{
  "index": "1987-A-4",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $P$ be a polynomial, with real coefficients, in three variables\nand $F$ be a function of two variables such that\n\\[\nP(ux, uy, uz) = u^2 F(y-x,z-x) \\quad \\mbox{for all real $x,y,z,u$},\n\\]\nand such that $P(1,0,0)=4$, $P(0,1,0)=5$, and $P(0,0,1)=6$. Also let\n$A,B,C$ be complex numbers with $P(A,B,C)=0$ and $|B-A|=10$. Find $|C-A|$.",
  "solution": "Solution. Letting \\( u=1 \\) and \\( x=0 \\), we have that \\( F(y, z)=P(0, y, z) \\) is a polynomial. Also, \\( F(u y, u z)=P(0, u y, u z)=u^{2} P(0, y, z)=u^{2} F(y, z) \\), so \\( F \\) is homogeneous of degree 2 . Therefore\n\\[\nP(x, y, z)=F(y-x, z-x)=a(y-x)^{2}+b(y-x)(z-x)+c(z-x)^{2}\n\\]\nfor some real \\( a, b, c \\). Then \\( 4=P(1,0,0)=a+b+c, 5=P(0,1,0)=a \\), and \\( 6=P(0,0,1)=c \\), so \\( b=-7 \\). Now\n\\[\n0=P(A, B, C)=5(B-A)^{2}-7(B-A)(C-A)+6(C-A)^{2},\n\\]\nso the number \\( m=(C-A) /(B-A) \\) satisfies \\( 5-7 m+6 m^{2}=0 \\). The zeros of \\( 6 m^{2}-7 m+5 \\) are complex conjugate with product \\( 5 / 6 \\), so \\( |m|=\\sqrt{5 / 6} \\). Thus \\( |C-A|=\\sqrt{5 / 6}|B-A|=(5 / 3) \\sqrt{30} \\).",
  "vars": [
    "x",
    "y",
    "z",
    "u",
    "m"
  ],
  "params": [
    "P",
    "F",
    "A",
    "B",
    "C",
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "varxaxis",
        "y": "varyaxis",
        "z": "varzaxis",
        "u": "varuscale",
        "m": "varmratio",
        "P": "polyfunc",
        "F": "auxfunc",
        "A": "constalpha",
        "B": "constbeta",
        "C": "constgamma",
        "a": "coeffaone",
        "b": "coeffbtwo",
        "c": "coeffcthree"
      },
      "question": "Let $polyfunc$ be a polynomial, with real coefficients, in three variables and $auxfunc$ be a function of two variables such that\n\\[\npolyfunc(varuscale varxaxis, varuscale varyaxis, varuscale varzaxis) = varuscale^{2} auxfunc(varyaxis-varxaxis,varzaxis-varxaxis) \\quad \\mbox{for all real $varxaxis,varyaxis,varzaxis,varuscale$},\n\\]\nand such that $polyfunc(1,0,0)=4$, $polyfunc(0,1,0)=5$, and $polyfunc(0,0,1)=6$. Also let $constalpha,constbeta,constgamma$ be complex numbers with $polyfunc(constalpha,constbeta,constgamma)=0$ and $|constbeta-constalpha|=10$. Find $|constgamma-constalpha|$.",
      "solution": "Solution. Letting \\( varuscale=1 \\) and \\( varxaxis=0 \\), we have that \\( auxfunc(varyaxis, varzaxis)=polyfunc(0, varyaxis, varzaxis) \\) is a polynomial. Also, \\( auxfunc(varuscale varyaxis, varuscale varzaxis)=polyfunc(0, varuscale varyaxis, varuscale varzaxis)=varuscale^{2} polyfunc(0, varyaxis, varzaxis)=varuscale^{2} auxfunc(varyaxis, varzaxis) \\), so \\( auxfunc \\) is homogeneous of degree 2 . Therefore\n\\[\npolyfunc(varxaxis, varyaxis, varzaxis)=auxfunc(varyaxis-varxaxis, varzaxis-varxaxis)=coeffaone(varyaxis-varxaxis)^{2}+coeffbtwo(varyaxis-varxaxis)(varzaxis-varxaxis)+coeffcthree(varzaxis-varxaxis)^{2}\n\\]\nfor some real \\( coeffaone, coeffbtwo, coeffcthree \\). Then \\( 4=polyfunc(1,0,0)=coeffaone+coeffbtwo+coeffcthree, 5=polyfunc(0,1,0)=coeffaone \\), and \\( 6=polyfunc(0,0,1)=coeffcthree \\), so \\( coeffbtwo=-7 \\). Now\n\\[\n0=polyfunc(constalpha, constbeta, constgamma)=5(constbeta-constalpha)^{2}-7(constbeta-constalpha)(constgamma-constalpha)+6(constgamma-constalpha)^{2},\n\\]\nso the number \\( varmratio=(constgamma-constalpha) /(constbeta-constalpha) \\) satisfies \\( 5-7 varmratio+6 varmratio^{2}=0 \\). The zeros of \\( 6 varmratio^{2}-7 varmratio+5 \\) are complex conjugate with product \\( 5 / 6 \\), so \\( |varmratio|=\\sqrt{5 / 6} \\). Thus \\( |constgamma-constalpha|=\\sqrt{5 / 6}|constbeta-constalpha|=(5 / 3) \\sqrt{30} \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lanterns",
        "y": "cupcakes",
        "z": "earrings",
        "u": "photonic",
        "m": "sandwich",
        "P": "asteroid",
        "F": "dinosaur",
        "A": "accordion",
        "B": "pineapple",
        "C": "backpack",
        "a": "galactic",
        "b": "fountain",
        "c": "butterfly"
      },
      "question": "Let $asteroid$ be a polynomial, with real coefficients, in three variables\nand $dinosaur$ be a function of two variables such that\n\\[\nasteroid(photonic lanterns, photonic cupcakes, photonic earrings) = photonic^2 dinosaur(cupcakes-lanterns,earrings-lanterns) \\quad \\mbox{for all real $lanterns,cupcakes,earrings,photonic$},\n\\]\nand such that $asteroid(1,0,0)=4$, $asteroid(0,1,0)=5$, and $asteroid(0,0,1)=6$. Also let\n$accordion,pineapple,backpack$ be complex numbers with $asteroid(accordion,pineapple,backpack)=0$ and $|pineapple-accordion|=10$. Find $|backpack-accordion|$.",
      "solution": "Solution. Letting \\( photonic=1 \\) and \\( lanterns=0 \\), we have that \\( dinosaur(cupcakes, earrings)=asteroid(0, cupcakes, earrings) \\) is a polynomial. Also, \\( dinosaur(photonic cupcakes, photonic earrings)=asteroid(0, photonic cupcakes, photonic earrings)=photonic^{2} asteroid(0, cupcakes, earrings)=photonic^{2} dinosaur(cupcakes, earrings) \\), so \\( dinosaur \\) is homogeneous of degree 2. Therefore\n\\[\nasteroid(lanterns, cupcakes, earrings)=dinosaur(cupcakes-lanterns, earrings-lanterns)=galactic(cupcakes-lanterns)^{2}+fountain(cupcakes-lanterns)(earrings-lanterns)+butterfly(earrings-lanterns)^{2}\n\\]\nfor some real \\( galactic, fountain, butterfly \\). Then \\( 4=asteroid(1,0,0)=galactic+fountain+butterfly,\\; 5=asteroid(0,1,0)=galactic \\), and \\( 6=asteroid(0,0,1)=butterfly \\), so \\( fountain=-7 \\). Now\n\\[\n0=asteroid(accordion, pineapple, backpack)=5(pineapple-accordion)^{2}-7(pineapple-accordion)(backpack-accordion)+6(backpack-accordion)^{2},\n\\]\nso the number \\( sandwich=(backpack-accordion)/(pineapple-accordion) \\) satisfies \\( 5-7\\,sandwich+6\\,sandwich^{2}=0 \\). The zeros of \\( 6\\,sandwich^{2}-7\\,sandwich+5 \\) are complex conjugate with product \\( 5/6 \\), so \\( |sandwich|=\\sqrt{5/6} \\). Thus \\( |backpack-accordion|=\\sqrt{5/6}\\,|pineapple-accordion|=(5/3)\\sqrt{30} \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "y": "steadyterm",
        "z": "unchchanged",
        "u": "staticcoef",
        "m": "divergent",
        "P": "chaosfunction",
        "F": "nonfunction",
        "A": "movepoint",
        "B": "shiftvalue",
        "C": "driftvalue",
        "a": "freemember",
        "b": "detached",
