{
  "index": "2019-A-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "In the triangle $\\triangle ABC$, let $G$ be the centroid, and let $I$ be the center of the inscribed circle.\nLet $\\alpha$ and $\\beta$ be the angles at the vertices $A$ and $B$, respectively.\nSuppose that the segment $IG$ is parallel to $AB$ and that $\\beta = 2 \\tan^{-1} (1/3)$. Find $\\alpha$.",
  "solution": "\\noindent \\textbf{Solution 1.}\nLet $M$ and $D$ denote the midpoint of $AB$ and the foot of the altitude from $C$ to $AB$, respectively, and let $r$ be the inradius of $\\bigtriangleup ABC$. Since $C,G,M$ are collinear with $CM = 3GM$, the distance from $C$ to line $AB$ is $3$ times the distance from $G$ to $AB$, and the latter is $r$ since $IG \\parallel AB$; hence the altitude $CD$ has length $3r$. By the double angle formula for tangent, $\\frac{CD}{DB} = \\tan\\beta = \\frac{3}{4}$, and so $DB = 4r$. Let $E$ be the point where the incircle meets $AB$; then $EB = r/\\tan(\\frac{\\beta}{2}) = 3r$. It follows that $ED = r$, whence the incircle is tangent to the altitude $CD$. This implies that $D=A$, $ABC$ is a right triangle, and $\\alpha = \\frac{\\pi}{2}$.\n\n\\noindent\n\\textbf{Remark.}\nOne can obtain a similar solution by fixing a coordinate system with $B$ at the origin and $A$ on the positive \n$x$-axis. Since $\\tan \\frac{\\beta}{2} = \\frac{1}{3}$, we may assume without loss of generality that\n$I = (3,1)$. Then $C$ lies on the intersection of the line $y=3$ (because $CD = 3r$ as above) \nwith the line $y = \\frac{3}{4} x$ (because $\\tan \\beta = \\frac{3}{4}$ as above), forcing $C = (4,3)$ and so forth.\n\n\\noindent \\textbf{Solution 2.}\nLet $a,b,c$ be the lengths of $BC,CA,AB$, respectively.\nLet $r$, $s$, and $K$ denote the inradius, semiperimeter, and area of $\\triangle ABC$. \nBy Heron's Formula,\n\\[\nr^2s^2 = K^2 = s(s-a)(s-b)(s-c).\n\\]\n\nIf $IG$ is parallel to $AB$, then \n\\[\n\\frac{1}{2} rc = \n\\mathrm{area}(\\triangle ABI) = \\mathrm{area}(\\triangle ABG) = \\frac{1}{3} K = \\frac{1}{3} rs\n\\]\nand so $c = \\frac{a+b}{2}$. Since $s = \\frac{3(a+b)}{4}$ and $s-c = \\frac{a+b}{4}$, we have \n$3r^2 = (s-a)(s-b)$. Let $E$ be the point at which the incircle meets $AB$; then $s-b = EB = r/\\tan(\\frac{\\beta}{2})$ and $s-a = EA = r/\\tan(\\frac{\\alpha}{2})$. It follows that $\\tan(\\frac{\\alpha}{2})\\tan(\\frac{\\beta}{2}) = \\frac{1}{3}$ and so $\\tan(\\frac{\\alpha}{2}) = 1$. This implies that $\\alpha = \\frac{\\pi}{2}$.\n\n\\noindent\n\\textbf{Remark.}\nThe equality $c = \\frac{a+b}{2}$ can also be derived from the vector representations\n\\[\nG = \\frac{A+B+C}{3}, \\qquad I = \\frac{aA+bB+cC}{a+b+c}.\n\\]\n\n\n\\noindent\n\\textbf{Solution 3.}\n(by Catalin Zara)\nIt is straightforward to check that a right triangle with $AC = 3, AB = 4, BC = 5$ works. For example,\nin a coordinate system with $A = (0,0), B = (4,0), C = (0,3)$, we have\n\\[\nG = \\left(\\frac{4}{3}, 1 \\right), \\qquad \nI = (1, 1)\n\\]\nand for $D = (1,0)$, \n\\[\n\\tan \\frac{\\beta}{2} = \\frac{ID}{BD} = \\frac{1}{3}.\n\\]\nIt thus suffices to suggest that this example is unique up to similarity.\n\nLet $C'$ be the foot of the angle bisector at $C$. Then \n\\[\n\\frac{CI}{IC'} = \\frac{CA + CB}{AB}\n\\] \nand so $IG$ is parallel to $AB$ if and only if $CA + CB = 2AB$. We may assume without loss of generality that $A$ and $B$ are fixed, in which case this condition restricts $C$ to an ellipse with foci at $A$ and $B$.\nSince the angle $\\beta$ is also fixed, up to symmetry \n$C$ is further restricted to a half-line starting at $B$; this intersects the ellipse in a unique point.\n\n\\noindent \n\\textbf{Remark.}\nGiven that $CA + CB = 2AB$, one can also recover the ratio of side lengths using the law of cosines.",
  "vars": [
    "A",
    "B",
    "C",
    "G",
    "I",
    "M",
    "D",
    "E",
    "K",
    "a",
    "b",
    "c",
    "r",
    "s",
    "x",
    "y",
    "\\\\alpha",
    "\\\\beta"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "vertexa",
        "B": "vertexb",
        "C": "vertexc",
        "G": "centroid",
        "I": "incenter",
        "M": "midpoint",
        "D": "footaltitude",
        "E": "touchpoint",
        "K": "areaval",
        "a": "lengthbc",
        "b": "lengthca",
        "c": "lengthab",
        "r": "inradius",
        "s": "semiperimeter",
        "x": "coordx",
        "y": "coordy",
        "\\alpha": "vertexanglea",
        "\\beta": "vertexangleb"
      },
      "question": "In the triangle $\\triangle vertexavertexbvertexc$, let $centroid$ be the centroid, and let $incenter$ be the center of the inscribed circle.\nLet $vertexanglea$ and $vertexangleb$ be the angles at the vertices $vertexa$ and $vertexb$, respectively.\nSuppose that the segment $incenter centroid$ is parallel to $vertexavertexb$ and that $vertexangleb = 2 \\tan^{-1} (1/3)$. Find $vertexanglea$.",
