PR 4 # 25
 

Given a point P on one side of a general triangle ABC, construct the line through P which will divide the area of the triangle into two equal halves.

[QUOTE=Wacky]Given a point P on one side of a general triangle ABC, construct the line through P which will divide the area of the triangle into two equal halves.[/QUOTE]
[spoiler]The line must pass through the center of gravity. This point lies 2/3 of the way on the line from any vetex to the mid-point of the opposite side. As all three such line must also pass through the CoG, that gives you an easy way of finding the 2/3 point.[/spoiler]

Paul,
I don't believe that this is correct.
The center of gravity is such that the area, weighted by its displacement is equal on the two sides. But our construct requires that the areas, with uniform weighting, be equal.

Consider the particular case of an isosceles triangle with its base aligned horizontally and choose the point P to be 1/3 of the way up one side.

As I understand your construction, the line of partition would be horizontal.

This gives us two similar triangles whose areas are proportional to the square of their height.
But the entire triangle is 3/2 times as tall as the upper portion. Therefore, the total area is 9/4 of the upper area. Thus, the lower portion has more area than the upper.

Similar problem
 

:smile:
Wacky this problem or a very similar one just like it has been given before and solved by Fetofs If not mistaken and it has an ingenious solution which I noted because of its originality.

However I lack the expertise or know how, how to track the thread and its URL.
It will be worth resurrecting it for all our benefit.

I have not really analysed Paul's solution but I dont think its as simple as being based on medians meeting at a point as he describes. But this is besides the point I am making.

I have enjoyed all your problems and followed them closely. Unfortunately I could not solve most of them and the ones I could I lacked the time to post them what with two daily matches of the World Cup and that too one past midnight IST. I am suffering from a sleep debt and even now with no match today its fifteen past midnight.

Keep up the good work Richard and keep us mystified and entertained forcing us to put our thinking caps on

Mally :coffee:

So what does the square root of two have to do with solving this one?

Fusion

PR4- 25
 

[QUOTE=Wacky]Paul,
I don't believe that this is correct.
The center of gravity is such that the area, weighted by its displacement is equal on the two sides. But our construct requires that the areas, with uniform weighting, be equal.

Consider the particular case of an isosceles triangle with its base aligned horizontally and choose the point P to be 1/3 of the way up one side.

As I understand your construction, the line of partition would be horizontal.

This gives us two similar triangles whose areas are proportional to the square of their height.
But the entire triangle is 3/2 times as tall as the upper portion. Therefore, the total area is 9/4 of the upper area. Thus, the lower portion has more area than the upper.[/QUOTE]
:smile:

Well Wacky, Paul has the general idea and so does your hint but not quite.

Dont restrict the point, so that the parallel line can come to a position in your particular case (isosceles triangle) where it is such that the area of the two similar triangles are equal. Use your theorem and equate the equations obtainable to 1/2 as the smaller is half the larger [thats how sq.rt 2 comes in Fusion power] as we are dealing in squares and it should work out.

Off hand the parallel should be of length around 6.3 if the base is put as 9.

Moreover I would dare say off hand that there is a theorem that says that the areas of similar triangles are not only proportionate to the squares ot their heights ( I presume you are right but no argument) but more generally to the squares of the any two similar sides by which you can tackle the general case. However I will have to check Euclid to vouch for it right now.

I also would say that your first para needs correction as the property is true for the medians of the triangle thru which the COG passes and the weight distribution is equal as Paul states.
The mistake here is that weight is proportional to area and area is in proportion to the squares of the sides hence the 1/ 3 height of the COG neither the midpoint height will do as at those two positions the areas are simply not equal as my equations will reveal It must be at 6.3 length as in my solution if the base is 9.

Mally :coffee:

Consider an Isosceles triangle. Draw a line through the triangle such that the triangle is divided into two areas of equal volume. Rotate the line through 360 degrees one degree at a time and maintaining the two areas of equal volume. For how many lines is the isosceles triangle divided into two similar but separate triangles? For how many lines is it divided into two similar but overlapping triangles? What figure will the centerpoint of the line inscribe as you rotate it through 360 degrees? What proof(s) can you derive from this?

