{
  "index": "1977-B-2",
  "type": "GEO",
  "tag": [
    "GEO"
  ],
  "difficulty": "",
  "question": "Problem B-2\nGiven a convex quadrilateral \\( A B C D \\) and a point \\( O \\) not in the plane of \\( A B C D \\), locate point \\( A^{\\prime} \\) on line \\( O A \\), point \\( B^{\\prime} \\) on line \\( O B \\), point \\( C^{\\prime} \\) on line \\( O C \\), and point \\( D^{\\prime} \\) on line \\( O D \\) so that \\( A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime} \\) is a parallelogram.",
  "solution": "B-2.\nLet \\( O^{\\prime} \\) be any point different from \\( O \\) on the line of intersection of planes \\( A O C \\) and \\( B O D \\), e.g., \\( O^{\\prime} \\) may be the intersection of lines \\( A C \\) and \\( B D \\). Let \\( A^{\\prime} \\) be the intersection of line \\( O A \\) with the line through \\( O^{\\prime} \\) and parallel to \\( O C \\). Let \\( C^{\\prime} \\) be the intersection of line \\( O C \\) with the line through \\( O^{\\prime} \\) which is parallel to \\( O A \\). Then \\( O A^{\\prime} O^{\\prime} C^{\\prime} \\) is a parallelogram and its diagonals \\( O O^{\\prime} \\) and \\( A^{\\prime} C^{\\prime} \\) bisect each other at a point \\( M \\). Choosing \\( B^{\\prime} \\) and \\( D^{\\prime} \\) in the same way, one obtains a parallelogram \\( O B^{\\prime} O^{\\prime} D^{\\prime} \\) whose diagonals \\( O O^{\\prime} \\) and \\( B^{\\prime} D^{\\prime} \\) also bisect each other at the midpoint \\( M \\) of segment \\( O O^{\\prime} \\). Hence segments \\( A^{\\prime} C^{\\prime} \\) and \\( B^{\\prime} D^{\\prime} \\) bisect each other (at \\( M \\) ) and \\( A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime} \\) is a parallelogram. (The parallelogram is not unique.)",
  "vars": [
    "M"
  ],
  "params": [
    "A",
    "B",
    "C",
    "D",
    "O"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "vertexa",
        "B": "vertexb",
        "C": "vertexc",
        "D": "vertexd",
        "O": "vertexo",
        "M": "midpoint"
      },
      "question": "Problem B-2\nGiven a convex quadrilateral \\( vertexa\\ vertexb\\ vertexc\\ vertexd \\) and a point \\( vertexo \\) not in the plane of \\( vertexa\\ vertexb\\ vertexc\\ vertexd \\), locate point \\( vertexa^{\\prime} \\) on line \\( vertexo\\ vertexa \\), point \\( vertexb^{\\prime} \\) on line \\( vertexo\\ vertexb \\), point \\( vertexc^{\\prime} \\) on line \\( vertexo\\ vertexc \\), and point \\( vertexd^{\\prime} \\) on line \\( vertexo\\ vertexd \\) so that \\( vertexa^{\\prime}\\ vertexb^{\\prime}\\ vertexc^{\\prime}\\ vertexd^{\\prime} \\) is a parallelogram.",
      "solution": "B-2.\nLet \\( vertexo^{\\prime} \\) be any point different from \\( vertexo \\) on the line of intersection of planes \\( vertexa\\ vertexo\\ vertexc \\) and \\( vertexb\\ vertexo\\ vertexd \\), e.g., \\( vertexo^{\\prime} \\) may be the intersection of lines \\( vertexa\\ vertexc \\) and \\( vertexb\\ vertexd \\). Let \\( vertexa^{\\prime} \\) be the intersection of line \\( vertexo\\ vertexa \\) with the line through \\( vertexo^{\\prime} \\) and parallel to \\( vertexo\\ vertexc \\). Let \\( vertexc^{\\prime} \\) be the intersection of line \\( vertexo\\ vertexc \\) with the line through \\( vertexo^{\\prime} \\) which is parallel to \\( vertexo\\ vertexa \\). Then \\( vertexo\\ vertexa^{\\prime}\\ vertexo^{\\prime}\\ vertexc^{\\prime} \\) is a parallelogram and its diagonals \\( vertexo\\ vertexo^{\\prime} \\) and \\( vertexa^{\\prime}\\ vertexc^{\\prime} \\) bisect each other at a point \\( midpoint \\). Choosing \\( vertexb^{\\prime} \\) and \\( vertexd^{\\prime} \\) in the same way, one obtains a parallelogram \\( vertexo\\ vertexb^{\\prime}\\ vertexo^{\\prime}\\ vertexd^{\\prime} \\) whose diagonals \\( vertexo\\ vertexo^{\\prime} \\) and \\( vertexb^{\\prime}\\ vertexd^{\\prime} \\) also bisect each other at the midpoint \\( midpoint \\) of segment \\( vertexo\\ vertexo^{\\prime} \\). Hence segments \\( vertexa^{\\prime}\\ vertexc^{\\prime} \\) and \\( vertexb^{\\prime}\\ vertexd^{\\prime} \\) bisect each other (at \\( midpoint \\) ) and \\( vertexa^{\\prime}\\ vertexb^{\\prime}\\ vertexc^{\\prime}\\ vertexd^{\\prime} \\) is a parallelogram. (The parallelogram is not unique.)"
