{
  "index": "1956-B-3",
  "type": "GEO",
  "tag": [
    "GEO"
  ],
  "difficulty": "",
  "question": "3. A sphere is inscribed in a tetrahedron and each point of contact of the sphere with the four faces is joined to the vertices of the face containing the point. Show that the four sets of three angles so formed are identical.",
  "solution": "Solution. Let the vertices of the tetrahedron be \\( P_{i}, i=1,2,3,4 \\), and let \\( Q_{i} \\) be the point of contact of the sphere with the face opposite \\( P_{i} \\). We adopt the convention that when \\( i, j, k, l \\) appear in a statement, they represent distinct elements of \\( \\{1,2,3,4\\} \\).\n\nNow \\( P_{i} Q_{i} \\) and \\( P_{i} Q_{k} \\) are tangents to the sphere from the same external point; hence \\( \\left|P_{i} Q_{i}\\right|=\\left|P_{i} Q_{k}\\right| \\). Similarly, \\( \\left|P_{i} Q_{i}\\right|=\\left|P_{1} Q_{k}\\right| \\). Hence, \\( \\Delta P_{1} Q_{l} P_{l} \\cong \\Delta P_{i} Q_{k} P_{l} \\) by s.s.s. Therefore, \\( \\angle P_{i} Q_{i} P_{l}=\\angle P_{i} Q_{k} P_{l} \\). We denote this angle by \\( |i l| \\). Clearly we have\n\\[\n|i l|=|l i| .\n\\]\n\nSince the angles at \\( Q_{i} \\) add to \\( 2 \\pi \\), we have\n\\[\n\\begin{array}{l}\n|23|+|34|+|42|=2 \\pi, \\\\\n|34|+|41|+|13|=2 \\pi, \\\\\n|41|+|12|+|24|=2 \\pi, \\\\\n|12|+|23|+|31|=2 \\pi .\n\\end{array}\n\\]\n\nIf we add the first two of these equations, subtract the third and fourth, and use (1), we obtain \\( 2 \\cdot|34|-2 \\cdot|12|=0 \\). Hence, \\( |12|=|34| \\), and by symmetry\n\\[\n|\\ddot{i}|=|k l| .\n\\]\n\nThe angles at \\( Q_{1} \\) are \\( |23|,|34| \\), and \\( |42| \\), and these are respectively equal to the angles at \\( Q_{2} \\), namely, \\( |41|,|34| \\), and \\( |13| \\), by (2). By symmetry, the central angles are the same in all four faces.\n\nRemark. The angles appear with the same orientation in each face. Thus, when viewed from outside the tetrahedron, the angles |23|, |34|, and \\( |42| \\) at \\( Q_{1} \\) appear in the same order (clockwise or counterclockwise) as the corresponding angles \\( |41|,|34| \\), and \\( |13| \\) at \\( Q_{2} \\). This is hard to visualize, but if it were not so, the faces of the tetrahedron would be asymmetrically partitioned into two classes, which is clearly impossible in a symmetrical situation.",
  "vars": [
    "P_i",
    "Q_i",
    "P_1",
    "Q_1",
    "P_l",
    "Q_k",
    "Q_l",
    "i",
    "j",
    "k",
    "l"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P_i": "vertexany",
        "Q_i": "contactany",
        "P_1": "vertexone",
        "Q_1": "contactone",
        "P_l": "vertexell",
        "Q_k": "contactkay",
        "Q_l": "contactell",
        "i": "indexi",
        "j": "indexj",
        "k": "indexk",
        "l": "indexl"
      },
      "question": "3. A sphere is inscribed in a tetrahedron and each point of contact of the sphere with the four faces is joined to the vertices of the face containing the point. Show that the four sets of three angles so formed are identical.",
      "solution": "Solution. Let the vertices of the tetrahedron be \\( vertexany, indexi=1,2,3,4 \\), and let \\( contactany \\) be the point of contact of the sphere with the face opposite \\( vertexany \\). We adopt the convention that when \\( indexi, indexj, indexk, indexl \\) appear in a statement, they represent distinct elements of \\{1,2,3,4\\}.\n\nNow \\( vertexany contactany \\) and \\( vertexany contactkay \\) are tangents to the sphere from the same external point; hence \\( \\left|vertexany contactany\\right|=\\left|vertexany contactkay\\right| \\). Similarly, \\( \\left|vertexany contactany\\right|=\\left|vertexone contactkay\\right| \\). Hence, \\( \\Delta vertexone contactell vertexell \\cong \\Delta vertexany contactkay vertexell \\) by s.s.s. Therefore, \\( \\angle vertexany contactany vertexell=\\angle vertexany contactkay vertexell \\). We denote this angle by \\( |indexi indexl| \\). Clearly we have\n\\[\n|indexi indexl|=|indexl indexi| .