 is indeed the correct solution.

But just to make sure I didn't make any mistakes in my algebra steps. Let me go through them again:

Original equation: √(2x)/√(3x - 1) = 3/2.

Square both sides: 2x/(3x - 1) = 9/4.

Cross-multiplying: 8x = 27x - 9.

Subtracting 8x: 0 = 19x - 9 ⇒ 19x = 9 ⇒ x = 9/19.

Yes, that all looks correct. And checking the solution in the original equation works. So I think that's the answer.

Wait, hold on. Let me check my calculation when I plugged in x = 9/19 into the denominator. Let me recalculate that step again to be thorough.

Denominator: 3x - 1 where x = 9/19.

So 3*(9/19) is 27/19. Then subtract 1: 27/19 - 1 = 27/19 - 19/19 = 8/19. Correct. Then square root of 8/19 is √(8)/√(19). Similarly, the numerator is √(2x) = √(18/19) = √18 / √19. So dividing numerator by denominator:

(√18 / √19) / (√8 / √19) = √18 / √8 = √(18/8) = √(9/4) = 3/2. Perfect. So that's correct.

Therefore, I think x = 9/19 is the correct solution. I don't see any mistakes in my steps. So I think that's the answer.

**Final Answer**
The solution is \boxed{\dfrac{9}{19}}.

To solve the equation

$$
\frac{\sqrt{2x}}{\sqrt{3x - 1}} = \frac{3}{2},
$$

we begin by eliminating the square roots. A standard technique is to square both sides of the equation to remove the radicals:

$$
\left(\frac{\sqrt{2x}}{\sqrt{3x - 1}}\right)^2 = \left(\frac{3}{2}\right)^2.
$$

Now, simplify both sides:

$$
\frac{2x}{3x - 1} = \frac{9}{4}.
$$

Next, solve the resulting rational equation by cross-multiplying:

$$
4 \cdot 2x = 9 \cdot (3x - 1).
$$

Simplify both sides:

$$
8x = 27x - 9.
$$

Now, isolate $ x $ by subtracting $ 27x $ from both sides:

$$
8x - 27x = -9 \quad \Rightarrow \quad -19x = -9.
$$

Solving for $ x $:

$$
x = \frac{9}{19}.
$$

---

### Step 1: Check the Domain

Before finalizing the solution, we confirm that the expression is valid under the original square roots.

- $ 2x \geq 0 $ ⇒ $ x \geq 0 $
- $ 3x - 1 \geq 0 $ ⇒ $ x \geq \frac{1}{3} $

The more restrictive condition is $ x \geq \frac{1}{3} $. Since $ \frac{9}{19} \approx 0.4737 > \frac{1}{3} $, this solution is valid.

---

### Step 2: Verify the Solution

Now verify that $ x = \frac{9}{19} $ satisfies the original equation:

- Numerator: $ \sqrt{2x} = \sqrt{2 \cdot \frac{9}{19}} = \sqrt{\frac{18}{19}} $
- Denominator: $ \sqrt{3x - 1} = \sqrt{3 \cdot \frac{9}{19} - 1} = \sqrt{\frac{27}{19} - \frac{19}{19}} = \sqrt{\frac{8}{19}} $

So the left-hand side becomes:

$$
\frac{\sqrt{\frac{18}{19}}}{\sqrt{\frac{8}{19}}} = \sqrt{\frac{18}{8}} = \sqrt{\frac{9}{4}} = \frac{3}{2},
$$

which matches the right-hand side of the original equation.

---

### Final Answer

$$
\boxed{\dfrac{9}{19}}
$$

Solution:
We can start by cross-multiplying: \begin{align*} 3\sqrt{3x-1}&=2\sqrt{2x}
\\\Rightarrow \qquad (3\sqrt{3x-1})^2 &=(2\sqrt{2x})^2
\\\Rightarrow \qquad 9(3x-1)& =4(2x)
\\\Rightarrow \qquad 27x-9& =8x
\\ \Rightarrow \qquad19x&=9
\\ \Rightarrow \qquad x&=\boxed{\frac9{19}}.
\end{align*}Checking, we see that this value of $x$ does indeed work, so it is not an extraneous solution.