# Tricky Geometry Problem involving rectangle rotation.

Problem: A rectangle is rolled along a straight line by rotations of $90\deg$ about each vertex in turn. One vertex traces out a series of arches where each arc is the union of three circular arcs. Show that the area of the region bounded by the straight line and one arch equals the area of the rectangle plus twice the area of the circumscribed circle.

My Attempt: I have assumed a rectangle with vertices on the following coordinates, $(a,0),(a,b),(0,b)$ and $(0,0)$. Thus the area of the rectangle is $A_1=ab$. The area of the circumscribed circle is clearly $$A_2=\frac{\pi(a^2+b^2)}{4}.$$ Now as far as I understand the rectangle is moving about the line $y=b/2.$ But I can't understand what exactly is meant by the words:

One vertex traces out a series of arches where each arc is the union of three circular arcs.

Kindly help me in understanding the problem. I am unable to figure out what areas of the arcs need to be calculated.

EDIT: After following the image given in the first answer, I am able to infer that one needs to find the area of the quarter circles that are described by the red arc. This means that the total area is $$A=\frac{\pi a^2}{4}+\frac{\pi a^2}{4}+\frac{\pi b^2}{4}.$$ But this is not equal to $ab+\frac{\pi (a^2+b^2)}{2}.$ So what am I missing now?

• You're missing the fact that you're supposed to ask one question, get an answer, and move on. If you need another question answered, you should ask another question. Chameleon questions are frowned on (by me, at least). – John Hughes May 8 '17 at 22:59

The answer to your question "So what am I missing now?" is that you have the incorrect radius for the second arc. The radius of that arc has length $\sqrt{a^2+b^2}$ because it is the diagonal of the rectangle.

Show that the area of the region bounded by the straight line and one arch equals the area of the rectangle plus twice the area of the circumscribed circle.

Here are my own "thousand words":

The arc-set (drawn in red) is made of three circle-arcs, one based at each of the blue, yellow and green points. The red vertex is the one travelling along this arc set.

Apologies that the red curve is not perfect -- PowerPoint doesn't have great curve-shaping tools. :(

• Worth a thousand words, I'm told. :) – John Hughes May 8 '17 at 11:25
• So we need to find the area of the blue, yellow and green circle? – Shrey Aryan May 8 '17 at 11:28
• I don;t think that's what it says. It says "the area between the line along which the rectangle is rolling and one arch"; that seems to me to be HALF the area of the union of the three circles. (Which is NOT the sum of the three circle-areas, because they overlap.) – John Hughes May 8 '17 at 11:30
• So it is the area below the red-blue-yellow-green-red line? – Shrey Aryan May 8 '17 at 11:33
• You have to find the area under "Flintstone's cycloid". The "cycloid" consists of three $90^\circ$ circular arcs, and the shape under the "cycloid" can be easily divided into simple pieces of total area as claimed. – Christian Blatter May 8 '17 at 13:12