<Question>
Rectangle $ABCD$ has center $O$ and $AB/AD=k$. A point is randomly chosen from the interior of rectangle $ABCD$. What is the probability that it is closer to $O$ than to any of the four vertices?  [asy]
size(200);
draw((-250,100)--(250,100)--(250,-100)--(-250,-100)--cycle);
dot((0,0));
label("$O$",(0,0),N);
label("$A$",(-250,100),NW); label("$B$",(250,100),NE); label("$C$",(250,-100),SE); label("$D$",(-250,-100),SW);[/asy]
</Question>

<think>
<Plan_1>
Alright, so I need to find the probability that a randomly chosen point inside rectangle ABCD is closer to the center O than to any of the four vertices. The rectangle has center O, and the ratio of AB to AD is k. Let me start by visualizing the problem.

<prompt_1.1> Visualize the problem by considering a coordinate system for the rectangle and identifying the regions where points are closer to the center than to the vertices. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let me set up a coordinate system. Let's assume the center O is at the origin (0,0). Since ABCD is a rectangle, the vertices can be defined based on the lengths of AB and AD. Let me denote AB as the horizontal side and AD as the vertical side. Given that AB/AD = k, so if I let AD = h, then AB = k*h. Therefore, the rectangle's vertices would be at A(-k*h/2, h/2), B(k*h/2, h/2), C(k*h/2, -h/2), and D(-k*h/2, -h/2). Wait, actually, since AB is the length and AD is the width, maybe I should adjust that.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
If AB is the horizontal side, then AB = k*AD. Let me set AD = a, so AB = k*a. Then, the rectangle extends from (-k*a/2, -a/2) to (k*a/2, a/2). Hmm, maybe it's simpler to set the rectangle such that AB is along the x-axis and AD along the y-axis. Let me define the rectangle with length AB = 2a and width AD = 2b, so that the center O is at (0,0), and the vertices are at A(-a, b), B(a, b), C(a, -b), and D(-a, -b). Then, the ratio AB/AD = 2a / 2b = a/b = k.

<prompt_2.1> Define the rectangle using a coordinate system with the center at the origin and calculate the probability that a point is closer to the center than to any vertex. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
So, a/b = k, which means b = a/k. That might be useful later.

Now, the problem is to find the probability that a randomly chosen point (x,y) inside the rectangle is closer to O(0,0) than to any of the four vertices A, B, C, D. The probability would be the area of the region where this condition holds divided by the area of the entire rectangle.

First, let's compute the area of the rectangle. Since AB = 2a and AD = 2b, the area is (2a)*(2b) = 4ab. But since b = a/k, substituting that gives 4a*(a/k) = 4a²/k. Alternatively, maybe it's easier to set specific values for a and b to simplify calculations. Let me see. If I set a = k and b = 1, then the rectangle would have length 2k and height 2.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Alternatively, maybe even set a = 1 and b = 1/k. But perhaps keeping it general is better for now.

The key is to find the region where the distance from (x,y) to O is less than the distance to any of the four vertices. The distance from (x,y) to O is sqrt(x² + y²). The distance to each vertex is sqrt((x ± a)^2 + (y ± b)^2). So, the condition is sqrt(x² + y²) < sqrt((x - a)^2 + (y - b)^2) for vertex A, and similarly for the other vertices.

<prompt_3.1> Set up the conditions for a point (x, y) to be closer to the center O than to any of the four vertices using distance inequalities. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
But since the rectangle is symmetric, maybe we can consider one quadrant and multiply the area by 4? Let me think.

Wait, the rectangle is symmetric with respect to both the x-axis and y-axis. Also, the problem is symmetric in the sense that the regions near each vertex would be similar. Therefore, perhaps the region where the point is closer to O than to any vertex is also symmetric in each quadrant. So maybe I can compute the area in one quadrant and multiply by 4. Let me confirm.

