{
  "index": "1982-B-1",
  "type": "GEO",
  "tag": [
    "GEO"
  ],
  "difficulty": "",
  "question": "Problem B-1\nLet \\( M \\) be the midpoint of side \\( B C \\) of a general \\( \\triangle A B C \\). Using the smallest possible \\( n \\), describe a method for cutting \\( \\triangle A M B \\) into \\( n \\) triangles which can be reassembled to form a triangle congruent to \\( \\triangle A M C \\).",
  "solution": "B-1.\nThe smallest \\( n \\) is 2 . Let \\( D \\) be the midpoint of side \\( A B \\). Cut \\( \\triangle A M B \\) along \\( D M \\). Then \\( \\triangle B M D \\) can be placed alongside \\( \\triangle A D M \\), with side \\( B D \\) atop side \\( A D \\), so as to form a triangle congruent to \\( \\triangle A M C \\). Since \\( \\triangle A M B \\) need not be congruent to \\( \\triangle A M C \\) in a general \\( \\triangle A B C \\), there is no method with \\( n=1 \\).",
  "vars": [
    "A",
    "B",
    "C",
    "M",
    "D"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "vertexa",
        "B": "vertexb",
        "C": "vertexc",
        "M": "midpointbc",
        "D": "midpointab",
        "n": "piececount"
      },
      "question": "Problem B-1\nLet \\( midpointbc \\) be the midpoint of side \\( vertexb vertexc \\) of a general \\( \\triangle vertexa vertexb vertexc \\). Using the smallest possible \\( piececount \\), describe a method for cutting \\( \\triangle vertexa midpointbc vertexb \\) into \\( piececount \\) triangles which can be reassembled to form a triangle congruent to \\( \\triangle vertexa midpointbc vertexc \\).",
      "solution": "B-1.\nThe smallest \\( piececount \\) is 2 . Let \\( midpointab \\) be the midpoint of side \\( vertexa vertexb \\). Cut \\( \\triangle vertexa midpointbc vertexb \\) along \\( midpointab midpointbc \\). Then \\( \\triangle vertexb midpointbc midpointab \\) can be placed alongside \\( \\triangle vertexa midpointab midpointbc \\), with side \\( vertexb midpointab \\) atop side \\( vertexa midpointab \\), so as to form a triangle congruent to \\( \\triangle vertexa midpointbc vertexc \\). Since \\( \\triangle vertexa midpointbc vertexb \\) need not be congruent to \\( \\triangle vertexa midpointbc vertexc \\) in a general \\( \\triangle vertexa vertexb vertexc \\), there is no method with \\( piececount=1 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "pebblestone",
        "B": "driftwood",
        "C": "marigolds",
        "M": "windchimes",
        "D": "flashlight",
        "n": "buttercup"
      },
      "question": "Problem B-1\nLet \\( windchimes \\) be the midpoint of side \\( driftwood marigolds \\) of a general \\( \\triangle pebblestone driftwood marigolds \\). Using the smallest possible \\( buttercup \\), describe a method for cutting \\( \\triangle pebblestone windchimes driftwood \\) into \\( buttercup \\) triangles which can be reassembled to form a triangle congruent to \\( \\triangle pebblestone windchimes marigolds \\).",
