{
  "index": "1946-A-3",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "3. A projectile in flight is observed simultaneously from four radar stations which are situated at the corners of a square of side \\( b \\). The distances of the projectile from the four stations, taken in order around the square, are found to be \\( \\boldsymbol{R}_{1}, \\boldsymbol{R}_{2}, \\boldsymbol{R}_{3}, \\boldsymbol{R}_{4} \\). Show that\n\\[\nR_{1}^{2}+R_{3}^{2}=R_{2}^{2}+R_{4}^{2}\n\\]\n\nShow also that the height \\( h \\) of the projectile above the ground is given by\n\\[\n\\begin{aligned}\nh^{2}= & -\\frac{1}{2} b^{2}+\\frac{1}{4}\\left(R_{1}^{2}+R_{2}^{2}+R_{3}^{2}+R_{4}^{2}\\right) \\\\\n& -\\frac{1}{8 b^{2}}\\left(R_{1}^{4}+R_{2}^{4}+R_{3}^{4}+R_{4}^{4}-2 R_{1}^{2} R_{3}^{2}-2 R_{2}^{2} R_{4}^{2}\\right)\n\\end{aligned}\n\\]",
  "solution": "Solution. Choose the diagonals of the square as axes in the plane. Then the four vertices are \\( ( \\pm a, 0),(0, \\pm a) \\), where \\( a^{2}=b^{2} / 2 \\), and we number them counterclockwise starting at \\( (a, 0) \\). If the projectile is above the point \\( (x, y) \\), then\n\\[\n\\begin{array}{ll}\nR_{1}^{2}=(x-a)^{2}+y^{2}+h^{2}, & R_{2}^{2}=x^{2}+(y-a)^{2}+h^{2} \\\\\nR_{3}^{2}=(x+a)^{2}+y^{2}+h^{2}, & R_{4}^{2}=x^{2}+(y+a)^{2}+h^{2}\n\\end{array}\n\\]\n\nHence, \\( R_{1}{ }^{2}+R_{3}{ }^{2}=R_{2}{ }^{2}+R_{4}{ }^{2} \\), as required, and\n\\[\nR_{1}^{2}+R_{2}^{2}+R_{3}^{2}+R_{4}^{2}-4 a^{2}-4 h^{2}=4\\left(x^{2}+y^{2}\\right)\n\\]\n\nTherefore,\n\\[\n\\begin{aligned}\nR_{1}^{4}+R_{2}^{4} & +R_{3}^{4}+R_{4}^{4}-2 R_{1}^{2} R_{3}^{2}-2 R_{2}^{2} R_{4}^{2} \\\\\n& =\\left(R_{3}^{2}-R_{1}^{2}\\right)^{2}+\\left(R_{4}^{2}-R_{2}^{2}\\right)^{2}=16 a^{2}\\left(x^{2}+y^{2}\\right) \\\\\n& =4 a^{2}\\left(R_{1}^{2}+R_{2}^{2}+R_{3}^{2}+R_{4}^{2}-4 a^{2}-4 h^{2}\\right)\n\\end{aligned}\n\\]\n\nHence,\n\\[\n\\begin{array}{c}\n16 a^{2} h^{2}=-16 a^{4}+4 a^{2}\\left(R_{1}^{2}+R_{2}^{2}+R_{3}^{2}+R_{4}^{2}\\right) \\\\\n-\\left(R_{1}^{4}+R_{2}^{4}+R_{3}^{4}+R_{4}^{4}-2 R_{1}^{2} R_{3}^{2}-2 R_{2}^{2} R_{4}^{2}\\right)\n\\end{array}\n\\]\n\nDividing through by \\( 16 a^{2}=8 b^{2} \\), we obtain the desired relation.\nRemark 1. To keep the algebra simple, it is essential to take advantage of the symmetry of the square. Another, almost equally good, choice of coordinates would make the vertices \\( ( \\pm c, \\pm c) \\), where \\( c=b / 2 \\).\n\nRemark 2. A point above a plane is determined by its distances from any three non-collinear points in the plane. Hence \\( h, x \\), and \\( y \\) are determined by any three of the \\( R \\) 's. This explains why there must be a nontrivial relation connecting the four R's.",
  "vars": [
    "x",
    "y",
    "h",
    "R_1",
    "R_2",
    "R_3",
    "R_4"
  ],
  "params": [
    "b",
    "a",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "coordx",
        "y": "coordy",
        "h": "heightp",
        "R_1": "rangeone",
        "R_2": "rangetwo",
        "R_3": "rangethree",
        "R_4": "rangefour",
        "b": "sidesquare",
        "a": "diaghalf",
        "c": "halfside"
      },
      "question": "3. A projectile in flight is observed simultaneously from four radar stations which are situated at the corners of a square of side \\( sidesquare \\). The distances of the projectile from the four stations, taken in order around the square, are found to be \\( \\boldsymbol{rangeone}, \\boldsymbol{rangetwo}, \\boldsymbol{rangethree}, \\boldsymbol{rangefour} \\). Show that\n\\[\nrangeone^{2}+rangethree^{2}=rangetwo^{2}+rangefour^{2}\n\\]\n\nShow also that the height \\( heightp \\) of the projectile above the ground is given by\n\\[\n\\begin{aligned}\nheightp^{2}= & -\\frac{1}{2} sidesquare^{2}+\\frac{1}{4}\\left(rangeone^{2}+rangetwo^{2}+rangethree^{2}+rangefour^{2}\\right) \\\\\n& -\\frac{1}{8 sidesquare^{2}}\\left(rangeone^{4}+rangetwo^{4}+rangethree^{4}+rangefour^{4}-2 rangeone^{2} rangethree^{2}-2 rangetwo^{2} rangefour^{2}\\right)\n\\end{aligned}\n\\]",