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "buttercup",
        "B": "cellophan",
        "C": "marigolds",
        "D": "lemontree",
        "O": "raincloud",
        "M": "gearshift"
      },
      "question": "Problem B-2\nGiven a convex quadrilateral \\( buttercup cellophan marigolds lemontree \\) and a point \\( raincloud \\) not in the plane of \\( buttercup cellophan marigolds lemontree \\), locate point \\( buttercup^{\\prime} \\) on line \\( raincloud buttercup \\), point \\( cellophan^{\\prime} \\) on line \\( raincloud cellophan \\), point \\( marigolds^{\\prime} \\) on line \\( raincloud marigolds \\), and point \\( lemontree^{\\prime} \\) on line \\( raincloud lemontree \\) so that \\( buttercup^{\\prime} cellophan^{\\prime} marigolds^{\\prime} lemontree^{\\prime} \\) is a parallelogram.",
      "solution": "B-2.\nLet \\( raincloud^{\\prime} \\) be any point different from \\( raincloud \\) on the line of intersection of planes \\( buttercup raincloud marigolds \\) and \\( cellophan raincloud lemontree \\), e.g., \\( raincloud^{\\prime} \\) may be the intersection of lines \\( buttercup marigolds \\) and \\( cellophan lemontree \\). Let \\( buttercup^{\\prime} \\) be the intersection of line \\( raincloud buttercup \\) with the line through \\( raincloud^{\\prime} \\) and parallel to \\( raincloud marigolds \\). Let \\( marigolds^{\\prime} \\) be the intersection of line \\( raincloud marigolds \\) with the line through \\( raincloud^{\\prime} \\) which is parallel to \\( raincloud buttercup \\). Then \\( raincloud buttercup^{\\prime} raincloud^{\\prime} marigolds^{\\prime} \\) is a parallelogram and its diagonals \\( raincloud raincloud^{\\prime} \\) and \\( buttercup^{\\prime} marigolds^{\\prime} \\) bisect each other at a point \\( gearshift \\). Choosing \\( cellophan^{\\prime} \\) and \\( lemontree^{\\prime} \\) in the same way, one obtains a parallelogram \\( raincloud cellophan^{\\prime} raincloud^{\\prime} lemontree^{\\prime} \\) whose diagonals \\( raincloud raincloud^{\\prime} \\) and \\( cellophan^{\\prime} lemontree^{\\prime} \\) also bisect each other at the midpoint \\( gearshift \\) of segment \\( raincloud raincloud^{\\prime} \\). Hence segments \\( buttercup^{\\prime} marigolds^{\\prime} \\) and \\( cellophan^{\\prime} lemontree^{\\prime} \\) bisect each other (at \\( gearshift \\) ) and \\( buttercup^{\\prime} cellophan^{\\prime} marigolds^{\\prime} lemontree^{\\prime} \\) is a parallelogram. (The parallelogram is not unique.)"
    },
    "descriptive_long_misleading": {
      "map": {
        "M": "endpoint",
        "A": "midpointalpha",
        "B": "midpointbeta",
        "C": "midpointgamma",
        "D": "midpointdelta",
        "O": "innerpoint"
      },
      "question": "Problem B-2\nGiven a convex quadrilateral \\( midpointalpha midpointbeta midpointgamma midpointdelta \\) and a point \\( innerpoint \\) not in the plane of \\( midpointalpha midpointbeta midpointgamma midpointdelta \\), locate point \\( midpointalpha^{\\prime} \\) on line \\( innerpoint midpointalpha \\), point \\( midpointbeta^{\\prime} \\) on line \\( innerpoint midpointbeta \\), point \\( midpointgamma^{\\prime} \\) on line \\( innerpoint midpointgamma \\), and point \\( midpointdelta^{\\prime} \\) on line \\( innerpoint midpointdelta \\) so that \\( midpointalpha^{\\prime} midpointbeta^{\\prime} midpointgamma^{\\prime} midpointdelta^{\\prime} \\) is a parallelogram.",
      "solution": "B-2.\nLet \\( innerpoint^{\\prime} \\) be any point different from \\( innerpoint \\) on the line of intersection of planes \\( midpointalpha innerpoint midpointgamma \\) and \\( midpointbeta innerpoint midpointdelta \\), e.g., \\( innerpoint^{\\prime} \\) may be the intersection of lines \\( midpointalpha midpointgamma \\) and \\( midpointbeta midpointdelta \\). Let \\( midpointalpha^{\\prime} \\) be the intersection of line \\( innerpoint midpointalpha \\) with the line through \\( innerpoint^{\\prime} \\) and parallel to \\( innerpoint midpointgamma \\). Let \\( midpointgamma^{\\prime} \\) be the intersection of line \\( innerpoint midpointgamma \\) with the line through \\( innerpoint^{\\prime} \\) which is parallel to \\( innerpoint midpointalpha \\). Then \\( innerpoint midpointalpha^{\\prime} innerpoint^{\\prime} midpointgamma^{\\prime} \\) is a parallelogram and its diagonals \\( innerpoint innerpoint^{\\prime} \\) and \\( midpointalpha^{\\prime} midpointgamma^{\\prime} \\) bisect each other at a point \\( endpoint \\). Choosing \\( midpointbeta^{\\prime} \\) and \\( midpointdelta^{\\prime} \\) in the same way, one obtains a parallelogram \\( innerpoint midpointbeta^{\\prime} innerpoint^{\\prime} midpointdelta^{\\prime} \\) whose diagonals \\( innerpoint innerpoint^{\\prime} \\) and \\( midpointbeta^{\\prime} midpointdelta^{\\prime} \\) also bisect each other at the midpoint \\( endpoint \\) of segment \\( innerpoint innerpoint^{\\prime} \\). Hence segments \\( midpointalpha^{\\prime} midpointgamma^{\\prime} \\) and \\( midpointbeta^{\\prime} midpointdelta^{\\prime} \\) bisect each other (at \\( endpoint \\) ) and \\( midpointalpha^{\\prime} midpointbeta^{\\prime} midpointgamma^{\\prime} midpointdelta^{\\prime} \\) is a parallelogram. (The parallelogram is not unique.)"
