{
  "index": "1954-A-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "4. A uniform rod of length \\( 2 k \\) and weight \\( w \\) rests with the end \\( A \\) against a smooth vertical wall, while to the lower end \\( B \\) is fastened a string \\( B C \\) of length \\( 2 b \\) coming from a point \\( C \\) in the wall directly above \\( A \\). If the system is in equilibrium, determine the angle \\( A B C \\).",
  "solution": "Solution. Obviously the system will be in equilibrium if the rod and the string are both vertical, i.e., if \\( B \\) is on the wall and \\( \\angle A B C=0 \\).\n\nAssume that the rod is in equilibrium in some non-vertical position, as pictured. Let \\( D \\) be the midpoint of \\( B C \\). The force of tension in the string and the force of gravity on the rod both act through the point \\( D \\) (since the latter force acts vertically through the midpoint of the rod). Hence for equilibrium the reaction of the wall must also act through \\( D \\). Since the wall is smooth this force is perpendicular to the wall. Hence we have \\( A D \\) perpendicular to \\( A C \\) and\n\\[\n|A C|^{2}+|A D|^{2}=|C D|^{2}=b^{2}\n\\]\n\nBy the law of cosines\n\\[\n\\begin{aligned}\n|A C|^{2} & =4 k^{2}+4 b^{2}-8 b k \\cos \\phi \\\\\n|A D|^{2} & =4 k^{2}+b^{2}-4 b k \\cos \\phi\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\cos \\phi=\\frac{2 k^{2}+b^{2}}{3 b k}=\\frac{2}{3} \\frac{k}{b}+\\frac{1}{3} \\frac{b}{k} .\n\\]\n\nThus\n\\[\n\\angle A B C=\\arccos \\left(\\frac{2}{3} \\frac{k}{b}+\\frac{1}{3} \\frac{b}{k}\\right)\n\\]\n\nSince angles \\( A D B \\) and \\( C A B \\) are obtuse, we must have \\( b<2 k<2 b \\). Since \\( 2 x / 3+1 / 3 x<1 \\) for \\( \\frac{1}{2}<x<1 \\), equation (1) determines a unique acute angle whenever \\( 1 / 2<k / b<1 \\). Otherwise there is no equilibrium position except with \\( B \\) on the wall.\n\nRemark. If we calculate the height of the center of gravity of the rod, keeping \\( A \\) directly below \\( C \\) on the wall, we find that the equilibrium position away from the wall is always unstable (i.e., a slight disturbance will tend to lower the center of gravity) while the vertical equilibrium is stable if and only if \\( b<2 k \\).",
  "vars": [
    "\\\\phi",
    "x"
  ],
  "params": [
    "k",
    "b",
    "w",
    "A",
    "B",
    "C",
    "D"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "\\phi": "anglephi",
        "x": "varunknown",
        "k": "halflength",
        "b": "halfstring",
        "w": "rodweight",
        "A": "pointa",
        "B": "pointb",
        "C": "pointc",
        "D": "pointd"
      },
      "question": "4. A uniform rod of length \\( 2 halflength \\) and weight \\( rodweight \\) rests with the end \\( pointa \\) against a smooth vertical wall, while to the lower end \\( pointb \\) is fastened a string \\( pointb pointc \\) of length \\( 2 halfstring \\) coming from a point \\( pointc \\) in the wall directly above \\( pointa \\). If the system is in equilibrium, determine the angle \\( pointa pointb pointc \\).",
      "solution": "Solution. Obviously the system will be in equilibrium if the rod and the string are both vertical, i.e., if \\( pointb \\) is on the wall and \\( \\angle pointa pointb pointc = 0 \\).\n\nAssume that the rod is in equilibrium in some non-vertical position, as pictured. Let \\( pointd \\) be the midpoint of \\( pointb pointc \\). The force of tension in the string and the force of gravity on the rod both act through the point \\( pointd \\) (since the latter force acts vertically through the midpoint of the rod). Hence for equilibrium the reaction of the wall must also act through \\( pointd \\). Since the wall is smooth this force is perpendicular to the wall. Hence we have \\( pointa pointd \\) perpendicular to \\( pointa pointc \\) and\n\\[\n|pointa pointc|^{2} + |pointa pointd|^{2} = |pointc pointd|^{2} = halfstring^{2}\n\\]\n\nBy the law of cosines\n\\[\n\\begin{aligned}\n|pointa pointc|^{2} & = 4\\, halflength^{2} + 4\\, halfstring^{2} - 8\\, halfstring\\, halflength \\cos anglephi \\\\\n|pointa pointd|^{2} & = 4\\, halflength^{2} + halfstring^{2} - 4\\, halfstring\\, halflength \\cos anglephi\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\cos anglephi = \\frac{2\\, halflength^{2} + halfstring^{2}}{3\\, halfstring\\, halflength} = \\frac{2}{3} \\frac{halflength}{halfstring} + \\frac{1}{3} \\frac{halfstring}{halflength}.