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Subject: "Cut Triangle in Half"     Previous Topic | Next Topic
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Conferences The CTK Exchange High school Topic #248
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Bractals
Member since Jun-9-03
Jul-08-03, 07:16 AM (EST)
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"Cut Triangle in Half"
 
   Hi,

I came up with the following construction for the given problem.
It checks out with Sketchpad. See what you think.

Bractals

PROBLEM:

Given an arbitrary point and triangle. Construct a line through the
point that divides the triangle into two polygons of equal area.

NOTATION:

^ denotes set intersection.

Let X' denote the midpoint of the side of the triangle opposite the vertex X.

If X and Y are distinct points, let

L(XY) denote the line determined by points X and Y,
R(XY) denote the ray starting at point X and passing through
point Y,
S(XY) denote the line segment determined by points X and Y, and
P(Z,XY) denote the line through point Z and parallel to L(XY).

L(YX) = L(XY), R(YX) != R(XY), S(YX) = S(XY)
S(XY) = R(XY) ^ R(YX)

CONSTRUCTION SCHEMA:

CIR(XYZ,r) : For fixed points Y and Z and fixed length r -
construct point X such that Y lies on S(XZ) and |XY| = r.

CONSTRUCTION:

If ( the point lies on L(XX') for some triangle vertex X ) {
L(XX') is the desired line.
} else {
Let P be the given point and G the centroid of the triangle.
Label the triangle ABC such that R(GP) intersects S(CB').
Construct point R such that R = P(P,AB) ^ L(AC).
Construct point S such that S = P(B',BR) ^ L(AB).

If ( P lies on the triangle ) {
L(PS) is the desired line.
} else {
Construct point T as the midpoint of S(AS).
CIR(QRP,|PR|).
CIR(DAT,|PQ|).
Construct a semicircle with diameter S(DT) on the opposite side
of S(DT) from vertex C.
Construct a line through vertex A, perpendicular to S(DT),
intersecting semicircle DT at point E.

If ( P is outside the triangle ) {
CIR(FTA,|TE|).
L(PF) is the desired line.
} else {
Construct point H such that |EH| = |AT| and H lies on S(AT).
CIR(KTH,|AH|).
L(PK) is the desired line.
}
}
}


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  Subject     Author     Message Date     ID  
Cut Triangle in Half Bractals Jul-08-03 TOP
  RE: Cut Triangle in Half Vladimir Jul-11-03 1
     RE: Cut Triangle in Half Bractals Jul-12-03 2
         RE: Cut Triangle in Half Vladimir Jul-12-03 3
             RE: Cut Triangle in Half Bractals Jul-13-03 5
             RE: Cut Triangle in Half Bractals Jul-15-03 12
                 RE: Cut Triangle in Half Golland Jul-15-03 13
                     RE: Cut Triangle in Half Bractals Jul-15-03 14
                         RE: Cut Triangle in Half Golland Jul-16-03 15
                             RE: Cut Triangle in Half Bractals Jul-16-03 16
                 RE: Cut Triangle in Half Vladimir Jul-17-03 17
         RE: Cut Triangle in Half golland Jul-13-03 4
             RE: Cut Triangle in Half Bractals Jul-14-03 6
             RE: Cut Triangle in Half Bractals Jul-14-03 7
                 RE: Cut Triangle in Half alexb Jul-14-03 8
                 RE: Cut Triangle in Half Vladimir Jul-18-03 18
                     RE: Cut Triangle in Half Bractals Jul-23-03 19
                         RE: Cut Triangle in Half Vladimir Jul-24-03 20
  RE: Cut Triangle in Half Bractals Jul-14-03 9
     RE: Cut Triangle in Half alexb Jul-14-03 10
         RE: Cut Triangle in Half Bractals Jul-14-03 11

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Vladimir
Member since Jun-22-03
Jul-11-03, 02:43 PM (EST)
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1. "RE: Cut Triangle in Half"
In response to message #0
 
   LAST EDITED ON Jul-17-03 AT 07:43 PM (EST)
 
I am having trouble reading your construction, so I checked only the part when the area bisector point P is on the triangle. That's just me, but you did not offer any proof of your construction, a serius defficiency, otherwise I like it. So let's prove the first part of the construction now:

Construct the triangle area bisector through a given point P on the triangle circumference: Connect the point P with the opposite vertex (B in our case). Draw a line parallel to this line (BP) through the median heel (Mb) on the same triangle side as the point P. The line intersects the triangle at another point S. PS is the triangle area bisector. See the attached drawing.

