{
  "index": "1942-A-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "1. A square of side \\( 2 a \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\) - and \\( Y \\)-axes respectively. Find the locus of the midpoint of the square.",
  "solution": "Solution. Let \\( A \\) and \\( B \\) be two consecutive vertices of the square lying on the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( C \\) be the center of the square and \\( D \\) and \\( E \\) the feet of perpendiculars from \\( C \\) to the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( O \\) be the origin.\n\nThen \\( \\angle A C B \\) is right angle, and \\( \\angle D C E \\) is a right angle and \\( \\triangle A C D \\) is congruent to \\( \\triangle B C E \\) (hypotenuse and acute angle). Hence the \\( x \\) and \\( y \\) coordinates of \\( C \\) are equal, and \\( O E C D \\) is a square. As the given square moves under the specified constraint, the center \\( C \\) moves back and forth along the segment from \\( (a, a) \\) to \\( (\\sqrt{2} a, \\sqrt{2} a) \\).\n\nRemark. A generalization of this problem appears as Problem P.M. 7 in this examination.",
  "vars": [
    "A",
    "B",
    "C",
    "D",
    "E",
    "x",
    "y"
  ],
  "params": [
    "a",
    "O"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "pointalpha",
        "B": "pointbeta",
        "C": "pointcenter",
        "D": "footdown",
        "E": "footleft",
        "x": "coordx",
        "y": "coordy",
        "a": "sidelength",
        "O": "originpnt"
      },
      "question": "1. A square of side \\( 2 sidelength \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\) - and \\( Y \\)-axes respectively. Find the locus of the midpoint of the square.",
      "solution": "Solution. Let \\( pointalpha \\) and \\( pointbeta \\) be two consecutive vertices of the square lying on the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( pointcenter \\) be the center of the square and \\( footdown \\) and \\( footleft \\) the feet of perpendiculars from \\( pointcenter \\) to the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( originpnt \\) be the origin.\n\nThen \\( \\angle pointalpha \\, pointcenter \\, pointbeta \\) is right angle, and \\( \\angle footdown \\, pointcenter \\, footleft \\) is a right angle and \\( \\triangle pointalpha \\, pointcenter \\, footdown \\) is congruent to \\( \\triangle pointbeta \\, pointcenter \\, footleft \\) (hypotenuse and acute angle). Hence the \\( coordx \\) and \\( coordy \\) coordinates of \\( pointcenter \\) are equal, and \\( originpnt \\, footleft \\, pointcenter \\, footdown \\) is a square. As the given square moves under the specified constraint, the center \\( pointcenter \\) moves back and forth along the segment from \\( (sidelength, sidelength) \\) to \\( (\\sqrt{2} sidelength, \\sqrt{2} sidelength) \\).\n\nRemark. A generalization of this problem appears as Problem P.M. 7 in this examination."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "blueberry",
        "B": "sailplane",
        "C": "nightfall",
        "D": "horsewhip",
        "E": "drumstick",
        "x": "windvessel",
        "y": "sandstorm",
        "a": "gemstone",
        "O": "floodgate"
      },
      "question": "1. A square of side \\( 2 gemstone \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\) - and \\( Y \\)-axes respectively. Find the locus of the midpoint of the square.",
