{
  "index": "1963-B-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "4. Let \\( C \\) be a closed plane curve that has a continuously turning tangent and bounds a convex region. If \\( T \\) is a triangle inscribed in \\( C \\) with maximum perimeter, show that the normal to \\( C \\) at each vertex of \\( T \\) bisects the angle of \\( T \\) at that vertex. If a triangle \\( T \\) has the property just described, does it necessarily have maximum perimeter? What is the situation if \\( C \\) is a circle? (A convex region is a connected open set such that the line segment joining any two points of the set lies entirely in the set.)",
  "solution": "Solution. As in Problem A.M. 2 of the Third Competition, many of the hypotheses in this problem tend to obscure the generality and the simplicity of the result. \\( C \\) need not bound a convex region and the continuous turning of the tangent is irrelevant.\n\nWe first prove the following general fact, from which the assertion of the problem will quickly follow.\n\nTheorem. If \\( C \\) is any set of points in the plane and \\( T \\) is a triangle with vertices in \\( C \\) whose perimeter is maximal among all such triangles, then \\( C \\) has at each vertex of \\( T \\) a support line perpendicular to the angle bisector of \\( T \\) at that vertex.\nProof. Suppose \\( P, Q, R \\) are the vertices of \\( T \\) and let \\( l \\) be the line through \\( P \\) perpendicular to the bisector of \\( \\angle Q P R \\). If \\( Q^{*} \\) is the reflection of \\( Q \\) in \\( l \\) then it is immediate that \\( R, P, Q^{*} \\) are collinear.\n\nIf \\( X \\) is any point on the \\( Q^{*} \\)-side of \\( l \\), then \\( R Q+R X+X Q>R Q+ \\) \\( R X+X Q^{*} \\geq R Q+R Q^{*}=R Q+R P+P Q \\). Thus \\( \\Delta R X Q \\) has greater perimeter than \\( \\triangle P Q R \\) and \\( X \\) is not in \\( C \\). By definition, then, \\( l \\) is a support line of \\( C \\) as claimed.\n\nIn the case where \\( C \\) is a differentiable closed curve, the support lines of \\( C \\) are precisely the tangents to \\( C \\). (This is clear, but more details of this argument are given in the solution to A.M. 2, Third Competition.)\n\nIf \\( C \\) is approximately an equilateral triangle with slightly rounded corners, then the triangle with its vertices at the midpoint of the sides of \\( C \\) has the property that the normals to \\( C \\) bisect its angles, but it is clearly not of maximum perimeter.\n\nIf \\( C \\) is a circle, then standard compactness arguments show that there is a triangle inscribed in \\( C \\) having largest perimeter. This triangle has its angle bisectors concurrent at the center \\( O \\) of \\( C \\). Hence, in the figure; \\( \\widehat{Q S}=\\widehat{R S} \\), so \\( P Q=\\widehat{P R} \\). By symmetry \\( \\widehat{P Q}=\\widehat{Q R}=\\hat{R P} \\), so \\( \\triangle P Q R \\) is equilateral. Therefore:\n\nOf all triangles inscribed in a given circle the equilateral triangles (and no others) have maximum perimeter.",
  "vars": [
    "C",
    "T",
    "P",
    "Q",
    "R",
    "l",
    "X",
    "S",
    "O",
    "\\\\Delta"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "C": "curvepoints",
        "T": "trianglemax",
        "P": "vertexp",
        "Q": "vertexq",
        "R": "vertexr",
        "l": "supportline",
        "X": "testpoint",
        "S": "sympoint",
        "O": "circlecent",
        "\\Delta": "triangle"
      },