Proof:

Denote x = SB/AB, y = PC/AC. Since the lines BP and SMb are parallel, DABP and DASMb are similar.
Consequently,

(1 - y)/(1/2) = 1/(1 - x) or y = 1 - 1/{2(1 - x)}

Consider yellow DASP and magenta DSBP. Their altitudes to the line AS or SB are the same. Therefore, their areas A(Y) and A(M) (for yellow and magenta) are in the ratio of their bases AS and SB:

A(Y)/A(M) = (1 - x)/x

Consider yellow & magenta DPAB and cyan DCPB. Again, their altitudes to the line PA or CP are the same. Therefore, their areas A(Y) + A(M) and A(C) (for yellow + magenta and cyan) are in the ratio of their bases PA and CP:

(A(Y) + A(M))/A(C) = (1 - y)/y

Let's calculate the area A(M) + A(C):

A(M) + A(C) = A(Y).x/(1 - x) + (A(Y) + A(M)).y/(1 - y) =

= A(Y).x/(1 - x) + {A(Y) + A(Y).x/(1 - x)}.y/(1 - y) =

= A(Y).(x + y - xy)/(1 - x)/(1 - y)

Substituting y = 1 - 1/{2(1-x)} we get A(M) + A(C) = A(Y) and the line PS is indeed the triangle area bisector.

Q.E.D.

Attachments
https://www.cut-the-knot.org/htdocs/dcforum/User_files/3f0e673c6fff5696.gif

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Bractals
Member since Jun-9-03
Jul-12-03, 11:10 AM (EST)
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2. "RE: Cut Triangle in Half"
In response to message #1
 
   Hi Vladimir,

I will give you an analysis of my construction if you first tell me how you got your figure ( looks like it came from Sketchpad ) into a gif format.

Using your figure ( with B' as the midpoint of AC - don't know how to do subscripts ) here is an easier proof for this case:

Triangles BAP and SAB' are similar. Therefore, AS/AB' = AB/AP.

AS*AP = AB*AB' = AB*AC/2
AS*AP*sin(<SAP)/2 = (AB*AC*sin(<BAC)/2)/2
Area(SAP) = Area(BAC)/2

Bractals


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Vladimir
Member since Jun-22-03
Jul-12-03, 10:10 PM (EST)
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3. "RE: Cut Triangle in Half"
In response to message #2
 
   It's a deal. There are many graphics programs and simple drawings like the triangle I do with the Windows Paint. That gives me a bitmap. Likewise, there are many graphics format converters, even as shareware. I use Adobe Photoshop (not shareware!). Simply copy the bitmap into Photoshop and export is as a GIF. If I want a transparent GIF (i.e., one color in the picture selected as transparent), I usually have to do it twice - the Photoshop does not let me do it in one step. First, I export as a normal GIF. Then I load this GIF file and export it as a transparent GIF.

To use subscripts, indexes, and greek symbols, you have to type your message as a kind of HTML code. For example, the triangle symbol is the capital greek delta: <font face="symbol">D</font>ABC, but you have to use rectangular parenthesis - DABC. Partial derivative is <font face="symbol"></font>P/<font face="symbol"></font>x - P/x. See the Windows accessory/system tool Character Map. Likewise, subscripts or superscripts are x<sub>i</sub> or x<sup>5</sup> - xi or x5. See the HTML reference help to the right of the window wher you type. It has some pitfalls: If you use <CODE> ... </CODE> and do not use line breaks, the message text does not wrap. The <ENTER> key or the HTML code <BR> do the line breaks.

I admit that your proof is simpler, you should have included it the first time. I guess I was stubbornly looking for some triangles with the same altitudes. When trying to read your construction, I encountered things like constructing a point that already existed, constructing a circle for no other purpose than to double a line segment and than using its length somewhere else, etc.

Regards, Vladimir


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Bractals
Member since Jun-9-03
Jul-13-03, 09:54 AM (EST)
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5. "RE: Cut Triangle in Half"
In response to message #3
 
   Hi Vladimir,

Please help me out. I created a square-bracketed html file and a gif file (7KB).

When I copied the html file into the post and did a preview everything looked okay except one place.

Source: "...end of line.#1#Start next line..."

Preview: "...end of line.#2#Start next line..."

where

#1# = {cr}{cr}{lsb}/br{rsb}

#2# = {lsb}P>{cr}

where

{cr} = carriage return

{lsb} = left square bracket

{rsb} = right square bracket

When I tried to attach the gif file:

1) I "Click here to choose your file"
2) I browsed my disk for the gif file
3) I selected it and clicked upload
4) It'said it uploaded the file
5) It said the name would be placed in the Attachment box
6) It lied.

Any help would be appreciated.

Bractals


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Bractals
Member since Jun-9-03
Jul-15-03, 12:39 PM (EST)
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12. "RE: Cut Triangle in Half"
In response to message #3
 
   Hi Vladimir,


Thanks for the Information. I hope the figure comes through OK.



