{
  "index": "1974-B-3",
  "type": "NT",
  "tag": [
    "NT",
    "ANA"
  ],
  "difficulty": "",
  "question": "B-3. Prove that if \\( \\alpha \\) is a real number such that\n\\[\n\\cos \\pi \\alpha=1 / 3,\n\\]\nthen \\( \\alpha \\) is irrational. (The angle \\( \\pi \\alpha \\) is in radians.)",
  "solution": "B-3.\nIf \\( \\alpha=r / s \\) with \\( r \\) and \\( s \\) integers and \\( s>0 \\), then \\( \\cos (n \\pi \\alpha) \\) takes on at most \\( 2 s \\) distinct values for integral choices of \\( n \\). When \\( \\cos \\pi \\alpha=1 / 3 \\), the formula \\( \\cos 2 \\theta=2 \\cos ^{2} \\theta-1 \\) and mathematical induction can be used to show that\n\\[\n\\cos \\left(2^{m} \\pi \\alpha\\right)=t / 3^{2^{m}} \\quad \\quad[m=1,2,3, \\cdots]\n\\]\nwith \\( t \\) an integer not divisible by 3 , and hence that these cosines form an infinite set of distinct values. Thus \\( \\alpha \\) is irrational.",
  "vars": [
    "\\\\alpha",
    "\\\\theta",
    "r",
    "s",
    "n",
    "m",
    "t"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "\\alpha": "anglevar",
        "\\theta": "auxangle",
        "r": "intnumer",
        "s": "intdenom",
        "n": "indexint",
        "m": "powindex",
        "t": "intconst"
      },
      "question": "B-3. Prove that if \\( anglevar \\) is a real number such that\n\\[\n\\cos \\pi anglevar=1 / 3,\n\\]\nthen \\( anglevar \\) is irrational. (The angle \\( \\pi anglevar \\) is in radians.)",
      "solution": "B-3.\nIf \\( anglevar=intnumer / intdenom \\) with \\( intnumer \\) and \\( intdenom \\) integers and \\( intdenom>0 \\), then \\( \\cos (indexint \\pi anglevar) \\) takes on at most \\( 2 intdenom \\) distinct values for integral choices of \\( indexint \\). When \\( \\cos \\pi anglevar=1 / 3 \\), the formula \\( \\cos 2 auxangle=2 \\cos ^{2} auxangle-1 \\) and mathematical induction can be used to show that\n\\[\n\\cos \\left(2^{powindex} \\pi anglevar\\right)=intconst / 3^{2^{powindex}} \\quad \\quad[powindex=1,2,3, \\cdots]\n\\]\nwith \\( intconst \\) an integer not divisible by 3 , and hence that these cosines form an infinite set of distinct values. Thus \\( anglevar \\) is irrational."
    },
    "descriptive_long_confusing": {
      "map": {
        "\\alpha": "marmalade",
        "\\theta": "wildberry",
        "r": "pinecone",
        "s": "driftwood",
        "n": "rainstorm",
        "m": "stonework",
        "t": "clifftops"
      },
      "question": "B-3. Prove that if \\( marmalade \\) is a real number such that\n\\[\n\\cos \\pi marmalade=1 / 3,\n\\]\nthen marmalade is irrational. (The angle \\( \\pi marmalade \\) is in radians.)",
      "solution": "B-3.\nIf \\( marmalade=pinecone / driftwood \\) with \\( pinecone \\) and \\( driftwood \\) integers and \\( driftwood>0 \\), then \\( \\cos (rainstorm \\pi marmalade) \\) takes on at most \\( 2 driftwood \\) distinct values for integral choices of \\( rainstorm \\). When \\( \\cos \\pi marmalade=1 / 3 \\), the formula \\( \\cos 2 wildberry=2 \\cos ^{2} wildberry-1 \\) and mathematical induction can be used to show that\n\\[\n\\cos \\left(2^{stonework} \\pi marmalade\\right)=clifftops / 3^{2^{stonework}} \\quad \\quad[stonework=1,2,3, \\cdots]\n\\]\nwith \\( clifftops \\) an integer not divisible by 3 , and hence that these cosines form an infinite set of distinct values. Thus \\( marmalade \\) is irrational."
