The arbelos An arbelos is the region enclosed by three mutuall...

Question

The arbelos An arbelos is the region enclosed by three mutually tangent semicircles; it is the region inside the larger semicircle and outside the two smaller semicircles (see figure).

b. Show that the area of the arbelos is the area of a circle whose diameter is the distance BD in the figure.

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The shaded region in the figure is called an arbelos, formed by three mutually tangent semicircles. The end circle is the largest circle that can fit inside the arbelos, and it is tangent to all three semicircles. If the largest semicircle has a diameter AB of 1, what should be the diameter X of one of the smaller semicircles to maximize the radius of the in circle R? Note that the radius of the end circle can be expressed in terms of the diameters of the two smaller semicircles as follows. R is equal to x multiplied by Y, all divided by 2 multiplied by X plus Y. Awesome, so it appears for this particular problem, we're asked to figure out what should be the diameter X of one of the smaller semicircles in order to maximize the radius of the end circle R. And we're also told to note that the radius of the end circle can be expressed in the terms of the diametters of the two smaller semicircles as shown in our formula, or rather our equation that's given to us, which is R equals x multiplied by Y, divided by 2 multiplied by X plus Y. Fantastic. So now that now that we know that we're ultimately trying to figure out what the diameter X of one of the smaller semi-circles to maximize the radius of the inner circle R is to be, now that we know that's what we're ultimately trying to solve for, let's take a moment here to examine our figure that's provided to us. As you can see, we have our shaded region represented in yellow, and that's our R bellows, and inside our R bellows, we have our inscribed in circle, which is represented in green, and it has a red dot in the center going out to the outermost edge of the green circle, which shows our radius. The in circle are. And as you can see, we have the two circles that are being cut out of this shaded area, and then we have the larger circle, which has the diameter, or rather, actually, yeah, the diameter which goes straight across AC, which is some value X. And then the smaller circle is CB, which is some distance Y, and then we have the whole circle, which has a diameter of 1, which is A, B, which is going straight across, including both circles. So now that we have a nice visualization of what's happening for this particular prompt, let's read off our multiple choice answers to see what our final answer might be, noting that they're all going to be in units of units, so just unit. So A is 1/4, B is 1/2, C is 3/4, and D is 4/5. Fantastic. So first off, we need to note that since we want to maximize the radius of the end circle, we must first differentiate R and set the derivative equal to 0 in order to help us find the critical points. So, with that said, let us first express R purely in terms of X. And once again we need to note that we are told that A, B is equal to 1. So from our figure, We can determine that the diameter of the largest semicircle will be equal to the sum of the diameters of the two smaller semicircles, meaning we can go ahead and state that A B is equal to X plus Y. So that means that we can take it a step further and say that X plus Y is equal to 1. So therefore, when we rearrange that expression to isolate in sulfur wise, when we take X plus Y equals 1, and we rearrange that equation to isolate sulfur Y, we will find that Y is equal to 1 minus X. So now we need to substitute this into our equation to solve for R, so we can go ahead and write that R is equal to X multiplied by 1 minus X, all divided by 2 multiplied by X + 1 minus X. So it's gonna be X + 1 minus X. Fantastic. And then this will be all equal to. 1/2 X multiplied by 1 minus X, which we could simplify again, and we could write that one, that it's gonna be equal to 1/2 x minus 1/2 X squared. Now we need to differentiate R with respect to X, so we need to take DR by DX is equal to 1/2 minus 2 multiplied by 1/2 X, which is equal to 1/2 minus X. And now we need to set our derivative equal to 0. So we need to take DR by DX is equal to 0. So we can go ahead and write that 1/2 minus X is equal to 0. So when we rearrange this expression to isolate and solve for X all by itself, we will find that X is equal to 1/2. So now we need to verify that this will indeed give us the maximum radius by finding the second derivative. So with that said, we now need to find the second derivative, so we need to take D2d of R divided by D X. By DX2. So D 2 R by DX2, which is gonna be equal to D by DX of 12 minus X, which is going to be equal to simply -1 when we perform the derivative. So now we need to note that since R to the power of N is less than 0, this will mean that the radius function is concave down at this point, which indicates that X equals 1/2 is indeed a local maximum. So we can say. is a maximum. I'll just put the word max. So that means once again since R to the power of N is less than 0, this will mean as a result the radius function will be concave down at this specific point, indicating that X equals 1/2 will be indeed a local maximum. So therefore, as a result, the radius of the end circle R would be maximum if the diameter X of one of the smaller semicircles is a 12 unit. So, since we have obtained Why, so once again, so since we have obtained that Y is equal to 1 minus X previously, this will mean that the other smaller semicircle would have to have a diameter of Y equals 1 minus 1/2, which is equal to 12 unit, and that is our final answer. So let's quickly recap here for a second. So since we have obtained Y is equal to 1 minus X previously, this will mean as a result, the other semis or rather the other smaller semicircle would have a diameter of Y is equal to 1. Minus 1/2, which is equal to 12 unit as well. So therefore, without a doubt, we can go ahead and say that the radius of the in circle is maximized if the diameters of the two smaller semicircles are equal, which in this case for a diameter of 12 unit. So that will mean when we go to look at our multiple choice answers, the correct answer. Without a doubt, has to be multiple choice answer letter B, which once again is one half unit. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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