Today's math puzzle!

[Ellen Forradalom]

I put a Nick's Mathematical Puzzles gadget on my Google toolbar, and will live to regret it.

Today's puzzle:

Find the area of the largest semicircle that can be inscribed in the unit square.

We're talking something like this:



The hint says to express the side of the square (unit square, so length = 1) in terms of the radius (unknown.)

[Suich]

1. Huh?

:dunce:

[Jimbo S]

2. r = 0.586

If the side of the square is 1, then
the coordiantes of the circle are
a = SQRT(2)-1 = 0.414
b = SQRT(2)-1 = 0.414

Draw an imaginary horizontal line through the circle's center.
You'll have a right triangle with base 0.414 and side also 0.414
The hypotonuse (sp?) is also the radius.

radius = SQRT(2)*

[Zavulon]

4. This was my first guess, too.

;)

[boston bean]

5. That's exactly what I came up with. But I went about it a little differently

I just copied your answer.:D

[Ellen Forradalom]

6. Kindly elaborate for the benefit of the geometrically rusty

Thank you

[Jimbo S]

8. My apologies

I'm at work and this puzzle was preventing me from going home, hence the lack of explanation. lol
No prob, I'll stick around.

Using analytic geometry
set the coordinate (a,b) equal to the circle's center.
the square's four corners are (0,0), (0,1), (1,1), (1,0) clockwise from the origin.

the equation for the circle is
(x-a)^2 + (y-b)^2 = r^2, as well as
x^2 + y^2 = r^2, the circle crosses the origin.

set
(x-a)^2 + (y-b)^2 = r^2
(x-a)^2 + (y-b)^2 = x^2 + y^2
x^2 + y^2 - 2*(ax + by) = 0

the circle's center lies in the line y = x

the apex of the circle is when y = 1, which is also when x = a

set y = 1 and x = a in the equation above, you get
a^2 + 2*a - 1 = 0

using Pethagarem's Theorum
a = (-2 +/- sqrt(2^2 - 4*1*1))/(2*1)
a = sqrt(2) - 1
a = 0.414

therefore, b = 0.414 as well.

and from above:
r^2 = a^2 + b^2
r = sqrt((sqrt(2) - 1)^2 + (sqrt(2) - 1)^2)
r = sqrt(2 * (sqrt(2) - 1)^2)
r = sqrt(2) * (sqrt(2) - 1)
r = 2 - sqrt(2)
r = 0.5858
on edit: this way of solving for r is easier than my earlier post, but it works either way.

It's been a while since I did a brain twister like this. Surprised I hadn't come across this one in my years of high school and college math. Nice to know this old fart still has it. :)

[Ellen Forradalom]

9. Thanks

This will keep me up tonight.

I don't have serious math chops but I do enjoy a nice puzzle.

[Blue-Jay]

3. .

[matcom]

7. i can barely type

i'm laughing so fucking hard at that :rofl:

oh and how do you get scotch out of a keyboard :spray:

[Ptah]

10. Without looking at any other responses, here's what I got

Edit to add:

The area of the semicircle would be: A = (pi * R^2)/2 ; A = .54 sq. units

[Ellen Forradalom]

16. Pretty picture

Kindly explain it.

[Ptah]

17. I'll try

[bertha katzenengel]

18. Okay. Somewhere early in my life - say, 18-30 - I went SERIOUSLY wrong... Why?

Because I effing understand this! And got a C in pre-algebra, and never went any farther. No algebra, calculus, geometry, nothing. :banghead:

OTOH, life is too short for regrets. I should've been a professional musician, should've gone into sports med, should've taught history, could have been a doctor if I'd known soon enough that I could become that.

But I'm a happy drone. I don't live to work. Life is too short for regrets.

Thanks for reminding me of this, L. :hug:

[Ptah]

19. I was surprised at how hard it was to explain something step-by-step.

I'm glad it made sense to you, B.

:hi:

:hug:

[hfojvt]

11. what a shame, nobody has the right answer yet

.539 :P

[Ptah]

12. See post 10

:hi:

[hfojvt]

13. and you beat me by five minutes

but I had only seen the unedited version before I posted

[Ptah]

14. But it took me nearly thirty minutes to solve it, so you probably beat me to it.

:hi:

And after my post I re-read to OP and saw
I was so focused on the radius, that I forgot the question was
looking for the area.

:blush:

[hfojvt]

15. nah, I admit, I just went from the radius in your answer

which was clearly correct. This after I had pondered the problem for about five minutes and then punted. My (incorrect) guess was that the circle had the same area as the square. My geometry has been too long ago. It took me a while to remember that the area of a circle is piRsquared rather than 2piR (which is, of course, the circumfrence). Since I could not (or did not want to put forth the effort to) solve it, I thought I would poke fun at everyone solving for R and then not finishing the slam dunk. Once R is known, the rest is just a victory lap - the marathon is over.