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Logophobic

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Posts posted by Logophobic

  2. ROLLO MKIII

    in Games

    Posted

    SHEEN

    SCARE

    If SHEEN = 1 then SH--- = 1 (EATEN, BEETS) and SCARE - 1 proves H, or 2 proves S (not A from SHADE - 2 or R from SHORE - 2 with SH--E - 2)

    If SHEEN = 0 then --ADE = --ORE = 2 so either ARE = 3 or ODE = 3: SCARE - 1 proves --ODE, 2 proves -CODE, 3 proves --ARE, or 4 proves -CARE

  7. Who should do what

    in New Logic/Math Puzzles

    Posted

    A different solution? Perhaps you should clarify: Was it intended that each person completes one task and each task is complete by one person, with the goal of finding the lowest-cost solution? If so, then for the given costs the solution I posted is the only solution with a cost of less than 19.

     

  9. ROLLO MKIII

    in Games

    Posted · Edited by Logophobic

    LOVES -- If 0 then R from LOVER-1 and I from LIVES, if 2 then -O--S (not L or V, and ---ES = 1), if 1 then E (but that's already covered)

  20. Making a spiral

    in New Logic/Math Puzzles

    Posted

    @bonanova It appears to me that you are way off:

    Spoiler

    To begin with, we have the angle between the walls fixed at 120 degrees. Label the point of intersection as point O. Measure a distance of 1 unit from point O on each line, and label the points A and B. Draw a horizontal line on each wall through point O. Draw vertical lines through A and B to find points C and D where the vertical lines intersect the horizontal. We now have right triangles ACO and BDO with AO =BO = 1, AC = BD = sin Theta, CO = DO = cos Theta. Note that the midpoint of AB coincides with the midpoint of CD, we'll call this point M. We now have right triangles CMO and AMO with angles COM = 120/2 =60 degrees and AOM = 137.5/2 = 68.75 degrees. We also have right triangle ACM, for which we can find the lengths of each side in terms of Theta.

    As stated above, from triangle ACO:

    AO = 1 (given)
    angle AOC = Theta (given)
    angle ACO = 90 (by construction)
    AC = sin Theta
    CO = cos Theta

    From triangle AMO:

    angle AOM = 68.75 (bisection of angle AOB, given to be 137.5)
    angle AMO = 90 (by construction)
    AM = AO * sin AOM = sin 68.75

    From triangle CMO:

    angle COM = 60 (bisection of angle COD, given to be 120)
    angle CMO = 90 (by construction)
    CM = CO * sin COM = cos Theta * sin 60

    From triangle ACM:

    angle ACM = 90 (AC is vertical, CM is horizontal)
    AC^2 + CM^2 = AM^2 (pythagorean theorem)

    (sin Theta)^2 + (cos Theta * sin 60)^2 = (sin 68.75)^2

    (sin Theta)^2 + (cos Theta)^2 * 0.75 = (sin 68.75)^2

    (sin Theta)^2 + (cos Theta)^2 * (1 - 0.25) = (sin 68.75)^2

    (sin Theta)^2 + (cos Theta)^2 - 0.25 * (cos Theta)^2 = (sin 68.75)^2

    1 - 0.25 * (cos Theta)^2 = (sin 68.75)^2

    -0.25 * (cos Theta)^2 = (sin 68.75)^2 - 1

    0.25 * (cos Theta)^2 = 1 - (sin 68.75)^2

    (cos Theta)^2 / 4 = (cos 68.75)^2

    (cos Theta) / 2 = cos 68.75

    cos Theta = 2 * cos 68.75

    Theta = 43.54...

     

  21. ROLLO MKIII

    in Games

    Posted

    R E G A L

    BORED - 0

    SILLY - 0

    HAPPY - 0

    HOURS - 0

    ZESTY - 1

    SALAD - 1 maurice +5

    HOSED - 0

    BELLY - 1 plainglazed +5

    REGAL - 5 maurice +20

    Logophobic - 86 + 9 = 95
    maurice - 41 + 25 = 66
    plainglazed - 60 + 5 = 65
    phaze - 18
    nana - 10

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