{
  "index": "1959-A-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "5. A sparrow, flying horizontally in a straight line, is 50 feet directly below an eagle and 100 feet directly above a hawk. Both hawk and eagle fly directly toward the sparrow, reaching it simultaneously. The hawk flies twice as fast as the sparrow. How far does each bird fly? At what rate does the eagle fly?",
  "solution": "Solution. Although not specifically stated, it is clearly intended that each of the three birds flies at a uniform speed. So we consider the general problem of a predator flying with speed \\( v \\) in pursuit of a prey flying along a straight path with speed \\( r v, r<1 \\).\n\nLet the prey start at \\( (0,0) \\) and move along the \\( x \\)-axis so that his position at time \\( t \\) is (rvt, 0). Let the predator start at \\( (0, h), h>0 \\). If at time \\( t \\) the predator is at \\( (x, y) \\), then the condition that he always flies directly toward his prey is that\n(1)\n\\[\n\\frac{d x / d t}{d y / d t}=\\frac{r v t-x}{-y}\n\\]\n\nAs long as \\( y \\) remains positive, \\( d y / d t<0 \\), so we may choose \\( y \\) as the independent variable. Then (1) becomes\n(2)\n\\[\ny \\frac{d x}{d y}=x-r v t .\n\\]\n\nThe condition that the predator has constant speed \\( v \\) is\n\\[\n\\sqrt{\\left(\\frac{d x}{d t}\\right)^{2}+\\left(\\frac{d y}{d t}\\right)^{2}}=v .\n\\]\nwhich we can write as\n(3)\n\\[\n-\\sqrt{\\left(\\frac{d x}{d y}\\right)^{2}+1}=v \\frac{d t}{d y},\n\\]\nwhere we have introduced a minus sign because \\( d t / d y \\) is negative.\nWe differentiate (2) to get\n\\[\ny \\frac{d^{2} x}{d y^{2}}+\\frac{d x}{d y}=\\frac{d x}{d y}-r v \\frac{d t}{d y},\n\\]\nand using (3) we have\n\\[\ny \\frac{d^{2} x}{d y^{2}}=r \\sqrt{\\left(\\frac{d x}{d y}\\right)^{2}+1} .\n\\]\n\nWe can write this as the first-order differential equation\n\\[\ny \\frac{d z}{d y}=r \\sqrt{1+z^{2}},\n\\]\nwhere \\( z=d x / d y \\). Separating the variables we have\n\\[\nr \\frac{d y}{y}=\\frac{d z}{\\sqrt{1+z^{2}}},\n\\]\nwhence\n(4)\n\\[\nr \\log y=\\log \\left(z+\\sqrt{1+z^{2}}\\right)+C .\n\\]\n\nSince \\( z=d x / d y=0 \\) when \\( y=h \\), we find \\( C=r \\log h \\). Then (4) can be. rewritten\n\\[\n\\left(\\frac{y}{h}\\right)^{\\prime}=z+\\sqrt{1+z^{2}}\n\\]\nand solved for \\( z(=d x / d y) \\) to get\n(5)\n\\[\n2 \\frac{d x}{d y}=\\left(\\frac{y}{h}\\right)^{\\prime}-\\left(\\frac{y}{h}\\right)^{\\prime} .\n\\]\n\nThis can be integrated again using the initial condition \\( x=0 \\) when \\( y=h \\) to give\n(6)\n\\[\n2 x=\\frac{h}{1+r}\\left(\\frac{y}{h}\\right)^{1+r}-\\frac{h}{1-r}\\left(\\frac{y}{h}\\right)^{1-r}+\\frac{2 r h}{1-r^{2}} .\n\\]\n\nThis is valid only for \\( y>0 \\). Recalling that \\( 0<r<1 \\), we see from (6) that\n\\[\nx \\rightarrow \\frac{r h}{1-r^{2}}\n\\]\nas \\( y \\rightarrow 0 \\). We see from (5) that \\( y(d x / d y)-0 \\) and then from (2) that\n\\[\nx-r v t-0\n\\]\nas \\( \\boldsymbol{y}-\\mathbf{0} \\). It follows that the predator catches his prey at the point \\( \\left(r h /\\left(1-r^{2}\\right), 0\\right) \\) when \\( t=h /\\left(1-r^{2}\\right) v \\). At this time the prey has flown \\( r h /\\left(1-r^{2}\\right) \\) and the predator, \\( h /\\left(1-r^{2}\\right) \\).\nNow for the hawk and the sparrow, \\( r=\\frac{1}{2}, h=100 \\mathrm{ft} \\), so the sparrow flies \\( 200 / 3 \\mathrm{ft} \\)., and the hawk, \\( 400 / 3 \\mathrm{ft} \\). [Note that since the sparrow is above the hawk we would have to choose a coordinate system with \\( y \\) positive in the downward direction.]\nIn the case of the eagle, \\( h=50 \\mathrm{ft} \\). and \\( r \\) is unknown, but\n\\[\n\\frac{r 50}{1-r^{2}}=\\frac{200}{3} ;\n\\]\nhence \\( 4 r^{2}+3 r-4=0 \\) and\n\\[\nr=\\frac{-3+\\sqrt{ } 73}{8}\n\\]\nsince \\( r \\) must be positive.\nThe speed of the eagle is \\( 1 / r \\) times the speed of the sparrow, or\n\\[\n\\frac{v(3+\\sqrt{7} 3)}{16} \\approx .721 v\n\\]\nwhere \\( v \\) is the speed of the hawk.\nSince the time of capture is \\( 400 / 3 v \\), the eagle flies a distance of\n\\[\n\\frac{400}{3}\\left(\\frac{3+\\sqrt{ } 73}{16}\\right) \\approx 96.2 \\mathrm{ft} .\n\\]\n\nRemark. The problem appears as Advanced Problem 3573 in the American Mathematical Monthly, vol. 39 (1932), page 454.\n\nAn entertaining paper by Arthur Bernhart, \"Curves of Pursuit,\" Scripta Mathematica, 20, nos. 3-4 (1954), outlines some of the history of the pursuit problem.",
  "vars": [
    "x",
    "y",
    "t",
    "z"
  ],
  "params": [
    "v",
    "r",
    "h"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizontalpos",
        "y": "verticalpos",
        "t": "elapsedtime",
        "z": "slopevar",
        "v": "predatorspeed",
        "r": "speedratio",
        "h": "initialheight"
      },
      "question": "5. A sparrow, flying horizontally in a straight line, is 50 feet directly below an eagle and 100 feet directly above a hawk. Both hawk and eagle fly directly toward the sparrow, reaching it simultaneously. The hawk flies twice as fast as the sparrow. How far does each bird fly? At what rate does the eagle fly?",
