Hmm, I came up with a purely geometrical argument. I imagined two line segments of length x (where x = 2 is the horizontal dimension of these rectangles) connected together by a pivot point that can otherwise move freely, and the right segment pinned to "the page" by another pivot point. The left segment must remain horizontal as the right segment can be swept upwards by an arbitrary angle, sweeping out an arc of radius = x. Thus one can manipulate this assembly to create the relevant aspects of the situation depicted in this problem. Because the left segment remains horizontal, the left edge of the left segment also sweeps out an arc of radius x. My initial question was, is the radius of the semicircle less than or greater than x? By visual inspection of how my system sweeps, I could see that if R > x or if R < x, there could never be a point "C" where the left end of the left segment intersects the semicircle. I can't really show this without a picture, but if you think about it, you'll see what I mean. Thus, R = x. Thus, the area of the semicircle is simply half the area of a circle of radius x = 2, which is pi*x^2 / 2 = 2 Pi.
This was fun being reminded of the chord-chord theorem. The first thing that struck me was that IF a solution exists, it must be 2*pi because x can be any length less than r. Therefore just take the degenerate case x = 0, and the diagram is simple. But how do we know if a solution exists that does not depend on x? Your proof works well. Actually, unless I'm missing something, there is a simpler visual proof you could do in your head without any algebra. Label the points as you did in your video. Call the point where the two rectangles touch F. Now draw a line parallel to FB through C. Call the point where this new line crosses the diameter O. CFBO is a parallelogram. In fact, it is a rhombus since CF = FB = 2. Therefore OB = 2 and OC = 2. There is a unique circle with center on AB that passes through B and C, which must be the (semi)circle in the diagram. Therefore, CO and OB are two radii of the semicircle. The radius of the semicircle is therefore 2, and the area is 2*pi.
Thanks for the detailed comment Scott! Glad you enjoyed the problem. I was able to follow along with the first half of your solution, but could you please explain how you determined that the unique circle was centered at point O?
@@intuitivemath1124 Yeah, I had to think about that one a bit too. I guess I should have explained better. Basically, the center of any circle that goes through C and B must lie on the perpendicular bisector of segment BC, because O is equidistant from all points on the circle. This perpendicular bisector intersects AB at only one point, so it must be O. Does that make sense?
Hmm, I came up with a purely geometrical argument. I imagined two line segments of length x (where x = 2 is the horizontal dimension of these rectangles) connected together by a pivot point that can otherwise move freely, and the right segment pinned to "the page" by another pivot point. The left segment must remain horizontal as the right segment can be swept upwards by an arbitrary angle, sweeping out an arc of radius = x. Thus one can manipulate this assembly to create the relevant aspects of the situation depicted in this problem. Because the left segment remains horizontal, the left edge of the left segment also sweeps out an arc of radius x. My initial question was, is the radius of the semicircle less than or greater than x? By visual inspection of how my system sweeps, I could see that if R > x or if R < x, there could never be a point "C" where the left end of the left segment intersects the semicircle. I can't really show this without a picture, but if you think about it, you'll see what I mean. Thus, R = x. Thus, the area of the semicircle is simply half the area of a circle of radius x = 2, which is pi*x^2 / 2 = 2 Pi.
This was fun being reminded of the chord-chord theorem. The first thing that struck me was that IF a solution exists, it must be 2*pi because x can be any length less than r. Therefore just take the degenerate case x = 0, and the diagram is simple.
But how do we know if a solution exists that does not depend on x? Your proof works well.
Actually, unless I'm missing something, there is a simpler visual proof you could do in your head without any algebra. Label the points as you did in your video. Call the point where the two rectangles touch F. Now draw a line parallel to FB through C. Call the point where this new line crosses the diameter O. CFBO is a parallelogram. In fact, it is a rhombus since CF = FB = 2. Therefore OB = 2 and OC = 2. There is a unique circle with center on AB that passes through B and C, which must be the (semi)circle in the diagram. Therefore, CO and OB are two radii of the semicircle. The radius of the semicircle is therefore 2, and the area is 2*pi.
Thanks for the detailed comment Scott! Glad you enjoyed the problem. I was able to follow along with the first half of your solution, but could you please explain how you determined that the unique circle was centered at point O?
@@intuitivemath1124 Yeah, I had to think about that one a bit too. I guess I should have explained better. Basically, the center of any circle that goes through C and B must lie on the perpendicular bisector of segment BC, because O is equidistant from all points on the circle. This perpendicular bisector intersects AB at only one point, so it must be O. Does that make sense?
@@ytseberle Got it, thanks for the clarification!
Just assume the height of the rectangle is 0 lol, the radius is 2 lol, the area is pi R^2/2 lol, the area is 4pi/2 lol, the area is 2pi lol