      "question": "4. Let \\( curvepoints \\) be a closed plane curve that has a continuously turning tangent and bounds a convex region. If \\( trianglemax \\) is a triangle inscribed in \\( curvepoints \\) with maximum perimeter, show that the normal to \\( curvepoints \\) at each vertex of \\( trianglemax \\) bisects the angle of \\( trianglemax \\) at that vertex. If a triangle \\( trianglemax \\) has the property just described, does it necessarily have maximum perimeter? What is the situation if \\( curvepoints \\) is a circle? (A convex region is a connected open set such that the line segment joining any two points of the set lies entirely in the set.)",
      "solution": "Solution. As in Problem A.M. 2 of the Third Competition, many of the hypotheses in this problem tend to obscure the generality and the simplicity of the result. \\( curvepoints \\) need not bound a convex region and the continuous turning of the tangent is irrelevant.\n\nWe first prove the following general fact, from which the assertion of the problem will quickly follow.\n\nTheorem. If \\( curvepoints \\) is any set of points in the plane and \\( trianglemax \\) is a triangle with vertices in \\( curvepoints \\) whose perimeter is maximal among all such triangles, then \\( curvepoints \\) has at each vertex of \\( trianglemax \\) a support line perpendicular to the angle bisector of \\( trianglemax \\) at that vertex.\n\nProof. Suppose \\( vertexp, vertexq, vertexr \\) are the vertices of \\( trianglemax \\) and let \\( supportline \\) be the line through \\( vertexp \\) perpendicular to the bisector of \\( \\angle vertexq \\; vertexp \\; vertexr \\). If \\( vertexq^{*} \\) is the reflection of \\( vertexq \\) in \\( supportline \\) then it is immediate that \\( vertexr, vertexp, vertexq^{*} \\) are collinear.\n\nIf \\( testpoint \\) is any point on the \\( vertexq^{*} \\)-side of \\( supportline \\), then \\( vertexr\\,vertexq + vertexr\\,testpoint + testpoint\\,vertexq > vertexr\\,vertexq + vertexr\\,testpoint + testpoint\\,vertexq^{*} \\geq vertexr\\,vertexq + vertexr\\,vertexq^{*} = vertexr\\,vertexq + vertexr\\,vertexp + vertexp\\,vertexq \\). Thus \\( triangle vertexr\\,testpoint\\,vertexq \\) has greater perimeter than \\( \\triangle vertexp\\,vertexq\\,vertexr \\) and \\( testpoint \\) is not in \\( curvepoints \\). By definition, then, \\( supportline \\) is a support line of \\( curvepoints \\) as claimed.\n\nIn the case where \\( curvepoints \\) is a differentiable closed curve, the support lines of \\( curvepoints \\) are precisely the tangents to \\( curvepoints \\). (This is clear, but more details of this argument are given in the solution to A.M. 2, Third Competition.)\n\nIf \\( curvepoints \\) is approximately an equilateral triangle with slightly rounded corners, then the triangle with its vertices at the midpoint of the sides of \\( curvepoints \\) has the property that the normals to \\( curvepoints \\) bisect its angles, but it is clearly not of maximum perimeter.\n\nIf \\( curvepoints \\) is a circle, then standard compactness arguments show that there is a triangle inscribed in \\( curvepoints \\) having largest perimeter. This triangle has its angle bisectors concurrent at the center \\( circlecent \\) of \\( curvepoints \\). Hence, in the figure; \\( \\widehat{vertexq\\,sympoint} = \\widehat{vertexr\\,sympoint} \\), so \\( vertexp vertexq = \\widehat{vertexp\\,vertexr} \\). By symmetry \\( \\widehat{vertexp\\,vertexq} = \\widehat{vertexq\\,vertexr} = \\hat{vertexr\\,vertexp} \\), so \\( \\triangle vertexp\\,vertexq\\,vertexr \\) is equilateral. Therefore:\n\nOf all triangles inscribed in a given circle the equilateral triangles (and no others) have maximum perimeter."