    },
    "descriptive_long_misleading": {
      "map": {
        "\\alpha": "omegasymbol",
        "\\theta": "straightline",
        "r": "irrational",
        "s": "numerator",
        "n": "fractional",
        "m": "constantvalue",
        "t": "divisible"
      },
      "question": "B-3. Prove that if \\( omegasymbol \\) is a real number such that\n\\[\n\\cos \\pi omegasymbol=1 / 3,\n\\]\nthen \\( omegasymbol \\) is irrational. (The angle \\( \\pi omegasymbol \\) is in radians.)",
      "solution": "B-3.\nIf \\( omegasymbol=irrational / numerator \\) with \\( irrational \\) and \\( numerator \\) integers and \\( numerator>0 \\), then \\( \\cos (fractional \\pi omegasymbol) \\) takes on at most \\( 2 numerator \\) distinct values for integral choices of \\( fractional \\). When \\( \\cos \\pi omegasymbol=1 / 3 \\), the formula \\( \\cos 2 straightline=2 \\cos ^{2} straightline-1 \\) and mathematical induction can be used to show that\n\\[\n\\cos \\left(2^{constantvalue} \\pi omegasymbol\\right)=divisible / 3^{2^{constantvalue}} \\quad \\quad[constantvalue=1,2,3, \\cdots]\n\\]\nwith \\( divisible \\) an integer not divisible by 3 , and hence that these cosines form an infinite set of distinct values. Thus \\( omegasymbol \\) is irrational."
    },
    "garbled_string": {
      "map": {
        "\\alpha": "\\qzxwvtnp",
        "\\theta": "\\hjgrksla",
        "r": "qprnjdks",
        "s": "xlmvczbt",
        "n": "zvyrhgpl",
        "m": "kdfsghtr",
        "t": "skvnjmwr"
      },
      "question": "B-3. Prove that if \\( \\qzxwvtnp \\) is a real number such that\n\\[\n\\cos \\pi \\qzxwvtnp = 1 / 3,\n\\]\nthen \\( \\qzxwvtnp \\) is irrational. (The angle \\( \\pi \\qzxwvtnp \\) is in radians.)",
      "solution": "B-3.\nIf \\( \\qzxwvtnp = qprnjdks / xlmvczbt \\) with \\( qprnjdks \\) and \\( xlmvczbt \\) integers and \\( xlmvczbt>0 \\), then \\( \\cos ( zvy rhgpl \\pi \\qzxwvtnp ) \\) takes on at most \\( 2 xlmvczbt \\) distinct values for integral choices of \\( zvyrhgpl \\). When \\( \\cos \\pi \\qzxwvtnp = 1 / 3 \\), the formula \\( \\cos 2 \\hjgrksla = 2 \\cos ^{2} \\hjgrksla - 1 \\) and mathematical induction can be used to show that\n\\[\n\\cos \\left(2^{ kdfsghtr } \\pi \\qzxwvtnp \\right)= skvnjmwr / 3^{ 2^{ kdfsghtr } } \\quad \\quad[ kdfsghtr = 1,2,3, \\cdots]\n\\]\nwith \\( skvnjmwr \\) an integer not divisible by 3, and hence that these cosines form an infinite set of distinct values. Thus \\( \\qzxwvtnp \\) is irrational."