Two very interesting numbers come into play, the first is .866 and the second is 1.414.


I knew the relationship of sqrt(2) to this triangle, was just putting it into the conversation to see what others would figure out about it without giving away the solution. The underlying problem is to divide a triangle into two parts of equal area which infers that (using Wacky's example) the line will be split based on sqrt(2).

Another interesting twist on this is to use a cone with a plane bisector.

Fusion

PR4#25
 

[QUOTE=mfgoode]:smile:

Well Wacky, Paul has the general idea and so does your hint but not quite.

Moreover I would dare say off hand that there is a theorem that says that the areas of similar triangles are not only proportionate to the squares ot their heights ( I presume you are right but no argument) but more generally to the squares of the any two similar sides by which you can tackle the general case. However I will have to check Euclid to vouch for it right now.

Mally :coffee:[/QUOTE]
:smile:
I hereby confirm that the theorem that I have stated is correct. By similar sides I meant corresponding sides so there's no ambiguity.
With this property any triangle( not necessarily the particular case of isosceles triangles or equilateral) for this problem can be solved.
Wacky kindly confirm this and I would like to know if you have any other method.
Thank you,
Mally :coffee:

PR#25
 

[QUOTE=Fusion_power]Consider an Isosceles triangle. Draw a line through the triangle such that the triangle is divided into two areas of equal volume. Rotate the line through 360 degrees one degree at a time and maintaining the two areas of equal volume. For how many lines is the isosceles triangle divided into two similar but separate triangles? For how many lines is it divided into two similar but overlapping triangles? What figure will the centerpoint of the line inscribe as you rotate it through 360 degrees? What proof(s) can you derive from this?

Two very interesting numbers come into play, the first is .866 and the second is 1.414.



Another interesting twist on this is to use a cone with a plane bisector.

Fusion[/QUOTE]
:smile:
You seem to be terribly confused ( to me anyway) because as stated it is ridiculous with an 'equally interesting twist'
Kindly read it and restate it to clarify what you exactly mean.
Mally :coffee:

I didn't quite understand mfgoode's explanation, it is probably correct, but my solution is:
[spoiler]
Suppose the point P is on the line AC, and is such that the length PC is a fraction x of the length AC, with 1/2 <= x <= 1 (relabel the triangle as necesary to satisfy these conditions. Then if Q is the point on the line BC such that PQ divides the triangle's area in half, then the length QC is a fraction y of the length BC, where xy = 1/2.
The only problem is that I don't know how (or if it's possible) to construct the point Q - anyone?
[/spoiler]

Chris

Equal areas
 

[QUOTE=Chris Card]I didn't quite understand mfgoode's explanation, it is probably correct, but my solution is:
[spoiler]
Suppose the point P is on the line AC, and is such that the length PC is a fraction x of the length AC, with 1/2 <= x <= 1 (relabel the triangle as necessary to satisfy these conditions. Then if Q is the point on the line BC such that PQ divides the triangle's area in half, then the length QC is a fraction y of the length BC, where xy = 1/2.
The only problem is that I don't know how (or if it's possible) to construct the point Q - anyone?
[/spoiler]

Chris[/QUOTE
smile:
Your solution is really a question as you have not solved anything rather confused yourself further.
Kindly read my solution again.
Without the aid of a diagram your problem seems to be that you must realise that you must get two similar triangles whatever the triangle you start with. Therefore PQ must be parallel to the side it does not intersect. Dont worry and waste time to where point Q will intersect BC. The // line PQ does the trick.
The ratio you have used is wrong. Go over the posts. Every post is important including the one on COG. The theorem is clearly and precisely stated more than once. You have used the wrong ratio and there fore not understood a gist of the solution because of lack of study of the solution.
You require more effort on your part and I cannot and refuse to spoon feed you any further.
And you have the audacity to state that my solution is 'probably correct'?
And you introduced a hairbrained solution of your own when you dont know the basic principles and expect to fox others with it?
Mally :coffee