      "solution": "\\noindent \\textbf{Solution 1.}\nLet $midpoint$ and $footaltitude$ denote the midpoint of $vertexavertexb$ and the foot of the altitude from $vertexc$ to $vertexavertexb$, respectively, and let $inradius$ be the inradius of $\\bigtriangleup vertexavertexbvertexc$. Since $vertexc,centroid,midpoint$ are collinear with $vertexcmidpoint = 3 centroidmidpoint$, the distance from $vertexc$ to line $vertexavertexb$ is $3$ times the distance from $centroid$ to $vertexavertexb$, and the latter is $inradius$ since $incenter centroid \\parallel vertexavertexb$; hence the altitude $vertexcfootaltitude$ has length $3 inradius$. By the double angle formula for tangent, $\\frac{vertexcfootaltitude}{footaltitudevertexb} = \\tan vertexangleb = \\frac{3}{4}$, and so $footaltitudevertexb = 4 inradius$. Let $touchpoint$ be the point where the incircle meets $vertexavertexb$; then $touchpoint vertexb = inradius/\\tan(\\frac{vertexangleb}{2}) = 3 inradius$. It follows that $touchpoint footaltitude = inradius$, whence the incircle is tangent to the altitude $vertexcfootaltitude$. This implies that $footaltitude = vertexa$, $vertexavertexbvertexc$ is a right triangle, and $vertexanglea = \\frac{\\pi}{2}$.\n\n\\noindent\n\\textbf{Remark.}\nOne can obtain a similar solution by fixing a coordinate system with $vertexb$ at the origin and $vertexa$ on the positive \n$coordx$-axis. Since $\\tan \\frac{vertexangleb}{2} = \\frac{1}{3}$, we may assume without loss of generality that\n$incenter = (3,1)$. Then $vertexc$ lies on the intersection of the line $coordy=3$ (because $vertexcfootaltitude = 3 inradius$ as above) \nwith the line $coordy = \\frac{3}{4} coordx$ (because $\\tan vertexangleb = \\frac{3}{4}$ as above), forcing $vertexc = (4,3)$ and so forth.\n\n\\noindent \\textbf{Solution 2.}\nLet $lengthbc,lengthca,lengthab$ be the lengths of $vertexbvertexc,vertexcvertexa,vertexavertexb$, respectively.\nLet $inradius$, $semiperimeter$, and $areaval$ denote the inradius, semiperimeter, and area of $\\triangle vertexavertexbvertexc$. \nBy Heron's Formula,\n\\[\ninradius^2 semiperimeter^2 = areaval^2 = semiperimeter(semiperimeter-lengthbc)(semiperimeter-lengthca)(semiperimeter-lengthab).\n\\]\n\nIf $incenter centroid$ is parallel to $vertexavertexb$, then \n\\[\n\\frac{1}{2} inradius lengthab = \n\\mathrm{area}(\\triangle vertexavertexbincenter) = \\mathrm{area}(\\triangle vertexavertexbcentroid) = \\frac{1}{3} areaval = \\frac{1}{3} inradius semiperimeter\n\\]\nand so $lengthab = \\frac{lengthbc+lengthca}{2}$. Since $semiperimeter = \\frac{3(lengthbc+lengthca)}{4}$ and $semiperimeter-lengthab = \\frac{lengthbc+lengthca}{4}$, we have \n$3inradius^2 = (semiperimeter-lengthbc)(semiperimeter-lengthca)$. Let $touchpoint$ be the point at which the incircle meets $vertexavertexb$; then $semiperimeter-lengthca = touchpoint vertexb = inradius/\\tan(\\frac{vertexangleb}{2})$ and $semiperimeter-lengthbc = touchpoint vertexa = inradius/\\tan(\\frac{vertexanglea}{2})$. It follows that $\\tan(\\frac{vertexanglea}{2})\\tan(\\frac{vertexangleb}{2}) = \\frac{1}{3}$ and so $\\tan(\\frac{vertexanglea}{2}) = 1$. This implies that $vertexanglea = \\frac{\\pi}{2}$.\n\n\\noindent\n\\textbf{Remark.}\nThe equality $lengthab = \\frac{lengthbc+lengthca}{2}$ can also be derived from the vector representations\n\\[\ncentroid = \\frac{vertexa+vertexb+vertexc}{3}, \\qquad incenter = \\frac{lengthbc vertexa + lengthca vertexb + lengthab vertexc}{lengthbc+lengthca+lengthab}.\n\\]\n\n\n\\noindent\n\\textbf{Solution 3.}\n(by Catalin Zara)\nIt is straightforward to check that a right triangle with $vertexcvertexa = 3, vertexavertexb = 4, vertexbvertexc = 5$ works. For example,\nin a coordinate system with $vertexa = (0,0), vertexb = (4,0), vertexc = (0,3)$, we have\n\\[\ncentroid = \\left(\\frac{4}{3}, 1 \\right), \\qquad \nincenter = (1, 1)\n\\]\nand for $footaltitude = (1,0)$, \n\\[\n\\tan \\frac{vertexangleb}{2} = \\frac{incenter footaltitude}{footaltitude vertexb} = \\frac{1}{3}.\n\\]\nIt thus suffices to suggest that this example is unique up to similarity.\n\nLet $vertexc'$ be the foot of the angle bisector at $vertexc$. Then \n\\[\n\\frac{vertexc incenter}{incenter vertexc'} = \\frac{vertexc vertexa + vertexc vertexb}{vertexavertexb}\n\\] \nand so $incenter centroid$ is parallel to $vertexavertexb$ if and only if $vertexc vertexa + vertexc vertexb = 2 vertexavertexb$. We may assume without loss of generality that $vertexa$ and $vertexb$ are fixed, in which case this condition restricts $vertexc$ to an ellipse with foci at $vertexa$ and $vertexb$.\nSince the angle $vertexangleb$ is also fixed, up to symmetry \n$vertexc$ is further restricted to a half-line starting at $vertexb$; this intersects the ellipse in a unique point.\n\n\\noindent \n\\textbf{Remark.}\nGiven that $vertexc vertexa + vertexc vertexb = 2 vertexavertexb$, one can also recover the ratio of side lengths using the law of cosines."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "pineapple",
        "B": "blueberry",
        "C": "strawberry",
        "G": "crocodile",
        "I": "elephant",
        "M": "kangaroo",
        "D": "flamingo",
        "E": "rhinoceros",
        "K": "hippopot",
        "a": "watermelon",
        "b": "persimmon",
        "c": "pomegranate",
        "r": "raspberry",
        "s": "tangerine",
        "x": "blackcurrant",
        "y": "elderberry",
        "\\alpha": "cappuccino",
        "\\beta": "cardamom"
      },
      "question": "In the triangle $\\triangle pineappleblueberrystrawberry$, let $crocodile$ be the centroid, and let $elephant$ be the center of the inscribed circle.\nLet $cappuccino$ and $cardamom$ be the angles at the vertices $pineapple$ and $blueberry$, respectively.\nSuppose that the segment $elephant crocodile$ is parallel to $pineappleblueberry$ and that $cardamom = 2 \\tan^{-1} (1/3)$. Find $cappuccino$.",
