{
  "index": "1957-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "3. \\( A \\) and \\( B \\) are real numbers and \\( k \\) a positive integer. Show that\n\\[\n\\left|\\frac{\\cos k B \\cos A-\\cos k A \\cos B}{\\cos B-\\cos A}\\right|<k^{2}-1\n\\]\nwhenever the left side is defined.",
  "solution": "Solution. Let \\( x=(A-B) / 2, y=(A+B) / 2 \\). The numerator of the expression on the left is\n\\[\n\\begin{array}{l}\n\\frac{1}{2}[\\cos (k B+A)+\\cos (k B-A)-\\cos (k A+B)-\\cos (k A-B)] \\\\\n=\\frac{1}{2}[\\cos (k B+A)-\\cos (k A+B)]+\\frac{1}{2}[\\cos (k B-A)-\\cos (k A-B)] \\\\\n=\\sin (k-1) x \\sin (k+1) y+\\sin (k+1) x \\sin (k-1) y .\n\\end{array}\n\\]\n\nThe denominator is \\( 2 \\sin x \\sin y \\), and we may assume this is not zero. Hence we have\n\\[\n\\begin{aligned}\n\\left|\\frac{\\cos k B \\cos A-\\cos k A \\cos B}{\\cos B-\\cos A}\\right| \\leq & \\frac{1}{2}\\left|\\frac{\\sin (k-1) x}{\\sin x}\\right| \\cdot\\left|\\frac{\\sin (k+1) y}{\\sin y}\\right| \\\\\n& +\\frac{1}{2}\\left|\\frac{\\sin (k+1) x}{\\sin x}\\right| \\cdot\\left|\\frac{\\sin (k-1) y}{\\sin y}\\right|\n\\end{aligned}\n\\]\n\nNow since \\( |\\sin n z| \\leq n|\\sin z| \\) for all \\( z \\) and all positive integers \\( n \\) with strict inequality unless \\( n=1 \\) or \\( \\sin z=0 \\) (this is proved below), both terms are less than \\( \\left(k^{2}-1\\right) / 2 \\) provided \\( k>1 \\). Therefore\n\\[\n\\left|\\frac{\\cos k B \\cos A-\\cos k A \\cos B}{\\cos B-\\cos A}\\right|<k^{2}-1\n\\]\nprovided \\( k>1 \\) and \\( \\cos B \\neq \\cos A \\). Obviously, we have equality if \\( k=1 \\), so the problem is not accurately phrased.\n\nWe now prove that \\( |\\sin n z| \\leq n|\\sin z| \\) for all positive integers \\( n \\). If \\( \\sin z=0 \\), equality holds for every integer \\( n \\), so we assume from now on that \\( \\sin z \\neq 0 \\); then \\( |\\cos z|<1 \\). We continue by induction on \\( n \\). Clearly there is equality if \\( n=1 \\). For \\( n=2 \\), we have \\( |\\sin 2 z|=2|\\cos z| \\cdot|\\sin z| \\) \\( <2|\\sin z| \\). Suppose we have strict inequality for \\( n=k \\). Then\n\\[\n\\begin{aligned}\n|\\sin (k+1) z| & =|\\sin k z \\cos z+\\cos k z \\sin z| \\\\\n& \\leq|\\sin k z|+|\\sin z|<(k+1)|\\sin z| .\n\\end{aligned}\n\\]\n\nThus we have strict inequality for \\( n=k+1 \\). Hence we have proved\n\\[\n|\\sin n z| \\leq|\\sin z|\n\\]\nfor all real \\( z \\) and all positive integers \\( n \\) with strict inequality unless \\( n=1 \\) or \\( \\sin z=0 \\).\n\nRemark. In the report on the Competition (American Mathematical Monthly, vol. 64 (1957), pages 649-654) the error in the statement of the problem was explained as follows. \"The questions committee originally had a ' \\( \\leq \\) ' in this question, but, in some manner, the ' \\( = \\) ' was lost before the question reached the director. The omission was discovered in reading the proofs, but the chairman of the committee decided that it was just as well to give the contestants the opportunity to discover the error and correct it.\"",
  "vars": [
    "x",
    "y",
    "z",
    "n"
  ],
  "params": [
    "A",
    "B",
    "k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "anglea",
        "B": "angleb",
        "k": "integerk",
        "x": "halfdiff",
        "y": "halfsum",
        "z": "variable",
        "n": "counter"
      },
      "question": "3. \\( anglea \\) and \\( angleb \\) are real numbers and \\( integerk \\) a positive integer. Show that\n\\[\n\\left|\\frac{\\cos integerk angleb \\cos anglea-\\cos integerk anglea \\cos angleb}{\\cos angleb-\\cos anglea}\\right|<integerk^{2}-1\n\\]\nwhenever the left side is defined.",
