Recent content by bob012345

Here is another way but it’s just a variation of @coquelicot above in post #7 but the lengths are not obvious so they must be calculated. Drop a perpendicular at point Q. Since point P is at the midpoint of the vertical line the distance PQ is $$\frac{1}{2\sqrt{2}}$$ Then the distance QM is...

Here are some more solutions; In each case, the triangle constructed in green is similar to the triangle highlighted in brown which has angle B and is added to angle A. The sum is equal to angle C.

This problem is another from Martin Gardner’s book Mathematical Circus but originates from a fourth grade extra credit question in a Moscow school long ago. It makes me wonder how America won the Space Race! The problem statement is simply; Using only elementary geometry (not even...

Here is another simple visual demonstration. The yellow is the original shaded region. The yellow plus blue is exactly half a circle. The yellow is less than half of a half of the circle which is less than a quarter circle. It’s obviously true but not quite the proof of post #5 without further...

The yellow area is half the original shaded area. Sorry, I should have made that more clear. I edited my post above.

Here is my solution;

You don’t need numbers but of course you can calculate things if you want.

Three identical circles intersect in such a way that each intersects the centers of the other two as in the figure. The problem is to determine if the shaded area is less than or greater than one quarter the area of one circle. Of course one can calculate the answer but the exercise is to show...

I believe that generally speaking, they would be the same. Especially when the author is a world renowned puzzlist.

Well I found several solutions I thought were clever and valid reasonable early but what took a while was that nagging feeling they weren’t what the author intended and I was looking for that solution so that when I looked at the book answer I would not be disappointed and it was worth the wait.

I understand but I think part of the game is to figure that out. It took me a few hours to get to the second assumption. It took only a few minutes once I realized it.

@DaveC426913 , while you are mathematically correct and from a purely mathematical point of view, you obeyed the rules, your solution exists along with a literal infinite number of similar solutions which should be a clue that is not the unique solution the author had in mind. These aren’t...

But isn’t there usually a class of trivial solutions which can be discounted? Maybe because they seem to violate the spirit of the puzzle.

I believe so, yes. How is that leaky though?