      "solution": "Solution. Choose the diagonals of the square as axes in the plane. Then the four vertices are \\( ( \\pm diaghalf, 0),(0, \\pm diaghalf) \\), where \\( diaghalf^{2}=sidesquare^{2} / 2 \\), and we number them counterclockwise starting at \\( (diaghalf, 0) \\). If the projectile is above the point \\( (coordx, coordy) \\), then\n\\[\n\\begin{array}{ll}\nrangeone^{2}=(coordx-diaghalf)^{2}+coordy^{2}+heightp^{2}, & rangetwo^{2}=coordx^{2}+(coordy-diaghalf)^{2}+heightp^{2} \\\\\nrangethree^{2}=(coordx+diaghalf)^{2}+coordy^{2}+heightp^{2}, & rangefour^{2}=coordx^{2}+(coordy+diaghalf)^{2}+heightp^{2}\n\\end{array}\n\\]\n\nHence, \\( rangeone^{2}+rangethree^{2}=rangetwo^{2}+rangefour^{2} \\), as required, and\n\\[\nrangeone^{2}+rangetwo^{2}+rangethree^{2}+rangefour^{2}-4 diaghalf^{2}-4 heightp^{2}=4\\left(coordx^{2}+coordy^{2}\\right)\n\\]\n\nTherefore,\n\\[\n\\begin{aligned}\nrangeone^{4}+rangetwo^{4} & +rangethree^{4}+rangefour^{4}-2 rangeone^{2} rangethree^{2}-2 rangetwo^{2} rangefour^{2} \\\\\n& =\\left(rangethree^{2}-rangeone^{2}\\right)^{2}+\\left(rangefour^{2}-rangetwo^{2}\\right)^{2}=16 diaghalf^{2}\\left(coordx^{2}+coordy^{2}\\right) \\\\\n& =4 diaghalf^{2}\\left(rangeone^{2}+rangetwo^{2}+rangethree^{2}+rangefour^{2}-4 diaghalf^{2}-4 heightp^{2}\\right)\n\\end{aligned}\n\\]\n\nHence,\n\\[\n\\begin{array}{c}\n16 diaghalf^{2} heightp^{2}=-16 diaghalf^{4}+4 diaghalf^{2}\\left(rangeone^{2}+rangetwo^{2}+rangethree^{2}+rangefour^{2}\\right) \\\\\n-\\left(rangeone^{4}+rangetwo^{4}+rangethree^{4}+rangefour^{4}-2 rangeone^{2} rangethree^{2}-2 rangetwo^{2} rangefour^{2}\\right)\n\\end{array}\n\\]\n\nDividing through by \\( 16 diaghalf^{2}=8 sidesquare^{2} \\), we obtain the desired relation.\nRemark 1. To keep the algebra simple, it is essential to take advantage of the symmetry of the square. Another, almost equally good, choice of coordinates would make the vertices \\( ( \\pm halfside, \\pm halfside) \\), where \\( halfside=sidesquare / 2 \\).\n\nRemark 2. A point above a plane is determined by its distances from any three non-collinear points in the plane. Hence \\( heightp, coordx \\), and \\( coordy \\) are determined by any three of the \\( R \\) 's. This explains why there must be a nontrivial relation connecting the four R's."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "almondtree",
        "y": "paintbrush",
        "h": "driftwood",
        "R_1": "earthquake",
        "R_2": "sugarcane",
        "R_3": "tidalwave",
        "R_4": "blacksmith",
        "b": "horseshoe",
        "a": "starflower",
        "c": "brainstorm"
      },
      "question": "3. A projectile in flight is observed simultaneously from four radar stations which are situated at the corners of a square of side \\( horseshoe \\). The distances of the projectile from the four stations, taken in order around the square, are found to be \\( \\boldsymbol{earthquake}, \\boldsymbol{sugarcane}, \\boldsymbol{tidalwave}, \\boldsymbol{blacksmith} \\). Show that\n\\[\nearthquake^{2}+tidalwave^{2}=sugarcane^{2}+blacksmith^{2}\n\\]\n\nShow also that the height \\( driftwood \\) of the projectile above the ground is given by\n\\[\n\\begin{aligned}\ndriftwood^{2}= & -\\frac{1}{2} horseshoe^{2}+\\frac{1}{4}\\left(earthquake^{2}+sugarcane^{2}+tidalwave^{2}+blacksmith^{2}\\right) \\\\\n& -\\frac{1}{8 horseshoe^{2}}\\left(earthquake^{4}+sugarcane^{4}+tidalwave^{4}+blacksmith^{4}-2 earthquake^{2} tidalwave^{2}-2 sugarcane^{2} blacksmith^{2}\\right)\n\\end{aligned}\n\\]",
      "solution": "Solution. Choose the diagonals of the square as axes in the plane. Then the four vertices are \\( ( \\pm starflower, 0),(0, \\pm starflower) \\), where \\( starflower^{2}=horseshoe^{2} / 2 \\), and we number them counterclockwise starting at \\( (starflower, 0) \\). If the projectile is above the point \\( (almondtree, paintbrush) \\), then\n\\[\n\\begin{array}{ll}\nearthquake^{2}=(almondtree-starflower)^{2}+paintbrush^{2}+driftwood^{2}, & sugarcane^{2}=almondtree^{2}+(paintbrush-starflower)^{2}+driftwood^{2} \\\\\ntidalwave^{2}=(almondtree+starflower)^{2}+paintbrush^{2}+driftwood^{2}, & blacksmith^{2}=almondtree^{2}+(paintbrush+starflower)^{2}+driftwood^{2}\n\\end{array}\n\\]\n\nHence, \\( earthquake{ }^{2}+tidalwave{ }^{2}=sugarcane{ }^{2}+blacksmith{ }^{2} \\), as required, and\n\\[\nearthquake^{2}+sugarcane^{2}+tidalwave^{2}+blacksmith^{2}-4 starflower^{2}-4 driftwood^{2}=4\\left(almondtree^{2}+paintbrush^{2}\\right)\n\\]\n\nTherefore,\n\\[\n\\begin{aligned}\nearthquake^{4}+sugarcane^{4} & +tidalwave^{4}+blacksmith^{4}-2 earthquake^{2} tidalwave^{2}-2 sugarcane^{2} blacksmith^{2} \\\\\n& =\\left(tidalwave^{2}-earthquake^{2}\\right)^{2}+\\left(blacksmith^{2}-sugarcane^{2}\\right)^{2}=16 starflower^{2}\\left(almondtree^{2}+paintbrush^{2}\\right) \\\\\n& =4 starflower^{2}\\left(earthquake^{2}+sugarcane^{2}+tidalwave^{2}+blacksmith^{2}-4 starflower^{2}-4 driftwood^{2}\\right)\n\\end{aligned}\n\\]\n\nHence,\n\\[\n\\begin{array}{c}\n16 starflower^{2} driftwood^{2}=-16 starflower^{4}+4 starflower^{2}\\left(earthquake^{2}+sugarcane^{2}+tidalwave^{2}+blacksmith^{2}\\right) \\\\\n-\\left(earthquake^{4}+sugarcane^{4}+tidalwave^{4}+blacksmith^{4}-2 earthquake^{2} tidalwave^{2}-2 sugarcane^{2} blacksmith^{2}\\right)\n\\end{array}\n\\]\n\nDividing through by \\( 16 starflower^{2}=8 horseshoe^{2} \\), we obtain the desired relation.