{
  "index": "2021-B-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "Suppose that the plane is tiled with an infinite checkerboard of unit squares. If another unit square is dropped on the plane at random with position and orientation independent of the checkerboard tiling, what is the probability that it does not cover any of the corners of the squares of the checkerboard?",
  "solution": "The probability is $2 - \\frac{6}{\\pi}$.\n\nSet coordinates so that the original tiling includes the (filled) square \n$S = \\{(x,y): 0 \\leq x,y \\leq 1 \\}$. It is then equivalent to choose the second square by first choosing a point uniformly at random in $S$ to be the center of the square, then choosing an angle of rotation uniformly at random from the interval $[0, \\pi/2]$.\n\nFor each $\\theta \\in [0, \\pi/2]$, circumscribe a square $S_\\theta$ around $S$ with angle of rotation $\\theta$ relative to $S$; this square has side length $\\sin \\theta + \\cos \\theta$. Inside $S_\\theta$, draw the smaller square $S_\\theta'$ consisting of points at distance greater than $1/2$ from each side of $S_\\theta$; this square has side length $\\sin \\theta + \\cos \\theta - 1$. \n\nWe now verify that a unit square with angle of rotation $\\theta$ fails to cover any corners of $S$ if and only if its center lies in the interior of $S_\\theta'$. In one direction, if one of the corners of $S$ is covered, then that corner lies on a side of $S_\\theta$ which meets the dropped square, so the center of the dropped square is at distance less than $1/2$ from that side of $S_\\theta$.\nTo check the converse, note that\nthere are two ways to dissect the square $S_\\theta$ into the square $S_\\theta'$ plus four $\\sin \\theta \\times \\cos \\theta$ rectangles. If $\\theta \\neq 0, \\pi/4$, then one of these dissections\nhas the property that each corner $P$ of $S$ appears as an interior point of a side (not a corner) of one of the rectangles $R$. \nIt will suffice to check that if the center of the dropped square is in $R$, then the dropped square covers $P$; this follows from the fact that $\\sin \\theta$ and $\\cos \\theta$ are both at most 1.\n\nIt follows that the conditional probability, given that the angle of rotation is chosen to be $\\theta$, that the dropped square does not cover any corners of $S$ is $(\\sin \\theta + \\cos \\theta - 1)^2$. We then compute the original probability as the integral\n\\begin{align*}\n&\\frac{2}{\\pi} \\int_0^{\\pi/2} (\\sin \\theta + \\cos \\theta - 1)^2\\,d\\theta \\\\\n&\\quad =\n\\frac{2}{\\pi} \\int_0^{\\pi/2} (2 + \\sin 2\\theta - 2\\sin \\theta - 2 \\cos \\theta)\\,d\\theta\\\\\n&\\quad = \\frac{2}{\\pi} \\left( 2 \\theta - \\frac{1}{2} \\cos 2\\theta + 2 \\cos \\theta - 2 \\sin \\theta \\right)_0^{\\pi/2} \\\\\n&\\quad = \\frac{2}{\\pi} \\left( \\pi + 1 - 2 - 2 \\right) = 2 - \\frac{6}{\\pi}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark:} Noam Elkies has some pictures illustrating this problem:\n\\href{https://abel.math.harvard.edu/~elkies/putnam_b1a.pdf}{image 1},\n\\href{https://abel.math.harvard.edu/~elkies/putnam_b1.pdf}{image 2}.",
  "vars": [
    "x",
    "y",
    "\\\\theta",
    "S",
    "S_\\\\theta",
    "R",
    "P"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horiznt",
        "y": "vertcl",
        "\\theta": "rotatn",
        "S": "basesq",
        "S_\\theta": "rotatsq",
        "R": "rectang",
        "P": "cornerp"
      },
      "question": "Suppose that the plane is tiled with an infinite checkerboard of unit squares. If another unit square is dropped on the plane at random with position and orientation independent of the checkerboard tiling, what is the probability that it does not cover any of the corners of the squares of the checkerboard?",
