- animeshsaxena
**Posts:**520**Joined:**

QuoteOriginally posted by: lindavincentI think I should share here one of the brainteaser which I'd enjoyed most - http://www.iqtestexperts.com/brainteasers/Which one on the site?The pot one is too kiddish!

QuoteOriginally posted by: outrunWhat are the radii of the two little circles expressed as a function of the width & height of the triangle?say the height=a, width=b, hypotenuse c=sqrt(a^2+b^2). relatively easy to show: sqrt(r/r1) = sqrt(b/r)+1, where r=ab/(a+b+c) is the radius of the incircle and r1 is the radius of the green circle. so r1= a*b/[(a+b+c)*(1+sqrt(a+b+c)/sqrt(a))^2]note the symmetry a<->b yields the radius of red circle r2

Last edited by wileysw on May 26th, 2011, 10:00 pm, edited 1 time in total.

QuoteOriginally posted by: wileyswoutrun, i think T4A & katastrofa both have valid points: i did assume it is simply a square, but the problem would work if it is a rectangle as well - the green circle is just the inscribed circle of the triangle. the same method applies, since the area and the circumference of the triangle are easy to calculate.your new problem seems to be related to the lower bound of integer complexity (is that the "C"?) and i believe the numbers listed here (1,2,3,7,15,63) are the only solutions (checked up to n=10,000. MCarreira would do a much better job). unfortunately the apparent bound 3*log3(n) does not seem to do the trick... will updateLooked at it, it seems valid to look only at 2^m-1 (2 excepted), and could not find anything for 7<=m<=20 (search was not complete though, used GA - which found {1, 2, 3, 7, 15, 63}).

- Traden4Alpha
**Posts:**23951**Joined:**

QuoteOriginally posted by: outrunWithout pen and paper...LOL! 20.

- Traden4Alpha
**Posts:**23951**Joined:**

QuoteOriginally posted by: outrunThis one is trickyHmmmm.. Tricky.... It can be done with algebra but that requires a bit of stack depth in the old (very old) noodle.

- Traden4Alpha
**Posts:**23951**Joined:**

37.If the height of the top row is an integer, then the unknown area is 7, 17, 27, 37, .... with respective heights of 5, 6, 7, 8, .... The first two values are clearly too small. The 27 & 7 solution looks close but is not right. The 37 & 8 solution gives a rectangle above the 29cm^2 one of 58cm^2, a height of 8, and a width of 7 1/4 which is the width of the lower rectangle.

- Traden4Alpha
**Posts:**23951**Joined:**

What's interesting is that some brainteasers have a general analytic/algebraic solution (e.g., the area of the gray triangle as a function of the five known numbers) but that solution can be cognitively hard (i.e., it needs paper and pencil). Yet these same brainteasers can have very easy solutions if one assumes the solution is highly constrained (e.g., integer unknown area, integer height, etc.). But the problem statement does not explicitly give this bit of information.It's almost as if there's an implicit bit of information in many brainteasers -- that an easy solution exists -- which constrains the solution process to the set of easy solution processes (e.g., trying a few specific numerical solutions rather than turning the crank on a general analytic solution).Yet the subset of real-world problems that have a priori easy solutions or tidy implicit constraints on the unknowns seems to be a very small to me. Your algebraic solution is superior and the brainteaser (with it's artificial simplification) is inferior.

QuoteOriginally posted by: outrunBtw these are Japanese high school problemsYes the best brainteaser are "ponder ponder ponder ... 2!" Or "not enought information, it can't be done! .. Oh wait a minute if it *has* to have a solution then ..."Somewhat in this category (hopefully not posted before here): http://www.nytimes.com/2015/04/15/scien ... .html?_r=0

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