\nRemark 1. To keep the algebra simple, it is essential to take advantage of the symmetry of the square. Another, almost equally good, choice of coordinates would make the vertices \\( ( \\pm brainstorm, \\pm brainstorm) \\), where \\( brainstorm=horseshoe / 2 \\).\n\nRemark 2. A point above a plane is determined by its distances from any three non-collinear points in the plane. Hence \\( driftwood, almondtree \\), and \\( paintbrush \\) are determined by any three of the \\( R \\) 's. This explains why there must be a nontrivial relation connecting the four R's."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "h": "depthvalue",
        "R_1": "closenessone",
        "R_2": "closenesstwo",
        "R_3": "closenessthree",
        "R_4": "closenessfour",
        "b": "narrowness",
        "a": "shortmeasure",
        "c": "microextent"
      },
      "question": "3. A projectile in flight is observed simultaneously from four radar stations which are situated at the corners of a square of side \\( narrowness \\). The distances of the projectile from the four stations, taken in order around the square, are found to be \\( \\boldsymbol{closenessone}, \\boldsymbol{closenesstwo}, \\boldsymbol{closenessthree}, \\boldsymbol{closenessfour} \\). Show that\n\\[\nclosenessone^{2}+closenessthree^{2}=closenesstwo^{2}+closenessfour^{2}\n\\]\n\nShow also that the height \\( depthvalue \\) of the projectile above the ground is given by\n\\[\n\\begin{aligned}\ndepthvalue^{2}= & -\\frac{1}{2} narrowness^{2}+\\frac{1}{4}\\left(closenessone^{2}+closenesstwo^{2}+closenessthree^{2}+closenessfour^{2}\\right) \\\\\n& -\\frac{1}{8 narrowness^{2}}\\left(closenessone^{4}+closenesstwo^{4}+closenessthree^{4}+closenessfour^{4}-2 closenessone^{2} closenessthree^{2}-2 closenesstwo^{2} closenessfour^{2}\\right)\n\\end{aligned}\n\\]",
      "solution": "Solution. Choose the diagonals of the square as axes in the plane. Then the four vertices are \\( ( \\pm shortmeasure, 0),(0, \\pm shortmeasure) \\), where \\( shortmeasure^{2}=narrowness^{2} / 2 \\), and we number them counterclockwise starting at \\( (shortmeasure, 0) \\). If the projectile is above the point \\( (verticalaxis, horizontalaxis) \\), then\n\\[\n\\begin{array}{ll}\nclosenessone^{2}=(verticalaxis-shortmeasure)^{2}+horizontalaxis^{2}+depthvalue^{2}, & closenesstwo^{2}=verticalaxis^{2}+(horizontalaxis-shortmeasure)^{2}+depthvalue^{2} \\\\\nclosenessthree^{2}=(verticalaxis+shortmeasure)^{2}+horizontalaxis^{2}+depthvalue^{2}, & closenessfour^{2}=verticalaxis^{2}+(horizontalaxis+shortmeasure)^{2}+depthvalue^{2}\n\\end{array}\n\\]\n\nHence, \\( closenessone^{2}+closenessthree^{2}=closenesstwo^{2}+closenessfour^{2} \\), as required, and\n\\[\nclosenessone^{2}+closenesstwo^{2}+closenessthree^{2}+closenessfour^{2}-4 shortmeasure^{2}-4 depthvalue^{2}=4\\left(verticalaxis^{2}+horizontalaxis^{2}\\right)\n\\]\n\nTherefore,\n\\[\n\\begin{aligned}\nclosenessone^{4}+closenesstwo^{4} & +closenessthree^{4}+closenessfour^{4}-2 closenessone^{2} closenessthree^{2}-2 closenesstwo^{2} closenessfour^{2} \\\\\n& =\\left(closenessthree^{2}-closenessone^{2}\\right)^{2}+\\left(closenessfour^{2}-closenesstwo^{2}\\right)^{2}=16 shortmeasure^{2}\\left(verticalaxis^{2}+horizontalaxis^{2}\\right) \\\\\n& =4 shortmeasure^{2}\\left(closenessone^{2}+closenesstwo^{2}+closenessthree^{2}+closenessfour^{2}-4 shortmeasure^{2}-4 depthvalue^{2}\\right)\n\\end{aligned}\n\\]\n\nHence,\n\\[\n\\begin{array}{c}\n16 shortmeasure^{2} depthvalue^{2}=-16 shortmeasure^{4}+4 shortmeasure^{2}\\left(closenessone^{2}+closenesstwo^{2}+closenessthree^{2}+closenessfour^{2}\\right) \\\\\n-\\left(closenessone^{4}+closenesstwo^{4}+closenessthree^{4}+closenessfour^{4}-2 closenessone^{2} closenessthree^{2}-2 closenesstwo^{2} closenessfour^{2}\\right)\n\\end{array}\n\\]\n\nDividing through by \\( 16 shortmeasure^{2}=8 narrowness^{2} \\), we obtain the desired relation.\nRemark 1. To keep the algebra simple, it is essential to take advantage of the symmetry of the square. Another, almost equally good, choice of coordinates would make the vertices \\( ( \\pm microextent, \\pm microextent) \\), where \\( microextent=narrowness / 2 \\).\n\nRemark 2. A point above a plane is determined by its distances from any three non-collinear points in the plane. Hence \\( depthvalue, verticalaxis \\), and \\( horizontalaxis \\) are determined by any three of the closeness 's. This explains why there must be a nontrivial relation connecting the four closeness's."