    },
    "garbled_string": {
      "map": {
        "M": "zpqmfkdl",
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "plomnbvc",
        "D": "sghrtuio",
        "O": "nmkeiyrz"
      },
      "question": "Problem B-2\nGiven a convex quadrilateral \\( qzxwvtnp hjgrksla plomnbvc sghrtuio \\) and a point \\( nmkeiyrz \\) not in the plane of \\( qzxwvtnp hjgrksla plomnbvc sghrtuio \\), locate point \\( qzxwvtnp^{\\prime} \\) on line \\( nmkeiyrz qzxwvtnp \\), point \\( hjgrksla^{\\prime} \\) on line \\( nmkeiyrz hjgrksla \\), point \\( plomnbvc^{\\prime} \\) on line \\( nmkeiyrz plomnbvc \\), and point \\( sghrtuio^{\\prime} \\) on line \\( nmkeiyrz sghrtuio \\) so that \\( qzxwvtnp^{\\prime} hjgrksla^{\\prime} plomnbvc^{\\prime} sghrtuio^{\\prime} \\) is a parallelogram.",
      "solution": "B-2.\nLet \\( nmkeiyrz^{\\prime} \\) be any point different from \\( nmkeiyrz \\) on the line of intersection of planes \\( qzxwvtnp nmkeiyrz plomnbvc \\) and \\( hjgrksla nmkeiyrz sghrtuio \\), e.g., \\( nmkeiyrz^{\\prime} \\) may be the intersection of lines \\( qzxwvtnp plomnbvc \\) and \\( hjgrksla sghrtuio \\). Let \\( qzxwvtnp^{\\prime} \\) be the intersection of line \\( nmkeiyrz qzxwvtnp \\) with the line through \\( nmkeiyrz^{\\prime} \\) and parallel to \\( nmkeiyrz plomnbvc \\). Let \\( plomnbvc^{\\prime} \\) be the intersection of line \\( nmkeiyrz plomnbvc \\) with the line through \\( nmkeiyrz^{\\prime} \\) which is parallel to \\( nmkeiyrz qzxwvtnp \\). Then \\( nmkeiyrz qzxwvtnp^{\\prime} nmkeiyrz^{\\prime} plomnbvc^{\\prime} \\) is a parallelogram and its diagonals \\( nmkeiyrz nmkeiyrz^{\\prime} \\) and \\( qzxwvtnp^{\\prime} plomnbvc^{\\prime} \\) bisect each other at a point \\( zpqmfkdl \\). Choosing \\( hjgrksla^{\\prime} \\) and \\( sghrtuio^{\\prime} \\) in the same way, one obtains a parallelogram \\( nmkeiyrz hjgrksla^{\\prime} nmkeiyrz^{\\prime} sghrtuio^{\\prime} \\) whose diagonals \\( nmkeiyrz nmkeiyrz^{\\prime} \\) and \\( hjgrksla^{\\prime} sghrtuio^{\\prime} \\) also bisect each other at the midpoint \\( zpqmfkdl \\) of segment \\( nmkeiyrz nmkeiyrz^{\\prime} \\). Hence segments \\( qzxwvtnp^{\\prime} plomnbvc^{\\prime} \\) and \\( hjgrksla^{\\prime} sghrtuio^{\\prime} \\) bisect each other (at \\( zpqmfkdl \\) ) and \\( qzxwvtnp^{\\prime} hjgrksla^{\\prime} plomnbvc^{\\prime} sghrtuio^{\\prime} \\) is a parallelogram. (The parallelogram is not unique.)"