\n\\]\n\nThus\n\\[\n\\angle pointa pointb pointc = \\arccos \\left( \\frac{2}{3} \\frac{halflength}{halfstring} + \\frac{1}{3} \\frac{halfstring}{halflength} \\right)\n\\]\n\nSince angles \\( pointa pointd pointb \\) and \\( pointc pointa pointb \\) are obtuse, we must have \\( halfstring < 2\\, halflength < 2\\, halfstring \\). Since \\( \\frac{2\\, varunknown}{3} + \\frac{1}{3}\\, varunknown < 1 \\) for \\( \\frac{1}{2} < varunknown < 1 \\), equation (1) determines a unique acute angle whenever \\( \\frac{1}{2} < \\frac{halflength}{halfstring} < 1 \\). Otherwise there is no equilibrium position except with \\( pointb \\) on the wall.\n\nRemark. If we calculate the height of the center of gravity of the rod, keeping \\( pointa \\) directly below \\( pointc \\) on the wall, we find that the equilibrium position away from the wall is always unstable (i.e., a slight disturbance will tend to lower the center of gravity) while the vertical equilibrium is stable if and only if \\( halfstring < 2\\, halflength \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "\\phi": "blueberry",
        "x": "marigolds",
        "k": "pavilion",
        "b": "necklaces",
        "w": "rainstorm",
        "A": "lighthouse",
        "B": "accordion",
        "C": "pineapple",
        "D": "snowflake"
      },
      "question": "4. A uniform rod of length \\( 2 pavilion \\) and weight \\( rainstorm \\) rests with the end \\( lighthouse \\) against a smooth vertical wall, while to the lower end \\( accordion \\) is fastened a string \\( accordion pineapple \\) of length \\( 2 necklaces \\) coming from a point \\( pineapple \\) in the wall directly above \\( lighthouse \\). If the system is in equilibrium, determine the angle \\( lighthouse accordion pineapple \\).",
      "solution": "Solution. Obviously the system will be in equilibrium if the rod and the string are both vertical, i.e., if \\( accordion \\) is on the wall and \\( \\angle lighthouse accordion pineapple=0 \\).\n\nAssume that the rod is in equilibrium in some non-vertical position, as pictured. Let \\( snowflake \\) be the midpoint of \\( accordion pineapple \\). The force of tension in the string and the force of gravity on the rod both act through the point \\( snowflake \\) (since the latter force acts vertically through the midpoint of the rod). Hence for equilibrium the reaction of the wall must also act through \\( snowflake \\). Since the wall is smooth this force is perpendicular to the wall. Hence we have \\( lighthouse snowflake \\) perpendicular to \\( lighthouse pineapple \\) and\n\\[\n|lighthouse pineapple|^{2}+|lighthouse snowflake|^{2}=|pineapple snowflake|^{2}=necklaces^{2}\n\\]\n\nBy the law of cosines\n\\[\n\\begin{aligned}\n|lighthouse pineapple|^{2} & =4 pavilion^{2}+4 necklaces^{2}-8 necklaces pavilion \\cos blueberry \\\\\n|lighthouse snowflake|^{2} & =4 pavilion^{2}+necklaces^{2}-4 necklaces pavilion \\cos blueberry\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\cos blueberry=\\frac{2 pavilion^{2}+necklaces^{2}}{3 necklaces pavilion}=\\frac{2}{3} \\frac{pavilion}{necklaces}+\\frac{1}{3} \\frac{necklaces}{pavilion} .\n\\]\n\nThus\n\\[\n\\angle lighthouse accordion pineapple=\\arccos \\left(\\frac{2}{3} \\frac{pavilion}{necklaces}+\\frac{1}{3} \\frac{necklaces}{pavilion}\\right)\n\\]\n\nSince angles \\( lighthouse snowflake accordion \\) and \\( pineapple lighthouse accordion \\) are obtuse, we must have \\( necklaces<2 pavilion<2 necklaces \\). Since \\( 2 marigolds / 3+1 / 3 marigolds<1 \\) for \\( \\frac{1}{2}<marigolds<1 \\), equation (1) determines a unique acute angle whenever \\( 1 / 2<pavilion / necklaces<1 \\). Otherwise there is no equilibrium position except with \\( accordion \\) on the wall.\n\nRemark. If we calculate the height of the center of gravity of the rod, keeping \\( lighthouse \\) directly below \\( pineapple \\) on the wall, we find that the equilibrium position away from the wall is always unstable (i.e., a slight disturbance will tend to lower the center of gravity) while the vertical equilibrium is stable if and only if \\( necklaces<2 pavilion \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "\\phi": "linearity",