      "solution": "Solution. Let \\( halfdiff=(anglea-angleb) / 2,\\, halfsum=(anglea+angleb) / 2 \\). The numerator of the expression on the left is\n\\[\n\\begin{array}{l}\n\\frac{1}{2}[\\cos (integerk angleb+anglea)+\\cos (integerk angleb-anglea)-\\cos (integerk anglea+angleb)-\\cos (integerk anglea-angleb)] \\\\\n=\\frac{1}{2}[\\cos (integerk angleb+anglea)-\\cos (integerk anglea+angleb)]+\\frac{1}{2}[\\cos (integerk angleb-anglea)-\\cos (integerk anglea-angleb)] \\\\\n=\\sin (integerk-1) halfdiff \\sin (integerk+1) halfsum+\\sin (integerk+1) halfdiff \\sin (integerk-1) halfsum .\n\\end{array}\n\\]\n\nThe denominator is \\( 2 \\sin halfdiff \\sin halfsum \\), and we may assume this is not zero. Hence we have\n\\[\n\\begin{aligned}\n\\left|\\frac{\\cos integerk angleb \\cos anglea-\\cos integerk anglea \\cos angleb}{\\cos angleb-\\cos anglea}\\right| \\leq & \\frac{1}{2}\\left|\\frac{\\sin (integerk-1) halfdiff}{\\sin halfdiff}\\right| \\cdot\\left|\\frac{\\sin (integerk+1) halfsum}{\\sin halfsum}\\right| \\\\\n& +\\frac{1}{2}\\left|\\frac{\\sin (integerk+1) halfdiff}{\\sin halfdiff}\\right| \\cdot\\left|\\frac{\\sin (integerk-1) halfsum}{\\sin halfsum}\\right|\n\\end{aligned}\n\\]\n\nNow since \\( |\\sin counter variable| \\leq counter|\\sin variable| \\) for all \\( variable \\) and all positive integers \\( counter \\) with strict inequality unless \\( counter=1 \\) or \\( \\sin variable=0 \\) (this is proved below), both terms are less than \\( \\left(integerk^{2}-1\\right) / 2 \\) provided \\( integerk>1 \\). Therefore\n\\[\n\\left|\\frac{\\cos integerk angleb \\cos anglea-\\cos integerk anglea \\cos angleb}{\\cos angleb-\\cos anglea}\\right|<integerk^{2}-1\n\\]\nprovided \\( integerk>1 \\) and \\( \\cos angleb \\neq \\cos anglea \\). Obviously, we have equality if \\( integerk=1 \\), so the problem is not accurately phrased.\n\nWe now prove that \\( |\\sin counter variable| \\leq counter|\\sin variable| \\) for all positive integers \\( counter \\). If \\( \\sin variable=0 \\), equality holds for every integer \\( counter \\), so we assume from now on that \\( \\sin variable \\neq 0 \\); then \\( |\\cos variable|<1 \\). We continue by induction on \\( counter \\). Clearly there is equality if \\( counter=1 \\). For \\( counter=2 \\), we have \\( |\\sin 2 variable|=2|\\cos variable| \\cdot|\\sin variable| <2|\\sin variable| \\). Suppose we have strict inequality for \\( counter=integerk \\). Then\n\\[\n\\begin{aligned}\n|\\sin (integerk+1) variable| & =|\\sin integerk variable \\cos variable+\\cos integerk variable \\sin variable| \\\\\n& \\leq|\\sin integerk variable|+|\\sin variable|<(integerk+1)|\\sin variable| .\n\\end{aligned}\n\\]\n\nThus we have strict inequality for \\( counter=integerk+1 \\). Hence we have proved\n\\[\n|\\sin counter variable| \\leq|\\sin variable|\n\\]\nfor all real \\( variable \\) and all positive integers \\( counter \\) with strict inequality unless \\( counter=1 \\) or \\( \\sin variable=0 \\).\n\nRemark. In the report on the Competition (American Mathematical Monthly, vol. 64 (1957), pages 649-654) the error in the statement of the problem was explained as follows. \"The questions committee originally had a ' \\( \\leq \\) ' in this question, but, in some manner, the ' \\( = \\) ' was lost before the question reached the director. The omission was discovered in reading the proofs, but the chairman of the committee decided that it was just as well to give the contestants the opportunity to discover the error and correct it.\""
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "sandcastle",
        "B": "moonflower",
        "k": "blackboard",
        "x": "riverstone",
        "y": "starbright",
        "z": "counterlid",
        "n": "bridgework"
      },
      "question": "3. \\( sandcastle \\) and \\( moonflower \\) are real numbers and \\( blackboard \\) a positive integer. Show that\n\\[\n\\left|\\frac{\\cos blackboard moonflower \\cos sandcastle-\\cos blackboard sandcastle \\cos moonflower}{\\cos moonflower-\\cos sandcastle}\\right|<blackboard^{2}-1\n\\]\nwhenever the left side is defined.",
