Recent content by bob012345

Expanding the original problem to be an ## nx1## rectangle, it turns out the angle B to add to angle A (defined by the ratio ##\large\frac{1}{n}##) to get ##\pi/4## is defined by the ratio ##\large\frac{n-1}{n+1}##. Here is an example by construction for ##n=4##. First, construct a circle of...

One can take the log of both side of the inequality and plot that. That plot looks identical but goes on exponentially further. The purple is blue overlapping red and boundary between blue and purple is where it’s failing due to large exponents. Also, it’s obvious but if you look at the...

The plot can also be generated easily on Desmos.

It can be shown that the yellow figure of the last graphic in post #8 can be completely contained in the blue figure. Take off an equivalent yellow area from the blue area yields only blue. That proves the original yellow shape was smaller than the blue thus less than a quarter circle.

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.