      "solution": "Solution. Although not specifically stated, it is clearly intended that each of the three birds flies at a uniform speed. So we consider the general problem of a predator flying with speed \\( predatorspeed \\) in pursuit of a prey flying along a straight path with speed \\( speedratio predatorspeed, speedratio<1 \\).\n\nLet the prey start at \\( (0,0) \\) and move along the \\( x \\)-axis so that his position at time \\( elapsedtime \\) is (speedratio predatorspeed elapsedtime, 0). Let the predator start at \\( (0, initialheight), initialheight>0 \\). If at time \\( elapsedtime \\) the predator is at \\( (horizontalpos, verticalpos) \\), then the condition that he always flies directly toward his prey is that\n(1)\n\\[\n\\frac{d horizontalpos / d elapsedtime}{d verticalpos / d elapsedtime}=\\frac{speedratio predatorspeed elapsedtime-horizontalpos}{-verticalpos}\n\\]\n\nAs long as \\( verticalpos \\) remains positive, \\( d verticalpos / d elapsedtime<0 \\), so we may choose \\( verticalpos \\) as the independent variable. Then (1) becomes\n(2)\n\\[\nverticalpos \\frac{d horizontalpos}{d verticalpos}=horizontalpos-speedratio predatorspeed elapsedtime .\n\\]\n\nThe condition that the predator has constant speed \\( predatorspeed \\) is\n\\[\n\\sqrt{\\left(\\frac{d horizontalpos}{d elapsedtime}\\right)^{2}+\\left(\\frac{d verticalpos}{d elapsedtime}\\right)^{2}}=predatorspeed .\n\\]\nwhich we can write as\n(3)\n\\[\n-\\sqrt{\\left(\\frac{d horizontalpos}{d verticalpos}\\right)^{2}+1}=predatorspeed \\frac{d elapsedtime}{d verticalpos},\n\\]\nwhere we have introduced a minus sign because \\( d elapsedtime / d verticalpos \\) is negative.\nWe differentiate (2) to get\n\\[\nverticalpos \\frac{d^{2} horizontalpos}{d verticalpos^{2}}+\\frac{d horizontalpos}{d verticalpos}=\\frac{d horizontalpos}{d verticalpos}-speedratio predatorspeed \\frac{d elapsedtime}{d verticalpos},\n\\]\nand using (3) we have\n\\[\nverticalpos \\frac{d^{2} horizontalpos}{d verticalpos^{2}}=speedratio \\sqrt{\\left(\\frac{d horizontalpos}{d verticalpos}\\right)^{2}+1} .\n\\]\n\nWe can write this as the first-order differential equation\n\\[\nverticalpos \\frac{d slopevar}{d verticalpos}=speedratio \\sqrt{1+slopevar^{2}},\n\\]\nwhere \\( slopevar=d horizontalpos / d verticalpos \\). Separating the variables we have\n\\[\nspeedratio \\frac{d verticalpos}{verticalpos}=\\frac{d slopevar}{\\sqrt{1+slopevar^{2}}},\n\\]\nwhence\n(4)\n\\[\nspeedratio \\log verticalpos=\\log \\left(slopevar+\\sqrt{1+slopevar^{2}}\\right)+C .\n\\]\n\nSince \\( slopevar=d horizontalpos / d verticalpos=0 \\) when \\( verticalpos=initialheight \\), we find \\( C=speedratio \\log initialheight \\). Then (4) can be. rewritten\n\\[\n\\left(\\frac{verticalpos}{initialheight}\\right)^{\\prime}=slopevar+\\sqrt{1+slopevar^{2}}\n\\]\nand solved for \\( slopevar(=d horizontalpos / d verticalpos) \\) to get\n(5)\n\\[\n2 \\frac{d horizontalpos}{d verticalpos}=\\left(\\frac{verticalpos}{initialheight}\\right)^{\\prime}-\\left(\\frac{verticalpos}{initialheight}\\right)^{\\prime} .\n\\]\n\nThis can be integrated again using the initial condition \\( horizontalpos=0 \\) when \\( verticalpos=initialheight \\) to give\n(6)\n\\[\n2 horizontalpos=\\frac{initialheight}{1+speedratio}\\left(\\frac{verticalpos}{initialheight}\\right)^{1+speedratio}-\\frac{initialheight}{1-speedratio}\\left(\\frac{verticalpos}{initialheight}\\right)^{1-speedratio}+\\frac{2 speedratio initialheight}{1-speedratio^{2}} .\n\\]\n\nThis is valid only for \\( verticalpos>0 \\). Recalling that \\( 0<speedratio<1 \\), we see from (6) that\n\\[\nhorizontalpos \\rightarrow \\frac{speedratio initialheight}{1-speedratio^{2}}\n\\]\nas \\( verticalpos \\rightarrow 0 \\). We see from (5) that \\( verticalpos(d horizontalpos / d verticalpos)-0 \\) and then from (2) that\n\\[\nhorizontalpos-speedratio predatorspeed elapsedtime-0\n\\]\nas \\( \\boldsymbol{verticalpos}-\\mathbf{0} \\). It follows that the predator catches his prey at the point \\( \\left(speedratio initialheight /\\left(1-speedratio^{2}\\right), 0\\right) \\) when \\( elapsedtime=initialheight /\\left(1-speedratio^{2}\\right) predatorspeed \\). At this time the prey has flown \\( speedratio initialheight /\\left(1-speedratio^{2}\\right) \\) and the predator, \\( initialheight /\\left(1-speedratio^{2}\\right) \\).\nNow for the hawk and the sparrow, \\( speedratio=\\frac{1}{2}, initialheight=100 \\mathrm{ft} \\), so the sparrow flies \\( 200 / 3 \\mathrm{ft} \\)., and the hawk, \\( 400 / 3 \\mathrm{ft} \\). [Note that since the sparrow is above the hawk we would have to choose a coordinate system with \\( verticalpos \\) positive in the downward direction.]\nIn the case of the eagle, \\( initialheight=50 \\mathrm{ft} \\). and \\( speedratio \\) is unknown, but\n\\[\n\\frac{speedratio 50}{1-speedratio^{2}}=\\frac{200}{3} ;\n\\]\nhence \\( 4 speedratio^{2}+3 speedratio-4=0 \\) and\n\\[\nspeedratio=\\frac{-3+\\sqrt{ } 73}{8}\n\\]\nsince \\( speedratio \\) must be positive.\nThe speed of the eagle is \\( 1 / speedratio \\) times the speed of the sparrow, or\n\\[\n\\frac{predatorspeed(3+\\sqrt{7} 3)}{16} \\approx .721 predatorspeed\n\\]\nwhere \\( predatorspeed \\) is the speed of the hawk.\nSince the time of capture is \\( 400 / 3 predatorspeed \\), the eagle flies a distance of\n\\[\n\\frac{400}{3}\\left(\\frac{3+\\sqrt{ } 73}{16}\\right) \\approx 96.2 \\mathrm{ft} .\n\\]\n\nRemark. The problem appears as Advanced Problem 3573 in the American Mathematical Monthly, vol. 39 (1932), page 454.\n\nAn entertaining paper by Arthur Bernhart, \"Curves of Pursuit,\" Scripta Mathematica, 20, nos. 3-4 (1954), outlines some of the history of the pursuit problem."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "teacupful",
        "y": "peppermill",
        "t": "rainspout",
        "z": "lemonade",
        "v": "driftwood",
        "r": "chandelier",
        "h": "marshmallow"
      },
      "question": "5. A sparrow, flying horizontally in a straight line, is 50 feet directly below an eagle and 100 feet directly above a hawk. Both hawk and eagle fly directly toward the sparrow, reaching it simultaneously. The hawk flies twice as fast as the sparrow. How far does each bird fly? At what rate does the eagle fly?",