      "solution": "The probability is $2 - \\frac{6}{\\pi}$. \n\nSet coordinates so that the original tiling includes the (filled) square \n$basesq = \\{(horiznt,vertcl): 0 \\leq horiznt,vertcl \\leq 1 \\}$. It is then equivalent to choose the second square by first choosing a point uniformly at random in $basesq$ to be the center of the square, then choosing an angle of rotation uniformly at random from the interval $[0, \\pi/2]$. \n\nFor each $\\rotatn \\in [0, \\pi/2]$, circumscribe a square $rotatsq$ around $basesq$ with angle of rotation $\\rotatn$ relative to $basesq$; this square has side length $\\sin \\rotatn + \\cos \\rotatn$. Inside $rotatsq$, draw the smaller square $rotatsq'$ consisting of points at distance greater than $1/2$ from each side of $rotatsq$; this square has side length $\\sin \\rotatn + \\cos \\rotatn - 1$. \n\nWe now verify that a unit square with angle of rotation $\\rotatn$ fails to cover any corners of $basesq$ if and only if its center lies in the interior of $rotatsq'$. In one direction, if one of the corners of $basesq$ is covered, then that corner lies on a side of $rotatsq$ which meets the dropped square, so the center of the dropped square is at distance less than $1/2$ from that side of $rotatsq$. To check the converse, note that there are two ways to dissect the square $rotatsq$ into the square $rotatsq'$ plus four $\\sin \\rotatn \\times \\cos \\rotatn$ rectangles. If $\\rotatn \\neq 0, \\pi/4$, then one of these dissections has the property that each corner $cornerp$ of $basesq$ appears as an interior point of a side (not a corner) of one of the rectangles $rectang$. It will suffice to check that if the center of the dropped square is in $rectang$, then the dropped square covers $cornerp$; this follows from the fact that $\\sin \\rotatn$ and $\\cos \\rotatn$ are both at most 1.\n\nIt follows that the conditional probability, given that the angle of rotation is chosen to be $\\rotatn$, that the dropped square does not cover any corners of $basesq$ is $(\\sin \\rotatn + \\cos \\rotatn - 1)^2$. We then compute the original probability as the integral\n\\begin{align*}\n&\\frac{2}{\\pi} \\int_0^{\\pi/2} (\\sin \\rotatn + \\cos \\rotatn - 1)^2\\,d\\rotatn \\\\\n&\\quad =\n\\frac{2}{\\pi} \\int_0^{\\pi/2} (2 + \\sin 2\\rotatn - 2\\sin \\rotatn - 2 \\cos \\rotatn)\\,d\\rotatn\\\\\n&\\quad = \\frac{2}{\\pi} \\left( 2 \\rotatn - \\frac{1}{2} \\cos 2\\rotatn + 2 \\cos \\rotatn - 2 \\sin \\rotatn \\right)_0^{\\pi/2} \\\\\n&\\quad = \\frac{2}{\\pi} \\left( \\pi + 1 - 2 - 2 \\right) = 2 - \\frac{6}{\\pi}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark:} Noam Elkies has some pictures illustrating this problem: \\href{https://abel.math.harvard.edu/~elkies/putnam_b1a.pdf}{image 1}, \\href{https://abel.math.harvard.edu/~elkies/putnam_b1.pdf}{image 2}."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marblecake",
        "y": "waterwheel",
        "\\theta": "goldfinch",
        "S": "lanternfly",
        "S_\\theta": "caterpillar",
        "R": "tangerine",
        "P": "windchime"
      },
      "question": "Suppose that the plane is tiled with an infinite checkerboard of unit squares. If another unit square is dropped on the plane at random with position and orientation independent of the checkerboard tiling, what is the probability that it does not cover any of the corners of the squares of the checkerboard?",
      "solution": "The probability is $2 - \\frac{6}{\\pi}$.  \n\nSet coordinates so that the original tiling includes the (filled) square \n$lanternfly = \\{(marblecake,waterwheel): 0 \\leq marblecake,waterwheel \\leq 1 \\}$. It is then equivalent to choose the second square by first choosing a point uniformly at random in $lanternfly$ to be the center of the square, then choosing an angle of rotation uniformly at random from the interval $[0, \\pi/2]$.  \n\nFor each $goldfinch \\in [0, \\pi/2]$, circumscribe a square $caterpillar$ around $lanternfly$ with angle of rotation $goldfinch$ relative to $lanternfly$; this square has side length $\\sin goldfinch + \\cos goldfinch$. Inside $caterpillar$, draw the smaller square $caterpillar'$ consisting of points at distance greater than $1/2$ from each side of $caterpillar$; this square has side length $\\sin goldfinch + \\cos goldfinch - 1$.  \n\nWe now verify that a unit square with angle of rotation $goldfinch$ fails to cover any corners of $lanternfly$ if and only if its center lies in the interior of $caterpillar'$. In one direction, if one of the corners of $lanternfly$ is covered, then that corner lies on a side of $caterpillar$ which meets the dropped square, so the center of the dropped square is at distance less than $1/2$ from that side of $caterpillar$. To check the converse, note that\nthere are two ways to dissect the square $caterpillar$ into the square $caterpillar'$ plus four $\\sin goldfinch \\times \\cos goldfinch$ rectangles. If $goldfinch \\neq 0, \\pi/4$, then one of these dissections\nhas the property that each corner $windchime$ of $lanternfly$ appears as an interior point of a side (not a corner) of one of the rectangles $tangerine$.  \nIt will suffice to check that if the center of the dropped square is in $tangerine$, then the dropped square covers $windchime$; this follows from the fact that $\\sin goldfinch$ and $\\cos goldfinch$ are both at most 1.\n\nIt follows that the conditional probability, given that the angle of rotation is chosen to be $goldfinch$, that the dropped square does not cover any corners of $lanternfly$ is $(\\sin goldfinch + \\cos goldfinch - 1)^2$. We then compute the original probability as the integral\n\\begin{align*}\n&\\frac{2}{\\pi} \\int_0^{\\pi/2} (\\sin goldfinch + \\cos goldfinch - 1)^2\\,d goldfinch \\\\\n&\\quad =\n\\frac{2}{\\pi} \\int_0^{\\pi/2} (2 + \\sin 2 goldfinch - 2\\sin goldfinch - 2 \\cos goldfinch)\\,d goldfinch\\\\\n&\\quad = \\frac{2}{\\pi} \\left( 2 goldfinch - \\frac{1}{2} \\cos 2 goldfinch + 2 \\cos goldfinch - 2 \\sin goldfinch \\right)_0^{\\pi/2} \\\\\n&\\quad = \\frac{2}{\\pi} \\left( \\pi + 1 - 2 - 2 \\right) = 2 - \\frac{6}{\\pi}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark:} Noam Elkies has some pictures illustrating this problem:\n\\href{https://abel.math.harvard.edu/~elkies/putnam_b1a.pdf}{image 1},\n\\href{https://abel.math.harvard.edu/~elkies/putnam_b1.pdf}{image 2}."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalcoord",
        "y": "horizontalcoord",
        "\\theta": "fixedangle",
        "S": "diskshape",
        "S_\\theta": "ellipseform",
        "R": "curvezone",
        "P": "areasite"
      },
      "question": "Suppose that the plane is tiled with an infinite checkerboard of unit squares. If another unit square is dropped on the plane at random with position and orientation independent of the checkerboard tiling, what is the probability that it does not cover any of the corners of the squares of the checkerboard?",
      "solution": "The probability is $2 - \\frac{6}{\\pi}$.\\n\\nSet coordinates so that the original tiling includes the (filled) square $diskshape = \\{(verticalcoord,horizontalcoord): 0 \\leq verticalcoord,horizontalcoord \\leq 1 \\}$. It is then equivalent to choose the second square by first choosing a point uniformly at random in $diskshape$ to be the center of the square, then choosing an angle of rotation uniformly at random from the interval $[0, \\pi/2]$.\\n\\nFor each $fixedangle \\in [0, \\pi/2]$, circumscribe a square $ellipseform$ around $diskshape$ with angle of rotation $fixedangle$ relative to $diskshape$; this square has side length $\\sin fixedangle + \\cos fixedangle$. Inside $ellipseform$, draw the smaller square $ellipseform'$ consisting of points at distance greater than $1/2$ from each side of $ellipseform$; this square has side length $\\sin fixedangle + \\cos fixedangle - 1$.