\n\\]\n\nSince the angles at \\( contactany \\) add to \\( 2 \\pi \\), we have\n\\[\n\\begin{array}{l}\n|23|+|34|+|42|=2 \\pi, \\\\ \n|34|+|41|+|13|=2 \\pi, \\\\ \n|41|+|12|+|24|=2 \\pi, \\\\ \n|12|+|23|+|31|=2 \\pi .\n\\end{array}\n\\]\n\nIf we add the first two of these equations, subtract the third and fourth, and use (1), we obtain \\( 2 \\cdot|34|-2 \\cdot|12|=0 \\). Hence, \\( |12|=|34| \\), and by symmetry\n\\[\n|\\ddot{indexi}|=|indexk indexl| .\n\\]\n\nThe angles at \\( contactone \\) are \\( |23|,|34| \\), and \\( |42| \\), and these are respectively equal to the angles at \\( Q_{2} \\), namely, \\( |41|,|34| \\), and \\( |13| \\), by (2). By symmetry, the central angles are the same in all four faces.\n\nRemark. The angles appear with the same orientation in each face. Thus, when viewed from outside the tetrahedron, the angles |23|, |34|, and \\( |42| \\) at \\( contactone \\) appear in the same order (clockwise or counterclockwise) as the corresponding angles \\( |41|,|34| \\), and \\( |13| \\) at \\( Q_{2} \\). This is hard to visualize, but if it were not so, the faces of the tetrahedron would be asymmetrically partitioned into two classes, which is clearly impossible in a symmetrical situation."
    },
    "descriptive_long_confusing": {
      "map": {
        "P_i": "blueprint",
        "Q_i": "gemstone",
        "P_1": "seashore",
        "Q_1": "blacksmith",
        "P_l": "nightfall",
        "Q_k": "sandalwood",
        "Q_l": "dawnlight",
        "i": "wallpaper",
        "j": "lemonades",
        "k": "brickwork",
        "l": "marshland"
      },
      "question": "3. A sphere is inscribed in a tetrahedron and each point of contact of the sphere with the four faces is joined to the vertices of the face containing the point. Show that the four sets of three angles so formed are identical.",
      "solution": "Solution. Let the vertices of the tetrahedron be \\( blueprint, wallpaper=1,2,3,4 \\), and let \\( gemstone \\) be the point of contact of the sphere with the face opposite \\( blueprint \\). We adopt the convention that when \\( wallpaper, lemonades, brickwork, marshland \\) appear in a statement, they represent distinct elements of \\( \\{1,2,3,4\\} \\).\n\nNow \\( blueprint gemstone \\) and \\( blueprint sandalwood \\) are tangents to the sphere from the same external point; hence \\( \\left| blueprint gemstone \\right|=\\left| blueprint sandalwood \\right| \\). Similarly, \\( \\left| blueprint gemstone \\right|=\\left| seashore sandalwood \\right| \\). Hence, \\( \\Delta seashore dawnlight nightfall \\cong \\Delta blueprint sandalwood nightfall \\) by s.s.s. Therefore, \\( \\angle blueprint gemstone nightfall = \\angle blueprint sandalwood nightfall \\). We denote this angle by \\( |wallpaper marshland| \\). Clearly we have\n\\[\n|wallpaper marshland| = |marshland wallpaper| .\n\\]\n\nSince the angles at \\( gemstone \\) add to \\( 2 \\pi \\), we have\n\\[\n\\begin{array}{l}\n|23|+|34|+|42|=2 \\pi, \\\\\n|34|+|41|+|13|=2 \\pi, \\\\\n|41|+|12|+|24|=2 \\pi, \\\\\n|12|+|23|+|31|=2 \\pi .\n\\end{array}\n\\]\n\nIf we add the first two of these equations, subtract the third and fourth, and use (1), we obtain \\( 2 \\cdot|34|-2 \\cdot|12|=0 \\). Hence, \\( |12|=|34| \\), and by symmetry\n\\[\n|\\ddot{wallpaper}| = |brickwork marshland| .\n\\]\n\nThe angles at \\( blacksmith \\) are \\( |23|,|34| \\), and \\( |42| \\), and these are respectively equal to the angles at \\( Q_{2} \\), namely, \\( |41|,|34| \\), and \\( |13| \\), by (2). By symmetry, the central angles are the same in all four faces.\n\nRemark. The angles appear with the same orientation in each face. Thus, when viewed from outside the tetrahedron, the angles |23|, |34|, and \\( |42| \\) at \\( blacksmith \\) appear in the same order (clockwise or counterclockwise) as the corresponding angles \\( |41|,|34| \\), and \\( |13| \\) at \\( Q_{2} \\). This is hard to visualize, but if it were not so, the faces of the tetrahedron would be asymmetrically partitioned into two classes, which is clearly impossible in a symmetrical situation."