      "solution": "B-1.\nThe smallest \\( buttercup \\) is 2 . Let \\( flashlight \\) be the midpoint of side \\( pebblestone driftwood \\). Cut \\( \\triangle pebblestone windchimes driftwood \\) along \\( flashlight windchimes \\). Then \\( \\triangle driftwood windchimes flashlight \\) can be placed alongside \\( \\triangle pebblestone flashlight windchimes \\), with side \\( driftwood flashlight \\) atop side \\( pebblestone flashlight \\), so as to form a triangle congruent to \\( \\triangle pebblestone windchimes marigolds \\). Since \\( \\triangle pebblestone windchimes driftwood \\) need not be congruent to \\( \\triangle pebblestone windchimes marigolds \\) in a general \\( \\triangle pebblestone driftwood marigolds \\), there is no method with \\( buttercup=1 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "abysspoint",
        "B": "gulfpoint",
        "C": "voidpoint",
        "M": "endpoint",
        "D": "edgepoint",
        "n": "infinitecount"
      },
      "question": "Problem B-1\nLet \\( endpoint \\) be the midpoint of side \\( gulfpoint voidpoint \\) of a general \\( \\triangle abysspoint gulfpoint voidpoint \\). Using the smallest possible \\( infinitecount \\), describe a method for cutting \\( \\triangle abysspoint endpoint gulfpoint \\) into \\( infinitecount \\) triangles which can be reassembled to form a triangle congruent to \\( \\triangle abysspoint endpoint voidpoint \\).",
      "solution": "B-1.\nThe smallest \\( infinitecount \\) is 2 . Let \\( edgepoint \\) be the midpoint of side \\( abysspoint gulfpoint \\). Cut \\( \\triangle abysspoint endpoint gulfpoint \\) along \\( edgepoint endpoint \\). Then \\( \\triangle gulfpoint endpoint edgepoint \\) can be placed alongside \\( \\triangle abysspoint edgepoint endpoint \\), with side \\( gulfpoint edgepoint \\) atop side \\( abysspoint edgepoint \\), so as to form a triangle congruent to \\( \\triangle abysspoint endpoint voidpoint \\). Since \\( \\triangle abysspoint endpoint gulfpoint \\) need not be congruent to \\( \\triangle abysspoint endpoint voidpoint \\) in a general \\( \\triangle abysspoint gulfpoint voidpoint \\), there is no method with \\( infinitecount=1 \\)."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "xvmnbeor",
        "M": "yplsdkqh",
        "D": "rcfgnuke",
        "n": "wqjmdaho"
      },
      "question": "Problem B-1\nLet \\( yplsdkqh \\) be the midpoint of side \\( hjgrksla xvmnbeor \\) of a general \\( \\triangle qzxwvtnp hjgrksla xvmnbeor \\). Using the smallest possible \\( wqjmdaho \\), describe a method for cutting \\( \\triangle qzxwvtnp yplsdkqh hjgrksla \\) into \\( wqjmdaho \\) triangles which can be reassembled to form a triangle congruent to \\( \\triangle qzxwvtnp yplsdkqh xvmnbeor \\).",
      "solution": "B-1.\nThe smallest \\( wqjmdaho \\) is 2 . Let \\( rcfgnuke \\) be the midpoint of side \\( qzxwvtnp hjgrksla \\). Cut \\( \\triangle qzxwvtnp yplsdkqh hjgrksla \\) along \\( rcfgnuke yplsdkqh \\). Then \\( \\triangle hjgrksla yplsdkqh rcfgnuke \\) can be placed alongside \\( \\triangle qzxwvtnp rcfgnuke yplsdkqh \\), with side \\( hjgrksla rcfgnuke \\) atop side \\( qzxwvtnp rcfgnuke \\), so as to form a triangle congruent to \\( \\triangle qzxwvtnp yplsdkqh xvmnbeor \\). Since \\( \\triangle qzxwvtnp yplsdkqh hjgrksla \\) need not be congruent to \\( \\triangle qzxwvtnp yplsdkqh xvmnbeor \\) in a general \\( \\triangle qzxwvtnp hjgrksla xvmnbeor \\), there is no method with \\( wqjmdaho=1 \\)."
