{
    "problem": "The orthocenter of triangle $ABC$ divides altitude $\\overline{CF}$ into segments with lengths $HF = 6$ and $HC = 15.$  Calculate $\\tan A \\tan B.$\n\n[asy]\nunitsize (1 cm);\n\npair A, B, C, D, E, F, H;\n\nA = (0,0);\nB = (5,0);\nC = (4,4);\nD = (A + reflect(B,C)*(A))/2;\nE = (B + reflect(C,A)*(B))/2;\nF = (C + reflect(A,B)*(C))/2;\nH = extension(A,D,B,E);\n\ndraw(A--B--C--cycle);\ndraw(C--F);\n\nlabel(\"$A$\", A, SW);\nlabel(\"$B$\", B, SE);\nlabel(\"$C$\", C, N);\nlabel(\"$F$\", F, S);\ndot(\"$H$\", H, W);\n[/asy]",
    "level": "Level 5",
    "type": "Precalculus",
    "solution": "Draw altitudes $\\overline{BE}$ and $\\overline{CF}.$\n\n[asy]\nunitsize (1 cm);\n\npair A, B, C, D, E, F, H;\n\nA = (0,0);\nB = (5,0);\nC = (4,4);\nD = (A + reflect(B,C)*(A))/2;\nE = (B + reflect(C,A)*(B))/2;\nF = (C + reflect(A,B)*(C))/2;\nH = extension(A,D,B,E);\n\ndraw(A--B--C--cycle);\ndraw(A--D);\ndraw(B--E);\ndraw(C--F);\n\nlabel(\"$A$\", A, SW);\nlabel(\"$B$\", B, SE);\nlabel(\"$C$\", C, N);\nlabel(\"$D$\", D, NE);\nlabel(\"$E$\", E, NW);\nlabel(\"$F$\", F, S);\nlabel(\"$H$\", H, NW, UnFill);\n[/asy]\n\nAs usual, let $a = BC,$ $b = AC,$ and $c = AB.$  From right triangle $AFC,$ $AF = b \\cos A.$  By the Extended Law of Sines, $b = 2R \\sin B,$ so\n\\[AF = 2R \\cos A \\sin B.\\]From right triangle $ADB,$ $\\angle DAB = 90^\\circ - B.$  Then $\\angle AHF = B,$ so\n\\[HF = \\frac{AF}{\\tan B} = \\frac{2R \\cos A \\sin B}{\\sin B/\\cos B} = 2R \\cos A \\cos B = 6.\\]Also from right triangle $AFC,$\n\\[CF = b \\sin A = 2R \\sin A \\sin B = 21.\\]Therefore,\n\\[\\tan A \\tan B = \\frac{2R \\sin A \\sin B}{2R \\cos A \\cos B} = \\frac{21}{6} = \\boxed{\\frac{7}{2}}.\\]"
}