    },
    "garbled_string": {
      "map": {
        "x": "tmprqzvn",
        "y": "kplhxsqt",
        "h": "jbvwlecm",
        "R_1": "qzxwvtnp",
        "R_2": "hjgrksla",
        "R_3": "vmclqser",
        "R_4": "dpnfrxug",
        "b": "snzptcwa",
        "a": "fvyqkdje",
        "c": "rslbpjmh"
      },
      "question": "3. A projectile in flight is observed simultaneously from four radar stations which are situated at the corners of a square of side \\( snzptcwa \\). The distances of the projectile from the four stations, taken in order around the square, are found to be \\( \\boldsymbol{qzxwvtnp}, \\boldsymbol{hjgrksla}, \\boldsymbol{vmclqser}, \\boldsymbol{dpnfrxug} \\). Show that\n\\[\nqzxwvtnp^{2}+vmclqser^{2}=hjgrksla^{2}+dpnfrxug^{2}\n\\]\n\nShow also that the height \\( jbvwlecm \\) of the projectile above the ground is given by\n\\[\n\\begin{aligned}\njbvwlecm^{2}= & -\\frac{1}{2} snzptcwa^{2}+\\frac{1}{4}\\left(qzxwvtnp^{2}+hjgrksla^{2}+vmclqser^{2}+dpnfrxug^{2}\\right) \\\\\n& -\\frac{1}{8 snzptcwa^{2}}\\left(qzxwvtnp^{4}+hjgrksla^{4}+vmclqser^{4}+dpnfrxug^{4}-2 qzxwvtnp^{2} vmclqser^{2}-2 hjgrksla^{2} dpnfrxug^{2}\\right)\n\\end{aligned}\n\\]\n",
      "solution": "Solution. Choose the diagonals of the square as axes in the plane. Then the four vertices are \\( ( \\pm fvyqkdje, 0),(0, \\pm fvyqkdje) \\), where \\( fvyqkdje^{2}=snzptcwa^{2} / 2 \\), and we number them counterclockwise starting at \\( (fvyqkdje, 0) \\). If the projectile is above the point \\( (tmprqzvn, kplhxsqt) \\), then\n\\[\n\\begin{array}{ll}\nqzxwvtnp^{2}=(tmprqzvn-fvyqkdje)^{2}+kplhxsqt^{2}+jbvwlecm^{2}, & hjgrksla^{2}=tmprqzvn^{2}+(kplhxsqt-fvyqkdje)^{2}+jbvwlecm^{2} \\\\\nvmclqser^{2}=(tmprqzvn+fvyqkdje)^{2}+kplhxsqt^{2}+jbvwlecm^{2}, & dpnfrxug^{2}=tmprqzvn^{2}+(kplhxsqt+fvyqkdje)^{2}+jbvwlecm^{2}\n\\end{array}\n\\]\n\nHence, \\( qzxwvtnp^{2}+vmclqser^{2}=hjgrksla^{2}+dpnfrxug^{2} \\), as required, and\n\\[\nqzxwvtnp^{2}+hjgrksla^{2}+vmclqser^{2}+dpnfrxug^{2}-4 fvyqkdje^{2}-4 jbvwlecm^{2}=4\\left(tmprqzvn^{2}+kplhxsqt^{2}\\right)\n\\]\n\nTherefore,\n\\[\n\\begin{aligned}\nqzxwvtnp^{4}+hjgrksla^{4} & +vmclqser^{4}+dpnfrxug^{4}-2 qzxwvtnp^{2} vmclqser^{2}-2 hjgrksla^{2} dpnfrxug^{2} \\\\\n& =\\left(vmclqser^{2}-qzxwvtnp^{2}\\right)^{2}+\\left(dpnfrxug^{2}-hjgrksla^{2}\\right)^{2}=16 fvyqkdje^{2}\\left(tmprqzvn^{2}+kplhxsqt^{2}\\right) \\\\\n& =4 fvyqkdje^{2}\\left(qzxwvtnp^{2}+hjgrksla^{2}+vmclqser^{2}+dpnfrxug^{2}-4 fvyqkdje^{2}-4 jbvwlecm^{2}\\right)\n\\end{aligned}\n\\]\n\nHence,\n\\[\n\\begin{array}{c}\n16 fvyqkdje^{2} jbvwlecm^{2}=-16 fvyqkdje^{4}+4 fvyqkdje^{2}\\left(qzxwvtnp^{2}+hjgrksla^{2}+vmclqser^{2}+dpnfrxug^{2}\\right) \\\\\n-\\left(qzxwvtnp^{4}+hjgrksla^{4}+vmclqser^{4}+dpnfrxug^{4}-2 qzxwvtnp^{2} vmclqser^{2}-2 hjgrksla^{2} dpnfrxug^{2}\\right)\n\\end{array}\n\\]\n\nDividing through by \\( 16 fvyqkdje^{2}=8 snzptcwa^{2} \\), we obtain the desired relation.\n\nRemark 1. To keep the algebra simple, it is essential to take advantage of the symmetry of the square. Another, almost equally good, choice of coordinates would make the vertices \\( ( \\pm rslbpjmh, \\pm rslbpjmh) \\), where \\( rslbpjmh=snzptcwa / 2 \\).\n\nRemark 2. A point above a plane is determined by its distances from any three non-collinear points in the plane. Hence \\( jbvwlecm, tmprqzvn \\), and \\( kplhxsqt \\) are determined by any three of the \\( R \\) 's. This explains why there must be a nontrivial relation connecting the four R's."
