[Not-so-daily puzzle] A fun geometry puzzle \ stacker news ~science

pull down to refresh

[Not-so-daily puzzle] A fun geometry puzzle

384 sats \ 11 comments \ @south_korea_ln 15 Jan science

The quarter circle and the rectangle coincide and/or are tangent at three points, as illustrated. Find the area of the shaded region.

I've posted many such a problem, but to me, this ended up being one of the more enjoyable geometry puzzles.

An easier geometry puzzle was posted in the previous iteration: #832245 (with the answer in #832640).

view all related items

138 sats \ 1 reply \ @Scroogey 15 Jan

It might help to assemble four quarter circles into a full one:

Let's say the doubly shaded right triangles have shorter sides

x \gt y

The outer square's side length is

2x+12

, area

4x^2+48x+144

We have to subtract the four white rectangles with area

1+x

each.

Further add the doubly shaded triangle areas

2xy

(they count twice).

Then the answer is obviously a fourth of that.

That's how far I got, but I couldn't get

x

and

y

0 sats \ 0 replies \ @south_korea_ln OP 15 Jan

I didn't proceed that way, but I imagine this may work too. I'll post the answer tomorrow.

1 sat \ 0 replies \ @south_korea_ln OP 16 Jan

The answer is 171.21.

Actual numbers are in the notes.

Main steps:

Realize that one can draw a vertical line from the top to the tangent point, due to it being a quarter circle.
Set the origin at the tangent point
Use point (5,1) = G being part of the circle to obtain the radius of the circle R by using

 $x^2 + (y-R)^2 = R^2$

With R = 13, we can use Pythagoras to find that the length of AB = 12.
We can find the angle t using the 5-12-13 triangle denoted by ABC corners.
Use the $\cos(t) = 12/13$ to find that $\sin(90-t)=12/13=a/12$ , hence $a=12$ .
Apply Pythagoras on triangle ADE to find that $13-b=5$ and thus $b=8$ .
Use the ratios between sides of similar triangles ABC and FHG to find that the length of FG equals $91/12$ .
You have all the information to find the area of the big rectangle (136) and the remaining triangle forming the rest of the shaded region ( $\sim 35.21$ ).
Add up the two values to find the final answer.

0 sats \ 1 reply \ @SimpleStacker 15 Jan

Hum ... this one looks hard.

30 sats \ 0 replies \ @south_korea_ln OP 15 Jan

It's from a page where they focus on "actually interesting math problems". Most are outside of my comfort zone, but this one felt accessible, for once. But it's not as simple as the one in the previous iteration, for sure.

0 sats \ 5 replies \ @mod 15 Jan freebie

🚩 This post might be more relevant and engaging in the ~gaming territory.