        "c": "isolated"
      },
      "question": "Let $chaosfunction$ be a polynomial, with real coefficients, in three variables\nand $nonfunction$ be a function of two variables such that\n\\[\nchaosfunction(staticcoefconstantvalue, staticcoefsteadyterm, staticcoefunchchanged) = staticcoef^2 nonfunction(steadyterm-constantvalue,unchchanged-constantvalue) \\quad \\mbox{for all real $constantvalue,steadyterm,unchchanged,staticcoef$},\n\\]\nand such that $chaosfunction(1,0,0)=4$, $chaosfunction(0,1,0)=5$, and $chaosfunction(0,0,1)=6$. Also let\n$movepoint,shiftvalue,driftvalue$ be complex numbers with $chaosfunction(movepoint,shiftvalue,driftvalue)=0$ and $|shiftvalue-movepoint|=10$. Find $|driftvalue-movepoint|$.",
      "solution": "Solution. Letting \\( staticcoef=1 \\) and \\( constantvalue=0 \\), we have that \\( nonfunction(steadyterm, unchchanged)=chaosfunction(0, steadyterm, unchchanged) \\) is a polynomial. Also, \\( nonfunction(staticcoef steadyterm, staticcoef unchchanged)=chaosfunction(0, staticcoef steadyterm, staticcoef unchchanged)=staticcoef^{2} chaosfunction(0, steadyterm, unchchanged)=staticcoef^{2} nonfunction(steadyterm, unchchanged) \\), so \\( nonfunction \\) is homogeneous of degree 2 . Therefore\n\\[\nchaosfunction(constantvalue, steadyterm, unchchanged)=nonfunction(steadyterm-constantvalue, unchchanged-constantvalue)=freemember(steadyterm-constantvalue)^{2}+detached(steadyterm-constantvalue)(unchchanged-constantvalue)+isolated(unchchanged-constantvalue)^{2}\n\\]\nfor some real \\( freemember, detached, isolated \\). Then \\( 4=chaosfunction(1,0,0)=freemember+detached+isolated, 5=chaosfunction(0,1,0)=freemember \\), and \\( 6=chaosfunction(0,0,1)=isolated \\), so \\( detached=-7 \\). Now\n\\[\n0=chaosfunction(movepoint, shiftvalue, driftvalue)=5(shiftvalue-movepoint)^{2}-7(shiftvalue-movepoint)(driftvalue-movepoint)+6(driftvalue-movepoint)^{2},\n\\]\nso the number \\( divergent=(driftvalue-movepoint) /(shiftvalue-movepoint) \\) satisfies \\( 5-7 divergent+6 divergent^{2}=0 \\). The zeros of \\( 6 divergent^{2}-7 divergent+5 \\) are complex conjugate with product \\( 5 / 6 \\), so \\( |divergent|=\\sqrt{5 / 6} \\). Thus \\( |driftvalue-movepoint|=\\sqrt{5 / 6}|shiftvalue-movepoint|=(5 / 3) \\sqrt{30} \\)."
    },
    "garbled_string": {
      "map": {
        "x": "frandolq",
        "y": "zibnexma",
        "z": "kotrenvi",
        "u": "psalmort",
        "m": "jugratis",
        "P": "tirgolanh",
        "F": "vecsundal",
        "A": "pobjarnik",
        "B": "wexlupor",
        "C": "mikzander",
        "a": "qusaplen",
        "b": "nithroqe",
        "c": "lombrastu"
      },
      "question": "Let $tirgolanh$ be a polynomial, with real coefficients, in three variables\nand $vecsundal$ be a function of two variables such that\n\\[\ntirgolanh(psalmort frandolq, psalmort zibnexma, psalmort kotrenvi) = psalmort^2 vecsundal(zibnexma-frandolq,kotrenvi-frandolq) \\quad \\mbox{for all real $frandolq,zibnexma,kotrenvi,psalmort$},\n\\]\nand such that $tirgolanh(1,0,0)=4$, $tirgolanh(0,1,0)=5$, and $tirgolanh(0,0,1)=6$. Also let\n$pobjarnik,wexlupor,mikzander$ be complex numbers with $tirgolanh(pobjarnik,wexlupor,mikzander)=0$ and $|wexlupor-pobjarnik|=10$. Find $|mikzander-pobjarnik|$.",