    },
    "kernel_variant": {
      "question": "Twelve synchronized beacons are fixed at the vertices  \nP₁ ,…, P₁₂ of a regular dodecagon which lies in the horizontal plane.  \nIts side–length is d > 0; the centre O coincides with the origin.  \nNumber the vertices counter-clockwise, starting with the one on the positive x-axis, so that  \n\n P_k = a exp(iθ_k) with θ_k = (k – 1)·π/6  (k = 1,…,12),\n\nwhere the circum-radius a is linked to the side-length by  \n\n d = 2a sin(π/12)   (a = d /(2 sin 15°) = ½d(√6+√2)).\n\nAt one instant an aircraft A = (x, y, h) with altitude h > 0 is observed.  \nPut z = x + iy and denote the simultaneous ranges\n\n L_k = |A P_k|  (k = 1,…,12).\n\nIntroduce the power sums  \n\n S = Σ_{k=1}^{12} L_k²,  \n Q = Σ_{k=1}^{12} L_k⁴ – 2 Σ_{k=1}^{6} L_k² L_{k+6}²  (indices mod 12).\n\n(a) (Discrete Fourier annihilation)  \nLet  \n\n \\tilde L_r := Σ_{k=1}^{12} e^{2π i r (k-1)/12}\\,L_k²  (r∈ℤ).\n\nShow that  \n\n \\tilde L_r = 0 whenever r (mod 12) ∉ {0,±1}.　　　(1)\n\nEquivalently, the only non-vanishing Fourier modes of the sequence {L_k²} occur for the\nfrequencies r≡0, 1, 11 (mod 12).\nIn particular\n\n L₁² – L₂² + L₃² – L₄² + ⋯ – L₁₂² = 0   (r = 6),  \n L₁²+L₃²+L₅²+L₇²+L₉²+L₁₁² = L₂²+L₄²+L₆²+L₈²+L₁₀²+L₁₂²  (r = 4),  \n L₁² – L₃² + L₅² – L₇² + L₉² – L₁₁² = 0  (r = 2)  \n\ntogether with three further independent alternating relations (r = 8, 10≡–2 and r = –4).\n\n(b) (Altitude formula)  \nProve that the aircraft altitude is determined by the ranges through  \n\n h² = –a² + S/12 – Q /(48 a²)  \n    = –d² /(4 sin² 15°) + S/12 – Q sin² 15° /(12 d²).\n\n(c) (Horizontal position and a built-in consistency check)  \nDefine the first two non-trivial Fourier coefficients\n\n F₁ = Σ_{k=1}^{12} e^{2π i (k-1)/12}\\,L_k²,  F₂ = Σ_{k=1}^{12} e^{4π i (k-1)/12}\\,L_k².\n\nShow that  \n\n F₂ = 0  and  z = –F₁ /(12 a).　　　　　　　　　　　(2)\n\nConsequently  \n\n |z|² = Q /(48 a²).　　　　　　　　　　　　　　　 (3)\n\nThus the triple (x, y, h) is uniquely determined by the twelve measured ranges,\nwhile the single identity F₂ = 0 supplies an immediate experimental consistency\ncheck for the measured data.",
      "solution": "Throughout write ζ := e^{2π i/12}; hence ζ¹² = 1 and ζ⁶ = –1.\n\n1. Expressing the squared ranges.  \nFor every k\n\n L_k² = |z – aζ^{k–1}|² + h²  \n    = |z|² + a² + h² – 2a Re( z ζ^{–(k–1)} ).\n\nIntroduce  \n\n ρ := |z|² + a² + h², u_k := Re( z ζ^{–(k–1)} ),　　　　(4)\n\nso that  \n\n L_k² = ρ – 2a u_k.　　　　　　　　　　　　　　　　(5)\n\nBecause the twelve roots ζ^{k–1} are uniformly distributed, we shall repeatedly use  \n\n Σ_{k=1}^{12} ζ^{m(k–1)} = 0  if m≢0 (mod 12).　　　　　　(6)\n\n2. Proof of (a).  \nLet r∈ℤ.  Compute the discrete Fourier coefficient\n\n \\tilde L_r = Σ_{k=1}^{12} ζ^{r(k–1)} L_k²  \n           = ρ Σ ζ^{r(k–1)} – 2a Σ u_k ζ^{r(k–1)}.　　　　　　(7)\n\nThe first sum vanishes unless r≡0 (mod 12).  \nFor the second one write u_k as half the sum of two conjugate exponentials,\n\n u_k = ½( z ζ^{–(k–1)} + \\bar z ζ^{k–1} ).\n\nHence\n\n Σ u_k ζ^{r(k–1)}  \n  = ½ z Σ ζ^{(r–1)(k–1)} + ½ \\bar z Σ ζ^{(r+1)(k–1)}.　　　　(8)\n\nEach geometric series on the right is non-zero only when r–1≡0 or r+1≡0 (mod 12), i.e. when r≡1 or r≡11.  Therefore\n\n \\tilde L_r ≠ 0 ⇔ r ≡ 0, 1, 11 (mod 12), proving (1).　　　　(9)\n\nParticular alternating identities.  \n• r = 6 → ζ^{6(k–1)} = (–1)^{k–1} gives the first displayed relation.  \n• r = 4 (or r = 8) separates odd from even indices, yielding the second.  \n• r = 2 (or r = 10) produces the third one cited in the statement; altogether six independent real relations arise from r = 2,4,6,8,10,–2.\n\n3. Global power sums S and Q.  \nSumming (5) immediately gives\n\n S = Σ L_k² = 12ρ.　　　　　　　　　　　　　　　　　　 (10)\n\nFor fourth powers we square (5):\n\n L_k⁴ = (ρ – 2a u_k)² = ρ² – 4aρ u_k + 4a²u_k².\n\nBecause Σ u_k = 0 (by (6) with m = 1),\n\n Σ L_k⁴ = 12ρ² + 4a²U, where U := Σ_{k=1}^{12} u_k².　　　　(11)\n\nOpposite beacons are antipodal: P_{k+6} = –P_k, hence u_{k+6} = –u_k and from (5)\n\n L_{k+6}² = ρ + 2a u_k.　　　　　　　　　　　　　　　　(12)\n\nTherefore for k = 1,…,6\n\n L_k² L_{k+6}² = (ρ – 2a u_k)(ρ + 2a u_k) = ρ² – 4a²u_k²,\n\nand\n\n Σ_{k=1}^{6} L_k² L_{k+6}² = 6ρ² – 2a²U.　　　　　　　　 (13)\n\nCombining (11) and (13) yields the corrected coefficient\n\n Q = Σ L_k⁴ – 2 Σ_{k=1}^{6} L_k² L_{k+6}²  \n   = (12ρ² + 4a²U) – 2(6ρ² – 2a²U) = 8a²U.　　　　　　 (14)\n\n4. Evaluation of U = Σ u_k².  \nWrite z = |z| e^{iφ}.  Then u_k = |z| cos(φ – θ_k).  \nBecause the twelve directions θ_k are equally spaced, the average of cos² over them is ½, whence\n\n U = Σ u_k² = 12·|z|² / 2 = 6|z|².　　　　　　　　　　　　(15)\n\n(One may also confirm this with roots of unity:  \nU = ¼(z² Σ ζ^{–2(k–1)} + \\bar z² Σ ζ^{2(k–1)} + 2|z|² Σ 1) = 6|z|².)\n\nSubstituting (15) into (14) gives\n\n Q = 48 a² |z|² ⇔ |z|² = Q /(48 a²).　　　　　　　　　 (16)\n\n5. Altitude formula (part (b)).  \nFrom (10) we have ρ = S/12.  By definition ρ = a² + h² + |z|², so\n\n h² = ρ – a² – |z|² = S/12 – a² – Q /(48 a²),　　　　　　(17)\n\nwhich is exactly the asserted expression.\n\n6. The first two Fourier coefficients (part (c)).  \nFirst coefficient:\n\n F₁ = Σ ζ^{k–1} L_k²  \n     = Σ ζ^{k–1}(ρ – 2a u_k)  \n     = –2a Σ u_k ζ^{k–1}  (because Σ ζ^{k–1}=0).　　　　(18)\n\nUsing (8) with r = 1 we obtain\n\n Σ u_k ζ^{k–1} = ½ z Σ ζ^{0} + ½ \\bar z Σ ζ^{2(k–1)} = 6z,　(19)\n\nsince Σ ζ^{0}=12 and Σ ζ^{2(k–1)}=0.  Hence\n\n F₁ = –12a z ⇔ z = –F₁ /(12 a).　　　　　　　　　　　 (20)\n\nSecond coefficient:\n\n F₂ = Σ ζ^{2(k–1)} L_k²  \n    = –2a Σ u_k ζ^{2(k–1)}.　　　　　　　　　　　　　　(21)\n\nInsert the decomposition of u_k again:\n\n Σ u_k ζ^{2(k–1)}  \n  = ½ z Σ ζ^{(2–1)(k–1)} + ½ \\bar z Σ ζ^{(2+1)(k–1)}  \n  = ½ z Σ ζ^{(k–1)} + ½ \\bar z Σ ζ^{3(k–1)} = 0,　　　　 (22)\n\nbecause both geometric sums vanish by (6).  Therefore\n\n F₂ = 0.　　　　　　　　　　　　　　　　　　　　　　 (23)\n\nEquations (20) and (16) complete part (c).  The identity F₂ = 0 furnishes an\nimmediate consistency test: any experimental data with F₂≠0 cannot come from\na single point in space.\n\nThus all three coordinates (x, y, h) are uniquely reconstructed from the twelve\nranges, and the corrected derivation is internally consistent.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.402304",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension in the sense of a 12-vertex configuration introduces six independent complex linear identities instead of one (square) or two (hexagon).  \n• The solution demands a discrete Fourier transform on the vertex set, not merely an alternating–sum trick.  Orthogonality of 12th roots of unity, manipulation of complex exponentials and symmetry arguments are indispensable.  \n• Altitude extraction now requires handling both fourth powers ΣL_k⁴ and the cross terms ΣL_k²L_{k+6}², together with an evaluation of Σu_k² via averaging of trigonometric squares—several more algebraic steps than in the hexagon case.  \n• Part (c) asks for the full horizontal coordinates, forcing the solver to connect Fourier coefficients with geometric data, something entirely absent from the original problem.  \nOverall the problem intertwines discrete Fourier analysis, classical geometry of regular polygons, and symmetric-polynomial elimination, making it markedly more sophisticated and lengthier than both the original square problem and the existing hexagon variant."