    },
    "descriptive_long_misleading": {
      "map": {
        "P_i": "voidvertex",
        "Q_i": "detachedpoint",
        "P_1": "voidvertexone",
        "Q_1": "detachedpointone",
        "P_l": "voidvertexl",
        "Q_k": "detachedpointk",
        "Q_l": "detachedpointl",
        "i": "constantin",
        "j": "immutable",
        "k": "stationary",
        "l": "changeless"
      },
      "question": "3. A sphere is inscribed in a tetrahedron and each point of contact of the sphere with the four faces is joined to the vertices of the face containing the point. Show that the four sets of three angles so formed are identical.",
      "solution": "Solution. Let the vertices of the tetrahedron be \\( voidvertex, constantin=1,2,3,4 \\), and let \\( detachedpoint \\) be the point of contact of the sphere with the face opposite \\( voidvertex \\). We adopt the convention that when \\( constantin, immutable, stationary, changeless \\) appear in a statement, they represent distinct elements of \\( \\{1,2,3,4\\} \\).\n\nNow \\( voidvertex detachedpoint \\) and \\( voidvertex detachedpointk \\) are tangents to the sphere from the same external point; hence \\( \\left|voidvertex detachedpoint\\right|=\\left|voidvertex detachedpointk\\right| \\). Similarly, \\( \\left|voidvertex detachedpoint\\right|=\\left|voidvertexone detachedpointk\\right| \\). Hence, \\( \\Delta voidvertexone detachedpointl voidvertexl \\cong \\Delta voidvertex detachedpointk voidvertexl \\) by s.s.s. Therefore, \\( \\angle voidvertex detachedpoint voidvertexl=\\angle voidvertex detachedpointk voidvertexl \\). We denote this angle by \\( |constantin changeless| \\). Clearly we have\n\\[\n|constantin changeless|=|changeless constantin| .\n\\]\n\nSince the angles at \\( detachedpoint \\) add to \\( 2 \\pi \\), we have\n\\[\n\\begin{array}{l}\n|23|+|34|+|42|=2 \\pi, \\\\\n|34|+|41|+|13|=2 \\pi, \\\\\n|41|+|12|+|24|=2 \\pi, \\\\\n|12|+|23|+|31|=2 \\pi .\n\\end{array}\n\\]\n\nIf we add the first two of these equations, subtract the third and fourth, and use (1), we obtain \\( 2 \\cdot|34|-2 \\cdot|12|=0 \\). Hence, \\( |12|=|34| \\), and by symmetry\n\\[\n|\\ddot{constantin}|=|stationary changeless| .\n\\]\n\nThe angles at \\( detachedpointone \\) are \\( |23|,|34| \\), and \\( |42| \\), and these are respectively equal to the angles at \\( detachedpoint \\), namely, \\( |41|,|34| \\), and \\( |13| \\), by (2). By symmetry, the central angles are the same in all four faces.\n\nRemark. The angles appear with the same orientation in each face. Thus, when viewed from outside the tetrahedron, the angles |23|, |34|, and \\( |42| \\) at \\( detachedpointone \\) appear in the same order (clockwise or counterclockwise) as the corresponding angles \\( |41|,|34| \\), and \\( |13| \\) at \\( detachedpoint \\). This is hard to visualize, but if it were not so, the faces of the tetrahedron would be asymmetrically partitioned into two classes, which is clearly impossible in a symmetrical situation."