Analysis of the construction:


Let the line through point P cut the sides AB and AC at points X and Y respectively.


Let c = |AB|, b = |AC|, x = |AX|, and y = |AY|. For the area of DXAY to be one-half the area of DBAC we need xy = bc/2 (*).


In the figure I show two cases: P1 inside the triangle and
P2 outside the triangle.


DXAY ~ DP1R1Y ~ DP2R2Y. Therefore,


|AX| / |AY| = |R1P1| / |R1Y| = |R2P2| / |R2Y|


|AX| / |AY| = |R1P1| / (|AY| - |AR1|) = |R2P2| / (|AR2| - |AY|)


x/y = u1/ (y - v1) = -u2/ (v2 - y) ,


where v = |AR| and |u| = |RP| with u < 0 if P is on the opposite side of line AC from the vertex B.


Therefore, x/y = u/(y - v) (**).


Eliminating y from (*) and (**) we get


2vx2 - bcx + bcu = 0.

Solving for x we get


x = w sqrt(w2 - 2uw) , where w = bc/(4v).


By labeling the triangle so that P lies in the yellow area (see the figure), we guarantee that w2 - 2uw > 0 and that X lies between B and C' if we use the plus sign ( I will let you verify this ). Hence,


x = w + sqrt(w2 - 2uw)


By letting t2 = (2|u|)w we have


x = w + sqrt(w2 + t2 ) for P outside the triangle and


x = w + sqrt(w2 - t2 ) for P inside the triangle.


I think you should be able to follow the construction from this. I hope my first attempt with HTML worked.


Bractals


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Golland
guest
Jul-15-03, 03:17 PM (EST)
 
13. "RE: Cut Triangle in Half"
In response to message #12
 
  
Hi Bractals,


I do not quite understand your construction: according to the

figure the points P1 and P2 are on the same line, but the solution

has x ( the length of AX ) different for P1 and P2.

Maybe it is the second solution for points P1 ( as we discussed )

but P1 outside of the "deltoid" but still inside the triangle has

only one solution.


Also the points in close proximity of centroid G clearly have 3

lines that cut area of the ABC in 2 equal parts.

The "deltoid" inside the ABC : is it a triangle?


How does your solution accounts for "deltoid" existance?

G


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Bractals
Member since Jun-9-03
Jul-15-03, 06:56 PM (EST)
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14. "RE: Cut Triangle in Half"
In response to message #13
 
   Hi Golland,

In calculating the x = w + sqrt(...) value one must use the u2,v2 values and for the x = w - sqrt(...) value one must use the u1,v1 values.

With respect to the deltoid, consider the following figure ( unlabeled - but consider the triangles labeled as in my previous figure ).

In the upper left corner the yellow area represents all the points P satisfying the inequality c/2 < x < c, where the x is the w + sqrt(...) value. In the upper right corner the blue area represents all the points P satisfying the same inequality c/2 < x < c, where the x is the w - sqrt(...) value.

When you overlay these figures you get the middle one with the small dark triangle representing points P which have two different lines cutting sides AB/AC as solutions.

About this dark triangle. The "side", which is not along a median of triangle ABC, is not a straight line. It is a hyperbolic arc concaved in towards the centroid. The equation for this arc is u = bc/(8v); which comes from setting the radicand equal to zero.

When you add in the points for the lines that cut sides AB/BC and AC/BC you add two more small triangles making up the deltoid.

After going through this further analysis - I want to modify my conjecture on the deltoid. All points in the interior have three solutions, all points in the exterior have one solution, and all points on the boundary have two solutions ( except for the deltoid vertices which have only one solution ). This can be seen by noting that a line cutting sides AB/BC can intersect the small dark triangle in my figure.

I hope this puts the problem to rest.

Bractals


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Golland
guest
Jul-16-03, 10:54 AM (EST)
 
15. "RE: Cut Triangle in Half"
In response to message #14
 
   Hello Bractals,

Thanks for the reply. I am still impressed.

There are 2 additional problems that have to be resolved before

we stop beating this horse: 1) draw a line perpendicular to this one

that divides each area in half again; 2) draw a line that divides

PERIMETER of the triangle in half.


Thanks.

G.


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Bractals
Member since Jun-9-03
Jul-16-03, 03:09 PM (EST)
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16. "RE: Cut Triangle in Half"
In response to message #15
 
   Hi Golland,

This deltoid problem, like Dracula, keeps coming back from the dead.

The following figure shows the deltoid for an equilateral triangle ABC ( not really - the arcs are circular ). Where G is the centroid and X* is the midpoint of the median through vertex X.