Consider the first quadrant where x ≥ 0 and y ≥ 0. In this quadrant, the vertex is B(a, b). So, in this quadrant, the condition for a point (x,y) to be closer to O than to B is sqrt(x² + y²) < sqrt((x - a)^2 + (y - b)^2). If I can find the area in this quadrant where this inequality holds, then multiplying by 4 (for all four quadrants) would give the total area where the point is closer to O than to any vertex.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Similarly, the distance conditions for the other vertices (A, C, D) would be similar in their respective quadrants. So yes, computing one quadrant and multiplying by 4 seems feasible.

So, let's focus on the first quadrant: x ≥ 0, y ≥ 0, with vertex B(a, b). The condition is sqrt(x² + y²) < sqrt((x - a)^2 + (y - b)^2). Let's square both sides to eliminate the square roots:

x² + y² < (x - a)² + (y - b)²

<prompt_4.1> Solve the inequality for the first quadrant to find the region where points are closer to the center than to vertex B. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Expanding the right side:

(x² - 2a x + a²) + (y² - 2b y + b²)

So, simplifying:

x² + y² < x² - 2a x + a² + y² - 2b y + b²

Subtract x² + y² from both sides:

0 < -2a x + a² - 2b y + b²

Rearranging:

2a x + 2b y < a² + b²

Divide both sides by 2:

a x + b y < (a² + b²)/2

So, in the first quadrant, the region where the point is closer to O than to vertex B is the set of (x, y) satisfying a x + b y < (a² + b²)/2, along with x ≥ 0, y ≥ 0, and also inside the rectangle (so x ≤ a, y ≤ b).

Therefore, the boundary in the first quadrant is the line a x + b y = (a² + b²)/2. The region we want is below this line.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Similarly, in other quadrants, the boundaries would be lines corresponding to the other vertices. For example, in the second quadrant (x ≤ 0, y ≥ 0), the vertex is A(-a, b), so the condition would be sqrt(x² + y²) < sqrt((x + a)^2 + (y - b)^2), leading to -a x + b y < (a² + b²)/2. Similarly, in the third quadrant (x ≤ 0, y ≤ 0), the vertex is D(-a, -b), so the condition would be -a x - b y < (a² + b²)/2, and in the fourth quadrant (x ≥ 0, y ≤ 0), the vertex is C(a, -b), leading to a x - b y < (a² + b²)/2.

Therefore, the overall region where a point is closer to O than to any vertex is the intersection of the regions defined by these four inequalities, along with the rectangle boundaries.

<prompt_5.1> Determine the shape of the region where points are closer to the center than to any vertex and plan to calculate its area. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
But actually, in each quadrant, the inequality is a linear inequality, so the intersection of all four regions would form a polygon. The shape of this region is likely a convex polygon, possibly a quadrilateral or a hexagon, depending on how the lines intersect. To find the total area, we need to find the area in each quadrant satisfying the respective inequality and sum them up.

But since the problem is symmetric, maybe the overall shape is a diamond or another rectangle rotated by 45 degrees, but clipped by the original rectangle. However, given the inequalities, it might be a more complex shape. Let me try to visualize or sketch this mentally.

Alternatively, maybe the region is a circle, but since we are dealing with distance to a point (O) versus distance to multiple points (vertices), the set of points closer to O than to any vertex is the intersection of the Voronoi regions around O with respect to the vertices. However, Voronoi diagrams usually partition the plane into regions based on proximity to each site (here, the vertices), but since O is a single point, the region closer to O than to any vertex would be the intersection of the half-planes defined by the perpendicular bisectors between O and each vertex.

Wait, exactly. The set of points closer to O than to any vertex is the intersection of the four regions defined by the perpendicular bisectors between O and each vertex. Each perpendicular bisector between O and a vertex is the set of points equidistant to O and that vertex. Therefore, the region where a point is closer to O than to any vertex is the intersection of the four half-planes lying on the side of the bisector closer to O.

Therefore, the boundary of this region is formed by the four perpendicular bisectors between O and each vertex. Each bisector is a line equidistant to O and a vertex, and the intersection of the four half-planes (closer to O) would form a convex polygon. Therefore, the region is a quadrilateral (possibly a rectangle or another shape) bounded by these four lines.

Therefore, to find the area, I need to find the equations of these four bisectors, find their intersection points, which would be the vertices of the region, and then compute the area of that polygon. Then, divide by the area of the rectangle to get the probability.