      "solution": "Solution. Letting \\( psalmort=1 \\) and \\( frandolq=0 \\), we have that \\( vecsundal(zibnexma, kotrenvi)=tirgolanh(0, zibnexma, kotrenvi) \\) is a polynomial. Also, \\( vecsundal(psalmort zibnexma, psalmort kotrenvi)=tirgolanh(0, psalmort zibnexma, psalmort kotrenvi)=psalmort^{2} tirgolanh(0, zibnexma, kotrenvi)=psalmort^{2} vecsundal(zibnexma, kotrenvi) \\), so \\( vecsundal \\) is homogeneous of degree 2 . Therefore\n\\[\ntirgolanh(frandolq, zibnexma, kotrenvi)=vecsundal(zibnexma-frandolq, kotrenvi-frandolq)=qusaplen(zibnexma-frandolq)^{2}+nithroqe(zibnexma-frandolq)(kotrenvi-frandolq)+lombrastu(kotrenvi-frandolq)^{2}\n\\]\nfor some real \\( qusaplen, nithroqe, lombrastu \\). Then \\( 4=tirgolanh(1,0,0)=qusaplen+nithroqe+lombrastu, 5=tirgolanh(0,1,0)=qusaplen \\), and \\( 6=tirgolanh(0,0,1)=lombrastu \\), so \\( nithroqe=-7 \\). Now\n\\[\n0=tirgolanh(pobjarnik, wexlupor, mikzander)=5(wexlupor-pobjarnik)^{2}-7(wexlupor-pobjarnik)(mikzander-pobjarnik)+6(mikzander-pobjarnik)^{2},\n\\]\nso the number \\( jugratis=(mikzander-pobjarnik) /(wexlupor-pobjarnik) \\) satisfies \\( 5-7 jugratis+6 jugratis^{2}=0 \\). The zeros of \\( 6 jugratis^{2}-7 jugratis+5 \\) are complex conjugate with product \\( 5 / 6 \\), so \\( |jugratis|=\\sqrt{5 / 6} \\). Thus \\( |mikzander-pobjarnik|=\\sqrt{5 / 6}|wexlupor-pobjarnik|=(5 / 3) \\sqrt{30} \\)."
    },
    "kernel_variant": {
      "question": "Let P be a polynomial with real coefficients in three variables and let F be a function of two variables such that  \n\n  P(ux, uy, uz) = u² F(y−x, z−x)  for every real x, y, z, u.  \n\nAssume further that  \n\n  P(1,0,0)=8, P(0,1,0)=3, P(0,0,1)=7.  \n\nComplex numbers A, B, C satisfy P(A,B,C)=0 and |B−A|=15.  Moreover  \n\n  Im ((C−A)/(B−A)) > 0.  \n\nAnswer the following questions:\n\n(a) Find |C−A|.  \n(b) Find |C−B|.  \n(c) Compute the area of the triangle ABC in the complex plane.  \n(d) Determine the radius R of the circumcircle of triangle ABC.",
      "solution": "Step 1.  Structure of P.  \nSetting u=1 in the defining identity gives P(x,y,z)=F(y−x, z−x); putting x=0 shows that F is a polynomial.  \nReplacing (y,z) by (uy,uz) shows F(uy,uz)=u²F(y,z), so F is homogeneous of degree 2.  \nHence for some real constants a,b,c we have  \n\n  P(x,y,z)=a(y−x)²+b(y−x)(z−x)+c(z−x)².  (1)\n\nStep 2.  Determining the coefficients.  \nSubstituting the three given values of P:\n\n•  P(0,1,0)=a=3;  \n•  P(0,0,1)=c=7;  \n•  P(1,0,0)=a+b+c=8 ⇒ b=8−a−c = 8−3−7 = −2.\n\nThus  \n\n  P(x,y,z)=3(y−x)²−2(y−x)(z−x)+7(z−x)².  (2)\n\nStep 3.  The fundamental quadratic for m=(C−A)/(B−A).  \nWrite s=B−A and t=C−A.  Then t = m s and P(A,B,C)=0 gives  \n\n  0 = 3s² − 2s·t + 7t²  \n  0 = 3 − 2m + 7m².  (3)\n\nEquation (3) is a quadratic in m:\n\n  7m² − 2m + 3 = 0.  (4)\n\nStep 4.  Solving (4) and recording useful data.  \nDiscriminant Δ = (−2)² − 4·7·3 = 4 − 84 = −80 < 0,  \nso m is non-real and the two roots are complex conjugates.  \nChoosing the one with positive imaginary part (by the hypothesis Im m>0),\n\n  m = (1 + i√20)/7.  (5)\n\nUseful numerical data:\n\n|m|² = (Re m)² + (Im m)² = (1/7)² + (√20/7)² = (1 + 20)/49 = 21/49 = 3/7 ⇒ |m| = √(3/7). (6)\n\nm−1 = −6/7 + i√20/7 ⇒ |m−1|² = 36/49 + 20/49 = 56/49 = 8/7 ⇒ |m−1| = √(8/7). (7)\n\ncos θ = Re m / |m| = (1/7)/√(3/7) = 1/√21,  so sin θ = √(1 − 1/21) = √(20/21). (8)\n\nHere θ = arg m = ∠BAC.\n\nStep 5.  Computing the requested quantities.\n\n(a) |C−A| = |m|·|B−A| = √(3/7)·15 = 15√(3/7).  \n\n(b) |C−B| = |t−s| = |(m−1)s| = |m−1|·|s| = √(8/7)·15 = 15√(8/7).  \n\n(c) Area of ΔABC.  In the complex plane, area = ½|B−A||C−A| sin θ.  \nUsing (a) and (8):\n\nArea = ½·15·15√(3/7)·√(20/21)  \n   = (225/2)·√[(3·20)/(7·21)]  \n   = (225/2)·√(60/147)  \n   = (225/2)·√(20/49)  \n   = (225/2)·(√20)/7  \n   = (225√5)/7.  \n\n(d) Circumradius R of ΔABC.  With side lengths  \n a = BC = 15√(8/7), b = AC = 15√(3/7), c = AB = 15,  \n\nthe classical formula R = abc / (4·Area) gives\n\nabc = 15·15√(3/7)·15√(8/7) = 15³√(24)/7 = 3375√24 / 7,  \n4·Area = 4·(225√5)/7 = 900√5 / 7,  \n\nso R = (3375√24 / 7) / (900√5 / 7) = (3375/900)·√(24/5)  \n   = (15/4)·√(24/5) = (15/4)·(2√6)/√5 = (15/2)·√(6/5).\n\nTherefore  \n\n |C−A| = 15√(3/7), |C−B| = 15√(8/7), Area(ΔABC) = (225√5)/7, R = (15/2)√(6/5).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.698670",
        "was_fixed": false,
        "difficulty_analysis": "•  The problem moves from merely finding a single distance to determining two distances, the signed angle, the area, and the circumradius—each requiring additional layers of computation.  \n•  Instead of stopping at the modulus of m, one must find Re m, Im m, and trigonometric functions of the argument; this obliges the competitor to solve the quadratic explicitly in the complex plane.  \n•  Part (c) demands converting algebraic information into geometric quantities (area via sine of an angle derived from complex data).  \n•  Part (d) introduces a further classical-geometry formula (circumradius) that combines all previous results; errors propagate if earlier work is not rigorous.  \n•  Altogether the solution chain involves functional equations, homogeneous polynomials, complex numbers, quadratic equations with negative discriminant, trigonometry, and classical Euclidean geometry, substantially raising both technical load and conceptual depth compared with the original single-answer problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let P be a polynomial with real coefficients in three variables and let F be a function of two variables such that  \n\n  P(ux, uy, uz) = u² F(y−x, z−x)  for every real x, y, z, u.  \n\nAssume further that  \n\n  P(1,0,0)=8, P(0,1,0)=3, P(0,0,1)=7.  \n\nComplex numbers A, B, C satisfy P(A,B,C)=0 and |B−A|=15.  Moreover  \n\n  Im ((C−A)/(B−A)) > 0.  \n\nAnswer the following questions:\n\n(a) Find |C−A|.  \n(b) Find |C−B|.  \n(c) Compute the area of the triangle ABC in the complex plane.  \n(d) Determine the radius R of the circumcircle of triangle ABC.",