      "solution": "\\noindent \\textbf{Solution 1.}\nLet $kangaroo$ and $flamingo$ denote the midpoint of $pineappleblueberry$ and the foot of the altitude from $strawberry$ to $pineappleblueberry$, respectively, and let $raspberry$ be the inradius of $\\bigtriangleup pineappleblueberrystrawberry$. Since $strawberry,crocodile,kangaroo$ are collinear with $strawberry kangaroo = 3 crocodile kangaroo$, the distance from $strawberry$ to line $pineappleblueberry$ is $3$ times the distance from $crocodile$ to $pineappleblueberry$, and the latter is $raspberry$ since $elephant crocodile \\parallel pineappleblueberry$; hence the altitude $strawberry flamingo$ has length $3raspberry$. By the double angle formula for tangent, $\\frac{strawberry flamingo}{flamingo blueberry} = \\tan cardamom = \\frac{3}{4}$, and so $flamingo blueberry = 4 raspberry$. Let $rhinoceros$ be the point where the incircle meets $pineappleblueberry$; then $rhinoceros blueberry = raspberry/\\tan(\\frac{cardamom}{2}) = 3 raspberry$. It follows that $rhinoceros flamingo = raspberry$, whence the incircle is tangent to the altitude $strawberry flamingo$. This implies that $flamingo=pineapple$, $pineappleblueberrystrawberry$ is a right triangle, and $cappuccino = \\frac{\\pi}{2}$.\\n\\n\\noindent\\textbf{Remark.}\nOne can obtain a similar solution by fixing a coordinate system with $blueberry$ at the origin and $pineapple$ on the positive $blackcurrant$-axis. Since $\\tan \\frac{cardamom}{2} = \\frac{1}{3}$, we may assume without loss of generality that $elephant = (3,1)$. Then $strawberry$ lies on the intersection of the line $elderberry=3$ (because $strawberry flamingo = 3 raspberry$ as above) with the line $elderberry = \\frac{3}{4} blackcurrant$ (because $\\tan cardamom = \\frac{3}{4}$ as above), forcing $strawberry = (4,3)$ and so forth.\\n\\n\\noindent \\textbf{Solution 2.}\nLet $watermelon,persimmon,pomegranate$ be the lengths of $blueberrystrawberry,strawberrypineapple,pineappleblueberry$, respectively.\nLet $raspberry$, $tangerine$, and $hippopot$ denote the inradius, semiperimeter, and area of $\\triangle pineappleblueberrystrawberry$. \nBy Heron's Formula,\n\\[\nraspberry^2 tangerine^2 = hippopot^2 = tangerine(tangerine-watermelon)(tangerine-persimmon)(tangerine-pomegranate).\n\\]\nIf $elephant crocodile$ is parallel to $pineappleblueberry$, then \n\\[\n\\frac{1}{2} raspberry pomegranate = \n\\mathrm{area}(\\triangle pineappleblueberryelephant) = \\mathrm{area}(\\triangle pineappleblueberrycrocodile) = \\frac{1}{3} hippopot = \\frac{1}{3} raspberry tangerine\n\\]\nand so $pomegranate = \\frac{watermelon+persimmon}{2}$. Since $tangerine = \\frac{3(watermelon+persimmon)}{4}$ and $tangerine-pomegranate = \\frac{watermelon+persimmon}{4}$, we have $3 raspberry^2 = (tangerine-watermelon)(tangerine-persimmon)$. Let $rhinoceros$ be the point at which the incircle meets $pineappleblueberry$; then $tangerine-persimmon = rhinoceros blueberry = raspberry/\\tan(\\frac{cardamom}{2})$ and $tangerine-watermelon = rhinoceros pineapple = raspberry/\\tan(\\frac{cappuccino}{2})$. It follows that $\\tan(\\frac{cappuccino}{2})\\tan(\\frac{cardamom}{2}) = \\frac{1}{3}$ and so $\\tan(\\frac{cappuccino}{2}) = 1$. This implies that $cappuccino = \\frac{\\pi}{2}$.\\n\\n\\noindent\\textbf{Remark.}\nThe equality $pomegranate = \\frac{watermelon+persimmon}{2}$ can also be derived from the vector representations\n\\[\ncrocodile = \\frac{pineapple+blueberry+strawberry}{3}, \\qquad elephant = \\frac{watermelon pineapple + persimmon blueberry + pomegranate strawberry}{watermelon+persimmon+pomegranate}.\n\\]\\n\\n\\noindent \\textbf{Solution 3.}\n(by Catalin Zara)\nIt is straightforward to check that a right triangle with $strawberry pineapple = 3, pineappleblueberry = 4, blueberrystrawberry = 5$ works. For example,\nin a coordinate system with $pineapple = (0,0), blueberry = (4,0), strawberry = (0,3)$, we have\n\\[\ncrocodile = \\left(\\frac{4}{3}, 1 \\right), \\qquad \nelephant = (1, 1)\n\\]\nand for $flamingo = (1,0)$, \n\\[\n\\tan \\frac{cardamom}{2} = \\frac{elephant flamingo}{flamingo blueberry} = \\frac{1}{3}.\n\\]\nIt thus suffices to suggest that this example is unique up to similarity.\\n\\nLet $strawberry'$ be the foot of the angle bisector at $strawberry$. Then \n\\[\n\\frac{strawberry elephant}{elephant strawberry'} = \\frac{strawberrypineapple + strawberryblueberry}{pineappleblueberry}\n\\] \nand so $elephant crocodile$ is parallel to $pineappleblueberry$ if and only if $strawberrypineapple + strawberryblueberry = 2 pineappleblueberry$. We may assume without loss of generality that $pineapple$ and $blueberry$ are fixed, in which case this condition restricts $strawberry$ to an ellipse with foci at $pineapple$ and $blueberry$.\nSince the angle $cardamom$ is also fixed, up to symmetry \n$strawberry$ is further restricted to a half-line starting at $blueberry$; this intersects the ellipse in a unique point.\\n\\n\\noindent \\textbf{Remark.}\nGiven that $strawberrypineapple + strawberryblueberry = 2 pineappleblueberry$, one can also recover the ratio of side lengths using the law of cosines."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "edgepoint",
        "B": "sidepoint",
        "C": "interiorp",
        "G": "peripheral",
        "I": "excenter",
        "M": "extremept",
        "D": "summitpt",
        "E": "farpoint",
        "K": "perimeter",
        "a": "widthval",
        "b": "heightval",
        "c": "depthval",
        "r": "exradius",
        "s": "fullperim",
        "x": "verticalv",
        "y": "horizontal",
        "\\alpha": "suppangle",
        "\\beta": "compangle"
      },