      "solution": "Solution. Let \\( riverstone=(sandcastle-moonflower) / 2, starbright=(sandcastle+moonflower) / 2 \\). The numerator of the expression on the left is\n\\[\n\\begin{array}{l}\n\\frac{1}{2}[\\cos (blackboard moonflower+sandcastle)+\\cos (blackboard moonflower-sandcastle)-\\cos (blackboard sandcastle+moonflower)-\\cos (blackboard sandcastle-moonflower)] \\\\\n=\\frac{1}{2}[\\cos (blackboard moonflower+sandcastle)-\\cos (blackboard sandcastle+moonflower)]+\\frac{1}{2}[\\cos (blackboard moonflower-sandcastle)-\\cos (blackboard sandcastle-moonflower)] \\\\\n=\\sin (blackboard-1) riverstone \\sin (blackboard+1) starbright+\\sin (blackboard+1) riverstone \\sin (blackboard-1) starbright .\n\\end{array}\n\\]\n\nThe denominator is \\( 2 \\sin riverstone \\sin starbright \\), and we may assume this is not zero. Hence we have\n\\[\n\\begin{aligned}\n\\left|\\frac{\\cos blackboard moonflower \\cos sandcastle-\\cos blackboard sandcastle \\cos moonflower}{\\cos moonflower-\\cos sandcastle}\\right| \\leq & \\frac{1}{2}\\left|\\frac{\\sin (blackboard-1) riverstone}{\\sin riverstone}\\right| \\cdot\\left|\\frac{\\sin (blackboard+1) starbright}{\\sin starbright}\\right| \\\\\n& +\\frac{1}{2}\\left|\\frac{\\sin (blackboard+1) riverstone}{\\sin riverstone}\\right| \\cdot\\left|\\frac{\\sin (blackboard-1) starbright}{\\sin starbright}\\right|\n\\end{aligned}\n\\]\n\nNow since \\( |\\sin bridgework counterlid| \\leq bridgework|\\sin counterlid| \\) for all \\( counterlid \\) and all positive integers \\( bridgework \\) with strict inequality unless \\( bridgework=1 \\) or \\( \\sin counterlid=0 \\) (this is proved below), both terms are less than \\( \\left(blackboard^{2}-1\\right) / 2 \\) provided \\( blackboard>1 \\). Therefore\n\\[\n\\left|\\frac{\\cos blackboard moonflower \\cos sandcastle-\\cos blackboard sandcastle \\cos moonflower}{\\cos moonflower-\\cos sandcastle}\\right|<blackboard^{2}-1\n\\]\nprovided \\( blackboard>1 \\) and \\( \\cos moonflower \\neq \\cos sandcastle \\). Obviously, we have equality if \\( blackboard=1 \\), so the problem is not accurately phrased.\n\nWe now prove that \\( |\\sin bridgework counterlid| \\leq bridgework|\\sin counterlid| \\) for all positive integers \\( bridgework \\). If \\( \\sin counterlid=0 \\), equality holds for every integer \\( bridgework \\), so we assume from now on that \\( \\sin counterlid \\neq 0 \\); then \\( |\\cos counterlid|<1 \\). We continue by induction on \\( bridgework \\). Clearly there is equality if \\( bridgework=1 \\). For \\( bridgework=2 \\), we have \\( |\\sin 2 counterlid|=2|\\cos counterlid| \\cdot|\\sin counterlid| \\)  \\(<2|\\sin counterlid| \\). Suppose we have strict inequality for \\( bridgework=blackboard \\). Then\n\\[\n\\begin{aligned}\n|\\sin (blackboard+1) counterlid| & =|\\sin blackboard counterlid \\cos counterlid+\\cos blackboard counterlid \\sin counterlid| \\\\\n& \\leq|\\sin blackboard counterlid|+|\\sin counterlid|<(blackboard+1)|\\sin counterlid| .\n\\end{aligned}\n\\]\n\nThus we have strict inequality for \\( bridgework=blackboard+1 \\). Hence we have proved\n\\[\n|\\sin bridgework counterlid| \\leq|\\sin counterlid|\n\\]\nfor all real \\( counterlid \\) and all positive integers \\( bridgework \\) with strict inequality unless \\( bridgework=1 \\) or \\( \\sin counterlid=0 \\).\n\nRemark. In the report on the Competition (American Mathematical Monthly, vol. 64 (1957), pages 649-654) the error in the statement of the problem was explained as follows. \"The questions committee originally had a ' \\( \\leq \\) ' in this question, but, in some manner, the ' \\( = \\) ' was lost before the question reached the director. The omission was discovered in reading the proofs, but the chairman of the committee decided that it was just as well to give the contestants the opportunity to discover the error and correct it.\""
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "additionsum",
        "y": "differencepart",
        "z": "constantvalue",
        "n": "fractionalpart",
        "A": "imaginaryvalue",
        "B": "complexnumber",
        "k": "negativefraction"
      },
