tag:blogger.com,1999:blog-5541500.post6290127405768057053..comments2024-03-28T18:35:54.237+09:00Comments on BigHominid's Hairy Chasms: another Math League challenge problem!Kevin Kimhttp://www.blogger.com/profile/01328790917314282058noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-5541500.post-15342281382226903272011-11-14T02:09:21.491+09:002011-11-14T02:09:21.491+09:00By the way, 3/4 is confirmed as correct. The orig...By the way, 3/4 is confirmed as correct. The original comment I wrote for this problem came to the correct solution, but I had forgotten to calculate area by using radii, not diameters. All my calculations were off by a factor of 2, which turned out not to matter since you lose the 2 once you establish the ratio.Kevin Kimhttps://www.blogger.com/profile/01328790917314282058noreply@blogger.comtag:blogger.com,1999:blog-5541500.post-76156053193474176782011-11-14T02:07:26.108+09:002011-11-14T02:07:26.108+09:00That's pretty much what I thought when I saw t...That's pretty much what I thought when I saw the black-and-white original.<br /><br />Will be posting the original soon; I brought the sheet home with me from school.Kevin Kimhttps://www.blogger.com/profile/01328790917314282058noreply@blogger.comtag:blogger.com,1999:blog-5541500.post-84744956906295132762011-11-13T21:19:17.724+09:002011-11-13T21:19:17.724+09:00Looks like the Korean flag went on a bender.
Yep,...Looks like the Korean flag went on a bender.<br /><br />Yep, that's all I've got.Charleshttp://www.liminality.orgnoreply@blogger.comtag:blogger.com,1999:blog-5541500.post-39986609405670861832011-11-13T10:10:50.689+09:002011-11-13T10:10:50.689+09:00I strained my brain for a few minutes after class ...I strained my brain for a few minutes after class trying to figure this one out... and then my colleague strolled over and told me how easy it was. Think of it this way: all the circles' centers lie on the same line, so you need to think in terms of semicircles.<br /><br />How do I find the area of the blue region, for example? I need to calculate the area of Circle C, which is the biggest circle. Then I need to calculate the respective areas of Circles A and B. Next, I need to subtract half the area of Circle B from half the area of Circle C, and I also need to add half the area of Circle A.<br /><br />Then we figure it the other way for red.<br /><br />Hell, let's do this.<br /><br />Based on the ratios given, we know the diameter of Circle C is 7. So...<br /><br />Area of C = (3.5^2)π = 12.25π<br />Area of B = (2^2)π = 4π<br />Area of A = (1.5^2)π = 2.25π<br /><br />Blue region's area:<br /><br />[(1/2)(12.25π)] - [(1/2)(4π)] + [(1/2)(2.25π)]<br /><br />= 5.25π<br /><br />Red region's area<br /><br />[(1/2)(12.25π)] + [(1/2)(4π)] - [(1/2)(2.25π)]<br /><br />= 7π<br /><br />The ratio of blue to red is thus<br /><br />5.25/7, or 21/28, or 3/4.<br /><br />That's a lot of work merely to confirm what I had initially suspected! Without doing any math-- and before my colleague had shown me the light-- I had thought the answer would be either 3/4 or 9/16, based purely on the info we were given about the respective diameters of Circles A and B, and my assumptions about the areas of those circles.<br /><br />So there we are. I'll need to confirm this answer, but I'm pretty sure I'm right.Kevin Kimhttps://www.blogger.com/profile/01328790917314282058noreply@blogger.com