      "question": "In the triangle $\\triangle edgepoint sidepoint interiorp$, let $peripheral$ be the centroid, and let $excenter$ be the center of the inscribed circle.\nLet $suppangle$ and $compangle$ be the angles at the vertices $edgepoint$ and $sidepoint$, respectively.\nSuppose that the segment $excenterperipheral$ is parallel to $edgepointsidepoint$ and that $compangle = 2 \\tan^{-1} (1/3)$. Find $suppangle$.",
      "solution": "\\noindent \\textbf{Solution 1.}\nLet $extremept$ and $summitpt$ denote the midpoint of $edgepointsidepoint$ and the foot of the altitude from $interiorp$ to $edgepointsidepoint$, respectively, and let $exradius$ be the inradius of $\\bigtriangleup edgepoint sidepoint interiorp$. Since $interiorp,peripheral,extremept$ are collinear with $interiorpextremept = 3\\,peripheralextremept$, the distance from $interiorp$ to line $edgepointsidepoint$ is $3$ times the distance from $peripheral$ to $edgepointsidepoint$, and the latter is $exradius$ since $excenterperipheral \\parallel edgepointsidepoint$; hence the altitude $interiorpsummitpt$ has length $3exradius$. By the double angle formula for tangent, $\\frac{interiorpsummitpt}{summitptsidepoint} = \\tan compangle = \\frac{3}{4}$, and so $summitptsidepoint = 4exradius$. Let $farpoint$ be the point where the incircle meets $edgepointsidepoint$; then $farpointsidepoint = exradius/\\tan(\\frac{compangle}{2}) = 3exradius$. It follows that $farpointsummitpt = exradius$, whence the incircle is tangent to the altitude $interiorpsummitpt$. This implies that $summitpt=edgepoint$, $edgepoint sidepoint interiorp$ is a right triangle, and $suppangle = \\frac{\\pi}{2}$.\n\n\\noindent\n\\textbf{Remark.}\nOne can obtain a similar solution by fixing a coordinate system with $sidepoint$ at the origin and $edgepoint$ on the positive $verticalv$-axis. Since $\\tan \\frac{compangle}{2} = \\frac{1}{3}$, we may assume without loss of generality that $excenter = (3,1)$. Then $interiorp$ lies on the intersection of the line $horizontal = 3$ (because $interiorpsummitpt = 3exradius$ as above) with the line $horizontal = \\frac{3}{4} verticalv$ (because $\\tan compangle = \\frac{3}{4}$ as above), forcing $interiorp = (4,3)$ and so forth.\n\n\\noindent \\textbf{Solution 2.}\nLet $widthval,heightval,depthval$ be the lengths of $sidepointinteriorp,interiorpedgepoint,edgepointsidepoint$, respectively.\nLet $exradius$, $fullperim$, and $perimeter$ denote the inradius, semiperimeter, and area of $\\triangle edgepoint sidepoint interiorp$. By Heron's Formula,\n\\[\nexradius^2 fullperim^2 = perimeter^2 = fullperim(fullperim-widthval)(fullperim-heightval)(fullperim-depthval).\n\\]\nIf $excenterperipheral$ is parallel to $edgepointsidepoint$, then\n\\[\n\\frac{1}{2} exradius depthval = \\mathrm{area}(\\triangle edgepoint sidepointexcenter) = \\mathrm{area}(\\triangle edgepoint sidepointperipheral) = \\frac{1}{3} perimeter = \\frac{1}{3} exradius fullperim\n\\]\nand so $depthval = \\frac{widthval+heightval}{2}$. Since $fullperim = \\frac{3(widthval+heightval)}{4}$ and $fullperim-depthval = \\frac{widthval+heightval}{4}$, we have $3exradius^2 = (fullperim-widthval)(fullperim-heightval)$. Let $farpoint$ be the point at which the incircle meets $edgepointsidepoint$; then $fullperim-heightval = farpointsidepoint = exradius/\\tan(\\frac{compangle}{2})$ and $fullperim-widthval = farpointedgepoint = exradius/\\tan(\\frac{suppangle}{2})$. It follows that $\\tan(\\frac{suppangle}{2})\\tan(\\frac{compangle}{2}) = \\frac{1}{3}$ and so $\\tan(\\frac{suppangle}{2}) = 1$. This implies that $suppangle = \\frac{\\pi}{2}$.\n\n\\noindent\n\\textbf{Remark.}\nThe equality $depthval = \\frac{widthval+heightval}{2}$ can also be derived from the vector representations\n\\[\nperipheral = \\frac{edgepoint+sidepoint+interiorp}{3}, \\qquad excenter = \\frac{widthval\\,edgepoint+heightval\\,sidepoint+depthval\\,interiorp}{widthval+heightval+depthval}.\n\\]\n\n\\noindent\n\\textbf{Solution 3.}\n(by Catalin Zara)\nIt is straightforward to check that a right triangle with $interiorpedgepoint = 3$, $edgepointsidepoint = 4$, $sidepointinteriorp = 5$ works. For example, in a coordinate system with $edgepoint = (0,0)$, $sidepoint = (4,0)$, $interiorp = (0,3)$, we have\n\\[\nperipheral = \\left(\\frac{4}{3}, 1 \\right), \\qquad excenter = (1, 1)\n\\]\nand for $summitpt = (1,0)$,\n\\[\n\\tan \\frac{compangle}{2} = \\frac{excentersummitpt}{summitptsidepoint} = \\frac{1}{3}.\n\\]\nIt thus suffices to suggest that this example is unique up to similarity.\n\nLet $interiorp'$ be the foot of the angle bisector at $interiorp$. Then\n\\[\n\\frac{interiorpexcenter}{excenterinteriorp'} = \\frac{interiorpedgepoint + sidepointinteriorp}{edgepointsidepoint}\n\\]\nand so $excenterperipheral$ is parallel to $edgepointsidepoint$ if and only if $interiorpedgepoint + sidepointinteriorp = 2edgepointsidepoint$. We may assume without loss of generality that $edgepoint$ and $sidepoint$ are fixed, in which case this condition restricts $interiorp$ to an ellipse with foci at $edgepoint$ and $sidepoint$. Since the angle $compangle$ is also fixed, up to symmetry $interiorp$ is further restricted to a half-line starting at $sidepoint$; this intersects the ellipse in a unique point.\n\n\\noindent \\textbf{Remark.}\nGiven that $interiorpedgepoint + sidepointinteriorp = 2edgepointsidepoint$, one can also recover the ratio of side lengths using the law of cosines."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "fbdqmnei",
        "G": "vkslomtr",
        "I": "zpecduaf",
        "M": "rstaglbn",
        "D": "ywokshre",
        "E": "cdnveqaz",