      "question": "3. \\( imaginaryvalue \\) and \\( complexnumber \\) are real numbers and \\( negativefraction \\) a positive integer. Show that\n\\[\n\\left|\\frac{\\cos negativefraction complexnumber \\cos imaginaryvalue-\\cos negativefraction imaginaryvalue \\cos complexnumber}{\\cos complexnumber-\\cos imaginaryvalue}\\right|<negativefraction^{2}-1\n\\]\nwhenever the left side is defined.",
      "solution": "Solution. Let \\( additionsum=(imaginaryvalue-complexnumber) / 2, differencepart=(imaginaryvalue+complexnumber) / 2 \\). The numerator of the expression on the left is\n\\[\n\\begin{array}{l}\n\\frac{1}{2}[\\cos (negativefraction complexnumber+imaginaryvalue)+\\cos (negativefraction complexnumber-imaginaryvalue)-\\cos (negativefraction imaginaryvalue+complexnumber)-\\cos (negativefraction imaginaryvalue-complexnumber)] \\\\\n=\\frac{1}{2}[\\cos (negativefraction complexnumber+imaginaryvalue)-\\cos (negativefraction imaginaryvalue+complexnumber)]+\\frac{1}{2}[\\cos (negativefraction complexnumber-imaginaryvalue)-\\cos (negativefraction imaginaryvalue-complexnumber)] \\\\\n=\\sin (negativefraction-1) additionsum \\sin (negativefraction+1) differencepart+\\sin (negativefraction+1) additionsum \\sin (negativefraction-1) differencepart .\n\\end{array}\n\\]\n\nThe denominator is \\( 2 \\sin additionsum \\sin differencepart \\), and we may assume this is not zero. Hence we have\n\\[\n\\begin{aligned}\n\\left|\\frac{\\cos negativefraction complexnumber \\cos imaginaryvalue-\\cos negativefraction imaginaryvalue \\cos complexnumber}{\\cos complexnumber-\\cos imaginaryvalue}\\right| \\leq & \\frac{1}{2}\\left|\\frac{\\sin (negativefraction-1) additionsum}{\\sin additionsum}\\right| \\cdot\\left|\\frac{\\sin (negativefraction+1) differencepart}{\\sin differencepart}\\right| \\\\\n& +\\frac{1}{2}\\left|\\frac{\\sin (negativefraction+1) additionsum}{\\sin additionsum}\\right| \\cdot\\left|\\frac{\\sin (negativefraction-1) differencepart}{\\sin differencepart}\\right|\n\\end{aligned}\n\\]\n\nNow since \\( |\\sin fractionalpart constantvalue| \\leq fractionalpart|\\sin constantvalue| \\) for all \\( constantvalue \\) and all positive integers \\( fractionalpart \\) with strict inequality unless \\( fractionalpart=1 \\) or \\( \\sin constantvalue=0 \\) (this is proved below), both terms are less than \\( \\left(negativefraction^{2}-1\\right) / 2 \\) provided \\( negativefraction>1 \\). Therefore\n\\[\n\\left|\\frac{\\cos negativefraction complexnumber \\cos imaginaryvalue-\\cos negativefraction imaginaryvalue \\cos complexnumber}{\\cos complexnumber-\\cos imaginaryvalue}\\right|<negativefraction^{2}-1\n\\]\nprovided \\( negativefraction>1 \\) and \\( \\cos complexnumber \\neq \\cos imaginaryvalue \\). Obviously, we have equality if \\( negativefraction=1 \\), so the problem is not accurately phrased.\n\nWe now prove that \\( |\\sin fractionalpart constantvalue| \\leq fractionalpart|\\sin constantvalue| \\) for all positive integers \\( fractionalpart \\). If \\( \\sin constantvalue=0 \\), equality holds for every integer \\( fractionalpart \\), so we assume from now on that \\( \\sin constantvalue \\neq 0 \\); then \\( |\\cos constantvalue|<1 \\). We continue by induction on \\( fractionalpart \\). Clearly there is equality if \\( fractionalpart=1 \\). For \\( fractionalpart=2 \\), we have \\( |\\sin 2 constantvalue|=2|\\cos constantvalue| \\cdot|\\sin constantvalue| \\) \\( <2|\\sin constantvalue| \\). Suppose we have strict inequality for \\( fractionalpart=negativefraction \\). Then\n\\[\n\\begin{aligned}\n|\\sin (negativefraction+1) constantvalue| & =|\\sin negativefraction constantvalue \\cos constantvalue+\\cos negativefraction constantvalue \\sin constantvalue| \\\\\n& \\leq|\\sin negativefraction constantvalue|+|\\sin constantvalue|<(negativefraction+1)|\\sin constantvalue| .\n\\end{aligned}\n\\]\n\nThus we have strict inequality for \\( fractionalpart=negativefraction+1 \\). Hence we have proved\n\\[\n|\\sin fractionalpart constantvalue| \\leq|\\sin constantvalue|\n\\]\nfor all real \\( constantvalue \\) and all positive integers \\( fractionalpart \\) with strict inequality unless \\( fractionalpart=1 \\) or \\( \\sin constantvalue=0 \\).\n\nRemark. In the report on the Competition (American Mathematical Monthly, vol. 64 (1957), pages 649-654) the error in the statement of the problem was explained as follows. \"The questions committee originally had a ' \\( \\leq \\) ' in this question, but, in some manner, the ' \\( = \\) ' was lost before the question reached the director. The omission was discovered in reading the proofs, but the chairman of the committee decided that it was just as well to give the contestants the opportunity to discover the error and correct it.\""