Alternatively, since in each quadrant the region is a triangle or a trapezoid, maybe it's easier to compute the area in one quadrant and multiply by four. Let me go back to the first quadrant.

In the first quadrant, the bisector between O and B(a, b) is the line a x + b y = (a² + b²)/2. The region closer to O is below this line. However, this line might intersect the sides of the rectangle. So, the area in the first quadrant is the area under the line a x + b y = (a² + b²)/2 within the first quadrant.

Similarly, in other quadrants, the bisectors will have similar equations but with signs adjusted for the quadrant. For example, in the second quadrant, the bisector between O and A(-a, b) is -a x + b y = (a² + b²)/2. The region closer to O would be where -a x + b y < (a² + b²)/2. But in the second quadrant, x ≤ 0 and y ≥ 0.

Therefore, perhaps the region closer to O is a polygon bounded by four lines, each in a different quadrant. The intersection points of these lines will give the vertices of the region. Let me try to find these intersection points.

First, let's consider the first quadrant. The bisector is a x + b y = (a² + b²)/2. Let's find where this line intersects the edges of the rectangle. The first quadrant part of the rectangle goes from x=0 to x=a and y=0 to y=b. So, the line a x + b y = (a² + b²)/2 will intersect the top edge y = b at some x, and the right edge x = a at some y.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Let me compute that.

<prompt_6.1> Find the intersection points of the bisector with the edges of the rectangle in the first quadrant. </prompt_6.1>
-----

Based on execution_5.1:
Let me verify that.

Suppose we have the line a x + b y = (a² + b²)/2.

<prompt_6.2> Verify the intersection points of the bisector with the edges of the rectangle and determine the region. </prompt_6.2>
</Plan_6>

<Execution_6>
<execution_6.1>
Setting y = b in the equation:

a x + b*b = (a² + b²)/2

=> a x = (a² + b²)/2 - b² = (a² - b²)/2

=> x = (a² - b²)/(2a)

Similarly, setting x = a:

a*a + b y = (a² + b²)/2

=> b y = (a² + b²)/2 - a² = (-a² + b²)/2

=> y = ( -a² + b² )/(2b )

But since we are in the first quadrant, x and y must be between 0 and a, 0 and b respectively. So, x = (a² - b²)/(2a). Depending on whether a > b or b > a, this x could be positive or negative. Similarly, y = (b² - a²)/(2b). So, if a > b, then x is positive, y is negative, but since we are in the first quadrant, y cannot be negative. Therefore, the intersection with the top edge y = b is at x = (a² - b²)/(2a), and the intersection with the right edge x = a is at y = (b² - a²)/(2b), but since y must be ≥ 0, if a > b, then y would be negative, which is outside the rectangle. Therefore, in the first quadrant, if a > b, the line intersects the top edge at x = (a² - b²)/(2a) and the right edge is outside the rectangle. Similarly, if a < b, the line would intersect the right edge at y = (b² - a²)/(2b) and the top edge is outside.
</execution_6.1>

<execution_6.2>
To find where it intersects the top edge y = b, set y = b:

a x + b*b = (a² + b²)/2

=> a x = (a² + b²)/2 - b² = (a² - b²)/2

=> x = (a² - b²)/(2a)

Similarly, to find where it intersects the right edge x = a:

a*a + b y = (a² + b²)/2

=> b y = (a² + b²)/2 - a² = (-a² + b²)/2

=> y = ( -a² + b² )/(2b )

So, the line intersects the top edge at ( (a² - b²)/(2a), b ) and the right edge at ( a, ( -a² + b² )/(2b ) ). However, we need to check if these intersection points lie within the edges.

For the top edge y = b, x ranges from 0 to a. The x-coordinate found is (a² - b²)/(2a). Let's compute this:

(a² - b²)/(2a) = (a/2) - (b²)/(2a)

Since a and b are positive, this value could be positive or negative depending on whether a² > b² or not. Similarly, for the right edge x = a, the y-coordinate is ( -a² + b² )/(2b ). Let's compute this:

( -a² + b² )/(2b ) = (b² - a²)/(2b ) = (b/2) - (a²)/(2b )

Again, this could be positive or negative.