\\n\\nWe now verify that a unit square with angle of rotation $fixedangle$ fails to cover any corners of $diskshape$ if and only if its center lies in the interior of $ellipseform'$. In one direction, if one of the corners of $diskshape$ is covered, then that corner lies on a side of $ellipseform$ which meets the dropped square, so the center of the dropped square is at distance less than $1/2$ from that side of $ellipseform$. To check the converse, note that there are two ways to dissect the square $ellipseform$ into the square $ellipseform'$ plus four $\\sin fixedangle \\times \\cos fixedangle$ rectangles. If $fixedangle \\neq 0, \\pi/4$, then one of these dissections has the property that each corner $areasite$ of $diskshape$ appears as an interior point of a side (not a corner) of one of the rectangles $curvezone$. It will suffice to check that if the center of the dropped square is in $curvezone$, then the dropped square covers $areasite$; this follows from the fact that $\\sin fixedangle$ and $\\cos fixedangle$ are both at most 1.\\n\\nIt follows that the conditional probability, given that the angle of rotation is chosen to be $fixedangle$, that the dropped square does not cover any corners of $diskshape$ is $(\\sin fixedangle + \\cos fixedangle - 1)^2$. We then compute the original probability as the integral\\n\\begin{align*}\\n&\\frac{2}{\\pi} \\int_0^{\\pi/2} (\\sin fixedangle + \\cos fixedangle - 1)^2\\,d fixedangle \\\\n&\\quad = \\frac{2}{\\pi} \\int_0^{\\pi/2} (2 + \\sin 2fixedangle - 2\\sin fixedangle - 2 \\cos fixedangle)\\,d fixedangle\\\\\n&\\quad = \\frac{2}{\\pi} \\left( 2 fixedangle - \\frac{1}{2} \\cos 2fixedangle + 2 \\cos fixedangle - 2 \\sin fixedangle \\right)_0^{\\pi/2} \\\\n&\\quad = \\frac{2}{\\pi} \\left( \\pi + 1 - 2 - 2 \\right) = 2 - \\frac{6}{\\pi}.\\n\\end{align*}\\n\\n\\noindent\\textbf{Remark:} Noam Elkies has some pictures illustrating this problem: \\href{https://abel.math.harvard.edu/~elkies/putnam_b1a.pdf}{image 1}, \\href{https://abel.math.harvard.edu/~elkies/putnam_b1.pdf}{image 2}."
    },
    "garbled_string": {
      "map": {
        "x": "kseuvlqp",
        "y": "qjtrsmda",
        "\\theta": "vkptnsha",
        "S": "yxpumzce",
        "S_\\theta": "hbzquwrn",
        "R": "jvthelmq",
        "P": "wcgznkia"
      },
      "question": "Suppose that the plane is tiled with an infinite checkerboard of unit squares. If another unit square is dropped on the plane at random with position and orientation independent of the checkerboard tiling, what is the probability that it does not cover any of the corners of the squares of the checkerboard?",
      "solution": "The probability is $2 - \\frac{6}{\\pi}$.  \n\nSet coordinates so that the original tiling includes the (filled) square \n$yxpumzce = \\{(kseuvlqp,qjtrsmda): 0 \\leq kseuvlqp,qjtrsmda \\leq 1 \\}$. It is then equivalent to choose the second square by first choosing a point uniformly at random in $yxpumzce$ to be the center of the square, then choosing an angle of rotation uniformly at random from the interval $[0, \\pi/2]$.  \n\nFor each $vkptnsha \\in [0, \\pi/2]$, circumscribe a square $hbzquwrn$ around $yxpumzce$ with angle of rotation $vkptnsha$ relative to $yxpumzce$; this square has side length $\\sin vkptnsha + \\cos vkptnsha$. Inside $hbzquwrn$, draw the smaller square $hbzquwrn'$ consisting of points at distance greater than $1/2$ from each side of $hbzquwrn$; this square has side length $\\sin vkptnsha + \\cos vkptnsha - 1$.  \n\nWe now verify that a unit square with angle of rotation $vkptnsha$ fails to cover any corners of $yxpumzce$ if and only if its center lies in the interior of $hbzquwrn'$. In one direction, if one of the corners of $yxpumzce$ is covered, then that corner lies on a side of $hbzquwrn$ which meets the dropped square, so the center of the dropped square is at distance less than $1/2$ from that side of $hbzquwrn$.  \nTo check the converse, note that\nthere are two ways to dissect the square $hbzquwrn$ into the square $hbzquwrn'$ plus four $\\sin vkptnsha \\times \\cos vkptnsha$ rectangles. If $vkptnsha \\neq 0, \\pi/4$, then one of these dissections\nhas the property that each corner $wcgznkia$ of $yxpumzce$ appears as an interior point of a side (not a corner) of one of the rectangles $jvthelmq$. \nIt will suffice to check that if the center of the dropped square is in $jvthelmq$, then the dropped square covers $wcgznkia$; this follows from the fact that $\\sin vkptnsha$ and $\\cos vkptnsha$ are both at most 1.\n\nIt follows that the conditional probability, given that the angle of rotation is chosen to be $vkptnsha$, that the dropped square does not cover any corners of $yxpumzce$ is $(\\sin vkptnsha + \\cos vkptnsha - 1)^2$. We then compute the original probability as the integral\n\\begin{align*}\n&\\frac{2}{\\pi} \\int_0^{\\pi/2} (\\sin vkptnsha + \\cos vkptnsha - 1)^2\\,dvkptnsha \\\\\n&\\quad =\n\\frac{2}{\\pi} \\int_0^{\\pi/2} (2 + \\sin 2 vkptnsha - 2\\sin vkptnsha - 2 \\cos vkptnsha)\\,dvkptnsha\\\\\n&\\quad = \\frac{2}{\\pi} \\left( 2 vkptnsha - \\frac{1}{2} \\cos 2 vkptnsha + 2 \\cos vkptnsha - 2 \\sin vkptnsha \\right)_0^{\\pi/2} \\\\\n&\\quad = \\frac{2}{\\pi} \\left( \\pi + 1 - 2 - 2 \\right) = 2 - \\frac{6}{\\pi}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark:} Noam Elkies has some pictures illustrating this problem:\n\\href{https://abel.math.harvard.edu/~elkies/putnam_b1a.pdf}{image 1},\n\\href{https://abel.math.harvard.edu/~elkies/putnam_b1.pdf}{image 2}."
    },
    "kernel_variant": {
      "question": "The plane is tiled by an infinite checkerboard whose squares all have side-length 3 and whose sides are parallel to the coordinate axes.  A second 3 × 3 square is dropped at random as follows.\n1.  Its centre (X ,Y) is chosen uniformly from the rectangle\n   R = { (x ,y) : 4 ≤ x ≤ 7 and −2 ≤ y ≤ 1 }\n   (area 9).\n2.  Independently, an angle θ is chosen uniformly from the interval\n   [π⁄8 , 5π⁄8] , and the square is then rotated counter-clockwise through the angle θ about its centre.\n\nWhat is the probability that the dropped square covers none of the vertices of the underlying checkerboard?",
      "solution": "Because all checkerboard squares are congruent and the tiling is invariant under translations by integer multiples of 3 and under 90° rotations, we may work inside the single 3 × 3 square\n   S₀ := { (x ,y) : 4 ≤ x ≤ 7 , −2 ≤ y ≤ 1 },\nand we may replace θ with its value modulo π⁄2.  In particular\n      φ := θ − π⁄8\nis uniformly distributed on [0 , π⁄2].  The problem therefore reduces to the following.\n\nGiven a fixed φ ∈ [0 , π⁄2] we place a 3 × 3 square having that relative angle with respect to the checkerboard; its centre is chosen uniformly in S₀.  What is the conditional probability (as a function of φ) that no corner of S₀ is covered?\n\nStep 1.  The bounding (axis-parallel) square.\nLet \\widehat S_φ be the smallest axis-parallel square that contains the 3 × 3 square obtained by rotating S₀ through φ about its centre.  A routine projection calculation shows that\n        side(\\widehat S_φ) = 3(\\sin φ + \\cos φ).         (1)\n\nStep 2.  The inner forbidden zone.\nInside \\widehat S_φ draw the concentric square\n        \\widehat S'_φ := {P ∈ \\widehat S_φ : dist(P,∂\\widehat S_φ) > 3⁄2}.\nIts side length is therefore\n        3(\\sin φ + \\cos φ) − 3 = 3(\\sin φ + \\cos φ − 1).      (2)\n\nStep 3.  Why centres that fall outside \\widehat S'_φ are bad.\nWe must show that\n  (i)  the dropped square misses every checkerboard vertex  ⇔  its centre lies in the interior of \\widehat S'_φ.   (∗)\n\nOne implication is easy: if the centre is inside \\widehat S'_φ then every side of the dropped square is at least 3⁄2 away from every side of \\widehat S_φ, hence at least 3⁄2 away from every checkerboard vertex; therefore no vertex can be covered.