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mnplqtrs",
        "n": "cskdfghj",
        "A": "vbxnmzpr",
        "B": "lkjhgfre",
        "k": "dfghjqwe"
      },
      "question": "3. \\( vbxnmzpr \\) and \\( lkjhgfre \\) are real numbers and \\( dfghjqwe \\) a positive integer. Show that\n\\[\n\\left|\\frac{\\cos dfghjqwe lkjhgfre \\cos vbxnmzpr-\\cos dfghjqwe vbxnmzpr \\cos lkjhgfre}{\\cos lkjhgfre-\\cos vbxnmzpr}\\right|<dfghjqwe^{2}-1\n\\]\nwhenever the left side is defined.",
      "solution": "Solution. Let \\( qzxwvtnp=(vbxnmzpr-lkjhgfre) / 2, hjgrksla=(vbxnmzpr+lkjhgfre) / 2 \\). The numerator of the expression on the left is\n\\[\n\\begin{array}{l}\n\\frac{1}{2}[\\cos (dfghjqwe lkjhgfre+vbxnmzpr)+\\cos (dfghjqwe lkjhgfre-vbxnmzpr)-\\cos (dfghjqwe vbxnmzpr+lkjhgfre)-\\cos (dfghjqwe vbxnmzpr-lkjhgfre)] \\\\\n=\\frac{1}{2}[\\cos (dfghjqwe lkjhgfre+vbxnmzpr)-\\cos (dfghjqwe vbxnmzpr+lkjhgfre)]+\\frac{1}{2}[\\cos (dfghjqwe lkjhgfre-vbxnmzpr)-\\cos (dfghjqwe vbxnmzpr-lkjhgfre)] \\\\\n=\\sin (dfghjqwe-1) qzxwvtnp \\sin (dfghjqwe+1) hjgrksla+\\sin (dfghjqwe+1) qzxwvtnp \\sin (dfghjqwe-1) hjgrksla .\n\\end{array}\n\\]\n\nThe denominator is \\( 2 \\sin qzxwvtnp \\sin hjgrksla \\), and we may assume this is not zero. Hence we have\n\\[\n\\begin{aligned}\n\\left|\\frac{\\cos dfghjqwe lkjhgfre \\cos vbxnmzpr-\\cos dfghjqwe vbxnmzpr \\cos lkjhgfre}{\\cos lkjhgfre-\\cos vbxnmzpr}\\right| \\leq & \\frac{1}{2}\\left|\\frac{\\sin (dfghjqwe-1) qzxwvtnp}{\\sin qzxwvtnp}\\right| \\cdot\\left|\\frac{\\sin (dfghjqwe+1) hjgrksla}{\\sin hjgrksla}\\right| \\\\\n& +\\frac{1}{2}\\left|\\frac{\\sin (dfghjqwe+1) qzxwvtnp}{\\sin qzxwvtnp}\\right| \\cdot\\left|\\frac{\\sin (dfghjqwe-1) hjgrksla}{\\sin hjgrksla}\\right|\n\\end{aligned}\n\\]\n\nNow since \\( |\\sin cskdfghj mnplqtrs| \\leq cskdfghj|\\sin mnplqtrs| \\) for all \\( mnplqtrs \\) and all positive integers \\( cskdfghj \\) with strict inequality unless \\( cskdfghj=1 \\) or \\( \\sin mnplqtrs=0 \\) (this is proved below), both terms are less than \\( \\left(dfghjqwe^{2}-1\\right) / 2 \\) provided \\( dfghjqwe>1 \\). Therefore\n\\[\n\\left|\\frac{\\cos dfghjqwe lkjhgfre \\cos vbxnmzpr-\\cos dfghjqwe vbxnmzpr \\cos lkjhgfre}{\\cos lkjhgfre-\\cos vbxnmzpr}\\right|<dfghjqwe^{2}-1\n\\]\nprovided \\( dfghjqwe>1 \\) and \\( \\cos lkjhgfre \\neq \\cos vbxnmzpr \\). Obviously, we have equality if \\( dfghjqwe=1 \\), so the problem is not accurately phrased.\n\nWe now prove that \\( |\\sin cskdfghj mnplqtrs| \\leq cskdfghj|\\sin mnplqtrs| \\) for all positive integers \\( cskdfghj \\). If \\( \\sin mnplqtrs=0 \\), equality holds for every integer \\( cskdfghj \\), so we assume from now on that \\( \\sin mnplqtrs \\neq 0 \\); then \\( |\\cos mnplqtrs|<1 \\). We continue by induction on \\( cskdfghj \\). Clearly there is equality if \\( cskdfghj=1 \\). For \\( cskdfghj=2 \\), we have \\( |\\sin 2 mnplqtrs|=2|\\cos mnplqtrs| \\cdot|\\sin mnplqtrs| \\) \\( <2|\\sin mnplqtrs| \\). Suppose we have strict inequality for \\( cskdfghj=dfghjqwe \\). Then\n\\[\n\\begin{aligned}\n|\\sin (dfghjqwe+1) mnplqtrs| & =|\\sin dfghjqwe mnplqtrs \\cos mnplqtrs+\\cos dfghjqwe mnplqtrs \\sin mnplqtrs| \\\\\n& \\leq|\\sin dfghjqwe mnplqtrs|+|\\sin mnplqtrs|<(dfghjqwe+1)|\\sin mnplqtrs| .\n\\end{aligned}\n\\]\n\nThus we have strict inequality for \\( cskdfghj=dfghjqwe+1 \\). Hence we have proved\n\\[\n|\\sin cskdfghj mnplqtrs| \\leq|\\sin mnplqtrs|\n\\]\nfor all real \\( mnplqtrs \\) and all positive integers \\( cskdfghj \\) with strict inequality unless \\( cskdfghj=1 \\) or \\( \\sin mnplqtrs=0 \\).