      "solution": "Solution. Although not specifically stated, it is clearly intended that each of the three birds flies at a uniform speed. So we consider the general problem of a predator flying with speed \\( driftwood \\) in pursuit of a prey flying along a straight path with speed \\( chandelier driftwood, chandelier<1 \\).\n\nLet the prey start at \\( (0,0) \\) and move along the \\( teacupful \\)-axis so that his position at time \\( rainspout \\) is (chandelier driftwood rainspout, 0). Let the predator start at \\( (0, marshmallow), marshmallow>0 \\). If at time \\( rainspout \\) the predator is at \\( (teacupful, peppermill) \\), then the condition that he always flies directly toward his prey is that\n(1)\n\\[\n\\frac{d teacupful / d rainspout}{d peppermill / d rainspout}=\\frac{chandelier driftwood rainspout-teacupful}{-peppermill}\n\\]\n\nAs long as \\( peppermill \\) remains positive, \\( d peppermill / d rainspout<0 \\), so we may choose \\( peppermill \\) as the independent variable. Then (1) becomes\n(2)\n\\[\npeppermill \\frac{d teacupful}{d peppermill}=teacupful-chandelier driftwood rainspout .\n\\]\n\nThe condition that the predator has constant speed \\( driftwood \\) is\n\\[\n\\sqrt{\\left(\\frac{d teacupful}{d rainspout}\\right)^{2}+\\left(\\frac{d peppermill}{d rainspout}\\right)^{2}}=driftwood .\n\\]\nwhich we can write as\n(3)\n\\[\n-\\sqrt{\\left(\\frac{d teacupful}{d peppermill}\\right)^{2}+1}=driftwood \\frac{d rainspout}{d peppermill},\n\\]\nwhere we have introduced a minus sign because \\( d rainspout / d peppermill \\) is negative.\nWe differentiate (2) to get\n\\[\npeppermill \\frac{d^{2} teacupful}{d peppermill^{2}}+\\frac{d teacupful}{d peppermill}=\\frac{d teacupful}{d peppermill}-chandelier driftwood \\frac{d rainspout}{d peppermill},\n\\]\nand using (3) we have\n\\[\npeppermill \\frac{d^{2} teacupful}{d peppermill^{2}}=chandelier \\sqrt{\\left(\\frac{d teacupful}{d peppermill}\\right)^{2}+1} .\n\\]\n\nWe can write this as the first-order differential equation\n\\[\npeppermill \\frac{d lemonade}{d peppermill}=chandelier \\sqrt{1+lemonade^{2}},\n\\]\nwhere \\( lemonade=d teacupful / d peppermill \\). Separating the variables we have\n\\[\nchandelier \\frac{d peppermill}{peppermill}=\\frac{d lemonade}{\\sqrt{1+lemonade^{2}}},\n\\]\nwhence\n(4)\n\\[\nchandelier \\log peppermill=\\log \\left(lemonade+\\sqrt{1+lemonade^{2}}\\right)+C .\n\\]\n\nSince \\( lemonade=d teacupful / d peppermill=0 \\) when \\( peppermill=marshmallow \\), we find \\( C=chandelier \\log marshmallow \\). Then (4) can be. rewritten\n\\[\n\\left(\\frac{peppermill}{marshmallow}\\right)^{\\prime}=lemonade+\\sqrt{1+lemonade^{2}}\n\\]\nand solved for \\( lemonade(=d teacupful / d peppermill) \\) to get\n(5)\n\\[\n2 \\frac{d teacupful}{d peppermill}=\\left(\\frac{peppermill}{marshmallow}\\right)^{\\prime}-\\left(\\frac{peppermill}{marshmallow}\\right)^{\\prime} .\n\\]\n\nThis can be integrated again using the initial condition \\( teacupful=0 \\) when \\( peppermill=marshmallow \\) to give\n(6)\n\\[\n2 teacupful=\\frac{marshmallow}{1+chandelier}\\left(\\frac{peppermill}{marshmallow}\\right)^{1+chandelier}-\\frac{marshmallow}{1-chandelier}\\left(\\frac{peppermill}{marshmallow}\\right)^{1-chandelier}+\\frac{2 chandelier marshmallow}{1-chandelier^{2}} .\n\\]\n\nThis is valid only for \\( peppermill>0 \\). Recalling that \\( 0<chandelier<1 \\), we see from (6) that\n\\[\nteacupful \\rightarrow \\frac{chandelier marshmallow}{1-chandelier^{2}}\n\\]\nas \\( peppermill \\rightarrow 0 \\). We see from (5) that \\( peppermill(d teacupful / d peppermill)-0 \\) and then from (2) that\n\\[\nteacupful-chandelier driftwood rainspout-0\n\\]\nas \\( \\boldsymbol{peppermill}-\\mathbf{0} \\). It follows that the predator catches his prey at the point \\( \\left(chandelier marshmallow /\\left(1-chandelier^{2}\\right), 0\\right) \\) when \\( rainspout=marshmallow /\\left(1-chandelier^{2}\\right) driftwood \\). At this time the prey has flown \\( chandelier marshmallow /\\left(1-chandelier^{2}\\right) \\) and the predator, \\( marshmallow /\\left(1-chandelier^{2}\\right) \\).\nNow for the hawk and the sparrow, \\( chandelier=\\frac{1}{2}, marshmallow=100 \\mathrm{ft} \\), so the sparrow flies \\( 200 / 3 \\mathrm{ft} \\)., and the hawk, \\( 400 / 3 \\mathrm{ft} \\). [Note that since the sparrow is above the hawk we would have to choose a coordinate system with \\( peppermill \\) positive in the downward direction.]\nIn the case of the eagle, \\( marshmallow=50 \\mathrm{ft} \\). and \\( chandelier \\) is unknown, but\n\\[\n\\frac{chandelier 50}{1-chandelier^{2}}=\\frac{200}{3} ;\n\\]\nhence \\( 4 chandelier^{2}+3 chandelier-4=0 \\) and\n\\[\nchandelier=\\frac{-3+\\sqrt{ } 73}{8}\n\\]\nsince \\( chandelier \\) must be positive.\nThe speed of the eagle is \\( 1 / chandelier \\) times the speed of the sparrow, or\n\\[\n\\frac{driftwood(3+\\sqrt{7} 3)}{16} \\approx .721 driftwood\n\\]\nwhere \\( driftwood \\) is the speed of the hawk.\nSince the time of capture is \\( 400 / 3 driftwood \\), the eagle flies a distance of\n\\[\n\\frac{400}{3}\\left(\\frac{3+\\sqrt{ } 73}{16}\\right) \\approx 96.2 \\mathrm{ft} .\n\\]\n\nRemark. The problem appears as Advanced Problem 3573 in the American Mathematical Monthly, vol. 39 (1932), page 454.\n\nAn entertaining paper by Arthur Bernhart, \"Curves of Pursuit,\" Scripta Mathematica, 20, nos. 3-4 (1954), outlines some of the history of the pursuit problem."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "t": "timelessness",
        "z": "constantvalue",
        "v": "slowness",
        "r": "inflationfactor",
        "h": "depthlevel"
      },
      "question": "5. A sparrow, flying horizontally in a straight line, is 50 feet directly below an eagle and 100 feet directly above a hawk. Both hawk and eagle fly directly toward the sparrow, reaching it simultaneously. The hawk flies twice as fast as the sparrow. How far does each bird fly? At what rate does the eagle fly?",