    },
    "kernel_variant": {
      "question": "Let α be a real number such that  \n  cos (π α)=2/5.\n\n(a)  Prove that α is irrational.\n\n(b)  Let p,s be positive integers with gcd(p,s)=1.  \n Determine exactly for which values of s the number  \n\n  cos(π p/s)\n\n is rational, and for every such s list all possible rational values of the cosine.\n\n(c)  Using your answer to (b) together with the special value cos (π α)=2/5, show that k α is irrational for every non–zero integer k.  \n(Throughout, angles are measured in radians.)",
      "solution": "Notation.  ζ_n:=e^{2π i/n}.  \nFor m≥0 let T_m be the Chebyshev polynomials of the first kind,\n T_0(x)=1, T_1(x)=x, T_{m+1}(x)=2xT_m(x)−T_{m−1}(x);  \nthey satisfy T_m(cos θ)=cos(m θ) and T_m∈ℤ[x].\n\n-------------------------------------------------\n(a) Irrationality of α\n\nAssume, for contradiction, that α=p/q with gcd(p,q)=1 and q>0.  \nPut  \n\n β:=2 cos(π α)=4/5∈ℚ. (1)\n\nWrite p=2^{ε}p₀ with ε∈{0,1} and p₀ odd.\n\n•  Case ε=0 (p odd).  \n Then e^{iπ α}=ζ_{2q}^{\\,p} is a primitive 2q-th root of unity, so let n:=2q.  \n Because β=ζ_n^{\\,p}+ζ_n^{−p}, we have β∈ℚ(ζ_n+ζ_n^{−1}).  \n We now show that β actually generates the whole real cyclotomic subfield.\n\n  Lemma 1.  If gcd(t,n)=1, then ζ_n^{\\,t}+ζ_n^{−t} is a primitive element of ℚ(ζ_n+ζ_n^{−1}); in particular  \n   ℚ(ζ_n^{\\,t}+ζ_n^{−t}) = ℚ(ζ_n+ζ_n^{−1}).  \n\n  Proof.  Let σ_t:ζ_n↦ζ_n^{\\,t}.  Since gcd(t,n)=1, σ_t is an automorphism of ℚ(ζ_n).  Hence  \n   σ_t(ζ_n+ζ_n^{−1}) = ζ_n^{\\,t}+ζ_n^{−t}.  \n  Thus ζ_n^{\\,t}+ζ_n^{−t} is Galois-conjugate to ζ_n+ζ_n^{−1}, and both elements therefore generate the same fixed field of complex conjugation, namely the real subfield. ∎\n\n Applying Lemma 1 with t=p gives  \n  [ℚ(β):ℚ]=[ℚ(ζ_n+ζ_n^{−1}):ℚ]=φ(n)/2   (n>2). (2)\n\n Because β∈ℚ, we must have φ(n)/2=1, i.e. φ(n)=2.  The positive integers n with φ(n)=2 are n=3,4,6.  Here n=2q is even and ≥4, hence n∈{4,6} giving q∈{2,3}.  \n  – If q=2 then α is a half–integer and cos(π α)∈{±1,0}.  \n  – If q=3 then α=p/3 with p odd and cos(π α)=±1/2.  \n Neither possibility equals 2/5, contradicting (1).\n\n•  Case ε=1 (p even).  \n Write p=2p₀ (p₀ odd); now q is odd.  Then e^{iπ α}=ζ_q^{\\,p₀} is primitive, so put n:=q.  \n Exactly as above, β generates ℚ(ζ_n+ζ_n^{−1}) and (2) holds.  With n odd, φ(n)=2 gives n=3 and q=3.  Then α=2p₀/3 and cos(π α)=cos(2π p₀/3)=±1/2, again contradicting (1).\n\nBoth cases are impossible; hence α is irrational.\n\n-------------------------------------------------\n(b) Rational angles yielding rational cosines\n\nLet θ:=p/s with gcd(p,s)=1, s>0, and set  \n\n γ:=2 cos(π p/s)=ζ_{2s}^{\\,p}+ζ_{2s}^{−p}. (3)\n\nWrite p=2^{ε}p₀ with ε∈{0,1} and p₀ odd, and define  \n\n m:=  { 2s if ε=0 (p odd),  \n     s  if ε=1 (p even).     (4)\n\nIf ε=1 then s is necessarily odd, so ζ_{2s}^{\\,p} becomes a primitive m-th root of unity; if ε=0 the primitivity is evident.  Thus γ=ζ_m^{\\,p₀}+ζ_m^{−p₀} with gcd(p₀,m)=1.\n\nBy Lemma 1, γ is a primitive element of the real cyclotomic subfield ℚ(ζ_m+ζ_m^{−1}).  