    },
    "descriptive_long_confusing": {
      "map": {
        "C": "lanterns",
        "T": "panorama",
        "P": "tapestry",
        "Q": "harboring",
        "R": "meadowed",
        "l": "copperart",
        "X": "journeys",
        "S": "galaxies",
        "O": "cascades",
        "\\Delta": "dragonfly"
      },
      "question": "4. Let \\( lanterns \\) be a closed plane curve that has a continuously turning tangent and bounds a convex region. If \\( panorama \\) is a triangle inscribed in \\( lanterns \\) with maximum perimeter, show that the normal to \\( lanterns \\) at each vertex of \\( panorama \\) bisects the angle of \\( panorama \\) at that vertex. If a triangle \\( panorama \\) has the property just described, does it necessarily have maximum perimeter? What is the situation if lanterns is a circle? (A convex region is a connected open set such that the line segment joining any two points of the set lies entirely in the set.)",
      "solution": "Solution. As in Problem A.M. 2 of the Third Competition, many of the hypotheses in this problem tend to obscure the generality and the simplicity of the result. \\( lanterns \\) need not bound a convex region and the continuous turning of the tangent is irrelevant.\n\nWe first prove the following general fact, from which the assertion of the problem will quickly follow.\n\nTheorem. If \\( lanterns \\) is any set of points in the plane and \\( panorama \\) is a triangle with vertices in \\( lanterns \\) whose perimeter is maximal among all such triangles, then \\( lanterns \\) has at each vertex of \\( panorama \\) a support line perpendicular to the angle bisector of \\( panorama \\) at that vertex.\nProof. Suppose \\( tapestry, harboring, meadowed \\) are the vertices of \\( panorama \\) and let \\( copperart \\) be the line through \\( tapestry \\) perpendicular to the bisector of \\( \\angle harboring\\ tapestry\\ meadowed \\). If \\( harboring^{*} \\) is the reflection of \\( harboring \\) in \\( copperart \\) then it is immediate that \\( meadowed, tapestry, harboring^{*} \\) are collinear.\n\nIf \\( journeys \\) is any point on the \\( harboring^{*} \\)-side of \\( copperart \\), then \\( meadowed harboring + meadowed journeys + journeys harboring > meadowed harboring + \\) \\( meadowed journeys + journeys harboring^{*} \\geq meadowed harboring + meadowed harboring^{*} = meadowed harboring + meadowed tapestry + tapestry harboring \\). Thus \\( dragonfly\\ meadowed\\ journeys\\ harboring \\) has greater perimeter than \\( \\triangle tapestry\\ harboring\\ meadowed \\) and \\( journeys \\) is not in \\( lanterns \\). By definition, then, \\( copperart \\) is a support line of \\( lanterns \\) as claimed.\n\nIn the case where \\( lanterns \\) is a differentiable closed curve, the support lines of \\( lanterns \\) are precisely the tangents to \\( lanterns \\). (This is clear, but more details of this argument are given in the solution to A.M. 2, Third Competition.)\n\nIf \\( lanterns \\) is approximately an equilateral triangle with slightly rounded corners, then the triangle with its vertices at the midpoint of the sides of \\( lanterns \\) has the property that the normals to \\( lanterns \\) bisect its angles, but it is clearly not of maximum perimeter.\n\nIf \\( lanterns \\) is a circle, then standard compactness arguments show that there is a triangle inscribed in \\( lanterns \\) having largest perimeter. This triangle has its angle bisectors concurrent at the center \\( cascades \\) of \\( lanterns \\). Hence, in the figure; \\( \\widehat{harboring\\ galaxies}=\\widehat{meadowed\\ galaxies} \\), so \\( tapestry harboring = \\widehat{tapestry meadowed} \\). By symmetry \\( \\widehat{tapestry harboring}=\\widehat{harboring meadowed}=\\hat{meadowed tapestry} \\), so \\( \\triangle tapestry\\ harboring\\ meadowed \\) is equilateral. Therefore:\n\nOf all triangles inscribed in a given circle the equilateral triangles (and no others) have maximum perimeter."