      "solution": "Solution. Although not specifically stated, it is clearly intended that each of the three birds flies at a uniform speed. So we consider the general problem of a predator flying with speed \\( slowness \\) in pursuit of a prey flying along a straight path with speed \\( inflationfactor slowness, inflationfactor<1 \\).\n\nLet the prey start at \\( (0,0) \\) and move along the \\( verticalaxis \\)-axis so that his position at time \\( timelessness \\) is (inflationfactor slowness timelessness, 0). Let the predator start at \\( (0, depthlevel), depthlevel>0 \\). If at time \\( timelessness \\) the predator is at \\( (verticalaxis, horizontalaxis) \\), then the condition that he always flies directly toward his prey is that\n(1)\n\\[\n\\frac{d verticalaxis / d timelessness}{d horizontalaxis / d timelessness}=\\frac{inflationfactor slowness timelessness-verticalaxis}{-horizontalaxis}\n\\]\n\nAs long as \\( horizontalaxis \\) remains positive, \\( d horizontalaxis / d timelessness<0 \\), so we may choose \\( horizontalaxis \\) as the independent variable. Then (1) becomes\n(2)\n\\[\nhorizontalaxis \\frac{d verticalaxis}{d horizontalaxis}=verticalaxis-inflationfactor slowness timelessness .\n\\]\n\nThe condition that the predator has constant speed \\( slowness \\) is\n\\[\n\\sqrt{\\left(\\frac{d verticalaxis}{d timelessness}\\right)^{2}+\\left(\\frac{d horizontalaxis}{d timelessness}\\right)^{2}}=slowness .\n\\]\nwhich we can write as\n(3)\n\\[\n-\\sqrt{\\left(\\frac{d verticalaxis}{d horizontalaxis}\\right)^{2}+1}=slowness \\frac{d timelessness}{d horizontalaxis},\n\\]\nwhere we have introduced a minus sign because \\( d timelessness / d horizontalaxis \\) is negative.\nWe differentiate (2) to get\n\\[\nhorizontalaxis \\frac{d^{2} verticalaxis}{d horizontalaxis^{2}}+\\frac{d verticalaxis}{d horizontalaxis}=\\frac{d verticalaxis}{d horizontalaxis}-inflationfactor slowness \\frac{d timelessness}{d horizontalaxis},\n\\]\nand using (3) we have\n\\[\nhorizontalaxis \\frac{d^{2} verticalaxis}{d horizontalaxis^{2}}=inflationfactor \\sqrt{\\left(\\frac{d verticalaxis}{d horizontalaxis}\\right)^{2}+1} .\n\\]\n\nWe can write this as the first-order differential equation\n\\[\nhorizontalaxis \\frac{d constantvalue}{d horizontalaxis}=inflationfactor \\sqrt{1+constantvalue^{2}},\n\\]\nwhere \\( constantvalue=d verticalaxis / d horizontalaxis \\). Separating the variables we have\n\\[\ninflationfactor \\frac{d horizontalaxis}{horizontalaxis}=\\frac{d constantvalue}{\\sqrt{1+constantvalue^{2}}},\n\\]\nwhence\n(4)\n\\[\ninflationfactor \\log horizontalaxis=\\log \\left(constantvalue+\\sqrt{1+constantvalue^{2}}\\right)+C .\n\\]\n\nSince \\( constantvalue=d verticalaxis / d horizontalaxis=0 \\) when \\( horizontalaxis=depthlevel \\), we find \\( C=inflationfactor \\log depthlevel \\). Then (4) can be. rewritten\n\\[\n\\left(\\frac{horizontalaxis}{depthlevel}\\right)^{\\prime}=constantvalue+\\sqrt{1+constantvalue^{2}}\n\\]\nand solved for \\( constantvalue(=d verticalaxis / d horizontalaxis) \\) to get\n(5)\n\\[\n2 \\frac{d verticalaxis}{d horizontalaxis}=\\left(\\frac{horizontalaxis}{depthlevel}\\right)^{\\prime}-\\left(\\frac{horizontalaxis}{depthlevel}\\right)^{\\prime} .\n\\]\n\nThis can be integrated again using the initial condition \\( verticalaxis=0 \\) when \\( horizontalaxis=depthlevel \\) to give\n(6)\n\\[\n2 verticalaxis=\\frac{depthlevel}{1+inflationfactor}\\left(\\frac{horizontalaxis}{depthlevel}\\right)^{1+inflationfactor}-\\frac{depthlevel}{1-inflationfactor}\\left(\\frac{horizontalaxis}{depthlevel}\\right)^{1-inflationfactor}+\\frac{2 inflationfactor depthlevel}{1-inflationfactor^{2}} .\n\\]\n\nThis is valid only for \\( horizontalaxis>0 \\). Recalling that \\( 0<inflationfactor<1 \\), we see from (6) that\n\\[\nverticalaxis \\rightarrow \\frac{inflationfactor depthlevel}{1-inflationfactor^{2}}\n\\]\nas \\( horizontalaxis \\rightarrow 0 \\). We see from (5) that \\( horizontalaxis(d verticalaxis / d horizontalaxis)-0 \\) and then from (2) that\n\\[\nverticalaxis-inflationfactor slowness timelessness-0\n\\]\nas \\( \\boldsymbol{horizontalaxis}-\\mathbf{0} \\). It follows that the predator catches his prey at the point \\( \\left(inflationfactor depthlevel /\\left(1-inflationfactor^{2}\\right), 0\\right) \\) when \\( timelessness=depthlevel /\\left(1-inflationfactor^{2}\\right) slowness \\). At this time the prey has flown \\( inflationfactor depthlevel /\\left(1-inflationfactor^{2}\\right) \\) and the predator, \\( depthlevel /\\left(1-inflationfactor^{2}\\right) \\).\nNow for the hawk and the sparrow, \\( inflationfactor=\\frac{1}{2}, depthlevel=100 \\mathrm{ft} \\), so the sparrow flies \\( 200 / 3 \\mathrm{ft} \\)., and the hawk, \\( 400 / 3 \\mathrm{ft} \\). [Note that since the sparrow is above the hawk we would have to choose a coordinate system with \\( horizontalaxis \\) positive in the downward direction.]\nIn the case of the eagle, \\( depthlevel=50 \\mathrm{ft} \\). and \\( inflationfactor \\) is unknown, but\n\\[\n\\frac{inflationfactor 50}{1-inflationfactor^{2}}=\\frac{200}{3} ;\n\\]\nhence \\( 4 inflationfactor^{2}+3 inflationfactor-4=0 \\) and\n\\[\ninflationfactor=\\frac{-3+\\sqrt{ } 73}{8}\n\\]\nsince \\( inflationfactor \\) must be positive.\nThe speed of the eagle is \\( 1 / inflationfactor \\) times the speed of the sparrow, or\n\\[\n\\frac{slowness(3+\\sqrt{7} 3)}{16} \\approx .721 slowness\n\\]\nwhere \\( slowness \\) is the speed of the hawk.\nSince the time of capture is \\( 400 / 3 slowness \\), the eagle flies a distance of\n\\[\n\\frac{400}{3}\\left(\\frac{3+\\sqrt{ } 73}{16}\\right) \\approx 96.2 \\mathrm{ft} .\n\\]\n\nRemark. The problem appears as Advanced Problem 3573 in the American Mathematical Monthly, vol. 39 (1932), page 454.\n\nAn entertaining paper by Arthur Bernhart, \"Curves of Pursuit,\" Scripta Mathematica, 20, nos. 3-4 (1954), outlines some of the history of the pursuit problem."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "t": "mndplrfq",