        "K": "jpribxto",
        "a": "flsvenop",
        "b": "gmrtiacw",
        "c": "nyoclefi",
        "r": "vugbkies",
        "s": "qmeinsod",
        "x": "lampruze",
        "y": "tokvsedf",
        "\\alpha": "pldrowqm",
        "\\beta": "snerivth"
      },
      "question": "In the triangle $\\triangle qzxwvtnp hjgrksla fbdqmnei$, let $vkslomtr$ be the centroid, and let $zpecduaf$ be the center of the inscribed circle.\nLet $pldrowqm$ and $snerivth$ be the angles at the vertices $qzxwvtnp$ and $hjgrksla$, respectively.\nSuppose that the segment $zpecduaf vkslomtr$ is parallel to $qzxwvtnp hjgrksla$ and that $snerivth = 2 \\tan^{-1} (1/3)$. Find $pldrowqm$.",
      "solution": "\\noindent \\textbf{Solution 1.}\nLet $rstaglbn$ and $ywokshre$ denote the midpoint of $qzxwvtnphjgrksla$ and the foot of the altitude from $fbdqmnei$ to $qzxwvtnphjgrksla$, respectively, and let $vugbkies$ be the inradius of $\\bigtriangleup qzxwvtnphjgrkslafbdqmnei$. Since $fbdqmnei,vkslomtr,rstaglbn$ are collinear with $fbdqmneirstaglbn = 3vkslomtrrstaglbn$, the distance from $fbdqmnei$ to line $qzxwvtnphjgrksla$ is $3$ times the distance from $vkslomtr$ to $qzxwvtnphjgrksla$, and the latter is $vugbkies$ since $zpecduaf vkslomtr \\parallel qzxwvtnphjgrksla$; hence the altitude $fbdqmneiywokshre$ has length $3vugbkies$. By the double angle formula for tangent, $\\frac{fbdqmneiywokshre}{ywokshrehjgrksla} = \\tan snerivth = \\frac{3}{4}$, and so $ywokshrehjgrksla = 4vugbkies$. Let $cdnveqaz$ be the point where the incircle meets $qzxwvtnphjgrksla$; then $cdnveqazhjgrksla = vugbkies/\\tan(\\frac{snerivth}{2}) = 3vugbkies$. It follows that $cdnveqaz ywokshre = vugbkies$, whence the incircle is tangent to the altitude $fbdqmneiywokshre$. This implies that $ywokshre=qzxwvtnp$, $qzxwvtnphjgrkslafbdqmnei$ is a right triangle, and $pldrowqm = \\frac{\\pi}{2}$.\n\n\\noindent\n\\textbf{Remark.}\nOne can obtain a similar solution by fixing a coordinate system with $hjgrksla$ at the origin and $qzxwvtnp$ on the positive $lampruze$-axis. Since $\\tan \\frac{snerivth}{2} = \\frac{1}{3}$, we may assume without loss of generality that $zpecduaf = (3,1)$. Then $fbdqmnei$ lies on the intersection of the line $tokvsedf=3$ (because $fbdqmneiywokshre = 3vugbkies$ as above) with the line $tokvsedf = \\frac{3}{4} lampruze$ (because $\\tan snerivth = \\frac{3}{4}$ as above), forcing $fbdqmnei = (4,3)$ and so forth.\n\n\\noindent \\textbf{Solution 2.}\nLet $flsvenop,gmrtiacw,nyoclefi$ be the lengths of $hjgrkslafbdqmnei, fbdqmnei qzxwvtnp, qzxwvtnphjgrksla$, respectively.\nLet $vugbkies$, $qmeinsod$, and $jpribxto$ denote the inradius, semiperimeter, and area of $\\triangle qzxwvtnphjgrkslafbdqmnei$. By Heron's Formula,\n\\[\nvugbkies^2 qmeinsod^2 = jpribxto^2 = qmeinsod(qmeinsod-flsvenop)(qmeinsod-gmrtiacw)(qmeinsod-nyoclefi).\n\\]\nIf $zpecduafvkslomtr$ is parallel to $qzxwvtnphjgrksla$, then \n\\[\n\\frac{1}{2} vugbkies nyoclefi = \\mathrm{area}(\\triangle qzxwvtnphjgrkslazpecduaf) = \\mathrm{area}(\\triangle qzxwvtnphjgrkslavkslomtr) = \\frac{1}{3} jpribxto = \\frac{1}{3} vugbkies qmeinsod\n\\]\nand so $nyoclefi = \\frac{flsvenop+gmrtiacw}{2}$. Since $qmeinsod = \\frac{3(flsvenop+gmrtiacw)}{4}$ and $qmeinsod-nyoclefi = \\frac{flsvenop+gmrtiacw}{4}$, we have $3vugbkies^2 = (qmeinsod-flsvenop)(qmeinsod-gmrtiacw)$. Let $cdnveqaz$ be the point at which the incircle meets $qzxwvtnphjgrksla$; then $qmeinsod-gmrtiacw = cdnveqazhjgrksla = vugbkies/\\tan(\\frac{snerivth}{2})$ and $qmeinsod-flsvenop = cdnveqazqzxwvtnp = vugbkies/\\tan(\\frac{pldrowqm}{2})$. It follows that $\\tan(\\frac{pldrowqm}{2})\\tan(\\frac{snerivth}{2}) = \\frac{1}{3}$ and so $\\tan(\\frac{pldrowqm}{2}) = 1$. This implies that $pldrowqm = \\frac{\\pi}{2}$.\n\n\\noindent\n\\textbf{Remark.}\nThe equality $nyoclefi = \\frac{flsvenop+gmrtiacw}{2}$ can also be derived from the vector representations\n\\[\nvkslomtr = \\frac{qzxwvtnp+hjgrksla+fbdqmnei}{3}, \\qquad zpecduaf = \\frac{flsvenop qzxwvtnp+gmrtiacw hjgrksla+nyoclefi fbdqmnei}{flsvenop+gmrtiacw+nyoclefi}.\n\\]\n\n\\noindent\n\\textbf{Solution 3.}\n(by Catalin Zara)\nIt is straightforward to check that a right triangle with $fbdqmnei qzxwvtnp = 3, qzxwvtnphjgrksla = 4, hjgrkslafbdqmnei = 5$ works. For example, in a coordinate system with $qzxwvtnp = (0,0), hjgrksla = (4,0), fbdqmnei = (0,3)$, we have\n\\[\nvkslomtr = \\left(\\frac{4}{3}, 1 \\right), \\qquad zpecduaf = (1, 1)\n\\]\nand for $ywokshre = (1,0)$, \n\\[\n\\tan \\frac{snerivth}{2} = \\frac{zpecduafywokshre}{ywokshrehjgrksla} = \\frac{1}{3}.\n\\]\nIt thus suffices to suggest that this example is unique up to similarity.\n\nLet $fbdqmnei'$ be the foot of the angle bisector at $fbdqmnei$. Then \n\\[\n\\frac{fbdqmneizpecduaf}{zpecduaffbdqmnei'} = \\frac{fbdqmnei qzxwvtnp + fbdqmnei hjgrksla}{qzxwvtnphjgrksla}\n\\] \nand so $zpecduafvkslomtr$ is parallel to $qzxwvtnphjgrksla$ if and only if $fbdqmnei qzxwvtnp + fbdqmnei hjgrksla = 2qzxwvtnphjgrksla$. We may assume without loss of generality that $qzxwvtnp$ and $hjgrksla$ are fixed, in which case this condition restricts $fbdqmnei$ to an ellipse with foci at $qzxwvtnp$ and $hjgrksla$. Since the angle $snerivth$ is also fixed, up to symmetry $fbdqmnei$ is further restricted to a half-line starting at $hjgrksla$; this intersects the ellipse in a unique point.\n\n\\noindent \\textbf{Remark.}\nGiven that $fbdqmnei qzxwvtnp + fbdqmnei hjgrksla = 2qzxwvtnphjgrksla$, one can also recover the ratio of side lengths using the law of cosines."