        "x": "difference",
        "k": "briefness",
        "b": "shortening",
        "w": "lightness",
        "A": "voidpoint",
        "B": "centerpoint",
        "C": "belowpoint",
        "D": "endpoint"
      },
      "question": "4. A uniform rod of length \\( 2 briefness \\) and weight \\( lightness \\) rests with the end \\( voidpoint \\) against a smooth vertical wall, while to the lower end \\( centerpoint \\) is fastened a string \\( centerpoint belowpoint \\) of length \\( 2 shortening \\) coming from a point \\( belowpoint \\) in the wall directly above \\( voidpoint \\). If the system is in equilibrium, determine the angle \\( voidpoint centerpoint belowpoint \\).",
      "solution": "Solution. Obviously the system will be in equilibrium if the rod and the string are both vertical, i.e., if \\( centerpoint \\) is on the wall and \\( \\angle voidpoint centerpoint belowpoint=0 \\).\n\nAssume that the rod is in equilibrium in some non-vertical position, as pictured. Let \\( endpoint \\) be the midpoint of \\( centerpoint belowpoint \\). The force of tension in the string and the force of gravity on the rod both act through the point \\( endpoint \\) (since the latter force acts vertically through the midpoint of the rod). Hence for equilibrium the reaction of the wall must also act through \\( endpoint \\). Since the wall is smooth this force is perpendicular to the wall. Hence we have \\( voidpoint endpoint \\) perpendicular to \\( voidpoint belowpoint \\) and\n\\[\n|voidpoint belowpoint|^{2}+|voidpoint endpoint|^{2}=|belowpoint endpoint|^{2}=shortening^{2}\n\\]\n\nBy the law of cosines\n\\[\n\\begin{aligned}\n|voidpoint\\ belowpoint|^{2} & =4 briefness^{2}+4 shortening^{2}-8 shortening briefness \\cos linearity \\\\\n|voidpoint\\ endpoint|^{2} & =4 briefness^{2}+shortening^{2}-4 shortening briefness \\cos linearity\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\cos linearity=\\frac{2 briefness^{2}+shortening^{2}}{3 shortening briefness}=\\frac{2}{3} \\frac{briefness}{shortening}+\\frac{1}{3} \\frac{shortening}{briefness} .\n\\]\n\nThus\n\\[\n\\angle voidpoint centerpoint belowpoint=\\arccos \\left(\\frac{2}{3} \\frac{briefness}{shortening}+\\frac{1}{3} \\frac{shortening}{briefness}\\right)\n\\]\n\nSince angles \\( voidpoint endpoint centerpoint \\) and \\( belowpoint voidpoint centerpoint \\) are obtuse, we must have \\( shortening<2 briefness<2 shortening \\). Since \\( 2 difference / 3+1 / 3 difference<1 \\) for \\( \\frac{1}{2}<difference<1 \\), equation (1) determines a unique acute angle whenever \\( 1 / 2<briefness / shortening<1 \\). Otherwise there is no equilibrium position except with \\( centerpoint \\) on the wall.\n\nRemark. If we calculate the height of the center of gravity of the rod, keeping \\( voidpoint \\) directly below \\( belowpoint \\) on the wall, we find that the equilibrium position away from the wall is always unstable (i.e., a slight disturbance will tend to lower the center of gravity) while the vertical equilibrium is stable if and only if \\( shortening<2 briefness \\)."
    },
    "garbled_string": {
      "map": {
        "\\phi": "lqsntbva",
        "x": "trqkmdse",
        "k": "vlgwzhrp",
        "b": "xhjfqmle",
        "w": "pzdfkqso",
        "A": "yknrslbt",
        "B": "oepmavcz",
        "C": "wjsrilkq",
        "D": "azdbqtnh"
      },
      "question": "4. A uniform rod of length \\( 2 vlgwzhrp \\) and weight \\( pzdfkqso \\) rests with the end \\( yknrslbt \\) against a smooth vertical wall, while to the lower end \\( oepmavcz \\) is fastened a string \\( oepmavcz wjsrilkq \\) of length \\( 2 xhjfqmle \\) coming from a point \\( wjsrilkq \\) in the wall directly above \\( yknrslbt \\). If the system is in equilibrium, determine the angle \\( yknrslbt oepmavcz wjsrilkq \\).",