    },
    "kernel_variant": {
      "question": "Let $\\Pi$ be an affine plane in the Euclidean space $\\mathbb R^{3}$ equipped with a chosen unit normal vector $n$.  \nFix the point  \n\n\\[\nO:=(0,0,0)\\qquad(\\text{the origin of }\\mathbb R^{3}),\n\\]\n\nso $O\\notin\\Pi$ by assumption.  \nChoose a second unit vector $\\hat r$ such that  \n\n(i) $\\hat r\\cdot n\\neq 0$ \\;(hence $\\hat r$ is not parallel to $\\Pi$),  \n\n(ii) $\\hat r\\times n\\neq 0$ \\;(so $\\hat r$ is not parallel to $n$).\n\nPut  \n\n\\[\nm:=\\hat r\\times n\n\\]\n\nand use the (in general non-orthonormal) ordered basis $(\\,\\hat r,n,m\\,)$ to write coordinates of vectors.  \nIf $v$ is any vector we denote by  \n\n\\[\nv=v_{r}\\,\\hat r+v_{n}\\,n+v_{m}\\,m\n\\]\n\nits coordinate triple $(v_{r},v_{n},v_{m})$ in this basis.\n\nFix four distinct points $A,B,C,D$ that lie in $\\Pi$ and write  \n\n\\[\na:=\\overrightarrow{OA},\\qquad\nb:=\\overrightarrow{OB},\\qquad\nc:=\\overrightarrow{OC},\\qquad\nd:=\\overrightarrow{OD}.\n\\]\n\nThroughout let  \n\n\\[\n\\rho:=\\hat r\\cdot n\\quad(\\rho\\neq 0),\\qquad\n\\Delta_{1}:=a_{r}c_{n}-a_{n}c_{r},\\qquad\n\\Delta_{2}:=b_{r}d_{n}-b_{n}d_{r}.\n\\]\n\nAssume the following algebraic side-conditions.\n\n(C1) $a_{m}=c_{m}=0$ \\;(the rays $OA,OC$ lie in $\\Sigma:=\\operatorname{span}\\{\\hat r,n\\}$).  \n\n(C2) $\\Delta_{1}\\neq 0$.  \n\n(C3) $a_{n}c_{r}+a_{r}c_{n}=0$.  \n\n(C4) $a_{r}>0$ and $c_{r}>0$.  \n\n(C5) $b_{n}d_{n}\\neq 0$ \\;(the rays $OB,OD$ meet $\\Pi$ transversally).  \n\n(C5$'$) $b_{m}^{\\,2}+d_{m}^{\\,2}\\neq 0$ \\;(at least one of $b,d$ does not lie in $\\Sigma$).  \n\n(C6) $\\Delta_{2}\\neq 0$.  \n\n(C7) $\\Theta_{r}:=\\rho\\bigl(b_{r}d_{n}+d_{r}b_{n}\\bigr)+2\\,b_{n}d_{n}=0$.  \n\n(C8) $\\Theta_{m}:=b_{m}d_{n}-d_{m}b_{n}=0$.  \n\n(C9) $d_{n}\\,\\Delta_{2}>0>\\;b_{n}\\,\\Delta_{2}$.\n\nFor every real number $t>0$ put  \n\n\\[\nM:=O+t\\,\\hat r .\n\\]\n\n(a) Show that for each $t>0$ there exist unique points  \n\n\\[\nA'\\in OA,\\qquad B'\\in OB,\\qquad C'\\in OC,\\qquad D'\\in OD\n\\]\n\nsuch that  \n\n(i) $A'B'C'D'$ is a parallelogram,  \n\n(ii) $M$ is the common midpoint of its two diagonals,  \n\n(iii) the diagonal $A'C'$ is perpendicular to $\\Pi$,  \n\n(iv) the other diagonal $B'D'$ is parallel to $\\Pi$ (equivalently $B'D'\\perp n$).\n\n(b) Prove that the parallelogram is non-degenerate (its four vertices are pairwise distinct) and that the length of the “vertical’’ diagonal depends linearly on $t$:\n\\[\n|A'C'|=\\kappa\\,t\n\\quad\\text{with}\\quad\n\\kappa:=\\dfrac{4\\,|a_{n}c_{n}|}{|\\Delta_{1}|}.\n\\]\n\n(c) Construct $A',B',C',D'$ explicitly in terms of $t$ and the given data, justify that the construction works for every $t>0$, and prove uniqueness.",
      "solution": "All coordinates are taken in the ordered basis $(\\hat r,n,m)$.  \nRecall $\\rho:=\\hat r\\cdot n\\neq 0$.\n\n\\textbf{Step 1.  Relations involving $A$ and $C$}.  \nBecause of (C1)\n\\[\na=a_{r}\\hat r+a_{n}n,\\qquad\nc=c_{r}\\hat r+c_{n}n .\n\\]\nConditions (C2)–(C3) give\n\\[\n\\Delta_{1}=a_{r}c_{n}-a_{n}c_{r}\n          =2\\,a_{r}c_{n}=-2\\,a_{n}c_{r}\\neq 0,\n\\]\nhence $a_{n}c_{n}\\neq 0$, and by (C4) $a_{r},c_{r}>0$.\n\n\\textbf{Step 2.  Construction of the “vertical’’ diagonal $A'C'$}.  \nSeek $\\alpha,\\gamma>0$ with  \n\\[\nA':=\\alpha\\,a\\in OA,\\qquad C':=\\gamma\\,c\\in OC\n\\]\nsuch that  \n\n(i$_{A}$) $\\alpha a+\\gamma c=2t\\hat r$  (makes $M$ the midpoint),  \n\n(ii$_{A}$) $\\alpha a-\\gamma c\\parallel n$  (so $A'C'$ is perpendicular to $\\Pi$).\n\nCondition (ii$_{A}$) forces the $\\hat r$-coordinate of $\\alpha a-\\gamma c$ to be $0$, and together with (i$_{A}$) yields the $2\\times 2$ system  \n\\[\na_{r}\\alpha+c_{r}\\gamma=2t,\\qquad\na_{r}\\alpha-c_{r}\\gamma=0.\n\\]\nSolving,\n\\[\n\\alpha=\\dfrac{t}{a_{r}}>0,\\qquad\n\\gamma=\\dfrac{t}{c_{r}}>0.\n\\]\nBecause of (C3)\n\\[\na_{n}\\alpha+c_{n}\\gamma\n     =t\\Bigl(\\dfrac{a_{n}}{a_{r}}+\\dfrac{c_{n}}{c_{r}}\\Bigr)=0,\n\\]\nso (i$_{A}$) is a vector identity and (ii$_{A}$) holds automatically.  