      "solution": "Step 1.  Structure of P.  \nSetting u=1 in the defining identity gives P(x,y,z)=F(y−x, z−x); putting x=0 shows that F is a polynomial.  \nReplacing (y,z) by (uy,uz) shows F(uy,uz)=u²F(y,z), so F is homogeneous of degree 2.  \nHence for some real constants a,b,c we have  \n\n  P(x,y,z)=a(y−x)²+b(y−x)(z−x)+c(z−x)².  (1)\n\nStep 2.  Determining the coefficients.  \nSubstituting the three given values of P:\n\n•  P(0,1,0)=a=3;  \n•  P(0,0,1)=c=7;  \n•  P(1,0,0)=a+b+c=8 ⇒ b=8−a−c = 8−3−7 = −2.\n\nThus  \n\n  P(x,y,z)=3(y−x)²−2(y−x)(z−x)+7(z−x)².  (2)\n\nStep 3.  The fundamental quadratic for m=(C−A)/(B−A).  \nWrite s=B−A and t=C−A.  Then t = m s and P(A,B,C)=0 gives  \n\n  0 = 3s² − 2s·t + 7t²  \n  0 = 3 − 2m + 7m².  (3)\n\nEquation (3) is a quadratic in m:\n\n  7m² − 2m + 3 = 0.  (4)\n\nStep 4.  Solving (4) and recording useful data.  \nDiscriminant Δ = (−2)² − 4·7·3 = 4 − 84 = −80 < 0,  \nso m is non-real and the two roots are complex conjugates.  \nChoosing the one with positive imaginary part (by the hypothesis Im m>0),\n\n  m = (1 + i√20)/7.  (5)\n\nUseful numerical data:\n\n|m|² = (Re m)² + (Im m)² = (1/7)² + (√20/7)² = (1 + 20)/49 = 21/49 = 3/7 ⇒ |m| = √(3/7). (6)\n\nm−1 = −6/7 + i√20/7 ⇒ |m−1|² = 36/49 + 20/49 = 56/49 = 8/7 ⇒ |m−1| = √(8/7). (7)\n\ncos θ = Re m / |m| = (1/7)/√(3/7) = 1/√21,  so sin θ = √(1 − 1/21) = √(20/21). (8)\n\nHere θ = arg m = ∠BAC.\n\nStep 5.  Computing the requested quantities.\n\n(a) |C−A| = |m|·|B−A| = √(3/7)·15 = 15√(3/7).  \n\n(b) |C−B| = |t−s| = |(m−1)s| = |m−1|·|s| = √(8/7)·15 = 15√(8/7).  \n\n(c) Area of ΔABC.  In the complex plane, area = ½|B−A||C−A| sin θ.  \nUsing (a) and (8):\n\nArea = ½·15·15√(3/7)·√(20/21)  \n   = (225/2)·√[(3·20)/(7·21)]  \n   = (225/2)·√(60/147)  \n   = (225/2)·√(20/49)  \n   = (225/2)·(√20)/7  \n   = (225√5)/7.  \n\n(d) Circumradius R of ΔABC.  With side lengths  \n a = BC = 15√(8/7), b = AC = 15√(3/7), c = AB = 15,  \n\nthe classical formula R = abc / (4·Area) gives\n\nabc = 15·15√(3/7)·15√(8/7) = 15³√(24)/7 = 3375√24 / 7,  \n4·Area = 4·(225√5)/7 = 900√5 / 7,  \n\nso R = (3375√24 / 7) / (900√5 / 7) = (3375/900)·√(24/5)  \n   = (15/4)·√(24/5) = (15/4)·(2√6)/√5 = (15/2)·√(6/5).\n\nTherefore  \n\n |C−A| = 15√(3/7), |C−B| = 15√(8/7), Area(ΔABC) = (225√5)/7, R = (15/2)√(6/5).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.545959",
        "was_fixed": false,
        "difficulty_analysis": "•  The problem moves from merely finding a single distance to determining two distances, the signed angle, the area, and the circumradius—each requiring additional layers of computation.  \n•  Instead of stopping at the modulus of m, one must find Re m, Im m, and trigonometric functions of the argument; this obliges the competitor to solve the quadratic explicitly in the complex plane.  \n•  Part (c) demands converting algebraic information into geometric quantities (area via sine of an angle derived from complex data).  \n•  Part (d) introduces a further classical-geometry formula (circumradius) that combines all previous results; errors propagate if earlier work is not rigorous.  \n•  Altogether the solution chain involves functional equations, homogeneous polynomials, complex numbers, quadratic equations with negative discriminant, trigonometry, and classical Euclidean geometry, substantially raising both technical load and conceptual depth compared with the original single-answer problem."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}