      "solution": "Solution. Obviously the system will be in equilibrium if the rod and the string are both vertical, i.e., if \\( oepmavcz \\) is on the wall and \\( \\angle yknrslbt oepmavcz wjsrilkq=0 \\).\n\nAssume that the rod is in equilibrium in some non-vertical position, as pictured. Let \\( azdbqtnh \\) be the midpoint of \\( oepmavcz wjsrilkq \\). The force of tension in the string and the force of gravity on the rod both act through the point \\( azdbqtnh \\) (since the latter force acts vertically through the midpoint of the rod). Hence for equilibrium the reaction of the wall must also act through \\( azdbqtnh \\). Since the wall is smooth this force is perpendicular to the wall. Hence we have \\( yknrslbt azdbqtnh \\) perpendicular to \\( yknrslbt wjsrilkq \\) and\n\\[\n|yknrslbt wjsrilkq|^{2}+|yknrslbt azdbqtnh|^{2}=|wjsrilkq azdbqtnh|^{2}=xhjfqmle^{2}\n\\]\n\nBy the law of cosines\n\\[\n\\begin{aligned}\n|yknrslbt wjsrilkq|^{2} & =4 vlgwzhrp^{2}+4 xhjfqmle^{2}-8 xhjfqmle vlgwzhrp \\cos lqsntbva \\\\\n|yknrslbt azdbqtnh|^{2} & =4 vlgwzhrp^{2}+xhjfqmle^{2}-4 xhjfqmle vlgwzhrp \\cos lqsntbva\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\cos lqsntbva=\\frac{2 vlgwzhrp^{2}+xhjfqmle^{2}}{3 xhjfqmle vlgwzhrp}=\\frac{2}{3} \\frac{vlgwzhrp}{xhjfqmle}+\\frac{1}{3} \\frac{xhjfqmle}{vlgwzhrp} .\n\\]\n\nThus\n\\[\n\\angle yknrslbt oepmavcz wjsrilkq=\\arccos \\left(\\frac{2}{3} \\frac{vlgwzhrp}{xhjfqmle}+\\frac{1}{3} \\frac{xhjfqmle}{vlgwzhrp}\\right)\n\\]\n\nSince angles \\( yknrslbt azdbqtnh oepmavcz \\) and \\( wjsrilkq yknrslbt oepmavcz \\) are obtuse, we must have \\( xhjfqmle<2 vlgwzhrp<2 xhjfqmle \\). Since \\( 2 trqkmdse / 3+1 / 3 trqkmdse<1 \\) for \\( \\frac{1}{2}<trqkmdse<1 \\), equation (1) determines a unique acute angle whenever \\( 1 / 2<vlgwzhrp / xhjfqmle<1 \\). Otherwise there is no equilibrium position except with \\( oepmavcz \\) on the wall.\n\nRemark. If we calculate the height of the center of gravity of the rod, keeping \\( yknrslbt \\) directly below \\( wjsrilkq \\) on the wall, we find that the equilibrium position away from the wall is always unstable (i.e., a slight disturbance will tend to lower the center of gravity) while the vertical equilibrium is stable if and only if \\( xhjfqmle<2 vlgwzhrp \\)."
    },
    "kernel_variant": {
      "question": "A right-angled vertical corner is formed by the two mutually perpendicular, perfectly smooth walls  \n\n   Π₁ : x = 0  and  Π₂ : y = 0 ,\n\nwhose common vertical edge is the z–axis Γ (x = y = 0).  \n\nInside the corner a uniform heavy beam AB of length 3 k  (k > 0) and weight P is placed so that  \n\n• its upper end A rests on the wall Π₁  (x_A = 0),  \n• its lower end B rests on the wall Π₂  (y_B = 0),  \n• the beam is not vertical (i.e. it is not parallel to Γ).\n\nWrite, a priori,\n\n  A = (0, y, a)     (y ≥ 0, a > 0),  \n  B = (x, 0, z)     (x > 0,  z < a);             (*)\n\n(the coordinate z may even be negative because no horizontal floor is present).  \n\nA light, inextensible string BC of fixed length 4 b (b > 0) joins the point B to a fixed point  \n\n   C ≡ (0, 0, h)   (h > 0)\n\non the edge Γ.  No other supports act.  The only forces on the system are  \n\n • P (the weight) acting vertically downward at the mid-point G of AB,  \n • T (the tension of the string) acting on B along the line CB,  \n • R₁ (the smooth reaction of Π₁) acting at A along the +eₓ–direction,  \n • R₂ (the smooth reaction of Π₂) acting at B along the +eᵧ–direction.  \n\nThe system is in static equilibrium and the beam is not parallel to either wall.  \n(N.B.  A line contained in a plane is not regarded as “parallel” to that plane.)\n\nIntroduce the positive parameters\n\n u := a − z  (the vertical rise of the beam),  d := h − a  (the extra rise of the string above A).   (**)\n\n(i) Starting exclusively from the six equilibrium equations\n\n  ΣF = 0,  ΣM_O = 0  (about the origin O),\n\nprove that  \n\n   y = 0  (so that A lies on the edge Γ),\n\nand that a non-vertical equilibrium is possible if and only if the two given lengths k, b satisfy the double inequality  \n\n   2 b < 3 k < 4 b,            (1)\n\nwhile the heights must obey the single relation  \n\n   3 d² = 16 b² − 9 k²  ( = 3 u² ).      (2)\n\nDeduce that relation (2) fixes only the vertical separation d (= u) between A and C, whereas the absolute height a (and hence h = a + d) may be chosen arbitrarily large, giving a one-parameter family of equilibrium positions.\n\n(ii) For any admissible triple (k, b, d) found in part (i) determine the acute angle  \n\n   φ = ∠ABC  \n\nand show that  \n\n   cos φ = (9 k² + 8 b²)/(18 k b).     (3)\n\n(iii) Using (2)–(3) obtain explicit expressions for the tension T and for the wall reactions R₁, R₂, proving that  \n\n   T  = (2 b P)/d = (2 b P √3)/√(16 b² − 9 k²),  \n   R₁ = (P/2) √{ (36 k² − 16 b²)/(16 b² − 9 k²) } > 0,  \n   R₂ ≡ 0.  \n\nDiscuss the behaviour of T and R₁ as k approaches the lower and the upper bounds in (1), and decide whether R₂ can ever be positive or negative.  Comment explicitly on the effect of a rigid vertical translation of the whole configuration.\n\n---------------------------------------------------------------",
      "solution": "We use throughout the orthonormal basis  \n\n eₓ = (1,0,0), eᵧ = (0,1,0), e_z = (0,0,1).\n\n--------------------------------------------------------------------\n0. Notation and purely geometric relations\n--------------------------------------------------------------------\nWith the points written as in (*) and the parameters defined in (**)\n\n u := a − z > 0,  d := h − a > 0,          (4)\n\nthe fixed lengths of the beam and of the string yield\n\n |AB|² = x² + y² + u² = (3 k)² = 9 k²,       (5a)  \n |BC|² = x² + (u + d)²     = (4 b)² = 16 b².    (5b)\n\nNo assumption about y has been made so far.\n\n--------------------------------------------------------------------\n1. Force equilibrium ΣF = 0\n--------------------------------------------------------------------\nThe forces at A and B are  \n\n at A: R₁ eₓ,  \n\n at B: R₂ eᵧ + T (−x eₓ + (u + d) e_z)/(4 b) − P e_z.\n\nHence\n\n ΣF_x : R₁ − T x/(4 b) = 0,            (6a)  \n ΣF_y : R₂ = 0,                 (6b)  \n ΣF_z : T(u + d)/(4 b) − P = 0.        (6c)\n\n--------------------------------------------------------------------\n2. Moment equilibrium ΣM_O = 0\n--------------------------------------------------------------------\nPosition vectors\n\n r_A = (0, y, a), r_B = (x, 0, z), r_G = (x/2, y/2, (a + z)/2).\n\n(i) Moment of R₁ : r_A × (R₁ eₓ) = (0, a R₁, −y R₁).  \n(ii) Moment of R₂ : r_B × (R₂ eᵧ) = (−z R₂, 0, x R₂).  \n(iii) Moment of T : CB = C − B = (−x, 0, u + d); so  \n\n r_B × [T CB/(4 b)] = (0, −T x(a + d)/(4 b), 0).  \n(iv) Moment of P : r_G × (−P e_z) = (−P y/2, P x/2, 0).\n\nAdding all four contributions and setting the result to zero yields  \n\n ΣM_x : −P y/2 − z R₂ = 0,           (7a)  \n ΣM_y : a R₁ − T x(a + d)/(4 b) + P x/2 = 0, (7b)  \n ΣM_z : −y R₁ + x R₂ = 0.           (7c)\n\n--------------------------------------------------------------------\n3. Point A lies on the edge Γ\n--------------------------------------------------------------------\nEquation (6b) already gives R₂ = 0.  Substituting this into (7a) and (7c) gives  \n\n −P y/2 = 0 and −y R₁ = 0.\n\nBecause P > 0, we must have y = 0.  Therefore  \n\n A = (0, 0, a) ∈ Γ.               (8)\n\nEquation (7b) now simplifies to  \n\n a R₁ − T x(a + d)/(4 b) + P x/2 = 0.     (9)\n\n--------------------------------------------------------------------\n4. Compatibility of the six equilibrium equations\n--------------------------------------------------------------------\nFrom (6c)  T = 4 b P /(u + d).          (10)\n\nInsert R₁ from (6a) into (9) and use (10).  Because x > 0 by hypothesis, we can divide by x:  \n\n a·T/(4 b) − T(a + d)/(4 b) + P/2 = 0  \n  ⇔ T(−d)/(4 b) + P/2 = 0.\n\nWith T > 0 this yields  \n\n T = 2 b P/d.                (11)\n\nEquating (10) and (11) gives  \n\n 2 b P/d = 4 b P/(u + d) ⇒ u = d.     (12)\n\nSince y = 0, eqn (5a) becomes\n\n x² + u² = 9 k².               (13)\n\nEliminating x² between (13) and (5b) with u = d gives\n\n d² = (16 b² − 9 k²)/3,   d > 0.     (14)\n\nPositivity of d supplies 3 k < 4 b, while x² > 0 from (13) gives d < 3 k, i.e. 3 k > 2 b.  Altogether\n\n 2 b < 3 k < 4 b,             (15)\n\nand, rewriting (14),\n\n 3 d² = 16 b² − 9 k²  (= 3 u²).     (16)\n\nThus (15) and (16) are the necessary and sufficient conditions for a non-vertical equilibrium.  Equation (16) fixes only the separation d ( = u ); the absolute height a is arbitrary, so the whole configuration can be translated vertically, providing a one-parameter family of equilibria.\n\n--------------------------------------------------------------------\n5. Angle between beam and string\n--------------------------------------------------------------------\nBecause u = d,\n\n BA = A − B = (−x, 0, d),  BC = C − B = (−x, 0, 2 d).\n\nHence\n\n BA·BC = x² + 2 d²,  |BA| = 3 k, |BC| = 4 b,\n\nso\n\n cos φ = (x² + 2 d²)/(12 k b).        (17)\n\nSubstituting x² = 9 k² − d² and d² from (16) gives\n\n cos φ = (9 k² + 8 b²)/(18 k b),      (18)\n\nwhich is formula (3).\n\n--------------------------------------------------------------------\n6. Magnitudes of the unknown forces\n--------------------------------------------------------------------\nThe two expressions (10) and (11) immediately yield\n\n T = (2 b P)/d = (2 b P √3)/√(16 b² − 9 k²). (19)\n\nUsing (6a) together with x² = 9 k² − d²,\n\n R₁ = T x/(4 b)  \n   = P x/(2 d)  \n   = (P/2) √{ (36 k² − 16 b²)/(16 b² − 9 k²) } > 0. (20)\n\nEquation (6b) has already given  \n\n R₂ ≡ 0.                 (21)\n\n--------------------------------------------------------------------\n7. Limiting behaviour and vertical translation invariance\n--------------------------------------------------------------------\nIntroduce r := 3 k/(4 b); the admissible range is ½ < r < 1.\n\n(i) As r ↓ ½ (i.e. 3 k → 2 b)\n\n d² → 4 b², d → 2 b, x² → 0,  \n R₁ → 0, T → P.\n\n(ii) As r ↑ 1 (i.e. 3 k → 4 b)\n\n d → 0⁺, T ≈ 2 b P/d → +∞,  \n R₁ ≈ (P/2)(2 x/d) → +∞.\n\nThroughout the admissible interval R₁ is strictly positive, while R₂ is identically zero; hence the wall Π₂ never exerts a positive or negative reaction—the contact at B is sustained purely by the string.\n\nFinally, replacing every position vector r by r′ = r + τ e_z (τ real) leaves each force unchanged and adds the same cancelling term τ e_z × ( • ) to every moment about O; therefore the equilibrium is invariant under any rigid vertical translation.  This corroborates that only the separation d ( = u ) is fixed.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.463112",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original planar “rod-and-string-against-one-wall” problem, the present variant is appreciably harder because\n\n1.  Three–dimensional setting  \n The rod touches two perpendicular walls; all unknowns (coordinates of B, length constraints, angles) live in R³ rather than R².\n\n2.  Four forces instead of three  \n Alongside weight and string tension there are two distinct wall reactions.  \n One must therefore invoke concurrency of non-parallel forces and justify why one reaction vanishes.\n\n3.  Non-linear elimination  \n The simultaneous equations (5)–(6)–(7) entail a quadratic substitution followed by an explicit non–negativity requirement, giving rise to the sharp double inequality (★).\n\n4.  Messy algebra in angle computation  \n Finding cos φ requires inserting derived expressions for x and z and carefully manipulating quartic terms; a naïve approach is error-prone.\n\n5.  Additional force analysis  \n Part (iii) demands resolution of forces in three orthogonal directions and discussion of the sign of R₂, something absent from the original problem.\n\nTaken together these features oblige the solver to blend 3-D geometry, statics (concurrency, force resolution), and substantial algebraic manipulation—well beyond the scope of simple pattern matching or of the original two-dimensional equilibrium exercise."