Wait, but in the first quadrant, x and y are both positive. So, if (a² - b²)/(2a) is positive, then the intersection is within the top edge. Similarly, if ( -a² + b² )/(2b ) is positive, then the intersection is within the right edge.

Therefore, we have two cases:

1. If a > b: Then, a² - b² > 0, so (a² - b²)/(2a) > 0. Similarly, if b > a, then ( -a² + b² )/(2b ) > 0. If a = b, then both intersections would be at the midpoint.
</execution_6.2>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Therefore, depending on whether a > b or b > a, the line intersects either the top edge or the right edge within the first quadrant.

Therefore, the region in the first quadrant where a x + b y < (a² + b²)/2 is a polygon bounded by:

<prompt_7.1> Analyze the intersection points and determine the shape of the region in the first quadrant. </prompt_7.1>
-----

Based on execution_6.2:
Let's analyze for a > b and a < b.

Case 1: a > b

In this case, (a² - b²)/(2a) is positive, so the line intersects the top edge at ( (a² - b²)/(2a), b ). However, the y-coordinate is b, which is the maximum y in the first quadrant. The other intersection with the right edge x = a is at y = ( -a² + b² )/(2b ), which is negative, so outside the first quadrant. Therefore, in the first quadrant, the line intersects the top edge at x = (a² - b²)/(2a), y = b, and the origin (0,0)?

<prompt_7.2> Analyze the region in the first quadrant for the case when a > b. </prompt_7.2>

Case 2: a < b

In this case, if we set x = a, then y = ( -a² + b² )/(2b ). Since a < b, b² - a² > 0, so y is positive. Therefore, the line intersects the right edge at ( a, (b² - a²)/(2b ) ). Similarly, setting y = b gives x = (a² - b²)/(2a ), which is negative, so outside the first quadrant. Therefore, in the first quadrant, the line intersects the right edge at x = a, y = (b² - a²)/(2b ) and perhaps the origin?

<prompt_7.3> Analyze the region in the first quadrant for the case when a < b. </prompt_7.3>

Case 3: a = b

If a = b, then the line becomes a x + a y = (a² + a²)/2 = a².

<prompt_7.4> Analyze the region in the first quadrant for the case when a = b. </prompt_7.4>
</Plan_7>

<Execution_7>
<execution_7.1>
- The origin (0,0)
- The intersection with the top edge, if within the edge
- The intersection with the right edge, if within the edge
- The point (a,0) or (0,b), depending on where the line intersects the edges

Wait, actually, the line a x + b y = (a² + b²)/2 may not necessarily pass through the origin. Let me check.

If x = 0, then b y = (a² + b²)/2 => y = (a² + b²)/(2b ). Similarly, if y = 0, then a x = (a² + b²)/2 => x = (a² + b²)/(2a ). So, unless (a² + b²)/(2b ) ≤ b and (a² + b²)/(2a ) ≤ a, the intercepts would be within the edges. Let's check:

For y-intercept: (a² + b²)/(2b ) ≤ b ?

=> (a² + b²) ≤ 2b²

=> a² ≤ b²

=> a ≤ b

Similarly, x-intercept: (a² + b²)/(2a ) ≤ a ?

=> (a² + b²) ≤ 2a²

=> b² ≤ a²

=> b ≤ a

Therefore, if a ≤ b, then the y-intercept is within the edge (since (a² + b²)/(2b ) ≤ b ), and the x-intercept is outside (since (a² + b²)/(2a ) > a ). Conversely, if a ≥ b, the x-intercept is within the edge and the y-intercept is outside. If a = b, both intercepts are equal to (a² + a²)/(2a ) = a, so the intercepts are at (a,0) and (0,a), which is the line x + y = a, passing through (a,0) and (0,a).

Therefore, in the first quadrant, depending on whether a is greater than or less than b, the line intersects either the top edge or the right edge, but not both. Moreover, the region closer to O is below the line. So, if a > b, the line intersects the top edge at ( (a² - b²)/(2a ), b ) and the origin (0,0), forming a triangle. Similarly, if a < b, the line intersects the right edge at (a, (b² - a²)/(2b )) and the origin, forming a triangle. If a = b, the line is x + y = a, forming a triangle with vertices at (0,0), (a,0), and (0,a).