    },
    "garbled_string": {
      "map": {
        "P_i": "mqtzwuvs",
        "Q_i": "nlrpfyga",
        "P_1": "sldfkjwe",
        "Q_1": "kwerjvbm",
        "P_l": "xcnvbrte",
        "Q_k": "plkshdub",
        "Q_l": "zbvtyqpo",
        "i": "qzxwvtnp",
        "j": "hjgrksla",
        "k": "mndryxqo",
        "l": "vpshakyt"
      },
      "question": "3. A sphere is inscribed in a tetrahedron and each point of contact of the sphere with the four faces is joined to the vertices of the face containing the point. Show that the four sets of three angles so formed are identical.",
      "solution": "Solution. Let the vertices of the tetrahedron be \\( mqtzwuvs, qzxwvtnp=1,2,3,4 \\), and let \\( nlrpfyga \\) be the point of contact of the sphere with the face opposite \\( mqtzwuvs \\). We adopt the convention that when \\( qzxwvtnp, hjgrksla, mndryxqo, vpshakyt \\) appear in a statement, they represent distinct elements of \\( \\{1,2,3,4\\} \\).\n\nNow \\( mqtzwuvs nlrpfyga \\) and \\( mqtzwuvs plkshdub \\) are tangents to the sphere from the same external point; hence \\( \\left|mqtzwuvs nlrpfyga\\right|=\\left|mqtzwuvs plkshdub\\right| \\). Similarly, \\( \\left|mqtzwuvs nlrpfyga\\right|=\\left|sldfkjwe plkshdub\\right| \\). Hence, \\( \\Delta sldfkjwe zbvtyqpo xcnvbrte \\cong \\Delta mqtzwuvs plkshdub xcnvbrte \\) by s.s.s. Therefore, \\( \\angle mqtzwuvs nlrpfyga xcnvbrte=\\angle mqtzwuvs plkshdub xcnvbrte \\). We denote this angle by \\( |qzxwvtnp vpshakyt| \\). Clearly we have\n\\[\n|qzxwvtnp vpshakyt|=|vpshakyt qzxwvtnp| .\n\\]\n\nSince the angles at \\( nlrpfyga \\) add to \\( 2 \\pi \\), we have\n\\[\n\\begin{array}{l}\n|23|+|34|+|42|=2 \\pi, \\\\\n|34|+|41|+|13|=2 \\pi, \\\\\n|41|+|12|+|24|=2 \\pi, \\\\\n|12|+|23|+|31|=2 \\pi .\n\\end{array}\n\\]\n\nIf we add the first two of these equations, subtract the third and fourth, and use (1), we obtain \\( 2 \\cdot|34|-2 \\cdot|12|=0 \\). Hence, \\( |12|=|34| \\), and by symmetry\n\\[\n|\\ddot{qzxwvtnp}|=|mndryxqo vpshakyt| .\n\\]\n\nThe angles at \\( kwerjvbm \\) are \\( |23|,|34| \\), and \\( |42| \\), and these are respectively equal to the angles at \\( Q_{2} \\), namely, \\( |41|,|34| \\), and \\( |13| \\), by (2). By symmetry, the central angles are the same in all four faces.\n\nRemark. The angles appear with the same orientation in each face. Thus, when viewed from outside the tetrahedron, the angles |23|, |34|, and \\( |42| \\) at \\( kwerjvbm \\) appear in the same order (clockwise or counterclockwise) as the corresponding angles \\( |41|,|34| \\), and \\( |13| \\) at \\( Q_{2} \\). This is hard to visualize, but if it were not so, the faces of the tetrahedron would be asymmetrically partitioned into two classes, which is clearly impossible in a symmetrical situation."
    },
    "kernel_variant": {
      "question": "A sphere is inscribed in a tetrahedron with vertices V₀ ,V₁ ,V₂ ,V₃.  For i = 0,1,2,3 let Tᵢ be the point where the sphere touches the face opposite Vᵢ.  From every Tᵢ draw the three straight segments joining Tᵢ to the vertices of its face; this partitions the face into three planar angles.  Angles are to be measured in gradians (400 g = one full revolution).  Prove that the **set** of three angles occurring on a face is the same for all four faces of the tetrahedron.",