      }
    },
    "original_kernel_variant": {
      "question": "A right-angled vertical corner is formed by the two mutually perpendicular, perfectly smooth walls  \n\n   Π₁ : x = 0  and  Π₂ : y = 0 ,\n\nwhose common vertical edge is the z–axis Γ (x = y = 0).  \n\nInside the corner a uniform heavy beam AB of length 3 k  (k > 0) and weight P is placed so that  \n\n• its upper end A rests on the wall Π₁  (x_A = 0),  \n• its lower end B rests on the wall Π₂  (y_B = 0),  \n• the beam is not vertical (i.e. it is not parallel to Γ).\n\nWrite, a priori,\n\n  A = (0, y, a)     (y ≥ 0, a > 0),  \n  B = (x, 0, z)     (x > 0,  z < a);             (*)\n\n(the coordinate z may even be negative because no horizontal floor is present).  \n\nA light, inextensible string BC of fixed length 4 b (b > 0) joins the point B to a fixed point  \n\n   C ≡ (0, 0, h)   (h > 0)\n\non the edge Γ.  No other supports act.  The only forces on the system are  \n\n • P (the weight) acting vertically downward at the mid-point G of AB,  \n • T (the tension of the string) acting on B along the line CB,  \n • R₁ (the smooth reaction of Π₁) acting at A along the +eₓ–direction,  \n • R₂ (the smooth reaction of Π₂) acting at B along the +eᵧ–direction.  \n\nThe system is in static equilibrium and the beam is not parallel to either wall.  \n(N.B.  A line contained in a plane is not regarded as “parallel” to that plane.)\n\nIntroduce the positive parameters\n\n u := a − z  (the vertical rise of the beam),  d := h − a  (the extra rise of the string above A).   (**)\n\n(i) Starting exclusively from the six equilibrium equations\n\n  ΣF = 0,  ΣM_O = 0  (about the origin O),\n\nprove that  \n\n   y = 0  (so that A lies on the edge Γ),\n\nand that a non-vertical equilibrium is possible if and only if the two given lengths k, b satisfy the double inequality  \n\n   2 b < 3 k < 4 b,            (1)\n\nwhile the heights must obey the single relation  \n\n   3 d² = 16 b² − 9 k²  ( = 3 u² ).      (2)\n\nDeduce that relation (2) fixes only the vertical separation d (= u) between A and C, whereas the absolute height a (and hence h = a + d) may be chosen arbitrarily large, giving a one-parameter family of equilibrium positions.\n\n(ii) For any admissible triple (k, b, d) found in part (i) determine the acute angle  \n\n   φ = ∠ABC  \n\nand show that  \n\n   cos φ = (9 k² + 8 b²)/(18 k b).     (3)\n\n(iii) Using (2)–(3) obtain explicit expressions for the tension T and for the wall reactions R₁, R₂, proving that  \n\n   T  = (2 b P)/d = (2 b P √3)/√(16 b² − 9 k²),  \n   R₁ = (P/2) √{ (36 k² − 16 b²)/(16 b² − 9 k²) } > 0,  \n   R₂ ≡ 0.  \n\nDiscuss the behaviour of T and R₁ as k approaches the lower and the upper bounds in (1), and decide whether R₂ can ever be positive or negative.  Comment explicitly on the effect of a rigid vertical translation of the whole configuration.\n\n---------------------------------------------------------------",
      "solution": "We use throughout the orthonormal basis  \n\n eₓ = (1,0,0), eᵧ = (0,1,0), e_z = (0,0,1).\n\n--------------------------------------------------------------------\n0. Notation and purely geometric relations\n--------------------------------------------------------------------\nWith the points written as in (*) and the parameters defined in (**)\n\n u := a − z > 0,  d := h − a > 0,          (4)\n\nthe fixed lengths of the beam and of the string yield\n\n |AB|² = x² + y² + u² = (3 k)² = 9 k²,       (5a)  \n |BC|² = x² + (u + d)²     = (4 b)² = 16 b².    (5b)\n\nNo assumption about y has been made so far.\n\n--------------------------------------------------------------------\n1. Force equilibrium ΣF = 0\n--------------------------------------------------------------------\nThe forces at A and B are  \n\n at A: R₁ eₓ,  \n\n at B: R₂ eᵧ + T (−x eₓ + (u + d) e_z)/(4 b) − P e_z.\n\nHence\n\n ΣF_x : R₁ − T x/(4 b) = 0,            (6a)  \n ΣF_y : R₂ = 0,                 (6b)  \n ΣF_z : T(u + d)/(4 b) − P = 0.        (6c)\n\n--------------------------------------------------------------------\n2. Moment equilibrium ΣM_O = 0\n--------------------------------------------------------------------\nPosition vectors\n\n r_A = (0, y, a), r_B = (x, 0, z), r_G = (x/2, y/2, (a + z)/2).\n\n(i) Moment of R₁ : r_A × (R₁ eₓ) = (0, a R₁, −y R₁).  \n(ii) Moment of R₂ : r_B × (R₂ eᵧ) = (−z R₂, 0, x R₂).  \n(iii) Moment of T : CB = C − B = (−x, 0, u + d); so  \n\n r_B × [T CB/(4 b)] = (0, −T x(a + d)/(4 b), 0).  \n(iv) Moment of P : r_G × (−P e_z) = (−P y/2, P x/2, 0).\n\nAdding all four contributions and setting the result to zero yields  \n\n ΣM_x : −P y/2 − z R₂ = 0,           (7a)  \n ΣM_y : a R₁ − T x(a + d)/(4 b) + P x/2 = 0, (7b)  \n ΣM_z : −y R₁ + x R₂ = 0.           (7c)\n\n--------------------------------------------------------------------\n3. Point A lies on the edge Γ\n--------------------------------------------------------------------\nEquation (6b) already gives R₂ = 0.  