\n\nRemark. In the report on the Competition (American Mathematical Monthly, vol. 64 (1957), pages 649-654) the error in the statement of the problem was explained as follows. \"The questions committee originally had a ' \\( \\leq \\) ' in this question, but, in some manner, the ' \\( = \\) ' was lost before the question reached the director. The omission was discovered in reading the proofs, but the chairman of the committee decided that it was just as well to give the contestants the opportunity to discover the error and correct it.\""
    },
    "kernel_variant": {
      "question": "Let k,m be positive integers with k>m≥1.  \nFor real numbers A,B satisfying cos A ≠ cos B define  \n\n  Φ_{k,m}(A,B)= (cos kB · cos mA – cos kA · cos mB)/(cos B – cos A).\n\n(a) Prove the sharp inequality  \n  |Φ_{k,m}(A,B)| < k² – m².  \n\n(b) Show that the constant k² – m² is best possible: for every ε>0 there exist real A,B with cos A≠cos B such that  \n\n  k² – m² – ε < |Φ_{k,m}(A,B)| < k² – m².  \n\nConsequently the strict inequality in part (a) is never an equality, yet the upper bound can be approached arbitrarily closely.",
      "solution": "Notation.  Put  \n\n x := (A–B)/2, y := (A+B)/2 so that A = y–x, B = y+x.  \n\nBecause cos A ≠ cos B we have  \n\n cos B – cos A = cos(y+x) – cos(y–x) = –2 sin x sin y ≠ 0. (1)\n\n\n\n1.  Trigonometric rewrites of the numerator  \n\nUsing the product–to–sum identities,\n\n cos kB cos mA = ½[cos(k(y+x)+m(y–x)) + cos(k(y+x)–m(y–x))],\n\n cos kA cos mB = ½[cos(k(y–x)+m(y+x)) + cos(k(y–x)–m(y+x))].\n\nSubtracting yields  \n\n N := cos kB cos mA – cos kA cos mB  \n\n  = –sin((k–m)x) sin((k+m)y) – sin((k+m)x) sin((k–m)y). (2)\n\n\n\n2.  Forming the quotient  \n\nFrom (1)–(2) and the triangle inequality we obtain  \n\n |Φ_{k,m}(A,B)| = |N| /(2|sin x sin y|)  \n\n  ≤ ½( |sin((k–m)x)|/|sin x|·|sin((k+m)y)|/|sin y|  \n\n    + |sin((k+m)x)|/|sin x|·|sin((k–m)y)|/|sin y| ). (3)\n\n\n\n3.  A classical majorisation of sine ratios  \n\nLemma 1.  For every integer n ≥ 1 and every real t,  \n\n |sin n t| ≤ n|sin t|,   \n\nwith equality iff sin t = 0 or n = 1.  \n\n(Proof: write sin n t = 2 sin t·T_{n–1}(cos t) and use |T_{n–1}(u)| ≤ n–1 for |u| ≤ 1.)\n\nApplying Lemma 1 termwise in (3) gives  \n\n |Φ_{k,m}(A,B)| ≤ ½[(k–m)(k+m) + (k+m)(k–m)] = k² – m². (4)\n\n\n\n4.  Why the inequality is strict  \n\nBecause sin x, sin y ≠ 0, at least one of the four ratios in (3) contains an integer n>1 (indeed k+m>1).  \nLemma 1 is then strict for that ratio, so (4) is strict:  \n\n |Φ_{k,m}(A,B)| < k² – m². Part (a) is proved.\n\n\n\n5.  Sharpness of the constant k²–m²  \n\nWe now construct a sequence (A_j,B_j) for which |Φ_{k,m}(A_j,B_j)| ↑ k²–m².\n\n5.1  A convenient one–parameter family  \n\nChoose x = y = δ with 0 < δ ≤ 1.  Then  \n\n A = 0, B = 2δ, sin x = sin y = sin δ > 0.  \n\nBoth terms inside the absolute value in (2) are negative, so the triangle inequality used in (3) is an equality; consequently  \n\n |Φ_{k,m}(A,B)| = |sin((k+m)δ)|/|sin δ| · |sin((k–m)δ)|/|sin δ|. (5)\n\n\n\n5.2  Introducing the deviation numbers ρ_n  \n\nFor n ≥ 1 and t ≠ 0 define  \n\n ρ_n(t) := n – |sin n t|/|sin t| ≥ 0. (6)\n\nWith n₁ := k+m, n₂ := k–m, equation (5) becomes  \n\n |Φ_{k,m}(A,B)| = (n₁ – ρ_{n₁}(δ))(n₂ – ρ_{n₂}(δ)). (7)\n\n\n\n5.3  A quantitative bound on ρ_n(t) valid up to t ≤ 1/n  \n\nLemma 2.  For every integer n ≥ 1 and |t| ≤ 1/n,  \n\n 0 ≤ ρ_n(t) ≤ (1/3) n³ t². (8)\n\nProof.  For |z| ≤ 1 the Taylor estimate  \n\n |sin z – z + z³/6| ≤ |z|⁵/120  \n\nimplies |sin z| ≥ |z| – |z|³/6.  Put z = n t with |t| ≤ 1/n so that |z| ≤ 1; then  \n\n |sin n t| ≥ n|t| – n³|t|³/6.  \n\nUsing |sin t| ≤ |t| gives  \n\n |sin n t|/|sin t| ≥ n – (n³ t²)/6,  \n\nwhence (8). ∎\n\n\n\n5.4  Estimating the distance to k²–m²  \n\nSubtract the expressions in (7):\n\n (k² – m²) – |Φ_{k,m}(A,B)|  \n  = n₁n₂ – (n₁ – ρ_{n₁})(n₂ – ρ_{n₂})  \n  = n₂ρ_{n₁}(δ) + n₁ρ_{n₂}(δ) – ρ_{n₁}(δ)ρ_{n₂}(δ)  \n  ≤ n₂ρ_{n₁}(δ) + n₁ρ_{n₂}(δ). (9)\n\n(The last, non–positive, product term is dropped.)