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  Jim Randell 8:13 am on 5 November 2023 Permalink | Reply
  Tags: by: Sir Brian Young   

  Brain-Teaser 935: Ears, noses & throats 

  From The Sunday Times, 22nd June 1980 [link]

  Our local hospital is a busy place, with a large department, luckily, to keep records of what is wrong with whom. Those who come to the ENT specialist to complain about their ears, sometimes complain about their nose or their throat.

  Of his 60 patients, the number complaining of ears and nose only is three times as great as the number complaining of ears and throat only.

  Another group consists of those who say that their ears are the only body part of the three where they are in good health; and this group is three times as large as the group which declares that the throat alone is healthy.

  It’s all very confusing, I know; but I can tell you that there are 110 complaints in all — if you count a complaint about two ears as one complaint.

  Everyone complains about at least one thing.

  How many patients come with complaints about one part of their anatomy (ears or nose or throat) only?

  This puzzle is included in the book The Sunday Times Book of Brainteasers (1994).

  [teaser935]

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    Jim Randell 8:14 am on 5 November 2023 Permalink | Reply

    We can label the areas of the Venn diagram we are interested in as: E, N, T, EN, ET, NT, ENT (this accounts for all patients, as each has at least one complaint).

    We have:

    E + N + T + EN + ET + NT + ENT = 60
    EN = 3 ET
    NT = 3 EN
    E + N + T + 2(EN + ET + NT) + 3 ENT = 110

    And we want to determine the value of E + N + T.

    So:

    EN = 3 ET
    NT = 3 EN = 9 ET

    Hence:

    EN + ET + NT = 3 ET + ET + 9 ET = 13 ET

    Writing: X = E + N + T, we get:

    X + 13 ET + ENT = 60
    X + 26 ET + 3 ENT = 110
    ⇒
    X = ENT + 10

    And so:

    2 ENT = 50 − 13 ET

    There are 2 possibilities:

    ET = 0 ⇒ ENT = 25, E+N+T = 35
    ET = 2 ⇒ ENT = 12, E+N+T = 22

    If we suppose the phrase: “three times as great/large” in the puzzle text precludes zero values, then only a single solution remains.

    Solution: 22 of the patients have a single complaint.

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  Jim Randell 9:14 am on 27 July 2021 Permalink | Reply
  Tags: by: Sir Brian Young   

  Brain-Teaser 817: Fish in the swim 

  From The Sunday Times, 20th March 1977 [link]

  The goldfish and the shubunkin swim in straight lines and at the same speed, and they always make right-angled turns when they do turn.

  From a point on the edge of the Round Lake (a perfect circle, of course) the goldfish swam due north and the shubunkin swam in a direction somewhere between north and east.

  After an hour the goldfish reached the edge of the lake and turned right; simultaneously the shubunkin turned right.

  Half an hour later the shubunkin reached the edge of the lake, and 15 minutes after that the goldfish again reached a point on the edge of the lake.

  The shubunkin, having rested for 15 minutes, wants the longest possible swim now without reaching a point on the edge of the lake.

  What course should the fish steer?

  This puzzle is included in the book The Sunday Times Book of Brain-Teasers: Book 1 (1980). The puzzle text above is taken from the book.

  [teaser817]

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    Jim Randell 9:14 am on 27 July 2021 Permalink | Reply

    

    The goldfish starts at A, swims due north to B (in 60 minutes), and then due east to C (in 45 minutes).

    If we use units of distance corresponding to how far a fish can travel in 15 minutes we get AB = 4, BC = 3.

    So, ABC is a (3, 4, 5) triangle, hence AC is a diameter of lake, and so the lake has radius 5/2.

    In 60 minutes, the shubunkin swam from A to X (AX = 4), and in 30 minutes swam from X to Y (XY = 2).

    The subunkin then wishes to make the longest possible straight line journey, which is the diameter of the lake YZ.

    We need to determine the direction of YZ (which is the same as the direction OZ).

    Labelling the angle BAC as φ and COZ as θ then the direction OZ (amount west of north) is given by:

    d = θ − φ

    From triangle ABC we have:

    cos(φ) = 4/5

    And using the cosine rule in triangle AOY to determine the angle AOY = θ, we have:

    4² + 2² = 2(5/2)² − 2(5/2)(5/2)cos(θ)
    cos(θ) = −3/5

    Hence the direction OZ is:

    d = acos(−3/5) − acos(4/5)

    And:

    d = acos(−3/5) − acos(4/5)
    d = asin(3/5) + asin(4/5)
    d = φ + (90° − φ)
    d = 90°

    Solution: The shubunkin should head due west.

    So a more accurate representation of the situation is:

    

    Where OXY is a (3/2, 4/2, 5/2) triangle.

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