Consequently  \n\n [ℚ(γ):ℚ]= { 1 if m≤2,  \n       φ(m)/2 if m≥3.    (5)\n\nTherefore γ∈ℚ exactly when m≤2 or φ(m)=2.  As before φ(m)=2 ⇔ m∈{3,4,6}.  Translating back to s via (4) gives\n\n m=1,2 ⇒ s=1  ⇒ cos(π p)=±1;  \n m=3  ⇒ ε=1, s=3 ⇒ cos(π p/3)=±1/2;  \n m=4  ⇒ ε=0, s=2 ⇒ cos(π p/2)=0;  \n m=6  ⇒ ε=0, s=3 ⇒ cos(π p/3)=±1/2.\n\nHence\n\n s∈{1,2,3} and cos(π p/s)∈{1,−1,0,1/2,−1/2}. (6)\n\nConversely each value in (6) is achieved for some p coprime to the corresponding s (for instance p≡1 mod 2,3), so the classification is complete.\n\n-------------------------------------------------\n(c) Every non–zero integral multiple of α is irrational\n\nFor any k∈ℤ we have  \n\n cos(k π α)=T_{|k|}(cos π α)=T_{|k|}(2/5)∈ℚ, (7)\n\nusing evenness of cosine to reduce to |k|.  Assume that kα=r/s with k≠0, gcd(r,s)=1, s>0.  \nThen cos(k π α)=cos(π r/s) is rational, so by (b)\n\n s∈{1,2,3} and cos(π r/s)∈{±1,0,±1/2}. (8)\n\nOn the other hand expand T_{|k|}:\n\n T_{|k|}(x)=∑_{j=0}^{⌊|k|/2⌋} a_{k,j}\\,x^{|k|−2j}, a_{k,j}∈ℤ, a_{k,0}=2^{|k|−1} (|k|≥1).\n\nPlugging x=2/5 gives  \n\n 5^{|k|}T_{|k|}(2/5)=2^{2|k|−1}+5⋅M, M∈ℤ. (9)\n\nThe leading term 2^{2|k|−1} is not divisible by 5, whereas every other term contains a factor 5.  Thus the reduced fraction T_{|k|}(2/5) has denominator exactly 5^{|k|}.  Consequently\n\n T_{|k|}(2/5)∉{±1,0,±1/2},  (|k|≥1). (10)\n\nThis contradicts (8).  Therefore no non-zero integral multiple of α can be rational; i.e. k α∉ℚ for all k≠0.\n\n-------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.614701",
        "was_fixed": false,
        "difficulty_analysis": "The original problem merely asked to show that a single specific value α with\ncos(π α)=1/3 is irrational, using elementary doubling identities.  The present\nvariant introduces three major layers of additional complexity:                \n\n1.  Cyclotomic‐field machinery and Euler’s totient function are required to\n    link rational angles to the degree of roots of unity, something entirely\n    absent from the original.                                                  \n\n2.  Part (b) demands a complete classification of all rational angles whose\n    cosine is rational, forcing the solver to solve φ(n)=2 and to understand\n    why only n∈{3,4,6} occur; this pushes the problem well beyond simple\n    trigonometric identities into algebraic-number theory and group theory.    \n\n3.  Part (c) couples the algebraic classification with explicit evaluations of\n    Chebyshev polynomials, requiring knowledge of their arithmetic properties\n    (denominators that are powers of 5) to exclude every historically allowed\n    rational value, thereby ruling out rational multiples of α.                \n\nThese added requirements make the problem significantly harder: the solver\nmust blend elementary trigonometry with cyclotomic fields, degree arguments,\nand polynomial arithmetic, whereas the original solution needed only an\ninduction on a double-angle formula."