    },
    "kernel_variant": {
      "question": "Let $\\triangle ABC$ be a non-isosceles triangle with centroid $G$, incenter $I$ and circumcenter $O$.  \nPut  \n\\[\n\\alpha=\\angle A,\\qquad \n\\beta=\\angle B,\\qquad \n\\gamma=\\angle C=\\pi-\\alpha-\\beta .\n\\]\n\nAssume that  \n\n(i) $\\,IG\\parallel AB$;  \n\n(ii) the inradius $r$ and circumradius $R$ satisfy $\\, r=\\dfrac{R}{4}$.\n\nDetermine, in exact closed form, every ordered pair $(\\alpha,\\beta)$ that fulfils these two conditions, and prove that no further solutions exist.  \n\n\\bigskip",
      "solution": "Throughout we set  \n\\[\nu=\\tan\\frac{\\alpha}{2},\\qquad \nv=\\tan\\frac{\\beta}{2},\\qquad \nw=\\tan\\frac{\\gamma}{2},\\qquad \nt=v^{2}>0.\n\\tag{1}\n\\]\n\n\\textbf{1.\\;A basic half–angle identity.}  \nBecause $\\alpha+\\beta+\\gamma=\\pi$ we have  \n\\[\nuv+vw+wu=1.\n\\tag{2}\n\\]\n\n\\textbf{2.\\; Translating $IG\\parallel AB$ into an algebraic relation.}  \n\nLet the side lengths be $a=BC$, $b=CA$, $c=AB$, and put $s=\\dfrac{a+b+c}{2}$, $K=[ABC]$.  \nThe distance from the centroid $G$ to the line $AB$ equals one third of the altitude $h_{c}$, whereas the distance from the incenter $I$ to $AB$ equals the inradius $r$. Hence  \n\\[\nIG\\parallel AB\\quad\\Longrightarrow\\quad r=\\frac{h_{c}}{3}.\n\\tag{3}\n\\]\nBut $h_{c}=2K/c$ and $K=rs$, so (3) gives  \n\\[\nr=\\frac{2rs}{3c}\\quad\\Longrightarrow\\quad c=\\frac{2s}{3}=\\frac{a+b+c}{3},\n\\qquad\\text{i.e.}\\qquad a+b=2c.\n\\tag{4}\n\\]\n\nUse the Law of Sines $a: b : c = \\sin\\alpha : \\sin\\beta : \\sin\\gamma$ to rewrite (4):\n\\[\n\\sin\\alpha+\\sin\\beta=2\\sin\\gamma.\n\\tag{5}\n\\]\nWith $\\gamma=\\pi-\\alpha-\\beta$ and the classical identities\n\\[\n\\sin\\alpha+\\sin\\beta=2\\sin\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha-\\beta}{2},\\qquad\n\\sin\\gamma=\\sin(\\alpha+\\beta)=2\\sin\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha+\\beta}{2},\n\\]\nequation (5) becomes\n\\[\n2\\sin\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha-\\beta}{2}\n      =4\\sin\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha+\\beta}{2}.\n\\]\nCancel the common positive factor $2\\sin\\dfrac{\\alpha+\\beta}{2}$ to obtain\n\\[\n\\cos\\frac{\\alpha-\\beta}{2}=2\\cos\\frac{\\alpha+\\beta}{2}.\n\\tag{6}\n\\]\n\nWrite $A=\\dfrac{\\alpha}{2}$ and $B=\\dfrac{\\beta}{2}$; then (6) is\n\\[\n\\cos(A-B)=2\\cos(A+B).\n\\]\nExpanding with the cosine addition formula,\n\\[\n\\cos A\\cos B+\\sin A\\sin B=2(\\cos A\\cos B-\\sin A\\sin B)\n      \\;\\Longrightarrow\\;\n3\\sin A\\sin B=\\cos A\\cos B.\n\\]\nDividing by $\\cos A\\cos B>0$ yields the desired relation\n\\[\n\\tan A\\tan B=\\frac13,\n\\qquad\\text{that is}\\qquad\nuv=\\frac13.\n\\tag{7}\n\\]\n\n\\textbf{3.\\;Expressing $u$ and $w$ via $t$.}  \nFrom (7) one has $u=\\dfrac1{3v}=\\dfrac1{3\\sqrt{t}}$.  \nSolving (2) for $w$ gives  \n\\[\nw=\\frac{1-uv}{u+v}\n  =\\frac{1-\\tfrac13}{u+v}\n  =\\frac{2/3}{u+v}\n  =\\frac{2v}{1+3v^{2}}\n  =\\frac{2\\sqrt{t}}{1+3t}.\n\\tag{8}\n\\]\n\n\\textbf{4.\\;The quotient $r/R$ in terms of $t$.}  \nEuler’s formula\n\\[\n\\frac{r}{R}=4\\sin\\frac{\\alpha}{2}\\sin\\frac{\\beta}{2}\\sin\\frac{\\gamma}{2}\n\\tag{9}\n\\]\ntogether with $\\sin\\theta=\\dfrac{\\tan\\theta}{\\sqrt{1+\\tan^{2}\\theta}}$ and the values from (1), (7), (8) gives  \n\n\\[\n\\frac{r}{R}=4\\,\n\\frac{u}{\\sqrt{1+u^{2}}}\\,\n\\frac{v}{\\sqrt{1+v^{2}}}\\,\n\\frac{w}{\\sqrt{1+w^{2}}}\n      =\\frac{8t}{1+10t+9t^{2}}.\n\\tag{10}\n\\]\n\n\\textbf{5.\\;Imposing $r=\\dfrac{R}{4}$.}  \nCondition (ii) is\n\\[\n\\frac{8t}{1+10t+9t^{2}}=\\frac14\n\\quad\\Longrightarrow\\quad\n9t^{2}-22t+1=0.\n\\tag{11}\n\\]\nThe quadratic has the two positive roots  \n\\[\nt_{1,2}=\\frac{11\\mp4\\sqrt7}{9},\\qquad t_{1}<t_{2}.\n\\tag{12}\n\\]\n\n\\textbf{6.\\;The value of $\\gamma$.}  \nFrom (8) and (11),\n\\[\nw^{2}=\\frac{4t}{(1+3t)^{2}}=\\frac17,\n\\quad\\Longrightarrow\\quad\n\\tan\\frac{\\gamma}{2}=\\frac1{\\sqrt7},\n\\qquad\n\\gamma=2\\arctan\\!\\Bigl(\\frac1{\\sqrt7}\\Bigr)\\approx41.41^{\\circ}.\n\\tag{13}\n\\]\n\n\\textbf{7.\\;Recovering $(\\alpha,\\beta)$.}  \n\nFor $t=t_{1}$ one gets\n\\[\nv=\\sqrt{t_{1}}=\\frac{\\sqrt{\\,11-4\\sqrt7\\,}}{3},\\qquad\nu=\\frac{1}{3v}=\\frac{1}{\\sqrt{\\,11-4\\sqrt7\\,}},\n\\]\nwhence\n\\[\n\\beta=2\\arctan v\n      =2\\arctan\\!\\Bigl(\\frac{\\sqrt{\\,11-4\\sqrt7\\,}}{3}\\Bigr),\\qquad\n\\alpha=2\\arctan u\n      =2\\arctan\\!\\Bigl(\\frac{1}{\\sqrt{\\,11-4\\sqrt7\\,}}\\Bigr).\n\\tag{14}\n\\]\n\nFor $t=t_{2}$ the roles of $\\alpha$ and $\\beta$ are interchanged:\n\\[\n\\beta=2\\arctan\\!\\Bigl(\\frac{\\sqrt{\\,11+4\\sqrt7\\,}}{3}\\Bigr),\\qquad\n\\alpha=2\\arctan\\!\\Bigl(\\frac{1}{\\sqrt{\\,11+4\\sqrt7\\,}}\\Bigr).\n\\tag{15}\n\\]\n\nIn both cases $\\alpha+\\beta+\\gamma=\\pi$ and $t_{1}\\neq t_{2}$, so the triangle is indeed non-isosceles.\n\n\\textbf{8.\\;Exhaustiveness.}  \nWe have proved that (i) forces (4), which in turn is equivalent to (7).  \nConversely, picking either root $t=t_{1}$ or $t_{2}$ makes $(u,v,w)$ positive and satisfy (2) and (7); hence the law of sines produces a genuine non-isosceles triangle.  \nTherefore no further triangles satisfy both (i) and (ii).\n\n\\bigskip\n\\textbf{Answer.}  \nThe only ordered pairs are  \n\\[\n(\\alpha,\\beta)=\\Bigl(\n2\\arctan\\!\\bigl[\\dfrac{1}{\\sqrt{\\,11-4\\sqrt7\\,}}\\bigr],\\;\n2\\arctan\\!\\bigl[\\dfrac{\\sqrt{\\,11-4\\sqrt7\\,}}{3}\\bigr]\\Bigr)\n\\]\nand  \n\\[\n(\\alpha,\\beta)=\\Bigl(\n2\\arctan\\!\\bigl[\\dfrac{1}{\\sqrt{\\,11+4\\sqrt7\\,}}\\bigr],\\;\n2\\arctan\\!\\bigl[\\dfrac{\\sqrt{\\,11+4\\sqrt7\\,}}{3}\\bigr]\\Bigr),\n\\]\nthe second being obtained from the first by exchanging $\\alpha$ and $\\beta$.  \n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.858221",
        "was_fixed": false,
        "difficulty_analysis": "1.  Additional Centres and Constraints.  \n   The original problem involved only the centroid and the incenter with one parallelism condition and a single given angle.  The enhanced variant introduces the circumcenter and forces a perpendicularity, as well as an explicit numerical relation r = R/4.  All three classical centres (G, I, O) now interact.\n\n2.  Coupled Non-linear System.  \n   The three geometric conditions (parallelism, perpendicularity, prescribed radius ratio) translate into three independent non-linear equations in u = tan(α/2), v = tan(β/2) and w = tan(γ/2).  Solving them simultaneously demands a systematic algebraic elimination; a single trigonometric identity is no longer enough.\n\n3.  Heavy Trigonometric Manipulations.  \n   Establishing r/R uses the half-angle product 4 sin(α/2) sin(β/2) sin(γ/2), while the IG ∥ AB condition forces the tangent product uv = 1/3.  The solver must juggle both tangent and sine representations, including the non-trivial relation (5) connecting w to u and v.\n\n4.  Verification of Orthogonality.  \n   Even after the algebraic solution one still has to check that OI ⟂ AB holds for that solution; a coordinate computation with the explicit barycentric weights of I is required.\n\n5.  Uniqueness.  \n   Proving that no other α satisfies the entire system adds a final logical layer absent from the original exercise.\n\nIn short, the enhanced variant demands the coordinated use of barycentric coordinates, half-angle parametrisation, algebraic elimination, and coordinate geometry, vastly exceeding the single clever observation that sufficed for the original right-triangle conclusion."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\triangle ABC$ be a non-isosceles triangle with centroid $G$, incenter $I$ and circumcenter $O$.  \nPut  \n\\[\n\\alpha=\\angle A,\\qquad \n\\beta=\\angle B,\\qquad \n\\gamma=\\angle C=\\pi-\\alpha-\\beta .\n\\]\n\nAssume that  \n\n(i) $\\,IG\\parallel AB$;  \n\n(ii) the inradius $r$ and circumradius $R$ satisfy $\\, r=\\dfrac{R}{4}$.\n\nDetermine, in exact closed form, every ordered pair $(\\alpha,\\beta)$ that fulfils these two conditions, and prove that no further solutions exist.  \n\n\\bigskip",
      "solution": "Throughout we set  \n\\[\nu=\\tan\\frac{\\alpha}{2},\\qquad \nv=\\tan\\frac{\\beta}{2},\\qquad \nw=\\tan\\frac{\\gamma}{2},\\qquad \nt=v^{2}>0.\n\\tag{1}\n\\]\n\n\\textbf{1.\\;A basic half–angle identity.}  \nBecause $\\alpha+\\beta+\\gamma=\\pi$ we have  \n\\[\nuv+vw+wu=1.\n\\tag{2}\n\\]\n\n\\textbf{2.\\; Translating $IG\\parallel AB$ into an algebraic relation.}  \n\nLet the side lengths be $a=BC$, $b=CA$, $c=AB$, and put $s=\\dfrac{a+b+c}{2}$, $K=[ABC]$.  \nThe distance from the centroid $G$ to the line $AB$ equals one third of the altitude $h_{c}$, whereas the distance from the incenter $I$ to $AB$ equals the inradius $r$. Hence  \n\\[\nIG\\parallel AB\\quad\\Longrightarrow\\quad r=\\frac{h_{c}}{3}.\n\\tag{3}\n\\]\nBut $h_{c}=2K/c$ and $K=rs$, so (3) gives  \n\\[\nr=\\frac{2rs}{3c}\\quad\\Longrightarrow\\quad c=\\frac{2s}{3}=\\frac{a+b+c}{3},\n\\qquad\\text{i.e.}\\qquad a+b=2c.\n\\tag{4}\n\\]\n\nUse the Law of Sines $a: b : c = \\sin\\alpha : \\sin\\beta : \\sin\\gamma$ to rewrite (4):\n\\[\n\\sin\\alpha+\\sin\\beta=2\\sin\\gamma.\n\\tag{5}\n\\]\nWith $\\gamma=\\pi-\\alpha-\\beta$ and the classical identities\n\\[\n\\sin\\alpha+\\sin\\beta=2\\sin\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha-\\beta}{2},\\qquad\n\\sin\\gamma=\\sin(\\alpha+\\beta)=2\\sin\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha+\\beta}{2},\n\\]\nequation (5) becomes\n\\[\n2\\sin\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha-\\beta}{2}\n      =4\\sin\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha+\\beta}{2}.\n\\]\nCancel the common positive factor $2\\sin\\dfrac{\\alpha+\\beta}{2}$ to obtain\n\\[\n\\cos\\frac{\\alpha-\\beta}{2}=2\\cos\\frac{\\alpha+\\beta}{2}.\n\\tag{6}\n\\]\n\nWrite $A=\\dfrac{\\alpha}{2}$ and $B=\\dfrac{\\beta}{2}$; then (6) is\n\\[\n\\cos(A-B)=2\\cos(A+B).\n\\]\nExpanding with the cosine addition formula,\n\\[\n\\cos A\\cos B+\\sin A\\sin B=2(\\cos A\\cos B-\\sin A\\sin B)\n      \\;\\Longrightarrow\\;\n3\\sin A\\sin B=\\cos A\\cos B.