        "z": "bxvskjtr",
        "v": "rtylqsmd",
        "r": "hpqsnwve",
        "h": "vnbglkcz"
      },
      "question": "5. A sparrow, flying horizontally in a straight line, is 50 feet directly below an eagle and 100 feet directly above a hawk. Both hawk and eagle fly directly toward the sparrow, reaching it simultaneously. The hawk flies twice as fast as the sparrow. How far does each bird fly? At what rate does the eagle fly?",
      "solution": "Solution. Although not specifically stated, it is clearly intended that each of the three birds flies at a uniform speed. So we consider the general problem of a predator flying with speed \\( rtylqsmd \\) in pursuit of a prey flying along a straight path with speed \\( hpqsnwve\\, rtylqsmd, hpqsnwve<1 \\).\n\nLet the prey start at \\( (0,0) \\) and move along the \\( x \\)-axis so that his position at time \\( mndplrfq \\) is (hpqsnwve rtylqsmd mndplrfq, 0). Let the predator start at \\( (0, vnbglkcz), vnbglkcz>0 \\). If at time \\( mndplrfq \\) the predator is at \\( (qzxwvtnp, hjgrksla) \\), then the condition that he always flies directly toward his prey is that\n(1)\n\\[\n\\frac{d qzxwvtnp / d mndplrfq}{d hjgrksla / d mndplrfq}=\\frac{hpqsnwve \\, rtylqsmd \\, mndplrfq-qzxwvtnp}{-hjgrksla}\n\\]\n\nAs long as \\( hjgrksla \\) remains positive, \\( d hjgrksla / d mndplrfq<0 \\), so we may choose \\( hjgrksla \\) as the independent variable. Then (1) becomes\n(2)\n\\[\nhjgrksla \\frac{d qzxwvtnp}{d hjgrksla}=qzxwvtnp-hpqsnwve \\, rtylqsmd \\, mndplrfq .\n\\]\n\nThe condition that the predator has constant speed \\( rtylqsmd \\) is\n\\[\n\\sqrt{\\left(\\frac{d qzxwvtnp}{d mndplrfq}\\right)^{2}+\\left(\\frac{d hjgrksla}{d mndplrfq}\\right)^{2}}=rtylqsmd .\n\\]\nwhich we can write as\n(3)\n\\[\n-\\sqrt{\\left(\\frac{d qzxwvtnp}{d hjgrksla}\\right)^{2}+1}=rtylqsmd \\frac{d mndplrfq}{d hjgrksla},\n\\]\nwhere we have introduced a minus sign because \\( d mndplrfq / d hjgrksla \\) is negative.\nWe differentiate (2) to get\n\\[\nhjgrksla \\frac{d^{2} qzxwvtnp}{d hjgrksla^{2}}+\\frac{d qzxwvtnp}{d hjgrksla}=\\frac{d qzxwvtnp}{d hjgrksla}-hpqsnwve \\, rtylqsmd \\frac{d mndplrfq}{d hjgrksla},\n\\]\nand using (3) we have\n\\[\nhjgrksla \\frac{d^{2} qzxwvtnp}{d hjgrksla^{2}}=hpqsnwve \\sqrt{\\left(\\frac{d qzxwvtnp}{d hjgrksla}\\right)^{2}+1} .\n\\]\n\nWe can write this as the first-order differential equation\n\\[\nhjgrksla \\frac{d bxvskjtr}{d hjgrksla}=hpqsnwve \\sqrt{1+bxvskjtr^{2}},\n\\]\nwhere \\( bxvskjtr=d qzxwvtnp / d hjgrksla \\). Separating the variables we have\n\\[\nhpqsnwve \\frac{d hjgrksla}{hjgrksla}=\\frac{d bxvskjtr}{\\sqrt{1+bxvskjtr^{2}}},\n\\]\nwhence\n(4)\n\\[\nhpqsnwve \\log hjgrksla=\\log \\left(bxvskjtr+\\sqrt{1+bxvskjtr^{2}}\\right)+C .\n\\]\n\nSince \\( bxvskjtr=d qzxwvtnp / d hjgrksla=0 \\) when \\( hjgrksla=vnbglkcz \\), we find \\( C=hpqsnwve \\log vnbglkcz \\). Then (4) can be. rewritten\n\\[\n\\left(\\frac{hjgrksla}{vnbglkcz}\\right)^{\\prime}=bxvskjtr+\\sqrt{1+bxvskjtr^{2}}\n\\]\nand solved for \\( bxvskjtr(=d qzxwvtnp / d hjgrksla) \\) to get\n(5)\n\\[\n2 \\frac{d qzxwvtnp}{d hjgrksla}=\\left(\\frac{hjgrksla}{vnbglkcz}\\right)^{\\prime}-\\left(\\frac{hjgrksla}{vnbglkcz}\\right)^{\\prime} .\n\\]\n\nThis can be integrated again using the initial condition \\( qzxwvtnp=0 \\) when \\( hjgrksla=vnbglkcz \\) to give\n(6)\n\\[\n2 qzxwvtnp=\\frac{vnbglkcz}{1+hpqsnwve}\\left(\\frac{hjgrksla}{vnbglkcz}\\right)^{1+hpqsnwve}-\\frac{vnbglkcz}{1-hpqsnwve}\\left(\\frac{hjgrksla}{vnbglkcz}\\right)^{1-hpqsnwve}+\\frac{2 hpqsnwve \\, vnbglkcz}{1-hpqsnwve^{2}} .\n\\]\n\nThis is valid only for \\( hjgrksla>0 \\). Recalling that \\( 0<hpqsnwve<1 \\), we see from (6) that\n\\[\nqzxwvtnp \\rightarrow \\frac{hpqsnwve \\, vnbglkcz}{1-hpqsnwve^{2}}\n\\]\nas \\( hjgrksla \\rightarrow 0 \\). We see from (5) that \\( hjgrksla(d qzxwvtnp / d hjgrksla)-0 \\) and then from (2) that\n\\[\nqzxwvtnp-hpqsnwve \\, rtylqsmd \\, mndplrfq-0\n\\]\nas \\( \\boldsymbol{hjgrksla}-\\mathbf{0} \\). It follows that the predator catches his prey at the point \\( \\left(hpqsnwve \\, vnbglkcz /\\left(1-hpqsnwve^{2}\\right), 0\\right) \\) when \\( mndplrfq=vnbglkcz /\\left(1-hpqsnwve^{2}\\right) rtylqsmd \\). At this time the prey has flown \\( hpqsnwve \\, vnbglkcz /\\left(1-hpqsnwve^{2}\\right) \\) and the predator, \\( vnbglkcz /\\left(1-hpqsnwve^{2}\\right) \\).\nNow for the hawk and the sparrow, \\( hpqsnwve=\\frac{1}{2}, vnbglkcz=100 \\mathrm{ft} \\), so the sparrow flies \\( 200 / 3 \\mathrm{ft} \\)., and the hawk, \\( 400 / 3 \\mathrm{ft} \\). [Note that since the sparrow is above the hawk we would have to choose a coordinate system with \\( hjgrksla \\) positive in the downward direction.]\nIn the case of the eagle, \\( vnbglkcz=50 \\mathrm{ft} .\\) and \\( hpqsnwve \\) is unknown, but\n\\[\n\\frac{hpqsnwve \\, 50}{1-hpqsnwve^{2}}=\\frac{200}{3} ;\n\\]\nhence \\( 4 hpqsnwve^{2}+3 hpqsnwve-4=0 \\) and\n\\[\nhpqsnwve=\\frac{-3+\\sqrt{ } 73}{8}\n\\]\nsince \\( hpqsnwve \\) must be positive.\nThe speed of the eagle is \\( 1 / hpqsnwve \\) times the speed of the sparrow, or\n\\[\n\\frac{rtylqsmd(3+\\sqrt{7} 3)}{16} \\approx .721 \\, rtylqsmd\n\\]\nwhere \\( rtylqsmd \\) is the speed of the hawk.\nSince the time of capture is \\( 400 / 3 \\, rtylqsmd \\), the eagle flies a distance of\n\\[\n\\frac{400}{3}\\left(\\frac{3+\\sqrt{ } 73}{16}\\right) \\approx 96.2 \\mathrm{ft} .\n\\]\n\nRemark. The problem appears as Advanced Problem 3573 in the American Mathematical Monthly, vol. 39 (1932), page 454.\n\nAn entertaining paper by Arthur Bernhart, \"Curves of Pursuit,\" Scripta Mathematica, 20, nos. 3-4 (1954), outlines some of the history of the pursuit problem."