Hence  \n\\[\n\\overrightarrow{A'C'}=\\alpha a-\\gamma c\n        =t\\Bigl(\\dfrac{a}{a_{r}}-\\dfrac{c}{c_{r}}\\Bigr)\n        =-2t\\,\\dfrac{c_{n}}{c_{r}}\\,n,\n\\]\nwhich is non-zero, orthogonal to $\\Pi$, and has length\n\\[\n|A'C'|\n     =2t\\,\\dfrac{|c_{n}|}{|c_{r}|}\n     =2t\\,\\dfrac{|a_{n}|}{|a_{r}|}\n     =\\dfrac{4\\,|a_{n}c_{n}|}{|\\Delta_{1}|}\\,t,\n\\]\nestablishing formula (b).\n\n\\textbf{Step 3.  Construction of the “horizontal’’ diagonal $B'D'$}.  \nWrite the full coordinates\n\\[\nb=b_{r}\\hat r+b_{n}n+b_{m}m,\\qquad\nd=d_{r}\\hat r+d_{n}n+d_{m}m.\n\\]\nLook for $\\lambda,\\mu>0$ such that  \n\\[\n\\lambda b+\\mu d=2t\\hat r,      \\tag{1}\n\\]\n\\[\n(\\mu d-\\lambda b)\\cdot n=0.   \\tag{2}\n\\]\nEquation (1) is equivalent to the three scalar equations  \n\\[\nb_{r}\\lambda+d_{r}\\mu=2t,\\qquad\nb_{n}\\lambda+d_{n}\\mu=0,\\qquad\nb_{m}\\lambda+d_{m}\\mu=0.   \\tag{3}\n\\]\n\nOnly the first two equalities of (3) are independent; the third is forced by (C8).  \nSolving the first two yields  \n\\[\n\\lambda=\\dfrac{2t\\,d_{n}}{\\Delta_{2}},\\qquad\n\\mu   =-\\dfrac{2t\\,b_{n}}{\\Delta_{2}},\n\\]\nwith $\\Delta_{2}\\neq 0$ by (C6).  Condition (C9) guarantees $\\lambda,\\mu>0$, so  \n\\[\nB':=\\lambda b\\in OB,\\qquad D':=\\mu d\\in OD .\n\\]\nSubstituting these values into the third line of (3) gives  \n\\[\nb_{m}\\lambda+d_{m}\\mu\n   =\\dfrac{2t}{\\Delta_{2}}\\,(b_{m}d_{n}-d_{m}b_{n})\n   =\\dfrac{2t}{\\Delta_{2}}\\;\\Theta_{m}=0,\n\\]\nthanks to (C8); hence (1) holds as a vector identity.  \n\nFor (2) we compute\n\\[\n(\\mu d-\\lambda b)\\cdot n\n   =\\rho(d_{r}\\mu-b_{r}\\lambda)+(d_{n}\\mu-b_{n}\\lambda)\n   =-\\dfrac{2t}{\\Delta_{2}}\\,\n      \\bigl[\\rho(b_{r}d_{n}+d_{r}b_{n})+2\\,b_{n}d_{n}\\bigr]\n   =-\\dfrac{2t}{\\Delta_{2}}\\;\\Theta_{r}=0\n\\]\nby (C7), so $B'D'\\perp n$, i.e.\\ $B'D'$ is parallel to $\\Pi$.\n\n\\textbf{Step 4.  Verification of (i)–(iv)}.  \nFrom the identities of Steps 2 and 3 we have  \n\\[\n\\overrightarrow{OM}=t\\hat r\n                 =\\tfrac12(\\lambda b+\\mu d)\n                 =\\tfrac12(\\alpha a+\\gamma c),\n\\]\nwhence $\\overrightarrow{OM}$ is the common midpoint vector of $A'C'$ and $B'D'$.  \nConsequently $A'B'C'D'$ is a parallelogram, proving (i) and (ii).  \nProperties (iii) and (iv) were ensured in Steps 2 and 3, respectively.\n\n\\textbf{Step 5.  Non-degeneracy}.  \nWe already know $A'\\neq C'$ because $\\overrightarrow{A'C'}\\neq 0$, and $B'\\neq D'$ because $\\Delta_{2}\\neq 0$.  \nTo show that \\emph{every} pair of distinct indices corresponds to distinct points, suppose, toward a contradiction, that two different vertices coincide.\n\n\\smallskip\n$\\bullet$ \\emph{Case $A'=B'$}.  \nThen $\\alpha a=\\lambda b$.  Because $\\alpha a$ lies in $\\Sigma$, this forces $b\\in\\Sigma$, hence $b_{m}=0$.  \nUsing $\\mu d=2t\\hat r-\\lambda b$, we also have $\\mu d\\in\\Sigma$ and therefore $d_{m}=0$, contradicting (C5$'$).\n\n$\\bullet$ \\emph{Case $A'=D'$}.  \nNow $\\alpha a=\\mu d$.  Arguing as above yields $d_{m}=0$, and\n\\[\nb_{m}\\lambda=(b_{m}\\lambda+d_{m}\\mu)-d_{m}\\mu=0,\n\\]\nso $b_{m}=0$, again contradicting (C5$'$).\n\n$\\bullet$ \\emph{Case $B'=C'$}.  \nHere $\\lambda b=\\gamma c$.  Because $c\\in\\Sigma$, we deduce $b_{m}=0$, which forces $d_{m}=0$ as before, a contradiction.\n\n$\\bullet$ \\emph{Case $C'=D'$}.  \nThen $\\gamma c=\\mu d$.  Since $c\\in\\Sigma$, this gives $d_{m}=0$; subsequently $b_{m}=0$ follows from (C8), contradicting (C5$'$).\n\n$\\bullet$ \\emph{Case $A'=C'$ or $B'=D'$}.  \nThese were already excluded.\n\nThus all four vertices are pairwise distinct; the parallelogram is non-degenerate.  \nEquation \\(|A'C'|=\\kappa t\\) was obtained in Step 2, completing part (b).\n\n\\textbf{Step 6.  Uniqueness and linear dependence on $t$}.  \nThe $2\\times2$ systems in Steps 2 and 3 have unique solutions, hence no other quadruple of points on the same rays can satisfy (i)–(iv) for the same $t$.  \nFormulae for $\\alpha,\\gamma,\\lambda,\\mu$ are linear in $t$, so the vertices and the length $|A'C'|$ vary linearly with $t$.  \nThis completes the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.632664",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra geometric requirements The original problem asks only for the existence of a parallelogram; the enhanced version forces  \n   • its centre to lie on a prescribed skew line r,  \n   • one diagonal to be perpendicular to a given plane Π,  \n   • the two diagonals to be mutually perpendicular.  \n   Fulfilling four simultaneous conditions instead of one makes the system highly constrained.\n\n2. Higher-dimensional reasoning One has to juggle three distinct affine objects (plane Π, skew lines ℓ and r) in ℝ³, keep track of orthogonality relations, and work in the orthogonal complement r⊥—all absent from the original task.\n\n3. Vector-parameter calculus Solving the problem naturally leads to introducing four unknown positive scalars λ , μ , α , γ and an auxiliary parameter κ, then cracking a coupled vector–scalar system (7)–(10).  This demands linear-algebra techniques in a geometric setting.\n\n4. Non-trivial uniqueness issues Besides showing existence, one must prove that the construction returns exactly one quadrilateral for every centre position M, and that every admissible parallelogram is produced—subtleties not present in the basic variant.\n\n5. Quantitative finale The last request (express |A′C′| in terms of t) turns the exercise into an analytic one, forcing the solver to translate the whole figure into explicit trigonometric form.\n\nAll these layers oblige the contestant to combine synthetic 3-space geometry, affine-vector methods, and a dose of analytic geometry—considerably more sophisticated than the single-plane, single-condition setting of the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\Pi$ be an affine plane in the Euclidean space $\\mathbb R^{3}$ equipped with a chosen unit normal vector $n$.  \nFix the point  \n\n\\[\nO:=(0,0,0)\\qquad(\\text{the origin of }\\mathbb R^{3}),\n\\]\n\nso $O\\notin\\Pi$ by assumption.  \nChoose a second unit vector $\\hat r$ such that  \n\n(i) $\\hat r\\cdot n\\neq 0$ \\;(hence $\\hat r$ is not parallel to $\\Pi$),  \n\n(ii) $\\hat r\\times n\\neq 0$ \\;(so $\\hat r$ is not parallel to $n$).\n\nPut  \n\n\\[\nm:=\\hat r\\times n\n\\]\n\nand use the (in general non-orthonormal) ordered basis $(\\,\\hat r,n,m\\,)$ to write coordinates of vectors.  \nIf $v$ is any vector we denote by  \n\n\\[\nv=v_{r}\\,\\hat r+v_{n}\\,n+v_{m}\\,m\n\\]\n\nits coordinate triple $(v_{r},v_{n},v_{m})$ in this basis.\n\nFix four distinct points $A,B,C,D$ that lie in $\\Pi$ and write  \n\n\\[\na:=\\overrightarrow{OA},\\qquad\nb:=\\overrightarrow{OB},\\qquad\nc:=\\overrightarrow{OC},\\qquad\nd:=\\overrightarrow{OD}.\n\\]\n\nThroughout let  \n\n\\[\n\\rho:=\\hat r\\cdot n\\quad(\\rho\\neq 0),\\qquad\n\\Delta_{1}:=a_{r}c_{n}-a_{n}c_{r},\\qquad\n\\Delta_{2}:=b_{r}d_{n}-b_{n}d_{r}.\n\\]\n\nAssume the following algebraic side-conditions.\n\n(C1) $a_{m}=c_{m}=0$ \\;(the rays $OA,OC$ lie in $\\Sigma:=\\operatorname{span}\\{\\hat r,n\\}$).  \n\n(C2) $\\Delta_{1}\\neq 0$.  \n\n(C3) $a_{n}c_{r}+a_{r}c_{n}=0$.  \n\n(C4) $a_{r}>0$ and $c_{r}>0$.  \n\n(C5) $b_{n}d_{n}\\neq 0$ \\;(the rays $OB,OD$ meet $\\Pi$ transversally).  \n\n(C5$'$) $b_{m}^{\\,2}+d_{m}^{\\,2}\\neq 0$ \\;(at least one of $b,d$ does not lie in $\\Sigma$).  \n\n(C6) $\\Delta_{2}\\neq 0$.  \n\n(C7) $\\Theta_{r}:=\\rho\\bigl(b_{r}d_{n}+d_{r}b_{n}\\bigr)+2\\,b_{n}d_{n}=0$.  \n\n(C8) $\\Theta_{m}:=b_{m}d_{n}-d_{m}b_{n}=0$.  \n\n(C9) $d_{n}\\,\\Delta_{2}>0>\\;b_{n}\\,\\Delta_{2}$.\n\nFor every real number $t>0$ put  \n\n\\[\nM:=O+t\\,\\hat r .