      }
    },
    "original_kernel_variant": {
      "question": "Let α be a real number such that  \n  cos (π α)=2/5.\n\n(a)  Prove that α is irrational.\n\n(b)  Let p,s be positive integers with gcd(p,s)=1.  \n Determine exactly for which values of s the number  \n\n  cos(π p/s)\n\n is rational, and for every such s list all possible rational values of the cosine.\n\n(c)  Using your answer to (b) together with the special value cos (π α)=2/5, show that k α is irrational for every non–zero integer k.  \n(Throughout, angles are measured in radians.)",
      "solution": "Notation.  ζ_n:=e^{2π i/n}.  \nFor m≥0 let T_m be the Chebyshev polynomials of the first kind,\n T_0(x)=1, T_1(x)=x, T_{m+1}(x)=2xT_m(x)−T_{m−1}(x);  \nthey satisfy T_m(cos θ)=cos(m θ) and T_m∈ℤ[x].\n\n-------------------------------------------------\n(a) Irrationality of α\n\nAssume, for contradiction, that α=p/q with gcd(p,q)=1 and q>0.  \nPut  \n\n β:=2 cos(π α)=4/5∈ℚ. (1)\n\nWrite p=2^{ε}p₀ with ε∈{0,1} and p₀ odd.\n\n•  Case ε=0 (p odd).  \n Then e^{iπ α}=ζ_{2q}^{\\,p} is a primitive 2q-th root of unity, so let n:=2q.  \n Because β=ζ_n^{\\,p}+ζ_n^{−p}, we have β∈ℚ(ζ_n+ζ_n^{−1}).  \n We now show that β actually generates the whole real cyclotomic subfield.\n\n  Lemma 1.  If gcd(t,n)=1, then ζ_n^{\\,t}+ζ_n^{−t} is a primitive element of ℚ(ζ_n+ζ_n^{−1}); in particular  \n   ℚ(ζ_n^{\\,t}+ζ_n^{−t}) = ℚ(ζ_n+ζ_n^{−1}).  \n\n  Proof.  Let σ_t:ζ_n↦ζ_n^{\\,t}.  Since gcd(t,n)=1, σ_t is an automorphism of ℚ(ζ_n).  Hence  \n   σ_t(ζ_n+ζ_n^{−1}) = ζ_n^{\\,t}+ζ_n^{−t}.  \n  Thus ζ_n^{\\,t}+ζ_n^{−t} is Galois-conjugate to ζ_n+ζ_n^{−1}, and both elements therefore generate the same fixed field of complex conjugation, namely the real subfield. ∎\n\n Applying Lemma 1 with t=p gives  \n  [ℚ(β):ℚ]=[ℚ(ζ_n+ζ_n^{−1}):ℚ]=φ(n)/2   (n>2). (2)\n\n Because β∈ℚ, we must have φ(n)/2=1, i.e. φ(n)=2.  The positive integers n with φ(n)=2 are n=3,4,6.  Here n=2q is even and ≥4, hence n∈{4,6} giving q∈{2,3}.  \n  – If q=2 then α is a half–integer and cos(π α)∈{±1,0}.  \n  – If q=3 then α=p/3 with p odd and cos(π α)=±1/2.  \n Neither possibility equals 2/5, contradicting (1).\n\n•  Case ε=1 (p even).  \n Write p=2p₀ (p₀ odd); now q is odd.  Then e^{iπ α}=ζ_q^{\\,p₀} is primitive, so put n:=q.  \n Exactly as above, β generates ℚ(ζ_n+ζ_n^{−1}) and (2) holds.  With n odd, φ(n)=2 gives n=3 and q=3.  Then α=2p₀/3 and cos(π α)=cos(2π p₀/3)=±1/2, again contradicting (1).\n\nBoth cases are impossible; hence α is irrational.\n\n-------------------------------------------------\n(b) Rational angles yielding rational cosines\n\nLet θ:=p/s with gcd(p,s)=1, s>0, and set  \n\n γ:=2 cos(π p/s)=ζ_{2s}^{\\,p}+ζ_{2s}^{−p}. (3)\n\nWrite p=2^{ε}p₀ with ε∈{0,1} and p₀ odd, and define  \n\n m:=  { 2s if ε=0 (p odd),  \n     s  if ε=1 (p even).     (4)\n\nIf ε=1 then s is necessarily odd, so ζ_{2s}^{\\,p} becomes a primitive m-th root of unity; if ε=0 the primitivity is evident.  Thus γ=ζ_m^{\\,p₀}+ζ_m^{−p₀} with gcd(p₀,m)=1.\n\nBy Lemma 1, γ is a primitive element of the real cyclotomic subfield ℚ(ζ_m+ζ_m^{−1}).  