    },
    "kernel_variant": {
      "question": "Let a tern leave the origin $O=(0,0,0)$ at time $t=0$ and fly with the constant speed $v>0$ along the horizontal line\n\n\\[\n\\ell:\\qquad \\mathbf r_{\\mathrm T}(t)=v\\,t\\,\\frac{(1,1,0)}{\\sqrt 2},\n\\qquad t\\ge 0 .\n\\]\n\nAt the same instant three hawks take off from the fixed points  \n\n\\[\n\\begin{aligned}\nH_{1}:&\\;A_{1}=(0,\\,96,\\,128),\\\\\nH_{2}:&\\;A_{2}=(-128,\\,-96,\\,-32),\\\\\nH_{3}:&\\;A_{3}=(160,\\,64,\\,96),\n\\end{aligned}\n\\]\n\nkeeping their \\emph{own} constant speeds and, during the whole flight, heading at the instantaneous position of the tern (pure pursuit).  \nIt is known that Hawk $H_{2}$ is exactly four times as fast as the tern: $\\lVert\\mathbf V_{2}\\rVert =4v$.  \nAll three predators reach the tern simultaneously.\n\n(i)  Determine the (constant) speeds of $H_{1}$ and $H_{3}$ expressed as multiples of $v$.\n\n(ii)  Find the capture­–time $T$ and the capture point $\\mathbf C$.\n\n(iii)  Compute the total length of the flight path of \\emph{each} bird up to the moment of capture.\n\n(iv)  Give \\emph{explicit} parametric equations, in ordinary Cartesian coordinates $(x,y,z)$, for the whole pursuit curve of $H_{1}$ from launch to capture.  \nYour parametrisation must be continuously differentiable on the closed parameter interval (so that the curve itself is $C^{1}$). No differentiability requirement is imposed on higher derivatives.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Notation.  \nPut \n\\[\n\\hat g=\\frac{(1,1,0)}{\\sqrt 2},\n\\qquad \n\\mathbf r_{\\mathrm T}(t)=v\\,t\\,\\hat g .\n\\]\n\nFor every hawk $H_{i}$ introduce  \n\n\\[\nx_{0i}=A_{i}\\!\\cdot\\!\\hat g,\\qquad\n\\mathbf a_{i}=A_{i}-x_{0i}\\hat g,\\qquad\nh_{i}=\\lVert\\mathbf a_{i}\\rVert ,\\qquad\n\\hat e_{i}=\\frac{\\mathbf a_{i}}{h_{i}},\\qquad\nV_{i}=k_{i}v \\quad(k_{i}>1),\n\\]\n\nso that the individual motion of $H_{i}$ is contained in the plane\n\\[\n\\Pi_{i}=\\operatorname{span}\\{\\hat g,\\hat e_{i}\\}.\n\\]\n\nA straightforward calculation gives  \n\n\\[\n\\begin{array}{c|ccc}\n & H_{1}&H_{2}&H_{3}\\\\ \\hline\nx_{0i}\\;(\\mathrm m)& 48\\sqrt 2 & -\\dfrac{224}{\\sqrt2}& 112\\sqrt2\\\\[6pt]\nh_{i}\\;(\\mathrm m)& 16\\sqrt{82}& 16\\sqrt6 & 48\\sqrt6 \\\\ \\hline\n\\end{array}\n\\]\nand\n\\[\n\\hat e_{1}=\\frac{(-3,\\,3,\\,8)}{\\sqrt{82}},\\qquad\n\\hat e_{2}=\\frac{(-1,\\,1,\\,-2)}{\\sqrt 6},\\qquad\n\\hat e_{3}=\\frac{(1,\\,-1,\\,2)}{\\sqrt 6}.\n\\]\n\n--------------------------------------------------------------------\n1.  A general planar capture–distance formula.  \n\nConsider in an arbitrary plane $\\Pi$ the classical pursuit setting:  \n\n•\tPrey moves on the positive $x$–axis with speed $v$, starting from the origin;  \n•\tPredator starts from $(x_{0},h)$, $h>0$, and heads continuously at the prey with constant speed $V=kv$ ($k>1$).  \n\nWith $r=1/k\\;(0<r<1)$ and $z=\\mathrm d x/\\mathrm d y$ the usual reduction (omitted) leads to  \n\n\\[\ny\\frac{\\mathrm d z}{\\mathrm d y}=r\\sqrt{1+z^{2}},\\qquad\nz+\\sqrt{1+z^{2}}=F\\Bigl(\\frac{y}{h}\\Bigr)^{r},\n\\qquad\nF:=z_{0}+\\sqrt{1+z_{0}^{2}},\\; z_{0}:=\\frac{x_{0}}{h},\n\\]\nand finally to the \\emph{capture–distance} identity  \n\\[\n\\boxed{\\;\nvT=\\frac{r\\,h}{2}\\Bigl[\\frac{F}{1+r}+\\frac{1}{F(1-r)}\\Bigr]\n\\;}\n\\tag{1}\n\\]\nvalid whenever $F>0$ (which is our case).\n\n--------------------------------------------------------------------\n2.  Capture time from the data of $H_{2}$.  \n\nFor $H_{2}$ one has  \n\n\\[\nh_{2}=16\\sqrt6,\\qquad z_{02}=-\\frac 7{\\sqrt3},\\qquad\nF_{2}=\\frac{-7+2\\sqrt{13}}{\\sqrt3},\\qquad\nr_{2}=\\frac14 .\n\\]\n\nSubstituting in (1) gives  \n\n\\[\n\\boxed{\\;\nvT=S:=8\\sqrt6\\Bigl[\\frac{F_{2}}{5}+\\frac{1}{3F_{2}}\\Bigr]\n\\;}\n\\quad\\Longrightarrow\\quad\nT=\\frac{S}{v}\\approx54.0769\\;\\mathrm s .\n\\tag{2}\n\\]\n\nNumerically  \n\n\\[\nS=vT\\approx54.0769\\;\\mathrm m .\n\\]\n\n--------------------------------------------------------------------\n3.  Solving for the unknown speed factors $k_{1}$ and $k_{3}$.  \n\nFor a second predator the same distance $S$ must result from (1).  Eliminating $T$ yields the quadratic\n\n\\[\n\\bigl[(1-F^{2})+2SF/h\\bigr]\\,r^{2}+(1+F^{2})\\,r-2SF/h=0,\n\\qquad r=\\frac1k\\in(0,1),\n\\tag{3}\n\\]\nwith $(h,F)\\in\\{(h_{1},F_{1}),\\,(h_{3},F_{3})\\}$.\n\n(a)  Hawk $H_{1}$  \n\n\\[\nh_{1}=16\\sqrt{82},\\qquad \nF_{1}=\\frac{3+5\\sqrt2}{\\sqrt{41}} ,\n\\]\ngives  \n\n\\[\nr_{1}\\approx0.34846,\\qquad\nk_{1}=\\frac1{r_{1}}\\approx2.87.\n\\]\n\n(b)  Hawk $H_{3}$  \n\n\\[\nh_{3}=48\\sqrt6,\\qquad \nF_{3}=\\frac{7+2\\sqrt{19}}{3\\sqrt3},\n\\]\ngives  \n\n\\[\nr_{3}\\approx0.33301,\\qquad\nk_{3}=\\frac1{r_{3}}\\approx3.00 .\n\\]\n\nAnswer to (i)  \n\n\\[\n\\boxed{\\,V_{1}\\approx2.87\\,v,\\qquad V_{3}\\approx3.00\\,v\\,}.\n\\]\n\n--------------------------------------------------------------------\n4.  Capture point (answer to (ii)).  \n\nDuring the time $T$ the tern travels the distance $S=vT$ along $\\hat g$, hence  \n\n\\[\n\\boxed{\\;\n\\mathbf C=S\\,\\hat g\n          \\;\\approx\\;(38.25\\;\\mathrm m,\\;38.25\\;\\mathrm m,\\;0)\\;}.\n\\]\n\n--------------------------------------------------------------------\n5.  Flight–path lengths (answer to (iii)).  \n\n\\[\n\\begin{aligned}\ns_{\\mathrm T}&=vT=S\\approx54.08\\ \\mathrm m,\\\\\ns_{1}&=k_{1}vT\\approx155.4\\ \\mathrm m,\\\\\ns_{2}&=4vT\\approx216.3\\ \\mathrm m,\\\\\ns_{3}&=k_{3}vT\\approx162.2\\ \\mathrm m.\n\\end{aligned}\n\\]\n\n--------------------------------------------------------------------\n6.  A \\emph{$C^{1}$} parametrisation of the whole pursuit curve of $H_{1}$ (answer to (iv)).  \n\nKeep the basis $(\\hat g,\\hat e_{1})$ in $\\Pi_{1}$ and put  \n\n\\[\nr_{1}\\approx0.34846,\\qquad\nh=h_{1}=16\\sqrt{82},\\qquad\nF=F_{1},\\qquad\nx_{01}=48\\sqrt2 .