\n\nAssume in addition that  \n\n δ ≤ 1/(k+m) = 1/n₁,  (10)\n\nso Lemma 2 applies to both ρ_{n₁}(δ) and ρ_{n₂}(δ) (note n₂ < n₁).\n\nCombining (8)–(10) with (9) we get  \n\n (k² – m²) – |Φ_{k,m}(A,B)|  \n  ≤ n₂·(1/3)n₁³δ² + n₁·(1/3)n₂³δ²  \n  = (1/3)n₁n₂(n₁² + n₂²)δ²  \n  ≤ (2/3)n₁³n₂ δ²  \n  ≤ (2/3)(k+m)⁴ δ². (11)\n\nDefine  \n\n C(k,m) := (2/3)(k+m)⁴. (12)\n\n\n\n5.5  Choosing δ in terms of ε  \n\nGiven an arbitrary ε>0 set  \n\n δ := min{ 1/(k+m), √(ε / C(k,m)) }. (13)\n\nCondition (10) is satisfied by construction, and (11) then yields  \n\n (k² – m²) – |Φ_{k,m}(A,B)| ≤ C(k,m)δ² ≤ ε.  \n\nSince δ>0, equality cannot occur.  Thus  \n\n k² – m² – ε < |Φ_{k,m}(A,B)| < k² – m². (14)\n\nBecause ε was arbitrary, |Φ_{k,m}| can be made as close to k²–m² as desired but never reaches it.  Part (b) is established. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.486654",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional parameters:  The original problem involved only one positive integer \\(k\\); the variant introduces a second independent integer \\(m\\) (with \\(k>m\\)), so two angular harmonics interact simultaneously.\n\n2. More intricate algebra:  The mixed products \\(\\cos kB\\,\\cos mA\\) and \\(\\cos kA\\,\\cos mB\\) force the use of two–parameter sum–to–product formulas, greatly lengthening the trigonometric manipulations (Step 2).\n\n3. Sharper bounding:  Establishing (1) now requires controlling expressions containing both \\(k+m\\) and \\(k-m\\).  Care is needed to keep track of these factors and to see that they combine to the single quadratic difference \\(k^{2}-m^{2}\\).\n\n4. Strictness and optimality:  One must analyse when the chain of inequalities can be tight, which is subtler here because four different sine factors appear, two of them involving \\(k+m\\) and \\(k-m\\).  Showing the bound is sharp (but unattainable) introduces an additional limiting argument.\n\nBecause of these new dimensions—two independent frequency parameters, lengthier algebra, and a finer optimality discussion—the enhanced variant requires substantially deeper insight and more extensive calculations than both the original problem and its existing kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let k,m be positive integers with k>m≥1.  \nFor real numbers A,B satisfying cos A ≠ cos B define  \n\n  Φ_{k,m}(A,B)= (cos kB · cos mA – cos kA · cos mB)/(cos B – cos A).\n\n(a) Prove the sharp inequality  \n  |Φ_{k,m}(A,B)| < k² – m².  \n\n(b) Show that the constant k² – m² is best possible: for every ε>0 there exist real A,B with cos A≠cos B such that  \n\n  k² – m² – ε < |Φ_{k,m}(A,B)| < k² – m².  \n\nConsequently the strict inequality in part (a) is never an equality, yet the upper bound can be approached arbitrarily closely.",
      "solution": "Notation.  Put  \n\n x := (A–B)/2, y := (A+B)/2 so that A = y–x, B = y+x.  \n\nBecause cos A ≠ cos B we have  \n\n cos B – cos A = cos(y+x) – cos(y–x) = –2 sin x sin y ≠ 0. (1)\n\n\n\n1.  Trigonometric rewrites of the numerator  \n\nUsing the product–to–sum identities,\n\n cos kB cos mA = ½[cos(k(y+x)+m(y–x)) + cos(k(y+x)–m(y–x))],\n\n cos kA cos mB = ½[cos(k(y–x)+m(y+x)) + cos(k(y–x)–m(y+x))].\n\nSubtracting yields  \n\n N := cos kB cos mA – cos kA cos mB  \n\n  = –sin((k–m)x) sin((k+m)y) – sin((k+m)x) sin((k–m)y). (2)\n\n\n\n2.  Forming the quotient  \n\nFrom (1)–(2) and the triangle inequality we obtain  \n\n |Φ_{k,m}(A,B)| = |N| /(2|sin x sin y|)  \n\n  ≤ ½( |sin((k–m)x)|/|sin x|·|sin((k+m)y)|/|sin y|  \n\n    + |sin((k+m)x)|/|sin x|·|sin((k–m)y)|/|sin y| ). (3)\n\n\n\n3.  A classical majorisation of sine ratios  \n\nLemma 1.  For every integer n ≥ 1 and every real t,  \n\n |sin n t| ≤ n|sin t|,   \n\nwith equality iff sin t = 0 or n = 1.  \n\n(Proof: write sin n t = 2 sin t·T_{n–1}(cos t) and use |T_{n–1}(u)| ≤ n–1 for |u| ≤ 1.)