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Math Help - Tough Word Problem

  1. #1
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    Tough Word Problem

    I posted this on another part of the forum too, since I found it to be an interesting puzzle. The problem is as follows, and I need a quick answer!:


    The coffee shop is 655 meters from the grocery store, 393 meters from the bookshop, and 314 meters from the bar. The bookshop is 236 meters from the bar and 524 meters from the grocery store.

    Assuming the distance between the bar and the grocery store is less than the distance between the coffee shop and the grocery store, what is the distance between the bar and the grocery store, rounded to the nearest meter?
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  2. #2
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    Hello, blacksuzaku!

    The coffee shop is 655 meters from the grocery store,
    393 meters from the bookshop, and 314 meters from the tavern.
    The bookshop is 236 meters from the tavern and 524 meters from the grocery.

    Assuming the distance between the bar and the grocery is less than
    the distance between the coffee shop and the grocery,
    what is the distance between the bar and the grocery,
    rounded to the nearest meter?

    If I interpret the question correctly,
    . . the tavern (T) is inside the triangle CGB.

    Since CB^2 + BG^2 \:=\:CG^2\;(393^2 + 524^2\:=\:655^2),\;\angle B = 90^o
    . . In fact, \Delta CGB is a 3-4-5 right triangle.
    Code:
          C
          *
          |**
          | * *
          |  *  *
          |   *   *
          |    *314 *
      393 |     *     *
          |      *      * 655
          |       *T      *
          |     *     *     *
          | α *236      x *   *
          | * θ               * *
        B *-----------------------* G
                      524

    Let \alpha = \angle CBT,\;\theta = \angle TBG,\;x = TG

    Use the Law of Cosines in \Delta CBT:
    . . \cos\alpha \:=\:\frac{393^2 + 236^2 - 314^2}{2(393)(236)} \:=\:0.601355285

    . . Hence: . \alpha \:\approx\:53^o\quad\Rightarrow\quad\theta \:\approx\:37^o


    Use the Law of Cosines in \Delta TBG:
    . . x^2\:=\:236^2 + 524^2 - 2(236)(524)\cos37^o\:=\:132747.0766

    Therefore: . x\:\approx\:364 m.

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  3. #3
    MHF Contributor
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    Quote Originally Posted by blacksuzaku View Post
    I posted this on another part of the forum too, since I found it to be an interesting puzzle. The problem is as follows, and I need a quick answer!:

    Here is one way.

    The coffee shop is 655 meters from the grocery store, 393 meters from the bookshop, and 314 meters from the bar. The bookshop is 236 meters from the bar and 524 meters from the grocery store.

    Assuming the distance between the bar and the grocery store is less than the distance between the coffee shop and the grocery store, what is the distance between the bar and the grocery store, rounded to the nearest meter?
    So I drew 3 concentric circles whose centers is the coffee shop (I called the coffee shop, point A).
    Radius for bar, Br, is 314 m,
    Radius for bookshop, Bk, is 393 m.
    Radius for grocery store, Gs, is 655 m.

    Then I drew a horizontal line segment from the center A (coffe shop) to circle of grocery store. Thus, the 3 radii Br, Bk, and Gs are all in this horizontal line segment. The end of radius Br I called point B. The end of radius Bk I called point C. The end of radius Gs I called point D.

    At the end of Br, or at point B, as center, I drew a circle whose radius is 236 m. One of the two intersections of this 4th circle and the circle of bookshop from A, I called point M.
    ABM is a triangle.

    Then from M, I drew a fifth circle, whose radius is 524 m. One of the two intersections, the one nearer to point D, I called point N.

    Now I drew by sketching only, so the radii, circles are not exact.

    Then I checked first if triangle ABM is a right triangle by the Pythagorean theorem.
    AM = Bk = 393 m.
    AB = Br = 314 m.
    BM = 236 m.
    So AM is the longest, thus, it could be the hypotenuse.
    So, (393)^2 =? (314)^2 +(236)^2
    154,449 =? 154,292
    Wow, almost equal! So triangle ABM is a right triangle--well, almost.

    Umm, could it be that triangle AMN is a right triangle also?
    If so, then point N could be point D also---because line ABN could be a straight line segment, then it is the straight line ABD. Then triangle AMN is the same or congruent to triangle AMD.
    So I checked triangle AMD. In triangle AMD,
    AD = 655 m.
    AM = 393 m.
    MD = 524 m.
    (655)^2 =? (393)^2 +(524)^2
    429,025 =? 429,025
    Yes!!
    Hence, MN = MD = 524 m.
    Therefore, point N is actually point D.

    Therefore, the distance between the bar and the grocery store, which is BD, is AD -AB = 655 -314 = 341 m. ----------answer.

    ---------
    May I add then that the coffee shop, bar and grocery store are in a straight line. The bookshop is not in line, but is perpendicular to that line. If you draw the perpendicular line from the bookshop to that line, then it will hit the bar.
    Last edited by ticbol; January 11th 2007 at 10:34 AM. Reason: additional something, the first N should have been M.
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