      "solution": "Solution. Let \\( blueberry \\) and \\( sailplane \\) be two consecutive vertices of the square lying on the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( nightfall \\) be the center of the square and \\( horsewhip \\) and \\( drumstick \\) the feet of perpendiculars from \\( nightfall \\) to the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( floodgate \\) be the origin.\n\nThen \\( \\angle blueberry nightfall sailplane \\) is right angle, and \\( \\angle horsewhip nightfall drumstick \\) is a right angle and \\( \\triangle blueberry nightfall horsewhip \\) is congruent to \\( \\triangle sailplane nightfall drumstick \\) (hypotenuse and acute angle). Hence the \\( windvessel \\) and \\( sandstorm \\) coordinates of \\( nightfall \\) are equal, and \\( floodgate drumstick nightfall horsewhip \\) is a square. As the given square moves under the specified constraint, the center \\( nightfall \\) moves back and forth along the segment from \\( (gemstone, gemstone) \\) to \\( (\\sqrt{2} gemstone, \\sqrt{2} gemstone) \\).\n\nRemark. A generalization of this problem appears as Problem P.M. 7 in this examination."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "basementpt",
        "B": "ceilingpt",
        "C": "boundarypt",
        "D": "headspot",
        "E": "crownspot",
        "x": "verticalco",
        "y": "horizontalco",
        "a": "megascale",
        "O": "infinitept"
      },
      "question": "1. A square of side \\( 2\\megascale \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\) - and \\( Y \\)-axes respectively. Find the locus of the midpoint of the square.",
      "solution": "Solution. Let \\( \\basementpt \\) and \\( \\ceilingpt \\) be two consecutive vertices of the square lying on the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( \\boundarypt \\) be the center of the square and \\( \\headspot \\) and \\( \\crownspot \\) the feet of perpendiculars from \\( \\boundarypt \\) to the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( \\infinitept \\) be the origin.\n\nThen \\( \\angle \\basementpt \\boundarypt \\ceilingpt \\) is right angle, and \\( \\angle \\headspot \\boundarypt \\crownspot \\) is a right angle and \\( \\triangle \\basementpt \\boundarypt \\headspot \\) is congruent to \\( \\triangle \\ceilingpt \\boundarypt \\crownspot \\) (hypotenuse and acute angle). Hence the \\( \\verticalco \\) and \\( \\horizontalco \\) coordinates of \\( \\boundarypt \\) are equal, and \\( \\infinitept \\crownspot \\boundarypt \\headspot \\) is a square. As the given square moves under the specified constraint, the center \\( \\boundarypt \\) moves back and forth along the segment from \\( (\\megascale, \\megascale) \\) to \\( (\\sqrt{2}\\,\\megascale, \\sqrt{2}\\,\\megascale) \\).\n\nRemark. A generalization of this problem appears as Problem P.M. 7 in this examination."
    },
    "garbled_string": {
      "map": {
        "A": "xjvmdpqa",
        "B": "tzknorhc",
        "C": "hmglsewu",
        "D": "fqubylzi",
        "E": "rpxgsnca",
        "x": "vzeqimoy",
        "y": "kldhptsu",
        "a": "wjkfrdbe",
        "O": "snylvoha"
      },
      "question": "1. A square of side \\( 2 wjkfrdbe \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\) - and \\( Y \\)-axes respectively. Find the locus of the midpoint of the square.",
      "solution": "Solution. Let \\( xjvmdpqa \\) and \\( tzknorhc \\) be two consecutive vertices of the square lying on the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( hmglsewu \\) be the center of the square and \\( fqubylzi \\) and \\( rpxgsnca \\) the feet of perpendiculars from \\( hmglsewu \\) to the \\( X \\) - and \\( Y \\)-axes, respectively. Let \\( snylvoha \\) be the origin.\n\nThen \\( \\angle xjvmdpqa hmglsewu tzknorhc \\) is right angle, and \\( \\angle fqubylzi hmglsewu rpxgsnca \\) is a right angle and \\( \\triangle xjvmdpqa hmglsewu fqubylzi \\) is congruent to \\( \\triangle tzknorhc hmglsewu rpxgsnca \\) (hypotenuse and acute angle). Hence the \\( vzeqimoy \\) and \\( kldhptsu \\) coordinates of \\( hmglsewu \\) are equal, and \\( snylvoha rpxgsnca hmglsewu fqubylzi \\) is a square. As the given square moves under the specified constraint, the center \\( hmglsewu \\) moves back and forth along the segment from \\( (wjkfrdbe, wjkfrdbe) \\) to \\( (\\sqrt{2} wjkfrdbe, \\sqrt{2} wjkfrdbe) \\).\n\nRemark. A generalization of this problem appears as Problem P.M. 7 in this examination."