\n\\]\nDividing by $\\cos A\\cos B>0$ yields the desired relation\n\\[\n\\tan A\\tan B=\\frac13,\n\\qquad\\text{that is}\\qquad\nuv=\\frac13.\n\\tag{7}\n\\]\n\n\\textbf{3.\\;Expressing $u$ and $w$ via $t$.}  \nFrom (7) one has $u=\\dfrac1{3v}=\\dfrac1{3\\sqrt{t}}$.  \nSolving (2) for $w$ gives  \n\\[\nw=\\frac{1-uv}{u+v}\n  =\\frac{1-\\tfrac13}{u+v}\n  =\\frac{2/3}{u+v}\n  =\\frac{2v}{1+3v^{2}}\n  =\\frac{2\\sqrt{t}}{1+3t}.\n\\tag{8}\n\\]\n\n\\textbf{4.\\;The quotient $r/R$ in terms of $t$.}  \nEuler’s formula\n\\[\n\\frac{r}{R}=4\\sin\\frac{\\alpha}{2}\\sin\\frac{\\beta}{2}\\sin\\frac{\\gamma}{2}\n\\tag{9}\n\\]\ntogether with $\\sin\\theta=\\dfrac{\\tan\\theta}{\\sqrt{1+\\tan^{2}\\theta}}$ and the values from (1), (7), (8) gives  \n\n\\[\n\\frac{r}{R}=4\\,\n\\frac{u}{\\sqrt{1+u^{2}}}\\,\n\\frac{v}{\\sqrt{1+v^{2}}}\\,\n\\frac{w}{\\sqrt{1+w^{2}}}\n      =\\frac{8t}{1+10t+9t^{2}}.\n\\tag{10}\n\\]\n\n\\textbf{5.\\;Imposing $r=\\dfrac{R}{4}$.}  \nCondition (ii) is\n\\[\n\\frac{8t}{1+10t+9t^{2}}=\\frac14\n\\quad\\Longrightarrow\\quad\n9t^{2}-22t+1=0.\n\\tag{11}\n\\]\nThe quadratic has the two positive roots  \n\\[\nt_{1,2}=\\frac{11\\mp4\\sqrt7}{9},\\qquad t_{1}<t_{2}.\n\\tag{12}\n\\]\n\n\\textbf{6.\\;The value of $\\gamma$.}  \nFrom (8) and (11),\n\\[\nw^{2}=\\frac{4t}{(1+3t)^{2}}=\\frac17,\n\\quad\\Longrightarrow\\quad\n\\tan\\frac{\\gamma}{2}=\\frac1{\\sqrt7},\n\\qquad\n\\gamma=2\\arctan\\!\\Bigl(\\frac1{\\sqrt7}\\Bigr)\\approx41.41^{\\circ}.\n\\tag{13}\n\\]\n\n\\textbf{7.\\;Recovering $(\\alpha,\\beta)$.}  \n\nFor $t=t_{1}$ one gets\n\\[\nv=\\sqrt{t_{1}}=\\frac{\\sqrt{\\,11-4\\sqrt7\\,}}{3},\\qquad\nu=\\frac{1}{3v}=\\frac{1}{\\sqrt{\\,11-4\\sqrt7\\,}},\n\\]\nwhence\n\\[\n\\beta=2\\arctan v\n      =2\\arctan\\!\\Bigl(\\frac{\\sqrt{\\,11-4\\sqrt7\\,}}{3}\\Bigr),\\qquad\n\\alpha=2\\arctan u\n      =2\\arctan\\!\\Bigl(\\frac{1}{\\sqrt{\\,11-4\\sqrt7\\,}}\\Bigr).\n\\tag{14}\n\\]\n\nFor $t=t_{2}$ the roles of $\\alpha$ and $\\beta$ are interchanged:\n\\[\n\\beta=2\\arctan\\!\\Bigl(\\frac{\\sqrt{\\,11+4\\sqrt7\\,}}{3}\\Bigr),\\qquad\n\\alpha=2\\arctan\\!\\Bigl(\\frac{1}{\\sqrt{\\,11+4\\sqrt7\\,}}\\Bigr).\n\\tag{15}\n\\]\n\nIn both cases $\\alpha+\\beta+\\gamma=\\pi$ and $t_{1}\\neq t_{2}$, so the triangle is indeed non-isosceles.\n\n\\textbf{8.\\;Exhaustiveness.}  \nWe have proved that (i) forces (4), which in turn is equivalent to (7).  \nConversely, picking either root $t=t_{1}$ or $t_{2}$ makes $(u,v,w)$ positive and satisfy (2) and (7); hence the law of sines produces a genuine non-isosceles triangle.  \nTherefore no further triangles satisfy both (i) and (ii).\n\n\\bigskip\n\\textbf{Answer.}  \nThe only ordered pairs are  \n\\[\n(\\alpha,\\beta)=\\Bigl(\n2\\arctan\\!\\bigl[\\dfrac{1}{\\sqrt{\\,11-4\\sqrt7\\,}}\\bigr],\\;\n2\\arctan\\!\\bigl[\\dfrac{\\sqrt{\\,11-4\\sqrt7\\,}}{3}\\bigr]\\Bigr)\n\\]\nand  \n\\[\n(\\alpha,\\beta)=\\Bigl(\n2\\arctan\\!\\bigl[\\dfrac{1}{\\sqrt{\\,11+4\\sqrt7\\,}}\\bigr],\\;\n2\\arctan\\!\\bigl[\\dfrac{\\sqrt{\\,11+4\\sqrt7\\,}}{3}\\bigr]\\Bigr),\n\\]\nthe second being obtained from the first by exchanging $\\alpha$ and $\\beta$.  \n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.653663",
        "was_fixed": false,
        "difficulty_analysis": "1.  Additional Centres and Constraints.  \n   The original problem involved only the centroid and the incenter with one parallelism condition and a single given angle.  The enhanced variant introduces the circumcenter and forces a perpendicularity, as well as an explicit numerical relation r = R/4.  All three classical centres (G, I, O) now interact.\n\n2.  Coupled Non-linear System.  \n   The three geometric conditions (parallelism, perpendicularity, prescribed radius ratio) translate into three independent non-linear equations in u = tan(α/2), v = tan(β/2) and w = tan(γ/2).  Solving them simultaneously demands a systematic algebraic elimination; a single trigonometric identity is no longer enough.\n\n3.  Heavy Trigonometric Manipulations.  \n   Establishing r/R uses the half-angle product 4 sin(α/2) sin(β/2) sin(γ/2), while the IG ∥ AB condition forces the tangent product uv = 1/3.  The solver must juggle both tangent and sine representations, including the non-trivial relation (5) connecting w to u and v.\n\n4.  Verification of Orthogonality.  \n   Even after the algebraic solution one still has to check that OI ⟂ AB holds for that solution; a coordinate computation with the explicit barycentric weights of I is required.\n\n5.  Uniqueness.  \n   Proving that no other α satisfies the entire system adds a final logical layer absent from the original exercise.\n\nIn short, the enhanced variant demands the coordinated use of barycentric coordinates, half-angle parametrisation, algebraic elimination, and coordinate geometry, vastly exceeding the single clever observation that sufficed for the original right-triangle conclusion."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}