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  • What is the value of b/a ?

    Kaustubh Katdare

    Administrator

    Updated: Oct 22, 2024
    Views: 1.1K
    Here is yet another puzzle from Archimedes Lab:

    [FONT=Verdana, Arial, Helvetica, sans-serif]Two intersected semi-circles inscribed in a rectangle are tangent to 5 discs (soccer balls) as depicted in the image. What is the value of b/a?

    [​IMG]

    Puzzle taken from #-Link-Snipped-#

    Get all mathy now! Happy Brainache 😁😁

    -The Big K-
    [/FONT]
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Replies

  • Rocker

    Member • Apr 24, 2007

    Looking at the diagram, my best guess is '1'. Is that the right answer??

  • Prasad Ajinkya

    Member • Apr 24, 2007

    I think the answer is sqrt(2)

    Construction: Let the centers of the two big semicircles be A and B. Let the semicircles cut each other at C and D. So, we have a quadrilateral ABCD.

    Proof: Since AC and AD are tangents to circle B, and BC and BD are tangents to circle A. We can conclude that ACB and BDA are right angles. Also, since all of the sides of this quadrilateral are radii or either circle A or circle B, which equal to a length of b/2. Thus, ABCD is a square of sides b/2. Diagonal AB is of the length sqrt(2)*b/2 = b/sqrt(2).

    Also, len(AB) = a = b/sqrt(2). Thus, b/a = sqrt(2). Ta-da!

  • Rocker

    Member • Apr 25, 2007

    kidakaka
    I think the answer is sqrt(2)

    Construction: Let the centers of the two big semicircles be A and B. Let the semicircles cut each other at C and D. So, we have a quadrilateral ABCD.

    Proof: Since AC and AD are tangents to circle B, and BC and BD are tangents to circle A. We can conclude that ACB and BDA are right angles. Also, since all of the sides of this quadrilateral are radii or either circle A or circle B, which equal to a length of b/2. Thus, ABCD is a square of sides b/2. Diagonal AB is of the length sqrt(2)*b/2 = b/sqrt(2).

    Also, len(AB) = a = b/sqrt(2). Thus, b/a = sqrt(2). Ta-da!
    well done kidakaka😁! I think you got the correct answer!😁😁😁

  • Kaustubh Katdare

    Administrator • Apr 25, 2007

    Right or Wrong, we'll have to wait for the right answer 😁!

    Others, what's your answer?

    -The Big K-

  • NiLaY

    Member • Apr 26, 2007

    kidakaka
    I think the answer is sqrt(2)

    Construction: Let the centers of the two big semicircles be A and B. Let the semicircles cut each other at C and D. So, we have a quadrilateral ABCD.

    Proof: Since AC and AD are tangents to circle B, and BC and BD are tangents to circle A. We can conclude that ACB and BDA are right angles. Also, since all of the sides of this quadrilateral are radii or either circle A or circle B, which equal to a length of b/2. Thus, ABCD is a square of sides b/2. Diagonal AB is of the length sqrt(2)*b/2 = b/sqrt(2).

    Also, len(AB) = a = b/sqrt(2). Thus, b/a = sqrt(2). Ta-da!
    Insufficient data to prove ABCD is a square. Agreed ACB and BDA are right angles, but CAD and DBC (well, this is my fav - Death By Chocolate!) being right angles is an unproved assumption.

  • Prasad Ajinkya

    Member • Apr 26, 2007

    NiLaY, its not an assumption, ABCD has to be a square. This is the logic -

    Since AC = AD = b/2 (radii of the circle) and BC = BD = b/2 (radii of the other circle)
    Thus, it is established that ABCD is a paralellogram (a rhombus infact)

    Now, its common sense that if two opposite angles of a parallelogram are right angles, the other two are also right angles (do the math 😀 ). Thus, our rhombus is a square.

    QED.

  • NiLaY

    Member • Apr 26, 2007

    Please refer to the image below (common sense). Though, the answer might be correct, I am afraid solution is not. 😕
    We need to consider the footballs in this puzzle.
    [​IMG]

  • crook

    Member • May 1, 2007

    Where did you get that image NiLaY 😉 The puzzle is turning out to be more difficult that I thought 😁

  • NiLaY

    Member • May 1, 2007

    I did not GET 😒 it, I combined a few autoshape circles/rectangles/lines in M$ Word to make this.
    By the way, anyone up with another solution?

  • Kaustubh Katdare

    Administrator • Jun 7, 2007

    Here we go -

    [​IMG]

    [FONT=Verdana, Arial, Helvetica, sans-serif]We can therefore solve this puzzle with the help of the theorem of Pytagora:
    AA'2 = AB2 + BA'2
    a2 = (b/2)2+ (b/2)2 = 2(b/2)2
    Simplifying: a = b[​IMG]2/2    [/FONT]
    [FONT=Verdana, Arial, Helvetica, sans-serif]So: b/a = b/(b[​IMG]2/2) = [​IMG]2[/FONT]


    [FONT=Verdana, Arial, Helvetica, sans-serif](as posted on #-Link-Snipped-# )[/FONT]


    [FONT=Verdana, Arial, Helvetica, sans-serif]Good job, Kidakaka !
    [/FONT]


    [FONT=Verdana, Arial, Helvetica, sans-serif]-The Big K-
    [/FONT]
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