    },
    "descriptive_long_misleading": {
      "map": {
        "C": "straightset",
        "T": "circlefig",
        "P": "planarobj",
        "Q": "voidspot",
        "R": "hollowpt",
        "l": "curvedarc",
        "X": "emptypos",
        "S": "distantpt",
        "O": "edgepoint",
        "\\\\Delta": "squaresym"
      },
      "question": "4. Let \\( straightset \\) be a closed plane curve that has a continuously turning tangent and bounds a convex region. If \\( circlefig \\) is a triangle inscribed in \\( straightset \\) with maximum perimeter, show that the normal to \\( straightset \\) at each vertex of \\( circlefig \\) bisects the angle of \\( circlefig \\) at that vertex. If a triangle \\( circlefig \\) has the property just described, does it necessarily have maximum perimeter? What is the situation if \\( straightset \\) is a circle? (A convex region is a connected open set such that the line segment joining any two points of the set lies entirely in the set.)",
      "solution": "Solution. As in Problem A.M. 2 of the Third Competition, many of the hypotheses in this problem tend to obscure the generality and the simplicity of the result. \\( straightset \\) need not bound a convex region and the continuous turning of the tangent is irrelevant.\n\nWe first prove the following general fact, from which the assertion of the problem will quickly follow.\n\nTheorem. If \\( straightset \\) is any set of points in the plane and \\( circlefig \\) is a triangle with vertices in \\( straightset \\) whose perimeter is maximal among all such triangles, then \\( straightset \\) has at each vertex of \\( circlefig \\) a support line perpendicular to the angle bisector of \\( circlefig \\) at that vertex.\nProof. Suppose \\( planarobj, voidspot, hollowpt \\) are the vertices of \\( circlefig \\) and let \\( curvedarc \\) be the line through \\( planarobj \\) perpendicular to the bisector of \\( \\angle voidspot planarobj hollowpt \\). If \\( voidspot^{*} \\) is the reflection of \\( voidspot \\) in \\( curvedarc \\) then it is immediate that \\( hollowpt, planarobj, voidspot^{*} \\) are collinear.\n\nIf \\( emptypos \\) is any point on the \\( voidspot^{*} \\)-side of \\( curvedarc \\), then \\( hollowpt\\ voidspot+hollowpt\\ emptypos+emptypos\\ voidspot>hollowpt\\ voidspot+ \\) \\( hollowpt\\ emptypos+emptypos\\ voidspot^{*} \\geq hollowpt\\ voidspot+hollowpt\\ voidspot^{*}=hollowpt\\ voidspot+hollowpt\\ planarobj+planarobj\\ voidspot \\). Thus \\( squaresym\\ hollowpt\\ emptypos\\ voidspot \\) has greater perimeter than \\( \\triangle planarobj\\ voidspot\\ hollowpt \\) and \\( emptypos \\) is not in \\( straightset \\). By definition, then, \\( curvedarc \\) is a support line of \\( straightset \\) as claimed.\n\nIn the case where \\( straightset \\) is a differentiable closed curve, the support lines of \\( straightset \\) are precisely the tangents to \\( straightset \\). (This is clear, but more details of this argument are given in the solution to A.M. 2, Third Competition.)\n\nIf \\( straightset \\) is approximately an equilateral triangle with slightly rounded corners, then the triangle with its vertices at the midpoint of the sides of \\( straightset \\) has the property that the normals to \\( straightset \\) bisect its angles, but it is clearly not of maximum perimeter.\n\nIf \\( straightset \\) is a circle, then standard compactness arguments show that there is a triangle inscribed in \\( straightset \\) having largest perimeter. This triangle has its angle bisectors concurrent at the center \\( edgepoint \\) of \\( straightset \\). Hence, in the figure; \\( \\widehat{voidspot\\ distantpt}=\\widehat{hollowpt\\ distantpt} \\), so \\( planarobj\\ voidspot=\\widehat{planarobj\\ hollowpt} \\). By symmetry \\( \\widehat{planarobj\\ voidspot}=\\widehat{voidspot\\ hollowpt}=\\hat{hollowpt\\ planarobj} \\), so \\( \\triangle planarobj\\ voidspot\\ hollowpt \\) is equilateral. Therefore:\n\nOf all triangles inscribed in a given circle the equilateral triangles (and no others) have maximum perimeter."