\n\\]\n\nStep 1 — reuse the height–based functions.  \nFor $0<y\\le h$ define  \n\n\\[\n\\begin{aligned}\nf(y)&=F\\Bigl(\\frac{y}{h}\\Bigr)^{r_{1}},\\\\[4pt]\nz(y)&=\\frac12\\!\\Bigl[f(y)-f(y)^{-1}\\Bigr],\\\\[4pt]\nx(y)&=x_{01}+\\frac{h}{2}\\!\n      \\left\\{\\frac{F}{1+r_{1}}\n              \\Bigl[\\Bigl(\\tfrac{y}{h}\\Bigr)^{1+r_{1}}-1\\Bigr]\n            -\\frac{1}{F(1-r_{1})}\n              \\Bigl[\\Bigl(\\tfrac{y}{h}\\Bigr)^{1-r_{1}}-1\\Bigr]\\right\\}.\n\\end{aligned}\n\\]\n\nWith these expressions the \\emph{geometric} curve\n\\[\n\\Gamma_{1}=\\bigl\\{\\,x(y)\\,\\hat g+y\\,\\hat e_{1}\\;\\bigm|\\;0\\le y\\le h\\bigr\\}\n\\]\nis correct but the parameter $y$ destroys $C^{1}$–regularity at $y=0$ because $\\lVert\\mathrm d\\Gamma_{1}/\\mathrm dy\\rVert\\to\\infty$.\n\nStep 2 — introduce a \\emph{regular} parameter.  \nSet\n\\[\n\\theta=\\Bigl(\\frac{y}{h}\\Bigr)^{1-r_{1}},\\qquad\n0\\le\\theta\\le 1,\n\\]\nand write $y(\\theta)=h\\,\\theta^{\\,1/(1-r_{1})}$.  \nBecause $0<r_{1}<1$, one has $\\theta\\mapsto y(\\theta)$ strictly decreasing\nwith finite derivative at $\\theta=0$.  \n\nDefine\n\\[\n\\boxed{\\;\n\\mathbf R_{1}(\\theta)=x\\bigl(y(\\theta)\\bigr)\\,\\hat g+\n                       y(\\theta)\\,\\hat e_{1},\n\\qquad 0\\le\\theta\\le 1\\;}.\n\\]\n\nStep 3 — check regularity.  \nFrom the chain rule,\n\\[\n\\frac{\\mathrm d\\mathbf R_{1}}{\\mathrm d\\theta}\n       =\\Bigl[z\\bigl(y(\\theta)\\bigr)\\,\\hat g+\\hat e_{1}\\Bigr]\\,\n         \\frac{\\mathrm d y}{\\mathrm d\\theta}.\n\\]\nNear $\\theta=0$ one has $z(y)\\sim C\\,y^{-r_{1}}$ while \n$\\dfrac{\\mathrm d y}{\\mathrm d\\theta}\\sim\n\\dfrac{h}{1-r_{1}}\\,\\theta^{r_{1}/(1-r_{1})}$, so the product tends to the\nfinite vector\n\\[\n\\frac{hC}{1-r_{1}}\\;\\hat g+\\mathbf 0 ,\n\\]\nand $\\mathbf R_{1}$ is indeed $C^{1}$ on $[0,1]$.\n\nEvaluation at the ends:\n\n\\[\n\\mathbf R_{1}(1)=A_{1},\\qquad\n\\mathbf R_{1}(0)=\\mathbf C .\n\\]\n\nThus the requirement of a continuously differentiable parametrisation for the\n\\emph{entire} pursuit path of $H_{1}$ is satisfied.\n\nAll numerical work was carried out to at least ten significant digits; only the final answers are rounded.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.513215",
        "was_fixed": false,
        "difficulty_analysis": "• Dimension increase:  the motion now occurs in ℝ³, not in a plane as in the original problem.  \n• Multiple predators:  a third pursuer introduces an additional unknown speed and an extra simultaneity condition.  \n• Geometric decomposition:  the solver must project initial data onto and perpendicular to the prey’s direction, recognise that each predator moves in its own distinct 2-D plane, and keep track of three different perpendicular offsets ρᵢ.  \n• Non-trivial algebra:  the equal–time requirement produces two quadratic equations with irrational coefficients, leading to non-integer speed factors.  \n• Explicit curve requirement:  part (iv) asks for the full parametric description of a space curve, so the solver must re-derive the Cauchy–Liouville formula in a rotated basis and translate it back to Cartesian coordinates.  \nAll of these additions demand deeper geometric insight, heavier computation, and a broader familiarity with pursuit-curve theory than either the original Monthly problem or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "A tern starts from the origin $O$ and flies with constant speed $v>0$ along the straight line  \n\n\\[\n\\ell:\\qquad \\mathbf r(t)=v\\,t\\,\\frac{(1,1,0)}{\\sqrt{2}},\\qquad t\\ge 0 .\n\\]\n\n(Thus its horizontal heading is north–east and $\\ell$ lies in the plane $z=0$.)\n\nAt the same instant three hawks take off from  \n\n\\[\n\\begin{aligned}\nH_{1}:&\\quad A_{1}=(0,\\,96,\\,128)\\quad &(96\\text{ m north,\\; }128\\text{ m above the tern}),\\\\\nH_{2}:&\\quad A_{2}=(-128,\\,-96,\\,-32)\\quad &(128\\text{ m west,\\;}96\\text{ m south,\\;}32\\text{ m below}),\\\\\nH_{3}:&\\quad A_{3}=(160,\\,64,\\,96)\\quad &(160\\text{ m east,\\;}64\\text{ m north,\\;}96\\text{ m above}).\n\\end{aligned}\n\\]\n\nEach hawk keeps a personal \\emph{constant} speed and continuously steers toward the instantaneous position of the tern (pure pursuit).  \nHawk $H_{2}$ is known to fly four times as fast as the tern, i.e.\\ $\\lVert\\mathbf V_{2}\\rVert =4v$.  \nAll three hawks intercept the tern at the same moment.\n\n(i)\\; Determine the constant speeds of $H_{1}$ and $H_{3}$, expressed as multiples of $v$.  \n(ii)\\; Compute the capture time and the coordinates of the capture point.  \n(iii)\\; Find the length of the flight path of every bird up to the moment of capture.  \n(iv)\\; Give explicit parametric equations (in ordinary Cartesian coordinates) for the entire pursuit curve of $H_{1}$ from launch to capture.",
      "solution": "\\[\n\\text{\\bf Notation}\n\\]\nPut \n\\[\n\\hat g=\\frac{(1,1,0)}{\\sqrt{2}},\\qquad \n\\mathbf r(t)=v\\,t\\,\\hat g .\n\\]\nFor each hawk $H_{i}$ let\n\\[\nx_{0i}=A_{i}\\!\\cdot\\!\\hat g\\quad\\text{(signed offset parallel to }\\hat g),\\qquad\nh_{i}=\\lVert A_{i}-x_{0i}\\hat g\\rVert\\quad\\text{(perpendicular offset)},\n\\]\nand denote its (unknown) speed by $V_{i}=k_{i}v$ with $k_{i}>1$.  \nIn the plane $\\Pi_{i}$ spanned by $\\hat g$ and $A_{i}$ we choose the orthonormal basis $(\\hat g,\\hat e_{i})$; Cartesian coordinates therein will be written $(x,y)$ with $y$ measured along $\\hat e_{i}$.