# Another @solvemymaths problem

Another geometry puzzle from @solvemymaths:

I enjoyed this one – no solution immediately jumped out at me, and I spend a great deal of time looking smugly at a way over-engineered circle theorems approach I can no longer remember.

Let’s label the apex of the triangle P, and the octagons points A to H, anticlockwise from the bottom left (so we’re looking for angle APD).

Now, APB is obviously $\piby 6$, and BPD is an isosceles triangle, so if we can find the angle PBD, we can easily finish the puzzle.

From there, it’s easy: HBDF is a square, and BDP is $\piby 3$, so PBD is $\piby 6$. (Another solution, by @remygarreau noted that AP and BD are parallel, which gives the same result.)

That makes the ‘base’ angles of the isosceles triangle each $\frac{5}{12}\pi$, and the angle required $\frac{7}{12}\pi$. Lovely!