    },
    "garbled_string": {
      "map": {
        "C": "qzxwvtnp",
        "T": "hjgrksla",
        "P": "mdfqzlwi",
        "Q": "rksugnye",
        "R": "blskjdqe",
        "l": "wthzearp",
        "X": "uvmijcka",
        "S": "ozgtrela",
        "O": "ncamdzfh",
        "\\Delta": "pquxrevo"
      },
      "question": "4. Let \\( qzxwvtnp \\) be a closed plane curve that has a continuously turning tangent and bounds a convex region. If \\( hjgrksla \\) is a triangle inscribed in \\( qzxwvtnp \\) with maximum perimeter, show that the normal to \\( qzxwvtnp \\) at each vertex of \\( hjgrksla \\) bisects the angle of \\( hjgrksla \\) at that vertex. If a triangle \\( hjgrksla \\) has the property just described, does it necessarily have maximum perimeter? What is the situation if \\( qzxwvtnp \\) is a circle? (A convex region is a connected open set such that the line segment joining any two points of the set lies entirely in the set.)",
      "solution": "Solution. As in Problem A.M. 2 of the Third Competition, many of the hypotheses in this problem tend to obscure the generality and the simplicity of the result. \\( qzxwvtnp \\) need not bound a convex region and the continuous turning of the tangent is irrelevant.\n\nWe first prove the following general fact, from which the assertion of the problem will quickly follow.\n\nTheorem. If \\( qzxwvtnp \\) is any set of points in the plane and \\( hjgrksla \\) is a triangle with vertices in \\( qzxwvtnp \\) whose perimeter is maximal among all such triangles, then \\( qzxwvtnp \\) has at each vertex of \\( hjgrksla \\) a support line perpendicular to the angle bisector of \\( hjgrksla \\) at that vertex.\n\nProof. Suppose \\( mdfqzlwi, rksugnye, blskjdqe \\) are the vertices of \\( hjgrksla \\) and let \\( wthzearp \\) be the line through \\( mdfqzlwi \\) perpendicular to the bisector of \\( \\angle rksugnye\\, mdfqzlwi\\, blskjdqe \\). If \\( rksugnye^{*} \\) is the reflection of \\( rksugnye \\) in \\( wthzearp \\) then it is immediate that \\( blskjdqe, mdfqzlwi, rksugnye^{*} \\) are collinear.\n\nIf \\( uvmijcka \\) is any point on the \\( rksugnye^{*} \\)-side of \\( wthzearp \\), then \\( blskjdqe rksugnye+blskjdqe uvmijcka+uvmijcka rksugnye>blskjdqe rksugnye+ \\) \\( blskjdqe uvmijcka+uvmijcka rksugnye^{*} \\geq blskjdqe rksugnye+blskjdqe rksugnye^{*}=blskjdqe rksugnye+blskjdqe mdfqzlwi+mdfqzlwi rksugnye \\). Thus \\( pquxrevo blskjdqe uvmijcka rksugnye \\) has greater perimeter than \\( \\triangle mdfqzlwi\\, rksugnye\\, blskjdqe \\) and \\( uvmijcka \\) is not in \\( qzxwvtnp \\). By definition, then, \\( wthzearp \\) is a support line of \\( qzxwvtnp \\) as claimed.\n\nIn the case where \\( qzxwvtnp \\) is a differentiable closed curve, the support lines of \\( qzxwvtnp \\) are precisely the tangents to \\( qzxwvtnp \\). (This is clear, but more details of this argument are given in the solution to A.M. 2, Third Competition.)\n\nIf \\( qzxwvtnp \\) is approximately an equilateral triangle with slightly rounded corners, then the triangle with its vertices at the midpoint of the sides of \\( qzxwvtnp \\) has the property that the normals to \\( qzxwvtnp \\) bisect its angles, but it is clearly not of maximum perimeter.\n\nIf \\( qzxwvtnp \\) is a circle, then standard compactness arguments show that there is a triangle inscribed in \\( qzxwvtnp \\) having largest perimeter. This triangle has its angle bisectors concurrent at the center \\( ncamdzfh \\) of \\( qzxwvtnp \\). Hence, in the figure; \\( \\widehat{rksugnye\\, ozgtrela}=\\widehat{blskjdqe\\, ozgtrela} \\), so \\( mdfqzlwi rksugnye=\\widehat{mdfqzlwi\\, blskjdqe} \\). By symmetry \\( \\widehat{mdfqzlwi\\, rksugnye}=\\widehat{rksugnye\\, blskjdqe}=\\hat{blskjdqe\\, mdfqzlwi} \\), so \\( \\triangle mdfqzlwi\\, rksugnye\\, blskjdqe \\) is equilateral. Therefore:\n\nOf all triangles inscribed in a given circle the equilateral triangles (and no others) have maximum perimeter."