Wait, let's verify this. If a > b, then the line intersects the top edge at ( (a² - b²)/(2a ), b ) and the right edge is outside. So, the region closer to O is bounded by the line from (0,0) to ( (a² - b²)/(2a ), b ). Similarly, if a < b, the line intersects the right edge at (a, (b² - a²)/(2b )) and the top edge is outside, so the region is bounded by (0,0) to (a, (b² - a²)/(2b )). If a = b, the line is x + y = a, which connects (a,0) and (0,a).
</execution_7.1>

<execution_7.2>
Wait, no. When a > b, the line intersects the top edge at x = (a² - b²)/(2a), y = b, and the right edge is outside. So, the region in the first quadrant is bounded by the line from the origin (0,0) to the intersection point ( (a² - b²)/(2a), b ). Wait, but the line equation is a x + b y = (a² + b²)/2. Let me check if the origin (0,0) satisfies the inequality a x + b y < (a² + b²)/2. Plugging in (0,0), we get 0 < (a² + b²)/2, which is true. Therefore, the region in the first quadrant is the area below the line, which includes the origin and extends up to the intersection point with the top edge.

But wait, if the line intersects the top edge at ( (a² - b²)/(2a), b ), then the region in the first quadrant is a polygon with vertices at (0,0), (0, b), ( (a² - b²)/(2a), b ). Wait, but (0, b) is a corner of the rectangle. Let me confirm.

Wait, the line a x + b y = (a² + b²)/2 in the first quadrant. At x = 0, y = (a² + b²)/(2b ). But since the rectangle's top edge is at y = b, and (a² + b²)/(2b ) compared to b: (a² + b²)/(2b ) = (a²)/(2b ) + b/2. Since a > b, (a²)/(2b ) > (b²)/(2b ) = b/2. Therefore, (a² + b²)/(2b ) > b/2 + b/2 = b. Therefore, at x = 0, the line is above the top edge of the rectangle. Therefore, the line does not intersect the left edge (x=0) within the rectangle, but intersects the top edge at ( (a² - b²)/(2a), b ). Similarly, at y = 0, x = (a² + b²)/(2a ). Since a > b, (a² + b²)/(2a ) = a/2 + b²/(2a ). Since a > b, b²/(2a ) < a/2, so x = (a² + b²)/(2a ) < a/2 + a/2 = a. Therefore, the line intersects the bottom edge y=0 at x = (a² + b²)/(2a ), which is less than a. Therefore, the line intersects the bottom edge at ( (a² + b²)/(2a ), 0 ) and the top edge at ( (a² - b²)/(2a ), b ). Therefore, in the first quadrant, the region closer to O is the quadrilateral bounded by (0,0), ( (a² + b²)/(2a ), 0 ), ( (a² - b²)/(2a ), b ), and (0, b ). Wait, but (0, b ) is a corner of the rectangle. Let me check if (0, b ) is on the line.

Wait, plugging x=0, y=b into the line equation: a*0 + b*b = b². The right-hand side is (a² + b²)/2. So, unless a = b, these are not equal. Therefore, (0, b ) is not on the line. Therefore, the line in the first quadrant connects ( (a² + b²)/(2a ), 0 ) to ( (a² - b²)/(2a ), b ). Therefore, the region closer to O is the area below this line, which is a quadrilateral with vertices at (0,0), ( (a² + b²)/(2a ), 0 ), ( (a² - b²)/(2a ), b ), and (0, b ). Wait, but (0, b ) is a corner, and ( (a² - b²)/(2a ), b ) is somewhere along the top edge.
</execution_7.2>

<execution_7.3>
Wait, if x = 0, then a*0 + b y = (a² + b²)/2 => y = (a² + b²)/(2b ). If a < b, then (a² + b²)/(2b ) = (b² + a²)/(2b ) = b/2 + a²/(2b ). Since a < b, a²/(2b ) < b/2, so total y = b/2 + something less than b/2, so total y < b. Therefore, when x = 0, y = (a² + b²)/(2b ) < b. Therefore, the line intersects the left edge y-axis at (0, (a² + b²)/(2b )) and the right edge at (a, (b² - a²)/(2b )). Wait, but let's check:

Wait, if a < b, then setting x = 0:

a*0 + b y = (a² + b²)/2 => y = (a² + b²)/(2b )

Since a < b, (a² + b²)/(2b ) = (b² + a²)/(2b ) = b/2 + a²/(2b ). Since a < b, a²/(2b ) < b/2, so y = b/2 + something less than b/2, so y < b. Therefore, the line intersects the left edge at (0, (a² + b²)/(2b )) and the right edge at (a, (b² - a²)/(2b )). Wait, but (b² - a²)/(2b ) is positive because a < b. So, in this case, the line in the first quadrant intersects the left edge at (0, (a² + b²)/(2b )) and the right edge at (a, (b² - a²)/(2b )). Therefore, the region in the first quadrant where a x + b y < (a² + b²)/2 is the area below the line connecting these two points.

But wait, when a < b, the line from (0, (a² + b²)/(2b )) to (a, (b² - a²)/(2b )). Let me compute these coordinates:

First point: (0, (a² + b²)/(2b )) = (0, b/2 + a²/(2b )).

Second point: (a, (b² - a²)/(2b )) = (a, b/2 - a²/(2b )).

So, connecting these two points in the first quadrant, the region below the line is a trapezoid or triangle? Let me see. If you connect (0, c) to (a, d), where c and d are positive, then the area below the line is a trapezoid with vertices at (0,0), (0, c), (a, d), and (a, 0). Wait, no. The inequality is a x + b y < (a² + b²)/2. So, the boundary is the line, and the region closer to O is below the line. So, in the first quadrant, the region is bounded by x ≥ 0, y ≥ 0, and a x + b y < (a² + b²)/2. Therefore, the vertices of this region would be (0,0), (0, c), (a, d), and (a,0) if the line intersects both the left and right edges. Wait, but if a < b, the line intersects the left edge at (0, c) and the right edge at (a, d), so the region is a quadrilateral with vertices at (0,0), (0, c), (a, d), and (a, 0). But actually, no. The inequality is a x + b y < (a² + b²)/2. So, for x=0, y can be from 0 to c. For y=0, x can be from 0 to a. The line connects (0, c) to (a, d). Therefore, the region is the area under the line, which is a quadrilateral with vertices at (0,0), (0, c), (a, d), and (a,0). Wait, but actually, when x=0, y can go up to c, but the line is only part of the boundary. Wait, perhaps it's a triangle with vertices at (0,0), (0, c), and (a, d). Let me check.

Suppose we have the line connecting (0, c) and (a, d). The region a x + b y < (a² + b²)/2 is below this line. So, in the first quadrant, this region is bounded by:

- From (0,0) to (0, c): along the y-axis.

- From (0, c) to (a, d): along the line.

- From (a, d) to (a, 0): along the x-axis.

Wait, but actually, the line connects (0, c) to (a, d), so the region below the line in the first quadrant would be a polygon with vertices at (0,0), (0, c), (a, d), and (a,0). But (a,0) is not on the line unless d = (a² + b²)/(2b ). Wait, no. Let me compute when x = a, y = (b² - a²)/(2b ). So, if a < b, then y is positive. So, the line connects (0, c) to (a, d), where c = (a² + b²)/(2b ) and d = (b² - a²)/(2b ). Therefore, the region below the line in the first quadrant is a quadrilateral with vertices at (0,0), (0, c), (a, d), and (a,0). However, the line from (0, c) to (a, d) is the boundary, so the area under the line would be the area of the trapezoid formed by these points. But actually, when x goes from 0 to a, y decreases from c to d. Therefore, the region is a trapezoid with bases of length c and d and height a. The area would be (c + d)/2 * a.