\n\nThe converse requires more work.  Fix φ ∈ (0 , π⁄2) \\ {π⁄4}; these exceptional values have probability 0 and can be ignored.  Without loss of generality assume 0 < φ < π⁄4 so that \\sin φ ≤ \\cos φ (the other case is similar).\n\nDissect \\widehat S_φ as follows (see the figure in the original Putnam solution).  Through each side of the rotated square draw the line that continues that side until it meets \\widehat S_φ.  This partitions \\widehat S_φ into the rotated square itself plus four congruent rectangles; each rectangle has side lengths 3\\sin φ and 3\\cos φ, the longer side of length 3\\cos φ lying along one side of \\widehat S_φ.\n\nBecause φ ≠ 0, π⁄4, every vertex of S₀ lies on the interior of the long side of exactly one of those rectangles.  Denote such a rectangle by R and the corresponding vertex by P.  Suppose the centre C of the dropped square lies outside \\widehat S'_φ; then C is within 3⁄2 of some side of \\widehat S_φ, hence C lies in the rectangle R that borders that side.\n\nInside R the distances in the two coordinate directions do not exceed half of its side lengths; therefore\n        |C_x − P_x| ≤ (3\\sin φ)/2 ≤ 3⁄2,\n        |C_y − P_y| ≤ (3\\cos φ)/2 ≤ 3⁄2,\nwhere we used \\sin φ,\\cos φ ≤ 1.  Consequently P lies inside, or on the boundary of, the axis-parallel square of side 3 centred at C, i.e. inside the dropped 3 × 3 square.  Thus a checkerboard vertex is covered whenever the centre is outside \\widehat S'_φ.\n\nWe have therefore established the equivalence (∗).\n\nStep 4.  The conditional probability.\nGiven φ, the centre is uniformly distributed in S₀, so the conditional probability of success (no vertex covered) equals the ratio of the areas of \\widehat S'_φ and S₀.  Using (1) and (2):\n        P_success(φ) = [3(\\sin φ + \\cos φ − 1)]² / 3²\n                      = (\\sin φ + \\cos φ − 1)².       (3)\nThe factor 3 cancels, showing that the answer is independent of the common side length 3, exactly as in the unit-square version.\n\nStep 5.  Averaging over φ.\nSince φ is uniform on [0 , π⁄2],\n   P = (2/π) ∫₀^{π/2} (\\sin φ + \\cos φ − 1)² dφ.        (4)\nExpanding and integrating,\n   (\\sin φ + \\cos φ − 1)² = 2 + \\sin 2φ − 2\\sin φ − 2\\cos φ,\nso\n   P = (2/π) ∫₀^{π/2} (2 + \\sin 2φ − 2\\sin φ − 2\\cos φ) dφ\n     = (2/π) [2φ − (1/2)\\cos 2φ + 2\\cos φ − 2\\sin φ]₀^{π/2}\n     = (2/π) (π − 3)\n     = 2 − 6/π.\n\nTherefore the probability that the dropped 3 × 3 square misses every vertex of the underlying checkerboard is\n        2 − 6/π ≈ 0.0897.",
      "_meta": {
        "core_steps": [
          "Exploit translational and 90° rotational symmetry: pick the square’s centre uniformly in one fundamental tile and the rotation angle θ uniformly on a length-π/2 interval.",
          "For fixed θ, draw the θ-rotated square S_θ that just contains the reference tile; its side length is sinθ + cosθ.",
          "Inset an inner square S_θ′ at distance (side/2) from each side of S_θ; its side length is sinθ + cosθ − 1.",
          "Show: the dropped square avoids every lattice corner  ⇔  its centre lies in the interior of S_θ′.",
          "Hence conditional probability = (sinθ + cosθ − 1)²; integrate this over θ with uniform density 2/π to get 2 − 6/π."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Common side length of both the checkerboard squares and the dropped square (a global scaling factor).",
            "original": "1"
          },
          "slot2": {
            "description": "Specific placement of the length-π/2 rotation interval used for θ (only the interval’s length matters).",
            "original": "[0, π/2]"
          },
          "slot3": {
            "description": "Choice of the reference fundamental tile in which the centre is taken to be uniform.",
            "original": "S = {(x,y): 0 ≤ x,y ≤ 1}"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}