The first thing to notice is that the answer from the person who tried to use "reductio ad absurdum" is totally bogus. The opposite vertices are actually part of the sides, and he has totally missed the possibility that they just never intersect the opposite side at all.
Here is a sketch of the proof:
Start with the diagonal AC, with an arrow on it pointing towards C. If this bisects the area of the pentagon you are done. If not, then one of the two parts it cuts the pentagon into is bigger. You can refer to these parts by the left and the right as follows:
Left
------------->-----------------
Right
The right is the side that falls to the right of the arrow.
Slide the diagonal from AC to AD. One of two things happens:
The same side (right or left) of the diagonal is still bigger, or the other side is bigger. If the other side is bigger, that means somewhere in between C & D there will be a place with equal area.
If the same side is bigger, you slide the diagonal from AD to BD and apply the same check again. From BD you slide it to BE and from BE to CE and from CE to CA.
The only way for you to never switch which side has a bigger area is for the bigger area to always be on the same side of the diagonal. However, in the process of rotating the diagonal between the points, it is now "flipped", so what the arrow is pointing from C to A. So if the bigger side was left at the start, it must now be right, and vise versa. So, somewhere along the way (at least once), you have switched which side had the bigger area, and thus there is a place where the bisector is on the side.
2006-07-04 01:31:33
·
answer #1
·
answered by fatal_flaw_death 3
·
0⤊
0⤋
It's a bit hard to prove geometric problems by explanations (without drawing) but I try my best to explain:
First, I use 'reductio ad absurdum' method; that is if the area-halving lines don't intersect opposite sides then they should pass through opposite vertices. So they are internal diameters of polygon. Now we must simply prove that it is not possible for all vertices:
Suppose vertex 'A'. According the above assumption, one of the diameters (let say AC) must bisects the pentagon equally. Now consider the other opposite vertex (here 'D'). There are two diameters available; DA and DB. It is obvious that if AC halves the pentagon then DA could not (because then the triangle ACD must have zero area) so the only choice could be DB. Now, consider the triangles ABC and BCD. Both should have area equal to half of pentagon's area; therefore they must have same areas. Since they have one common side (BC) then we find out that the line AD (which is a diameter of pentagon) must be parallel to BC. Now, continue this method for other vertices and we find out that all opposite diameters and sides must be parallel (that is AB || CE, BC || AD, CD || BE, DE || AC, EA || BD). But the only pentagon that have such specification is an equilateral pentagon (and in this case also no diameter will halve the pentagon too). It is obvious that since we assume our pentagon a general one therefore our logic resulted to wrong statement. So the assumption must be false; that is some of the lines must intersect opposite sides.
2006-07-03 20:34:52
·
answer #2
·
answered by fredy1969 3
·
0⤊
0⤋
nicely, if the total form of steps is x, and man or woman A has climbed 1/2 of them (a million/2 x) and man or woman B has climbed a third of them (a million/3 x) and between them they have climbed 100 and sixty steps, then: a million/2 x + a million/3 x = 100 and sixty rewriting so both fractions have a undemanding denominator (the bottom variety) 3/6 x + 2/6 x = 100 and sixty 5/6 x = 100 and sixty dividing each and every aspect by utilizing 5/6 x = 192 therefore the total form of steps is 192. this signifies that man or woman A has run up ninety six steps and man or woman B has run up sixty 4 steps. we may be able to verify this somewhat a million/2(192) + a million/3(192) = 100 and sixty ninety six + sixty 4 = 100 and sixty
2016-10-14 02:39:14
·
answer #3
·
answered by ? 4
·
0⤊
0⤋
Draw a line in the middle to divide it in half. It's a pretty simple geometry problem.
2006-07-03 19:25:42
·
answer #4
·
answered by Anonymous
·
0⤊
0⤋
Draw the diagram and fold the paper in half. Hold it up to the light, and you'll find your answer: it's just that simple.
2006-07-03 19:26:35
·
answer #5
·
answered by Mr. Wizard 7
·
0⤊
0⤋
How can a pentagon be convex?
2006-07-03 19:35:08
·
answer #6
·
answered by Foolhardysage 2
·
0⤊
0⤋
Bumblebee?
2006-07-03 19:27:22
·
answer #7
·
answered by Anonymous
·
0⤊
0⤋
If you can't answer this question then you should concentrate in doing some business..!
2006-07-03 19:27:45
·
answer #8
·
answered by Anonymous
·
0⤊
0⤋
Wildraft is wrong, blue is the answer.
2006-07-03 19:27:29
·
answer #9
·
answered by Anonymous
·
0⤊
0⤋
oh man1 a lot of smart people around!
2006-07-04 00:33:31
·
answer #10
·
answered by pinkcloud2015 5
·
0⤊
0⤋