    },
    "kernel_variant": {
      "question": "A square of fixed side–length 4 c (with c > 0) moves in the plane subject to the following conditions.\n• Every point of the square satisfies x ≤ 0 and y ≤ 0; that is, the whole square always lies in the closed third quadrant, including the two negative coordinate axes.\n• At every instant two consecutive vertices of the square are situated on the negative X– and negative Y–axes, respectively.\nDetermine the locus of the centre of the square.",
      "solution": "Let the two consecutive vertices constrained to the axes be\n A = (–a , 0) (on the negative X–axis), B = (0 , –b) (on the negative Y–axis),\nwith a ≥ 0, b ≥ 0 and (a , b) ≠ (0 , 0).  Because AB is a side of the square of length 4c,\n AB = √[(0 + a)² + (–b – 0)²] = √(a² + b²) = 4c,               (1)\nso a² + b² = 16c².\n\n1. Orientation that keeps the square in the third quadrant\n---------------------------------------------------------\nThe vector AB is v = B – A = (a , –b).  A quarter–turn of v of the same length is a vector w with |w| = 4c and v·w = 0.  Up to sign there are two possibilities:\n w₁ = (–b , –a), w₂ = ( b ,  a).\nChoosing w₂ would send the next vertex A + w₂ = (–a + b , a) into the half‐plane y > 0 whenever a > 0, which is forbidden.  Hence we must take\n w = (–b , –a).\nWhen either a = 0 or b = 0 this choice still leaves all coordinates non–positive:\n A + w = (–a – b , –a), B + w = (–b , –a – b),\nboth lying in the closed third quadrant.  Thus the entire square satisfies the required positional constraint for all admissible (a , b).\n\n2. Coordinates of the centre\n----------------------------\nThe centre C is the midpoint of either diagonal.  Using the diagonal joining A to B + w,\n C = ½[ A + (B + w) ]\n   = ½[ (–a , 0) + (0 , –b) + (–b , –a) ]\n   = ½[ –(a + b) , –(a + b) ]\n   = ( –(a + b)/2 , –(a + b)/2 ).                                (2)\nHence the centre always lies on the bisector y = x in the third quadrant.  Write\n C = ( –t , –t ) with t = (a + b)/2 ≥ 0.                        (3)\n\n3. The range of t\n-----------------\nSet S = a + b.  From (1), a² + b² = 16c².  For non–negative a, b we have the classical bounds\n S² = (a + b)²  ≤  2(a² + b²) = 32c²             ⇒ S ≤ 4√2 c,   (upper bound)\n S² = (a + b)²  ≥  a² + b² = 16c²        ⇒ S ≥ 4c,      (lower bound)\nwhere equality S = 4c occurs exactly when one of a or b equals 0.  No geometric obstruction forbids this situation (the square merely touches one axis at the origin), so the lower bound is attainable.  Equality S = 4√2 c is attained when a = b = 4c/√2.\n\nTherefore\n 4c ≤ S ≤ 4√2 c and, by (3), 2c ≤ t ≤ 2√2 c.                (4)\n\n4. Locus of the centre\n----------------------\nCombining the description (3) with the range (4) we find that the centre of the square moves exactly on the closed line segment\n { ( –t , –t ) :  2c ≤ t ≤ 2√2 c }.\nThe endpoint ( –2c , –2c ) is realised, for example, when a = 0, b = 4c (the side AB coincides with the negative Y–axis), while ( –2√2 c , –2√2 c ) is obtained when a = b = 4c/√2.\n\nHence the required locus is the closed segment on the bisector y = x joining the points ( –2c , –2c ) and ( –2√2 c , –2√2 c ).",
      "_meta": {
        "core_steps": [
          "Place consecutive vertices A (on X-axis) and B (on Y-axis); denote the square’s center by C and drop perpendiculars from C to the axes (points D,E).",
          "Use the fact that square diagonals are perpendicular: ∠ACB = 90°, while ∠DCE = 90° because the axes are perpendicular.",
          "Right-triangle congruence (ΔACD ≅ ΔBCE) gives CD = CE, so the coordinates of C satisfy x = y; thus C lies on the line y = x.",
          "Determine the extreme positions of C: when a side of the square rests on an axis C is at (a,a); when a vertex touches both axes C is at (√2 a, √2 a).",
          "Hence the locus of the midpoint is the segment joining (a,a) and (√2 a, √2 a) on the line y = x."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Exact numerical value used for the side length; scaling the side from 2 a to any positive constant leaves the argument unchanged (the locus simply scales).",
            "original": "2 a"
          },
          "slot2": {
            "description": "The specific quadrant (here ‘first quadrant’) that forces the coordinates of all points to have the same sign; any other fixed quadrant or sign-restriction works equally well.",
            "original": "first quadrant"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}