Consequently  \n\n [ℚ(γ):ℚ]= { 1 if m≤2,  \n       φ(m)/2 if m≥3.    (5)\n\nTherefore γ∈ℚ exactly when m≤2 or φ(m)=2.  As before φ(m)=2 ⇔ m∈{3,4,6}.  Translating back to s via (4) gives\n\n m=1,2 ⇒ s=1  ⇒ cos(π p)=±1;  \n m=3  ⇒ ε=1, s=3 ⇒ cos(π p/3)=±1/2;  \n m=4  ⇒ ε=0, s=2 ⇒ cos(π p/2)=0;  \n m=6  ⇒ ε=0, s=3 ⇒ cos(π p/3)=±1/2.\n\nHence\n\n s∈{1,2,3} and cos(π p/s)∈{1,−1,0,1/2,−1/2}. (6)\n\nConversely each value in (6) is achieved for some p coprime to the corresponding s (for instance p≡1 mod 2,3), so the classification is complete.\n\n-------------------------------------------------\n(c) Every non–zero integral multiple of α is irrational\n\nFor any k∈ℤ we have  \n\n cos(k π α)=T_{|k|}(cos π α)=T_{|k|}(2/5)∈ℚ, (7)\n\nusing evenness of cosine to reduce to |k|.  Assume that kα=r/s with k≠0, gcd(r,s)=1, s>0.  \nThen cos(k π α)=cos(π r/s) is rational, so by (b)\n\n s∈{1,2,3} and cos(π r/s)∈{±1,0,±1/2}. (8)\n\nOn the other hand expand T_{|k|}:\n\n T_{|k|}(x)=∑_{j=0}^{⌊|k|/2⌋} a_{k,j}\\,x^{|k|−2j}, a_{k,j}∈ℤ, a_{k,0}=2^{|k|−1} (|k|≥1).\n\nPlugging x=2/5 gives  \n\n 5^{|k|}T_{|k|}(2/5)=2^{2|k|−1}+5⋅M, M∈ℤ. (9)\n\nThe leading term 2^{2|k|−1} is not divisible by 5, whereas every other term contains a factor 5.  Thus the reduced fraction T_{|k|}(2/5) has denominator exactly 5^{|k|}.  Consequently\n\n T_{|k|}(2/5)∉{±1,0,±1/2},  (|k|≥1). (10)\n\nThis contradicts (8).  Therefore no non-zero integral multiple of α can be rational; i.e. k α∉ℚ for all k≠0.\n\n-------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.491859",
        "was_fixed": false,
        "difficulty_analysis": "The original problem merely asked to show that a single specific value α with\ncos(π α)=1/3 is irrational, using elementary doubling identities.  The present\nvariant introduces three major layers of additional complexity:                \n\n1.  Cyclotomic‐field machinery and Euler’s totient function are required to\n    link rational angles to the degree of roots of unity, something entirely\n    absent from the original.                                                  \n\n2.  Part (b) demands a complete classification of all rational angles whose\n    cosine is rational, forcing the solver to solve φ(n)=2 and to understand\n    why only n∈{3,4,6} occur; this pushes the problem well beyond simple\n    trigonometric identities into algebraic-number theory and group theory.    \n\n3.  Part (c) couples the algebraic classification with explicit evaluations of\n    Chebyshev polynomials, requiring knowledge of their arithmetic properties\n    (denominators that are powers of 5) to exclude every historically allowed\n    rational value, thereby ruling out rational multiples of α.                \n\nThese added requirements make the problem significantly harder: the solver\nmust blend elementary trigonometry with cyclotomic fields, degree arguments,\nand polynomial arithmetic, whereas the original solution needed only an\ninduction on a double-angle formula."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}