      }
    },
    "original_kernel_variant": {
      "question": "Twelve synchronized beacons are fixed at the vertices  \nP₁ ,…, P₁₂ of a regular dodecagon which lies in the horizontal plane.  \nIts side–length is d > 0; the centre O coincides with the origin.  \nNumber the vertices counter-clockwise, starting with the one on the positive x-axis, so that  \n\n P_k = a exp(iθ_k) with θ_k = (k – 1)·π/6  (k = 1,…,12),\n\nwhere the circum-radius a is linked to the side-length by  \n\n d = 2a sin(π/12)   (a = d /(2 sin 15°) = ½d(√6+√2)).\n\nAt one instant an aircraft A = (x, y, h) with altitude h > 0 is observed.  \nPut z = x + iy and denote the simultaneous ranges\n\n L_k = |A P_k|  (k = 1,…,12).\n\nIntroduce the power sums  \n\n S = Σ_{k=1}^{12} L_k²,  \n Q = Σ_{k=1}^{12} L_k⁴ – 2 Σ_{k=1}^{6} L_k² L_{k+6}²  (indices mod 12).\n\n(a) (Discrete Fourier annihilation)  \nLet  \n\n \\tilde L_r := Σ_{k=1}^{12} e^{2π i r (k-1)/12}\\,L_k²  (r∈ℤ).\n\nShow that  \n\n \\tilde L_r = 0 whenever r (mod 12) ∉ {0,±1}.　　　(1)\n\nEquivalently, the only non-vanishing Fourier modes of the sequence {L_k²} occur for the\nfrequencies r≡0, 1, 11 (mod 12).\nIn particular\n\n L₁² – L₂² + L₃² – L₄² + ⋯ – L₁₂² = 0   (r = 6),  \n L₁²+L₃²+L₅²+L₇²+L₉²+L₁₁² = L₂²+L₄²+L₆²+L₈²+L₁₀²+L₁₂²  (r = 4),  \n L₁² – L₃² + L₅² – L₇² + L₉² – L₁₁² = 0  (r = 2)  \n\ntogether with three further independent alternating relations (r = 8, 10≡–2 and r = –4).\n\n(b) (Altitude formula)  \nProve that the aircraft altitude is determined by the ranges through  \n\n h² = –a² + S/12 – Q /(48 a²)  \n    = –d² /(4 sin² 15°) + S/12 – Q sin² 15° /(12 d²).\n\n(c) (Horizontal position and a built-in consistency check)  \nDefine the first two non-trivial Fourier coefficients\n\n F₁ = Σ_{k=1}^{12} e^{2π i (k-1)/12}\\,L_k²,  F₂ = Σ_{k=1}^{12} e^{4π i (k-1)/12}\\,L_k².\n\nShow that  \n\n F₂ = 0  and  z = –F₁ /(12 a).　　　　　　　　　　　(2)\n\nConsequently  \n\n |z|² = Q /(48 a²).　　　　　　　　　　　　　　　 (3)\n\nThus the triple (x, y, h) is uniquely determined by the twelve measured ranges,\nwhile the single identity F₂ = 0 supplies an immediate experimental consistency\ncheck for the measured data.",
      "solution": "Throughout write ζ := e^{2π i/12}; hence ζ¹² = 1 and ζ⁶ = –1.\n\n1. Expressing the squared ranges.  \nFor every k\n\n L_k² = |z – aζ^{k–1}|² + h²  \n    = |z|² + a² + h² – 2a Re( z ζ^{–(k–1)} ).\n\nIntroduce  \n\n ρ := |z|² + a² + h², u_k := Re( z ζ^{–(k–1)} ),　　　　(4)\n\nso that  \n\n L_k² = ρ – 2a u_k.　　　　　　　　　　　　　　　　(5)\n\nBecause the twelve roots ζ^{k–1} are uniformly distributed, we shall repeatedly use  \n\n Σ_{k=1}^{12} ζ^{m(k–1)} = 0  if m≢0 (mod 12).　　　　　　(6)\n\n2. Proof of (a).  \nLet r∈ℤ.  Compute the discrete Fourier coefficient\n\n \\tilde L_r = Σ_{k=1}^{12} ζ^{r(k–1)} L_k²  \n           = ρ Σ ζ^{r(k–1)} – 2a Σ u_k ζ^{r(k–1)}.　　　　　　(7)\n\nThe first sum vanishes unless r≡0 (mod 12).  \nFor the second one write u_k as half the sum of two conjugate exponentials,\n\n u_k = ½( z ζ^{–(k–1)} + \\bar z ζ^{k–1} ).\n\nHence\n\n Σ u_k ζ^{r(k–1)}  \n  = ½ z Σ ζ^{(r–1)(k–1)} + ½ \\bar z Σ ζ^{(r+1)(k–1)}.　　　　(8)\n\nEach geometric series on the right is non-zero only when r–1≡0 or r+1≡0 (mod 12), i.e. when r≡1 or r≡11.  Therefore\n\n \\tilde L_r ≠ 0 ⇔ r ≡ 0, 1, 11 (mod 12), proving (1).　　　　(9)\n\nParticular alternating identities.  \n• r = 6 → ζ^{6(k–1)} = (–1)^{k–1} gives the first displayed relation.  \n• r = 4 (or r = 8) separates odd from even indices, yielding the second.  \n• r = 2 (or r = 10) produces the third one cited in the statement; altogether six independent real relations arise from r = 2,4,6,8,10,–2.\n\n3. Global power sums S and Q.  \nSumming (5) immediately gives\n\n S = Σ L_k² = 12ρ.　　　　　　　　　　　　　　　　　　 (10)\n\nFor fourth powers we square (5):\n\n L_k⁴ = (ρ – 2a u_k)² = ρ² – 4aρ u_k + 4a²u_k².\n\nBecause Σ u_k = 0 (by (6) with m = 1),\n\n Σ L_k⁴ = 12ρ² + 4a²U, where U := Σ_{k=1}^{12} u_k².　　　　(11)\n\nOpposite beacons are antipodal: P_{k+6} = –P_k, hence u_{k+6} = –u_k and from (5)\n\n L_{k+6}² = ρ + 2a u_k.　　　　　　　　　　　　　　　　(12)\n\nTherefore for k = 1,…,6\n\n L_k² L_{k+6}² = (ρ – 2a u_k)(ρ + 2a u_k) = ρ² – 4a²u_k²,\n\nand\n\n Σ_{k=1}^{6} L_k² L_{k+6}² = 6ρ² – 2a²U.　　　　　　　　 (13)\n\nCombining (11) and (13) yields the corrected coefficient\n\n Q = Σ L_k⁴ – 2 Σ_{k=1}^{6} L_k² L_{k+6}²  \n   = (12ρ² + 4a²U) – 2(6ρ² – 2a²U) = 8a²U.　　　　　　 (14)\n\n4. Evaluation of U = Σ u_k².  \nWrite z = |z| e^{iφ}.  Then u_k = |z| cos(φ – θ_k).  \nBecause the twelve directions θ_k are equally spaced, the average of cos² over them is ½, whence\n\n U = Σ u_k² = 12·|z|² / 2 = 6|z|².　　　　　　　　　　　　(15)\n\n(One may also confirm this with roots of unity:  \nU = ¼(z² Σ ζ^{–2(k–1)} + \\bar z² Σ ζ^{2(k–1)} + 2|z|² Σ 1) = 6|z|².)\n\nSubstituting (15) into (14) gives\n\n Q = 48 a² |z|² ⇔ |z|² = Q /(48 a²).　　　　　　　　　 (16)\n\n5. Altitude formula (part (b)).  \nFrom (10) we have ρ = S/12.  By definition ρ = a² + h² + |z|², so\n\n h² = ρ – a² – |z|² = S/12 – a² – Q /(48 a²),　　　　　　(17)\n\nwhich is exactly the asserted expression.\n\n6. The first two Fourier coefficients (part (c)).  \nFirst coefficient:\n\n F₁ = Σ ζ^{k–1} L_k²  \n     = Σ ζ^{k–1}(ρ – 2a u_k)  \n     = –2a Σ u_k ζ^{k–1}  (because Σ ζ^{k–1}=0).　　　　(18)\n\nUsing (8) with r = 1 we obtain\n\n Σ u_k ζ^{k–1} = ½ z Σ ζ^{0} + ½ \\bar z Σ ζ^{2(k–1)} = 6z,　(19)\n\nsince Σ ζ^{0}=12 and Σ ζ^{2(k–1)}=0.  Hence\n\n F₁ = –12a z ⇔ z = –F₁ /(12 a).　　　　　　　　　　　 (20)\n\nSecond coefficient:\n\n F₂ = Σ ζ^{2(k–1)} L_k²  \n    = –2a Σ u_k ζ^{2(k–1)}.　　　　　　　　　　　　　　(21)\n\nInsert the decomposition of u_k again:\n\n Σ u_k ζ^{2(k–1)}  \n  = ½ z Σ ζ^{(2–1)(k–1)} + ½ \\bar z Σ ζ^{(2+1)(k–1)}  \n  = ½ z Σ ζ^{(k–1)} + ½ \\bar z Σ ζ^{3(k–1)} = 0,　　　　 (22)\n\nbecause both geometric sums vanish by (6).  Therefore\n\n F₂ = 0.　　　　　　　　　　　　　　　　　　　　　　 (23)\n\nEquations (20) and (16) complete part (c).  The identity F₂ = 0 furnishes an\nimmediate consistency test: any experimental data with F₂≠0 cannot come from\na single point in space.\n\nThus all three coordinates (x, y, h) are uniquely reconstructed from the twelve\nranges, and the corrected derivation is internally consistent.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.343842",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension in the sense of a 12-vertex configuration introduces six independent complex linear identities instead of one (square) or two (hexagon).  \n• The solution demands a discrete Fourier transform on the vertex set, not merely an alternating–sum trick.  Orthogonality of 12th roots of unity, manipulation of complex exponentials and symmetry arguments are indispensable.  \n• Altitude extraction now requires handling both fourth powers ΣL_k⁴ and the cross terms ΣL_k²L_{k+6}², together with an evaluation of Σu_k² via averaging of trigonometric squares—several more algebraic steps than in the hexagon case.  \n• Part (c) asks for the full horizontal coordinates, forcing the solver to connect Fourier coefficients with geometric data, something entirely absent from the original problem.  \nOverall the problem intertwines discrete Fourier analysis, classical geometry of regular polygons, and symmetric-polynomial elimination, making it markedly more sophisticated and lengthier than both the original square problem and the existing hexagon variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}