\n\n--------------------------------------------------------------------\nStep 1 – reduction to three planar pursuit problems  \n\n\\[\n\\begin{array}{lccc}\n\\hline\n& H_{1}&H_{2}&H_{3}\\\\\n\\hline\nx_{0i}\\;(\\text{m}) & 48\\sqrt{2}&-224/\\sqrt{2}&112\\sqrt{2}\\\\[2pt]\nh_{i}\\;(\\text{m}) & 16\\sqrt{82}&16\\sqrt{6}&48\\sqrt{6}\\\\[2pt]\nz_{0i}=x_{0i}/h_{i} &\\dfrac{3}{\\sqrt{41}} & -\\dfrac{7}{\\sqrt{3}} &\\dfrac{7}{3\\sqrt{3}}\\\\\n\\hline\n\\end{array}\n\\]\n\n--------------------------------------------------------------------\nStep 2 – general capture–time formula in an arbitrary planar setting  \n\nLet the prey move along the positive $x$–axis with speed $v$, the predator start at $(x_{0},h)$ with speed $V=kv$ ($k>1$) and pursue according to\n\\[\n\\frac{{\\rm d}x/{\\rm d}t}{{\\rm d}y/{\\rm d}t}=\\frac{v\\,t-x}{-y},\n\\qquad\n\\lVert({\\rm d}x/{\\rm d}t,{\\rm d}y/{\\rm d}t)\\rVert=kv .\n\\]\nPutting $r=1/k<1$ and $z={\\rm d}x/{\\rm d}y$ one obtains  \n\\[\ny\\frac{{\\rm d}z}{{\\rm d}y}=r\\sqrt{1+z^{2}},\\qquad\ny\\frac{{\\rm d}x}{{\\rm d}y}=x-vt .\n\\]\nWith $z_{0}=x_{0}/h$ and $F=z_{0}+\\sqrt{1+z_{0}^{2}}>0$ the first ODE integrates to  \n\\[\nz+\\sqrt{1+z^{2}}=F\\Bigl(\\frac{y}{h}\\Bigr)^{r}.\n\\]\nAfter rewriting $z$ and $\\sqrt{1+z^{2}}$ in terms of $f(y)=F(y/h)^{r}$ and performing a second quadrature one finds  \n\n\\[\n\\boxed{\\;\nvT=\\frac{r\\,h}{2}\\left[ \\frac{F}{1+r}+\\frac{1}{F(1-r)}\\right]\n\\;}.\n\\tag{1}\n\\]\n\n--------------------------------------------------------------------\nStep 3 – capture time from the data of $H_{2}$  \n\nFor $H_{2}$ we have $h_{2}=16\\sqrt{6}$, $z_{02}=-7/\\sqrt{3}$, hence\n\\[\nF_{2}=\\frac{-7+2\\sqrt{13}}{\\sqrt{3}},\\qquad r_{2}=\\frac14.\n\\]\nInsertion into (1) gives  \n\\[\nvT=8\\sqrt{6}\\left[\\frac{F_{2}}{5}+\\frac{1}{3F_{2}}\\right]\n=8\\sqrt{6}\\,(2.75863\\ldots)\n\\approx 54.0769\\;\\text{m}.\n\\tag{2}\n\\]\n\n--------------------------------------------------------------------\nStep 4 – solving for $k_{1}$ and $k_{3}$  \n\nPut $S:=vT$ from (2).  For arbitrary $(h,F)$ the requirement $vT=S$ is equivalent, via (1), to the quadratic\n\n\\[\n\\bigl[(1-F^{2})+2SF/h\\bigr]\\,r^{2}+(1+F^{2})\\,r-2SF/h=0,\\qquad 0<r<1,\\;r=1/k .\n\\tag{3}\n\\]\n\n(a) $H_{1}$: $h_{1}=16\\sqrt{82}$, $F_{1}=(3+5\\sqrt{2})/\\sqrt{41}$  \n\n\\[\n\\begin{aligned}\nA_{1}&=(1-F_{1}^{2})+2SF_{1}/h_{1}\\approx-0.30068,\\\\\nB_{1}&=1+F_{1}^{2}\\approx3.47562,\\\\\nC_{1}&=-2SF_{1}/h_{1}\\approx-1.17495.\n\\end{aligned}\n\\]\nThe root of (3) in $(0,1)$ is  \n\\[\nr_{1}=\\frac{-B_{1}+\\sqrt{B_{1}^{2}-4A_{1}C_{1}}}{2A_{1}}\\approx0.3485,\n\\qquad k_{1}=\\frac1{r_{1}}\\approx2.87 .\n\\]\n\n(b) $H_{3}$: $h_{3}=48\\sqrt{6}$, $F_{3}=[\\,7+2\\sqrt{19}\\,]/(3\\sqrt{3})$\n\n\\[\n\\begin{aligned}\nA_{3}&\\approx-5.360,\\\\\nB_{3}&\\approx10.132,\\\\\nC_{3}&\\approx-2.774,\\\\\nr_{3}&\\approx0.3330,\\qquad\nk_{3}=1/r_{3}\\approx3.00 .\n\\end{aligned}\n\\]\n\nAnswer to (i):\n\\[\n\\boxed{\\,V_{1}\\approx2.87\\,v,\\qquad V_{3}\\approx3.00\\,v\\,}.\n\\]\n\n--------------------------------------------------------------------\nStep 5 – capture point (answer to (ii))  \n\nDuring the time $T$ the tern travels a distance $vT$ along $\\hat g$, hence  \n\n\\[\n\\mathbf C\n   =\\frac{vT}{\\sqrt 2}(1,1,0)\n   =\\frac{S}{\\sqrt 2}(1,1,0)\n   =8\\sqrt 3\\left[\\frac{F_{2}}{5}+\\frac{1}{3F_{2}}\\right](1,1,0)\n   \\approx(38.25\\text{ m},\\,38.25\\text{ m},\\,0).\n\\]\n\n--------------------------------------------------------------------\nStep 6 – flight–path lengths (answer to (iii))  \n\n\\[\n\\begin{aligned}\ns_{0}&=vT\\approx54.08\\text{ m}\\quad\\text{(tern)},\\\\\ns_{1}&=k_{1}vT\\approx155.4\\text{ m}\\quad\\text{($H_{1}$)},\\\\\ns_{2}&=4vT\\approx216.3\\text{ m}\\quad\\text{($H_{2}$)},\\\\\ns_{3}&=k_{3}vT\\approx162.2\\text{ m}\\quad\\text{($H_{3}$)}.\n\\end{aligned}\n\\]\n\n--------------------------------------------------------------------\nStep 7 – explicit pursuit curve of $H_{1}$ (answer to (iv))  \n\nWithin $\\Pi_{1}$ keep the basis $(\\hat g,\\hat e_{1})$ and set  \n\\[\nr_{1}\\approx0.3485,\\qquad h=h_{1}=16\\sqrt{82},\\qquad\nF=F_{1}.\n\\]\nFor $0\\le y\\le h$ define  \n\n\\[\n\\begin{aligned}\nf(y)&=F\\Bigl(\\frac{y}{h}\\Bigr)^{r_{1}},\\\\[4pt]\nz(y)&=\\frac12\\bigl[f(y)-f(y)^{-1}\\bigr],\\\\[4pt]\nx(y)&=x_{01}+\\frac{h}{2}\n\\Bigl\\{\\frac{F}{1+r_{1}}\\Bigl[\\Bigl(\\frac{y}{h}\\Bigr)^{1+r_{1}}-1\\Bigr]\n-\\frac{1}{F(1-r_{1})}\\Bigl[\\Bigl(\\frac{y}{h}\\Bigr)^{1-r_{1}}-1\\Bigr]\\Bigr\\}.\n\\end{aligned}\n\\]\n\nThe space curve is therefore  \n\n\\[\n\\boxed{\\;\n\\mathbf R_{1}(y)=x(y)\\,\\hat g+y\\,\\hat e_{1},\\qquad 0\\le y\\le h_{1}\\;},\n\\]\nwhich starts at $A_{1}$ ($y=h_{1}$) and terminates smoothly at the capture point $\\mathbf C$ ($y=0$).\n\n--------------------------------------------------------------------\nAll intermediate calculations were carried with at least eight significant digits; the displayed decimals are rounded only at the end.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.428676",
        "was_fixed": false,
        "difficulty_analysis": "• Dimension increase:  the motion now occurs in ℝ³, not in a plane as in the original problem.  \n• Multiple predators:  a third pursuer introduces an additional unknown speed and an extra simultaneity condition.  \n• Geometric decomposition:  the solver must project initial data onto and perpendicular to the prey’s direction, recognise that each predator moves in its own distinct 2-D plane, and keep track of three different perpendicular offsets ρᵢ.  \n• Non-trivial algebra:  the equal–time requirement produces two quadratic equations with irrational coefficients, leading to non-integer speed factors.  \n• Explicit curve requirement:  part (iv) asks for the full parametric description of a space curve, so the solver must re-derive the Cauchy–Liouville formula in a rotated basis and translate it back to Cartesian coordinates.  \nAll of these additions demand deeper geometric insight, heavier computation, and a broader familiarity with pursuit-curve theory than either the original Monthly problem or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}