    },
    "kernel_variant": {
      "question": "Let γ be a simple closed curve in the plane.  Among all triangles whose three vertices lie on γ, let ΔPQR be one having the largest possible perimeter.\n\na)  Fix a vertex, say P, and let m be the internal bisector of ∠QPR.  Show that the line ℓ through P that is perpendicular to m is a support-line of γ.  Consequently, if γ is differentiable at P, its inward normal coincides with the bisector m.\n\nb)  Suppose an inscribed triangle has the property that at each of its vertices the inward normal to γ bisects the corresponding interior angle.  Must such a triangle necessarily have maximal perimeter?  Either prove it must, or give a counter-example.\n\nc)  Answer parts (a) and (b) when γ is a circle.",
      "solution": "Throughout we call a line ℓ a support-line of γ if γ lies entirely in one of the two closed half–planes determined by ℓ and meets ℓ at least at one point.\n\nPart (a)\nExistence of ΔPQR.  Because γ is compact, the function (X,Y,Z)↦|XY|+|YZ|+|ZX| attains a maximum on γ×γ×γ.  Let (P,Q,R) realise that maximum and write α = ∠QPR.\n\nConstruction and elementary geometry.  Let m be the internal bisector of α, and let ℓ be the line through P perpendicular to m.  Reflect Q in ℓ and denote the image by Q*.  By symmetry ∠Q*PR equals ∠RPQ, so Q* lies on the ray PR beyond P – in fact R,P,Q* are collinear and P is between R and Q*.  (A quick way to see this is to notice that the composition “reflection in ℓ followed by reflection in the line PQ” is the rotation that carries the ray PQ onto PR; hence Q* lies on PR.)  Moreover PQ* = PQ.\n\nWhy ℓ is a support line.  Suppose, to obtain a contradiction, that γ meets the open half-plane H on the Q*–side of ℓ.  Choose X ∈ γ ∩ H.  Because ℓ is the perpendicular bisector of QQ* we have XQ > XQ*.  The triangle inequality then gives\n  |RX| + |XQ| + |RQ| > |RX| + |XQ*| + |RQ| ≥ |RR| + |RQ*| + |RQ| = |PR| + |PQ| + |RQ|,\nso the perimeter of ΔRXQ is strictly larger than that of ΔPQR – a contradiction.  Thus γ is contained in the closed half-plane that does not meet H, and ℓ is indeed a support-line of γ at P.\n\nIf γ is differentiable at P, it has a unique tangent there.  Because any support-line through a differentiable point must coincide with that tangent, ℓ is the tangent, and the inward unit normal is the bisector m.\n\nPart (b)\nThe converse is false.\n\nConstruction of γ.  Start with an equilateral triangle A  B  C of side length 1.  On each side replace a small open middle segment – say a segment of length 1/10 – by an outward circular arc of very large radius ρ ≫ 1 whose endpoints coincide with the deleted segment’s endpoints and whose centre lies on the inward normal to that side.  Join the three resulting arcs by the untouched portions of the original sides.  If the three radii are chosen equal, the final curve γ is C¹-smooth, simple, and arbitrarily close to the original equilateral triangle.\n\nThe special inscribed triangle.  