Substituting this into (7a) and (7c) gives  \n\n −P y/2 = 0 and −y R₁ = 0.\n\nBecause P > 0, we must have y = 0.  Therefore  \n\n A = (0, 0, a) ∈ Γ.               (8)\n\nEquation (7b) now simplifies to  \n\n a R₁ − T x(a + d)/(4 b) + P x/2 = 0.     (9)\n\n--------------------------------------------------------------------\n4. Compatibility of the six equilibrium equations\n--------------------------------------------------------------------\nFrom (6c)  T = 4 b P /(u + d).          (10)\n\nInsert R₁ from (6a) into (9) and use (10).  Because x > 0 by hypothesis, we can divide by x:  \n\n a·T/(4 b) − T(a + d)/(4 b) + P/2 = 0  \n  ⇔ T(−d)/(4 b) + P/2 = 0.\n\nWith T > 0 this yields  \n\n T = 2 b P/d.                (11)\n\nEquating (10) and (11) gives  \n\n 2 b P/d = 4 b P/(u + d) ⇒ u = d.     (12)\n\nSince y = 0, eqn (5a) becomes\n\n x² + u² = 9 k².               (13)\n\nEliminating x² between (13) and (5b) with u = d gives\n\n d² = (16 b² − 9 k²)/3,   d > 0.     (14)\n\nPositivity of d supplies 3 k < 4 b, while x² > 0 from (13) gives d < 3 k, i.e. 3 k > 2 b.  Altogether\n\n 2 b < 3 k < 4 b,             (15)\n\nand, rewriting (14),\n\n 3 d² = 16 b² − 9 k²  (= 3 u²).     (16)\n\nThus (15) and (16) are the necessary and sufficient conditions for a non-vertical equilibrium.  Equation (16) fixes only the separation d ( = u ); the absolute height a is arbitrary, so the whole configuration can be translated vertically, providing a one-parameter family of equilibria.\n\n--------------------------------------------------------------------\n5. Angle between beam and string\n--------------------------------------------------------------------\nBecause u = d,\n\n BA = A − B = (−x, 0, d),  BC = C − B = (−x, 0, 2 d).\n\nHence\n\n BA·BC = x² + 2 d²,  |BA| = 3 k, |BC| = 4 b,\n\nso\n\n cos φ = (x² + 2 d²)/(12 k b).        (17)\n\nSubstituting x² = 9 k² − d² and d² from (16) gives\n\n cos φ = (9 k² + 8 b²)/(18 k b),      (18)\n\nwhich is formula (3).\n\n--------------------------------------------------------------------\n6. Magnitudes of the unknown forces\n--------------------------------------------------------------------\nThe two expressions (10) and (11) immediately yield\n\n T = (2 b P)/d = (2 b P √3)/√(16 b² − 9 k²). (19)\n\nUsing (6a) together with x² = 9 k² − d²,\n\n R₁ = T x/(4 b)  \n   = P x/(2 d)  \n   = (P/2) √{ (36 k² − 16 b²)/(16 b² − 9 k²) } > 0. (20)\n\nEquation (6b) has already given  \n\n R₂ ≡ 0.                 (21)\n\n--------------------------------------------------------------------\n7. Limiting behaviour and vertical translation invariance\n--------------------------------------------------------------------\nIntroduce r := 3 k/(4 b); the admissible range is ½ < r < 1.\n\n(i) As r ↓ ½ (i.e. 3 k → 2 b)\n\n d² → 4 b², d → 2 b, x² → 0,  \n R₁ → 0, T → P.\n\n(ii) As r ↑ 1 (i.e. 3 k → 4 b)\n\n d → 0⁺, T ≈ 2 b P/d → +∞,  \n R₁ ≈ (P/2)(2 x/d) → +∞.\n\nThroughout the admissible interval R₁ is strictly positive, while R₂ is identically zero; hence the wall Π₂ never exerts a positive or negative reaction—the contact at B is sustained purely by the string.\n\nFinally, replacing every position vector r by r′ = r + τ e_z (τ real) leaves each force unchanged and adds the same cancelling term τ e_z × ( • ) to every moment about O; therefore the equilibrium is invariant under any rigid vertical translation.  This corroborates that only the separation d ( = u ) is fixed.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.392626",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original planar “rod-and-string-against-one-wall” problem, the present variant is appreciably harder because\n\n1.  Three–dimensional setting  \n The rod touches two perpendicular walls; all unknowns (coordinates of B, length constraints, angles) live in R³ rather than R².\n\n2.  Four forces instead of three  \n Alongside weight and string tension there are two distinct wall reactions.  \n One must therefore invoke concurrency of non-parallel forces and justify why one reaction vanishes.\n\n3.  Non-linear elimination  \n The simultaneous equations (5)–(6)–(7) entail a quadratic substitution followed by an explicit non–negativity requirement, giving rise to the sharp double inequality (★).\n\n4.  Messy algebra in angle computation  \n Finding cos φ requires inserting derived expressions for x and z and carefully manipulating quartic terms; a naïve approach is error-prone.\n\n5.  Additional force analysis  \n Part (iii) demands resolution of forces in three orthogonal directions and discussion of the sign of R₂, something absent from the original problem.\n\nTaken together these features oblige the solver to blend 3-D geometry, statics (concurrency, force resolution), and substantial algebraic manipulation—well beyond the scope of simple pattern matching or of the original two-dimensional equilibrium exercise."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}