    },
    "kernel_variant": {
      "question": "Let \\(\\triangle PQR\\) be an arbitrary non-isosceles triangle and let \\(M\\) be the midpoint of side \\(QR\\).  How many triangles are needed, in the smallest possible dissection, to cut the half-triangle \\(\\triangle PMQ\\) into pieces that can be reassembled by rigid motions to form a triangle congruent to the complementary half-triangle \\(\\triangle PMR\\)?  Determine the least positive integer \\(n\\) and describe explicitly a dissection with exactly \\(n\\) pieces that achieves this minimum.",
      "solution": "Answer: \\(n = 2\\).\n\n1.  Why one piece is impossible.\n   In a general triangle the two halves \\(\\triangle PMQ\\) and \\(\\triangle PMR\\) are not congruent ( for example \\(\\angle QPM \\neq \\angle RPM\\) unless \\(PQ = PR\\) ).  Hence no single rigid motion can transform one into the other, so \\(n=1\\) is impossible.\n\n2.  Choosing an auxiliary midpoint and making the cut.\n   Let \\(D\\) be the midpoint of side \\(PQ\\).  Join \\(D\\) to \\(M\\); the segment \\(DM\\) lies entirely inside \\(\\triangle PMQ\\).  Cutting along \\(DM\\) divides \\(\\triangle PMQ\\) into the two smaller triangles\n   \\[\n        \\triangle PDM \\quad\\text{and}\\quad \\triangle QMD.\n   \\]\n   Because \\(D\\) and \\(M\\) are midpoints we have\n   \\[\n       PD = DQ \\quad\\text{and}\\quad DM \\parallel PR \\quad\\text{(mid-segment theorem).}\n   \\]\n\n3.  A half-turn of \\(\\triangle QMD\\).\n   Perform a half-turn (a rotation through \\(180^{\\circ}\\)) about the point \\(D\\).  Under this isometry\n   \\[\n       Q \\mapsto P,\\qquad M \\mapsto R',\\qquad D \\mapsto D,     \\tag{1}\n   \\]\n   where \\(R'\\) is the image of \\(M\\).  Because the rotation is an isometry, the image of \\(\\triangle QMD\\) is the triangle \\(\\triangle PR'D\\).\n\n4.  What the two pieces now look like.\n   Leaving \\(\\triangle PDM\\) unchanged and replacing \\(\\triangle QMD\\) by its image \\(\\triangle PR'D\\), the two pieces fit together along their common side \\(PD\\) (this side is exactly where they overlap after the rotation), producing the single triangle\n   \\[\n       \\triangle PMR'.\n   \\]\n   Indeed the boundary of the combined figure is the broken line\n   \\(P\\to M\\to R'\\to P\\), so the union is precisely \\(\\triangle PMR'.\\)\n\n5.  Showing that \\(\\triangle PMR' \\cong \\triangle PMR\\).\n   After the half-turn we have \\(D\\) as the midpoint of \\(MR'\\) (because \\(D\\) is the centre of the half-turn), so\n   \\[\n       MR' = 2\\,DM.   \\tag{2}\n   \\]\n   Since \\(DM\\parallel PR\\) and, from the mid-segment theorem, \\(DM = \\tfrac12\\,PR\\), equation (2) yields\n   \\[\n       MR' = PR\\quad\\text{and}\\quad MR'\\parallel PR. \\tag{3}\n   \\]\n   Now compare \\(\\triangle PMR'\\) with \\(\\triangle PMR\\):\n   • they share the side \\(PM\\);\n   • by (3) the sides \\(MR'\\) and \\(PR\\) are equal and parallel; hence the angle between \\(PM\\) and \\(MR'\\) equals the angle between \\(PM\\) and \\(PR\\).\n\n   Therefore the two triangles are congruent by the SAS criterion (common side \\(PM\\), equal side \\(MR' = PR\\), and equal included angles).  Consequently the two dissected pieces of \\(\\triangle PMQ\\) really do reassemble to a triangle congruent to \\(\\triangle PMR\\).\n\n6.  Minimality of \\(n=2\\).\n   We have shown that two pieces suffice, and step 1 proved that one piece does not.  Hence the least possible number of pieces is \\(n = 2\\).\n\nExplicit construction.\n   Cut \\(\\triangle PMQ\\) once along the segment joining the midpoints \\(D\\) of \\(PQ\\) and \\(M\\) of \\(QR\\).  Rotate the piece \\(\\triangle QMD\\) through \\(180^{\\circ}\\) about \\(D\\) and place it alongside \\(\\triangle PDM\\); the two fit exactly along \\(PD\\) and form a triangle congruent to \\(\\triangle PMR\\).",
      "_meta": {
        "core_steps": [
          "Note n = 1 is impossible because △AMB is not generally congruent to △AMC.",
          "Pick the midpoint D of a second side (here AB).",
          "Cut △AMB along DM, the segment that joins the two midpoints M and D.",
          "Use the Midpoint Theorem (DM ∥ AC and AD = DB) to ensure the two pieces match edge-to-edge.",
          "Glue △BMD to △ADM along AD (=DB) to obtain a triangle congruent to △AMC, so the minimal n is 2."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Side of the original triangle whose midpoint is named M.",
            "original": "BC"
          },
          "slot2": {
            "description": "Second side from which a midpoint is taken to create the cut segment.",
            "original": "AB"
          },
          "slot3": {
            "description": "Letter labels assigned to the triangle’s vertices (pure notation).",
            "original": "A, B, C"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}