\n\nApplying Lemma 1 termwise in (3) gives  \n\n |Φ_{k,m}(A,B)| ≤ ½[(k–m)(k+m) + (k+m)(k–m)] = k² – m². (4)\n\n\n\n4.  Why the inequality is strict  \n\nBecause sin x, sin y ≠ 0, at least one of the four ratios in (3) contains an integer n>1 (indeed k+m>1).  \nLemma 1 is then strict for that ratio, so (4) is strict:  \n\n |Φ_{k,m}(A,B)| < k² – m². Part (a) is proved.\n\n\n\n5.  Sharpness of the constant k²–m²  \n\nWe now construct a sequence (A_j,B_j) for which |Φ_{k,m}(A_j,B_j)| ↑ k²–m².\n\n5.1  A convenient one–parameter family  \n\nChoose x = y = δ with 0 < δ ≤ 1.  Then  \n\n A = 0, B = 2δ, sin x = sin y = sin δ > 0.  \n\nBoth terms inside the absolute value in (2) are negative, so the triangle inequality used in (3) is an equality; consequently  \n\n |Φ_{k,m}(A,B)| = |sin((k+m)δ)|/|sin δ| · |sin((k–m)δ)|/|sin δ|. (5)\n\n\n\n5.2  Introducing the deviation numbers ρ_n  \n\nFor n ≥ 1 and t ≠ 0 define  \n\n ρ_n(t) := n – |sin n t|/|sin t| ≥ 0. (6)\n\nWith n₁ := k+m, n₂ := k–m, equation (5) becomes  \n\n |Φ_{k,m}(A,B)| = (n₁ – ρ_{n₁}(δ))(n₂ – ρ_{n₂}(δ)). (7)\n\n\n\n5.3  A quantitative bound on ρ_n(t) valid up to t ≤ 1/n  \n\nLemma 2.  For every integer n ≥ 1 and |t| ≤ 1/n,  \n\n 0 ≤ ρ_n(t) ≤ (1/3) n³ t². (8)\n\nProof.  For |z| ≤ 1 the Taylor estimate  \n\n |sin z – z + z³/6| ≤ |z|⁵/120  \n\nimplies |sin z| ≥ |z| – |z|³/6.  Put z = n t with |t| ≤ 1/n so that |z| ≤ 1; then  \n\n |sin n t| ≥ n|t| – n³|t|³/6.  \n\nUsing |sin t| ≤ |t| gives  \n\n |sin n t|/|sin t| ≥ n – (n³ t²)/6,  \n\nwhence (8). ∎\n\n\n\n5.4  Estimating the distance to k²–m²  \n\nSubtract the expressions in (7):\n\n (k² – m²) – |Φ_{k,m}(A,B)|  \n  = n₁n₂ – (n₁ – ρ_{n₁})(n₂ – ρ_{n₂})  \n  = n₂ρ_{n₁}(δ) + n₁ρ_{n₂}(δ) – ρ_{n₁}(δ)ρ_{n₂}(δ)  \n  ≤ n₂ρ_{n₁}(δ) + n₁ρ_{n₂}(δ). (9)\n\n(The last, non–positive, product term is dropped.)\n\nAssume in addition that  \n\n δ ≤ 1/(k+m) = 1/n₁,  (10)\n\nso Lemma 2 applies to both ρ_{n₁}(δ) and ρ_{n₂}(δ) (note n₂ < n₁).\n\nCombining (8)–(10) with (9) we get  \n\n (k² – m²) – |Φ_{k,m}(A,B)|  \n  ≤ n₂·(1/3)n₁³δ² + n₁·(1/3)n₂³δ²  \n  = (1/3)n₁n₂(n₁² + n₂²)δ²  \n  ≤ (2/3)n₁³n₂ δ²  \n  ≤ (2/3)(k+m)⁴ δ². (11)\n\nDefine  \n\n C(k,m) := (2/3)(k+m)⁴. (12)\n\n\n\n5.5  Choosing δ in terms of ε  \n\nGiven an arbitrary ε>0 set  \n\n δ := min{ 1/(k+m), √(ε / C(k,m)) }. (13)\n\nCondition (10) is satisfied by construction, and (11) then yields  \n\n (k² – m²) – |Φ_{k,m}(A,B)| ≤ C(k,m)δ² ≤ ε.  \n\nSince δ>0, equality cannot occur.  Thus  \n\n k² – m² – ε < |Φ_{k,m}(A,B)| < k² – m². (14)\n\nBecause ε was arbitrary, |Φ_{k,m}| can be made as close to k²–m² as desired but never reaches it.  Part (b) is established. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.407257",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional parameters:  The original problem involved only one positive integer \\(k\\); the variant introduces a second independent integer \\(m\\) (with \\(k>m\\)), so two angular harmonics interact simultaneously.\n\n2. More intricate algebra:  The mixed products \\(\\cos kB\\,\\cos mA\\) and \\(\\cos kA\\,\\cos mB\\) force the use of two–parameter sum–to–product formulas, greatly lengthening the trigonometric manipulations (Step 2).\n\n3. Sharper bounding:  Establishing (1) now requires controlling expressions containing both \\(k+m\\) and \\(k-m\\).  Care is needed to keep track of these factors and to see that they combine to the single quadratic difference \\(k^{2}-m^{2}\\).\n\n4. Strictness and optimality:  One must analyse when the chain of inequalities can be tight, which is subtler here because four different sine factors appear, two of them involving \\(k+m\\) and \\(k-m\\).  Showing the bound is sharp (but unattainable) introduces an additional limiting argument.\n\nBecause of these new dimensions—two independent frequency parameters, lengthier algebra, and a finer optimality discussion—the enhanced variant requires substantially deeper insight and more extensive calculations than both the original problem and its existing kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}