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# What is the value of b/a ?

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

Puzzle taken from https://www.archimedes-lab.org

Get all mathy now! Happy Brainache 😁😁

[/FONT]

[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**?Puzzle taken from https://www.archimedes-lab.org

Get all mathy now! Happy Brainache 😁😁

**-The Big K-**[/FONT]

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

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!

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!😁😁😁kidakakaI 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!

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

Others, what's your answer?

Others, what's your answer?

**-The Big K-**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.kidakakaI 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!

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.

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.

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.

We need to consider the footballs in this puzzle.

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

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?

By the way, anyone up with another solution?

Here we go -

[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

Simplifying:

[FONT=Verdana, Arial, Helvetica, sans-serif]So:

[FONT=Verdana, Arial, Helvetica, sans-serif]

[FONT=Verdana, Arial, Helvetica, sans-serif]

[FONT=Verdana, Arial, Helvetica, sans-serif]

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[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

*a*2 = (*b*/2)2+ (*b*/2)2 = 2(*b*/2)2Simplifying:

*a*=*b*2/2[/FONT][FONT=Verdana, Arial, Helvetica, sans-serif]So:

*b*/*a*=*b*/(*b*2/2) = 2[/FONT][FONT=Verdana, Arial, Helvetica, sans-serif]

*(as posted on www.archimedes-lab.org )*[/FONT][FONT=Verdana, Arial, Helvetica, sans-serif]

*Good job,*

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

**-The Big K-**[/FONT]