But let's verify. The area under the line in the first quadrant can be calculated as the integral from x=0 to x=a of y(x) dx, where y(x) = [ (a² + b²)/2 - a x ] / b. So:

Area = ∫ from 0 to a [ (a² + b²)/(2b ) - (a x)/b ] dx

= [ (a² + b²)/(2b ) * x - (a / (2b )) x² ] from 0 to a

= [ (a² + b²)/(2b ) * a - (a / (2b )) * a² ] - 0

= [ a(a² + b²)/(2b ) - a³/(2b ) ]

= [ (a³ + a b² )/(2b ) - a³/(2b ) ]

= (a³ + a b² - a³ )/(2b )

= (a b² )/(2b )

= a b / 2

Wait, that's interesting. The area in the first quadrant where a x + b y < (a² + b²)/2 is (a b)/2. But if a < b, then this would be (a b)/2 regardless. Wait, but in the case when a = b, this gives (a²)/2, which is a square with side a, area a², but (a²)/2 would be half of that. Wait, but the total area of the rectangle when a = b is 4a², so the total region would be 4*(a²/2) = 2a², and the probability would be 2a² / 4a² = 1/2. But that seems high. Wait, if a = b, the rectangle is a square, and the region closer to the center than to any vertex would be the intersection of the four regions defined by the perpendicular bisectors. Let me think. If the rectangle is a square, then the Voronoi regions around the center would be the set of points closer to the center than to any vertex. In a square, the region closer to the center than to the vertices would be another square rotated by 45 degrees, whose diagonals are equal to the distance from the center to the sides. Wait, but in this calculation, we got that the area in each quadrant is (a b)/2, so total area would be 4*(a b)/2 = 2a b. But if a = b, then 2a². The area of the square is (2a)*(2a) = 4a². Then, the probability would be 2a² /4a² = 1/2. But is that correct?

Let me think. If the square has side length 2a (from -a to a in both x and y), then the center is at (0,0). The region closer to the center than to any vertex (which are at (a,a), (-a,a), etc. ) would be the set of points (x,y) such that sqrt(x² + y²) < sqrt((x - a)^2 + (y - a)^2 ), and similarly for the other vertices. Solving one inequality: sqrt(x² + y²) < sqrt((x - a)^2 + (y - a)^2 ). Squaring both sides:

x² + y² < (x - a)^2 + (y - a)^2

Expanding RHS: x² - 2a x + a² + y² - 2a y + a²

So:

x² + y² < x² - 2a x + a² + y² - 2a y + a²

Simplify:

0 < -2a x + 2a² - 2a y

Divide by 2a:

0 < -x + a - y

So, x + y < a

Similarly, for the other vertices, we get:

From vertex (-a, a): -x + y < a

From vertex (a, -a): x - y < a

From vertex (-a, -a): -x - y < a

Therefore, the region closer to O than to any vertex is the intersection of:

x + y < a

-x + y < a

x - y < a

-x - y < a

Which is a diamond (rotated square) with vertices at (a,0), (0,a), (-a,0), (0,-a). The area of this diamond is 2a². Since the area of the square is (2a)^2 = 4a², the probability is 2a² /4a² = 1/2. So, the calculation above for a = b giving probability 1/2 is correct. Therefore, the formula (a b)/2 per quadrant, summing to 2a b, is correct. Therefore, in the case a = b, the area is 2a², which is half the area of the square, which makes sense. Therefore, the general formula seems to hold.
-----

Based on execution_7.1:
**Final Answer**
The probability that a randomly chosen point from the interior of rectangle \(ABCD\) is closer to \(O\) than to any of the four vertices is \boxed{\dfrac{k}{2}}.
</execution_7.3>

<execution_7.4>
=> x + y = a.

So, in this case, the line is x + y = a, which intersects the top edge y = a at x = 0 and the right edge x = a at y = 0. So, in this case, the region in the first quadrant is the triangle bounded by (0,0), (a,0), and (0,a).
</execution_7.4>
</Execution_7>

<Plan_8>
Based on execution_7:
**Final Answer**
The probability is \boxed{\frac{k}{2}}.
</Plan_8>

<Execution_8>
No prompt to execute.
</Execution_8>

<Final_answer>
Therefore, the probability that a randomly chosen point from the interior of rectangle \(ABCD\) is closer to the center \(O\) than to any of the four vertices is \boxed{\frac{k}{2}}.
</Final_answer>
</think>