      "solution": "Throughout we measure angles in gradians; thus a right angle is 100 g and a full revolution 400 g.  Whenever i, j, k, ℓ occur together they are distinct elements of {0,1,2,3}.\n\n1.  Definition of the bracket  ⟨ij⟩\n   ------------------------------------------------\n   For two distinct indices i, j choose any k ≠ i, j and put\n          ⟨ij⟩  :=  ∠ Vᵢ T_k Vⱼ             (1)\n   (the angle at the point of contact situated on the face not containing Vᵢ and Vⱼ).\n\n2.  Well–definition of ⟨ij⟩\n   ------------------------------------------------\n   Let k, ℓ be the two indices different from i, j.  Because VᵢT_k and VᵢT_ℓ are tangents drawn from the same external point Vᵢ to the inscribed sphere we have |VᵢT_k| = |VᵢT_ℓ|.  The same argument from Vⱼ gives |VⱼT_k| = |VⱼT_ℓ|, and the side VᵢVⱼ is common to the two triangles ΔVᵢT_kVⱼ and ΔVᵢT_ℓVⱼ.  By the side–side–side criterion the two triangles are congruent, hence ∠VᵢT_kVⱼ = ∠VᵢT_ℓVⱼ.  The right–hand side of (1) is therefore independent of the auxiliary index and ⟨ij⟩ is well defined.  Evidently ⟨ij⟩ = ⟨ji⟩.\n\n3.  A 400 g relation\n   ------------------------------------------------\n   Fix k.  At the contact point T_k the three angles around that point lie in the face VᵢVⱼV_ℓ and add up to one full revolution:\n          ⟨ij⟩ + ⟨jℓ⟩ + ⟨ℓi⟩ = 400 g.        (2)\n   Writing (2) for k = 0,1,2,3 we obtain the system\n      ⟨12⟩+⟨23⟩+⟨31⟩ = 400 g,\n      ⟨23⟩+⟨30⟩+⟨02⟩ = 400 g,\n      ⟨30⟩+⟨01⟩+⟨13⟩ = 400 g,\n      ⟨01⟩+⟨12⟩+⟨20⟩ = 400 g.              (3)\n\n4.  Equalities for opposite edges\n   ------------------------------------------------\n   Add the first two and subtract the last two equations in (3):\n          2⟨23⟩ − 2⟨01⟩ = 0 ⟹ ⟨01⟩ = ⟨23⟩.\n   By cyclic relabelling we further get\n          ⟨02⟩ = ⟨13⟩,     ⟨03⟩ = ⟨12⟩.      (4)\n   Thus the angles subtended by opposite edges are equal.\n\n5.  The three angles on the four faces\n   ------------------------------------------------\n   •  Face V₁V₂V₃ (contact point T₀).\n      Its three angles are\n            A₁ = ⟨12⟩,   A₂ = ⟨23⟩,   A₃ = ⟨31⟩.      (5)\n\n   •  Face V₀V₂V₃ (contact point T₁).\n      By definition these angles are\n            B₁ = ⟨02⟩,   B₂ = ⟨23⟩,   B₃ = ⟨30⟩.\n      Using (4) we have ⟨02⟩ = ⟨31⟩ and ⟨30⟩ = ⟨12⟩, so\n            {B₁,B₂,B₃} = {⟨31⟩,⟨23⟩,⟨12⟩} = {A₁,A₂,A₃}.      (6)\n\n   •  Face V₀V₁V₃ (contact point T₂).\n      Its angles are C₁ = ⟨01⟩, C₂ = ⟨13⟩, C₃ = ⟨30⟩.\n      From (4) we substitute ⟨01⟩ = ⟨23⟩, ⟨13⟩ = ⟨02⟩ = ⟨31⟩, and ⟨30⟩ = ⟨12⟩; hence\n            {C₁,C₂,C₃} = {⟨23⟩,⟨31⟩,⟨12⟩} = {A₁,A₂,A₃}.      (7)\n\n   •  Face V₀V₁V₂ (contact point T₃).\n      Its angles are D₁ = ⟨01⟩, D₂ = ⟨12⟩, D₃ = ⟨20⟩.\n      Again by (4) we have ⟨01⟩ = ⟨23⟩ and ⟨20⟩ = ⟨03⟩ = ⟨12⟩, so\n            {D₁,D₂,D₃} = {⟨23⟩,⟨12⟩,⟨31⟩} = {A₁,A₂,A₃}.      (8)\n\n   In every case the unordered triple of angles is {⟨12⟩,⟨23⟩,⟨31⟩}.  Therefore each face of the tetrahedron is dissected into three angles of exactly the same magnitudes.\n\n6.  Conclusion\n   ------------------------------------------------\n   The three planar angles determined on any face by the lines joining the point of tangency to the three vertices form the same multiset {⟨12⟩,⟨23⟩,⟨31⟩}.  Consequently the four faces of a tangential tetrahedron are partitioned into congruent triples of angles, as was to be shown.",
      "_meta": {
        "core_steps": [
          "Equal-tangent lemma: from the same vertex, all tangent segments to the inscribed sphere have equal length.",
          "SSS-congruence of the triangles formed by two such tangents and the edge between their far ends gives equal subtended angles |il|.",
          "Symmetry fact: angle |il| = |li|, so only unordered pairs of indices matter.",
          "Angle-sum at each contact point Qi is 2π, giving four linear equations in the six unknown angles |ij|.",
          "Solving these linear relations (add first two, subtract last two) forces all opposite-pair angles equal, so every face receives the same ordered triple of angles."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Units used for the angle sum around a point.",
            "original": "2π radians (could be 360°)"
          },
          "slot2": {
            "description": "Labels and index set for vertices/contact points.",
            "original": "Vertices P1…P4, contact points Q1…Q4, indices {1,2,3,4}"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}