how to know value of half circle's total length on X axis?

Question

Imagine there is half circle and it has N lines in its inside. But those line's X axis's length is different in every single lines. And question is what is value of those line's X axis's length sum? Help me if i won't find value, my teacher going to kill me

Answer 1

Hello Bilguun,



Questions like this are designed to make you think and analyze a situation. I know that at times it feels like the questions are designed to drive your crazy.



From the question, it seems a discrete situation, rather than continuous. The answer will change on the value of N. This seems complex. Lets look at a simpler problem and then see how it could be adapted to your question. I also do not know how your instructor wants you to distribute the lines, by even intervals along the y-axis, of by angle measure.



For evenly along the y-axis:



For simplicity assume the radius is 1 -- later you can generalize by simply multiplying by R.



Lets assume the semi circle is centered at the origin and the lines are evenly distributed along the y-axis. That means for N=1, you have a single line at y=1/2; for N=2, there are two lines at y=1/3 and y=2/3; etc. (In a case like this, decimals are not your friends. Fractions will best show you the patterns. (You may want to sketch this a couple times to see the pattern.) Can you see that in your head? If not draw a picture.



Still too complex -- look only at the first quadrant -- a quarter circle. You can generalize back to your problem by doubling everything.



Really, you are talking about the "floor" of a triangle defined by a central angle, right? Do you see it?

The side opposite the central angle will have lengths that varies on N.

N=1 side opposite has length 1/2

N=2 side opposite has lengths 1/3 and 2/3

N=3 side opposite has lengths 1/4, 2/4, 3/4

N=4 side opposite has lengths 1/5, 2/5, 3/5, 4,5



For each case calculate (in fraction form) the x-lengths and add them. Look at how you have to calculate each summation. You will have to use summa notation (The huge summation thingy) to express the values in general in terms of N.



Keep in mind that was for a quarter circle or radius 1. Double it and multiply by R for your general case.



In the limit as N tends to infinity you can use the integral you suggested, (except you are only looking at one quadrant so what do you have to do?)



For angle measure:



This one is EASIER. Use the assumtpions above, except asssume evenly distributed by angle measure. You will be summing cosines.

N=1 cos(pi/4)

N=2 cos(pi/6), cos(2pi/6)

N=3 cos(pi/8), cos(2pi/8), cos(3pi/8)

N=4 cos(pi/10, cos(2pi/10), cos(3pi/10), cos(4pi/10)



Do you see the pattern? Look at the denominators. WHY are they what they are?



Again, summa notation, double and multiply by R.



I hope this helps you. You are more familiar with what your teacher said and what he/she will expect. Depending on the level of class angle-based is the easiest, y-axis based would be more advanced.



Good luck with the problem -- and your class. I hope you can crack this problem and hunt down the answer you seek.

Answer 2

I'll count on that you want a non-trivial dissection into acute isoceles triangles that aren't equilateral triangles, in view that in any other case the solution is a dissection into four equilateral triangles, for the minimal. I'm going to look more into this, however offhand, first the equilateral triangle can be dissected into one acute isosceles triangle and a pair of equal obtuse isosceles triangles. Every of the obtuse isosceles triangles can then be dissected into an irregular cyclic pentagon and two acute isosceles triangles, wherein that irregular cyclic pentagon can also be extra dissected into 5 acute isosceles triangles. Consequently, the total number of acute isosceles triangles that the equilateral triangle will also be dissected into is 1+ 2(2+5) = 15. I will provide a graph later. Edit: folks must cool it with the TUs earlier than I've had the threat to compare this in element. An answer with 15 acute isosceles, none of them equilateral triangles, would not look feasible. As a substitute, I ought to first break up the two obtuse triangles into 2 more acute isosceles triangles, and then dissect the rest obtuse triangles into 7 acute isosceles triangles, in order that the complete is 17, no longer 15. Maintain this query open unless I find the time to verify that a 17-acute-isosceles triangle dissection is feasible, none of which are equilateral triangles. Ugh. Edit 2: yes, just barely, there is a 17-acute-isosceles triangle dissection viable, none of which might be equilateral triangles. I'm going to work out the numbers later, now not tonight. Edit 3: here is the hyperlink to the graph of the 17-acute isosceles triangles dissection (with the proper facet uncompleted, but is mirror photograph of left). Probably gianlino does have a way to do that with 15, but i'll want extra time to compare his reply. Notes: to find the center of the circumscribed circle of the cyclic irregular pentagon, find intersection of bisectors of two angles of the long-established triangle. For that reason, it's constantly possible to dissect a triangle into 7 isosceles triangles. Nevertheless, the that they all be acute isosceles is what usually require that the common triangle have an additional acute isosceles triangle first whittled off. In this case, all of the isosceles triangles making up the cyclic irregular pentagon are acute, some just barely. Edit 4: okay, i have a difficult sketch of what I feel gianlino's solution is, and it does seem like it will have to allow a valid one. But i have not yet worked out the details on his answer. That's, making definite all 15 are acute isosceles. Edit 5: Ditto with gianlino's modern day solution with 13 (maybe) acute isosceles triangles. It can be the last isosceles triangle at the bottom that i'm having main issue verifying, and time is walking out and that i gotta handle every other matters. Edit 6: In a parallel universe i've discovered gianlino's resolution with 10 acute isosceles triangles, but on this universe I don't have any time to even watching at how this might even be viable. Nonetheless, I must agree that his thirteen acute isoscoles answer is completely possible, even though I've now not but demonstrated it myself, so maybe he will have to get the BA.

Answer 3

diameter