\n\\]\n\n(a) Show that for each $t>0$ there exist unique points  \n\n\\[\nA'\\in OA,\\qquad B'\\in OB,\\qquad C'\\in OC,\\qquad D'\\in OD\n\\]\n\nsuch that  \n\n(i) $A'B'C'D'$ is a parallelogram,  \n\n(ii) $M$ is the common midpoint of its two diagonals,  \n\n(iii) the diagonal $A'C'$ is perpendicular to $\\Pi$,  \n\n(iv) the other diagonal $B'D'$ is parallel to $\\Pi$ (equivalently $B'D'\\perp n$).\n\n(b) Prove that the parallelogram is non-degenerate (its four vertices are pairwise distinct) and that the length of the “vertical’’ diagonal depends linearly on $t$:\n\\[\n|A'C'|=\\kappa\\,t\n\\quad\\text{with}\\quad\n\\kappa:=\\dfrac{4\\,|a_{n}c_{n}|}{|\\Delta_{1}|}.\n\\]\n\n(c) Construct $A',B',C',D'$ explicitly in terms of $t$ and the given data, justify that the construction works for every $t>0$, and prove uniqueness.",
      "solution": "All coordinates are taken in the ordered basis $(\\hat r,n,m)$.  \nRecall $\\rho:=\\hat r\\cdot n\\neq 0$.\n\n\\textbf{Step 1.  Relations involving $A$ and $C$}.  \nBecause of (C1)\n\\[\na=a_{r}\\hat r+a_{n}n,\\qquad\nc=c_{r}\\hat r+c_{n}n .\n\\]\nConditions (C2)–(C3) give\n\\[\n\\Delta_{1}=a_{r}c_{n}-a_{n}c_{r}\n          =2\\,a_{r}c_{n}=-2\\,a_{n}c_{r}\\neq 0,\n\\]\nhence $a_{n}c_{n}\\neq 0$, and by (C4) $a_{r},c_{r}>0$.\n\n\\textbf{Step 2.  Construction of the “vertical’’ diagonal $A'C'$}.  \nSeek $\\alpha,\\gamma>0$ with  \n\\[\nA':=\\alpha\\,a\\in OA,\\qquad C':=\\gamma\\,c\\in OC\n\\]\nsuch that  \n\n(i$_{A}$) $\\alpha a+\\gamma c=2t\\hat r$  (makes $M$ the midpoint),  \n\n(ii$_{A}$) $\\alpha a-\\gamma c\\parallel n$  (so $A'C'$ is perpendicular to $\\Pi$).\n\nCondition (ii$_{A}$) forces the $\\hat r$-coordinate of $\\alpha a-\\gamma c$ to be $0$, and together with (i$_{A}$) yields the $2\\times 2$ system  \n\\[\na_{r}\\alpha+c_{r}\\gamma=2t,\\qquad\na_{r}\\alpha-c_{r}\\gamma=0.\n\\]\nSolving,\n\\[\n\\alpha=\\dfrac{t}{a_{r}}>0,\\qquad\n\\gamma=\\dfrac{t}{c_{r}}>0.\n\\]\nBecause of (C3)\n\\[\na_{n}\\alpha+c_{n}\\gamma\n     =t\\Bigl(\\dfrac{a_{n}}{a_{r}}+\\dfrac{c_{n}}{c_{r}}\\Bigr)=0,\n\\]\nso (i$_{A}$) is a vector identity and (ii$_{A}$) holds automatically.  Hence  \n\\[\n\\overrightarrow{A'C'}=\\alpha a-\\gamma c\n        =t\\Bigl(\\dfrac{a}{a_{r}}-\\dfrac{c}{c_{r}}\\Bigr)\n        =-2t\\,\\dfrac{c_{n}}{c_{r}}\\,n,\n\\]\nwhich is non-zero, orthogonal to $\\Pi$, and has length\n\\[\n|A'C'|\n     =2t\\,\\dfrac{|c_{n}|}{|c_{r}|}\n     =2t\\,\\dfrac{|a_{n}|}{|a_{r}|}\n     =\\dfrac{4\\,|a_{n}c_{n}|}{|\\Delta_{1}|}\\,t,\n\\]\nestablishing formula (b).\n\n\\textbf{Step 3.  Construction of the “horizontal’’ diagonal $B'D'$}.  \nWrite the full coordinates\n\\[\nb=b_{r}\\hat r+b_{n}n+b_{m}m,\\qquad\nd=d_{r}\\hat r+d_{n}n+d_{m}m.\n\\]\nLook for $\\lambda,\\mu>0$ such that  \n\\[\n\\lambda b+\\mu d=2t\\hat r,      \\tag{1}\n\\]\n\\[\n(\\mu d-\\lambda b)\\cdot n=0.   \\tag{2}\n\\]\nEquation (1) is equivalent to the three scalar equations  \n\\[\nb_{r}\\lambda+d_{r}\\mu=2t,\\qquad\nb_{n}\\lambda+d_{n}\\mu=0,\\qquad\nb_{m}\\lambda+d_{m}\\mu=0.   \\tag{3}\n\\]\n\nOnly the first two equalities of (3) are independent; the third is forced by (C8).  \nSolving the first two yields  \n\\[\n\\lambda=\\dfrac{2t\\,d_{n}}{\\Delta_{2}},\\qquad\n\\mu   =-\\dfrac{2t\\,b_{n}}{\\Delta_{2}},\n\\]\nwith $\\Delta_{2}\\neq 0$ by (C6).  Condition (C9) guarantees $\\lambda,\\mu>0$, so  \n\\[\nB':=\\lambda b\\in OB,\\qquad D':=\\mu d\\in OD .\n\\]\nSubstituting these values into the third line of (3) gives  \n\\[\nb_{m}\\lambda+d_{m}\\mu\n   =\\dfrac{2t}{\\Delta_{2}}\\,(b_{m}d_{n}-d_{m}b_{n})\n   =\\dfrac{2t}{\\Delta_{2}}\\;\\Theta_{m}=0,\n\\]\nthanks to (C8); hence (1) holds as a vector identity.  \n\nFor (2) we compute\n\\[\n(\\mu d-\\lambda b)\\cdot n\n   =\\rho(d_{r}\\mu-b_{r}\\lambda)+(d_{n}\\mu-b_{n}\\lambda)\n   =-\\dfrac{2t}{\\Delta_{2}}\\,\n      \\bigl[\\rho(b_{r}d_{n}+d_{r}b_{n})+2\\,b_{n}d_{n}\\bigr]\n   =-\\dfrac{2t}{\\Delta_{2}}\\;\\Theta_{r}=0\n\\]\nby (C7), so $B'D'\\perp n$, i.e.\\ $B'D'$ is parallel to $\\Pi$.\n\n\\textbf{Step 4.  Verification of (i)–(iv)}.  \nFrom the identities of Steps 2 and 3 we have  \n\\[\n\\overrightarrow{OM}=t\\hat r\n                 =\\tfrac12(\\lambda b+\\mu d)\n                 =\\tfrac12(\\alpha a+\\gamma c),\n\\]\nwhence $\\overrightarrow{OM}$ is the common midpoint vector of $A'C'$ and $B'D'$.  \nConsequently $A'B'C'D'$ is a parallelogram, proving (i) and (ii).  \nProperties (iii) and (iv) were ensured in Steps 2 and 3, respectively.\n\n\\textbf{Step 5.  Non-degeneracy}.  \nWe already know $A'\\neq C'$ because $\\overrightarrow{A'C'}\\neq 0$, and $B'\\neq D'$ because $\\Delta_{2}\\neq 0$.  \nTo show that \\emph{every} pair of distinct indices corresponds to distinct points, suppose, toward a contradiction, that two different vertices coincide.\n\n\\smallskip\n$\\bullet$ \\emph{Case $A'=B'$}.  \nThen $\\alpha a=\\lambda b$.  Because $\\alpha a$ lies in $\\Sigma$, this forces $b\\in\\Sigma$, hence $b_{m}=0$.  \nUsing $\\mu d=2t\\hat r-\\lambda b$, we also have $\\mu d\\in\\Sigma$ and therefore $d_{m}=0$, contradicting (C5$'$).\n\n$\\bullet$ \\emph{Case $A'=D'$}.  \nNow $\\alpha a=\\mu d$.  Arguing as above yields $d_{m}=0$, and\n\\[\nb_{m}\\lambda=(b_{m}\\lambda+d_{m}\\mu)-d_{m}\\mu=0,\n\\]\nso $b_{m}=0$, again contradicting (C5$'$).\n\n$\\bullet$ \\emph{Case $B'=C'$}.  \nHere $\\lambda b=\\gamma c$.  Because $c\\in\\Sigma$, we deduce $b_{m}=0$, which forces $d_{m}=0$ as before, a contradiction.\n\n$\\bullet$ \\emph{Case $C'=D'$}.  \nThen $\\gamma c=\\mu d$.  Since $c\\in\\Sigma$, this gives $d_{m}=0$; subsequently $b_{m}=0$ follows from (C8), contradicting (C5$'$).\n\n$\\bullet$ \\emph{Case $A'=C'$ or $B'=D'$}.  \nThese were already excluded.\n\nThus all four vertices are pairwise distinct; the parallelogram is non-degenerate.  \nEquation \\(|A'C'|=\\kappa t\\) was obtained in Step 2, completing part (b).\n\n\\textbf{Step 6.  Uniqueness and linear dependence on $t$}.  \nThe $2\\times2$ systems in Steps 2 and 3 have unique solutions, hence no other quadruple of points on the same rays can satisfy (i)–(iv) for the same $t$.  \nFormulae for $\\alpha,\\gamma,\\lambda,\\mu$ are linear in $t$, so the vertices and the length $|A'C'|$ vary linearly with $t$.  \nThis completes the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.504096",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra geometric requirements The original problem asks only for the existence of a parallelogram; the enhanced version forces  \n   • its centre to lie on a prescribed skew line r,  \n   • one diagonal to be perpendicular to a given plane Π,  \n   • the two diagonals to be mutually perpendicular.  \n   Fulfilling four simultaneous conditions instead of one makes the system highly constrained.\n\n2. Higher-dimensional reasoning One has to juggle three distinct affine objects (plane Π, skew lines ℓ and r) in ℝ³, keep track of orthogonality relations, and work in the orthogonal complement r⊥—all absent from the original task.\n\n3. Vector-parameter calculus Solving the problem naturally leads to introducing four unknown positive scalars λ , μ , α , γ and an auxiliary parameter κ, then cracking a coupled vector–scalar system (7)–(10).  This demands linear-algebra techniques in a geometric setting.\n\n4. Non-trivial uniqueness issues Besides showing existence, one must prove that the construction returns exactly one quadrilateral for every centre position M, and that every admissible parallelogram is produced—subtleties not present in the basic variant.\n\n5. Quantitative finale The last request (express |A′C′| in terms of t) turns the exercise into an analytic one, forcing the solver to translate the whole figure into explicit trigonometric form.\n\nAll these layers oblige the contestant to combine synthetic 3-space geometry, affine-vector methods, and a dose of analytic geometry—considerably more sophisticated than the single-plane, single-condition setting of the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}