Let M₁,M₂,M₃ be the midpoints of the three circular arcs; each Mi is the point of the i-th arc farthest from the original side.  By symmetry the tangents to γ at the three midpoints are still parallel to the corresponding original sides, so the inward normals are perpendicular to those sides.  Consequently the inward normal at Mi bisects ∠M_{i+1}MiM_{i−1}; in other words ΔM₁M₂M₃ satisfies the “normal bisects the angle’’ condition.\n\nWhy ΔM₁M₂M₃ is not maximal.  Because the arcs bulge outward, every Mi has been moved a distance ε ≈ ρ⁻¹ away from the side it replaced, whereas the endpoints of the same arc (which also lie on γ) have been moved by a far smaller amount (≈ ε²ρ/2) and remain much closer to the original vertices A,B,C.  Hence by sliding each vertex Mi slightly along its inward normal until it reaches one of the endpoints of the corresponding arc we obtain another inscribed triangle whose perimeter is strictly larger than that of ΔM₁M₂M₃.  Therefore ΔM₁M₂M₃ cannot be perimeter-maximising even though it enjoys the required normal–bisects–the–angle property.\n\nPart (c)  (γ a circle)\nLet γ be the circle with centre O and radius r.  For every point P ∈ γ the support-line is the tangent at P, which is perpendicular to OP.  Part (a) then implies that in any perimeter-maximising ΔPQR the internal bisector of each angle goes through O.  Hence all three bisectors concur at O; the triangle is therefore equiangular and so equilateral.  Conversely, an inscribed equilateral triangle plainly satisfies the normal-bisects-the-angle condition and, by the argument just given, must have maximal perimeter.  Thus:\n\nAmong all triangles inscribed in a circle, the equilateral ones – and only those – maximise the perimeter.",
      "_meta": {
        "core_steps": [
          "At a vertex P of a perimeter–maximizing triangle, draw the line ℓ through P perpendicular to the interior angle-bisector; reflect the adjacent vertex Q in ℓ, obtaining Q* with R, P, Q* collinear.",
          "Any point X of C lying on the Q*-side of ℓ would make triangle R-X-Q longer than triangle P-Q-R, contradicting maximality; hence C contains no such points and ℓ is a support line of C.",
          "Thus every vertex of a longest-perimeter triangle has a support line perpendicular to its angle-bisector.",
          "For a smooth curve, support lines coincide with tangents; therefore the tangent is perpendicular to the bisector and the normal (the line through the vertex perpendicular to the tangent) lies along the bisector.",
          "Circle case: tangents are perpendicular to radii, so all bisectors meet at the center; equal central angles give an equilateral triangle, while a rounded-triangle example shows the bisector/normal condition alone does not guarantee maximal perimeter."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The requirement that C bound a convex region; convexity is never used in the proof.",
            "original": "bounds a convex region"
          },
          "slot2": {
            "description": "The assumption that the tangent to C turns continuously; smoothness is unnecessary except where tangents are explicitly invoked.",
            "original": "has a continuously turning tangent"
          },
          "slot3": {
            "description": "The statement that C is closed; compactness is only needed to guarantee existence of a maximizing triangle, not for the support-line argument itself.",
            "original": "closed plane curve"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}