*****************************************************
My final conjecture (until next week - haha) about the number of line solutions for a point:

3 - The centroid G and the interior of the "triangles" A*B*G, B*C*G, and C*A*G
2 - Arcs A*B*, B*C*, and C*A* excluding the endpoints
1 - All other points
*****************************************************

See if you can find 3 solution lines for any point on the open line segment (A*G).

As for your two problems, let me think about them.

With respect to the area problem, would you agree that solving the problem for an equilateral triangle is enough?

Will be gone for a week.

Bractals


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Vladimir
Member since Jun-22-03
Jul-17-03, 06:52 PM (EST)
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17. "RE: Cut Triangle in Half"
In response to message #12
 
   LAST EDITED ON Jul-23-03 AT 05:02 PM (EST)
 
Hi, Bractals

I can follow your construction now. Your analysis is an excellent post. I am also presenting my (independent) analysis. It melts down to the same thing - solving a quadratic equation.

Draw a line from a point Q outside the DABC such that it cuts the triangle area in half.

The extended medians of a triangle divide the plane outside the triangle into 6 segments (see the attached drawing). If the point Q is on one of the extended medians, the problem is solved. Suppose the point Q is between the extended medians ma and mb. Put the origin of the xy coordinate system into the triangle vertex C and the y-axis along the side a (i.e., through the vertex B). If the triangle ABC is labeled counterclockwise, the y-coordinate of the vertex B is positive. If the triangle DABC is labeled clockwise, the y-coordinate of the vertex B is negative. In that case, reflect both the DABC and the point Q in the x-axis, reducing the problem to the former case. Select the unit length a = CB = 1.

The line cutting the triangle area in half intersect the triangle in 2 points: P (on the triangle side a = CB) and S (on the triangle side b = CA). The equation of the line QP is

y - yP = (yQ - yP) / (xQ - xP) * (x - xP) = (yQ - yP) / xQ * x

because xP = 0. Since the points Q, P, S are colinear (on a line), we have the following equation for their coordinates:

(1) yS - yP = (yQ - yP) / xQ * xS

Since the point S and the triangle vertex A are on a line through the coordinate origin,

(2) tan(g) = yA/xA = yS/xS

Finally, since the area of DSPC is half of the area of DABC

1/2 CS * CP sin(g) = 1/2 (1/2 CA * CB sin(g))

CS * CP = 1/2 CA * CB

Writing this using the xy coordinates of the points P, S, and A and using the unit length CB = 1, we get

(3) yP (xS2 + yS2) = 1/2 (xA2 + yA2)

We have 3 equations for 3 unknowns yP, yS, and xS. Simplifying the 3rd equation using the 2nd equation:

yP (xS2 + yS2) = yP.xS (1 + (yS/xS)2) = yP xS (1 + (yA/xA)2) = yP xS/xA (xA2 + yA2) = 1/2 (xA2 + yA2)

yP xS/xA = 1/2

Combining this with the 2nd equation we get

xS = xA/(2yP)
yS = xS yA/xA = xA/(2yP) yA/xA = yA/(2yP)

Substituting to the 1st equation we get the following quadratic equation

yP2 - xA/(2xQ) yP - xA/2 (yA/xA - yQ/xQ) = 0

The equations of medians ma = AGa and mb = BGb are

ma: y = (yA - 1/2) * x/xA + 1/2
mb: y = (yA - 1) * x/xA + 1

Since the point Q is between these 2 medians,

(yA - 1/2) * xQ/xA + 1/2 < yQ < (yA - 1) * xQ/xA + 1

Since xQ < 0

-(yA - 1/2)/xA - 1/(2xQ) < -yQ/xQ < -(yA - 1)/xA - 1/xQ

1/(2xA) - 1/(2xQ) < yA/xA - yQ/xQ < 1/xA - 1/xQ

1/4 - xA/(4xQ) < xA/2 (yQ/xQ - yA/xA) < 1/2 - xA/(2xQ)

Again, since xQ < 0 and xA > 0, both the left and right side of this inequality are positive and we can denote

u = -xA/(2xQ) > 0

v2 = xA/2 (yQ/xQ - yA/xA) > 0

and write our quadratic equation as

yP2 + uyP - v2 = 0

The solution is yP = -u/2 (u2/4 + v2). The root with minus sign before the square root is negative and cannot be on the side a = CB of our DABC.

To perform the construction, we have to construct the lengths u > 0 and v > 0. The square root in the formula for yP will be a hypothenuse of a right triangle with sides u/2 and v. Finaly, we will subtract u/2 from the length of this hypothenuse to get the length of yP (see the attached drawing).

1. Draw a unit circle (radius a = CB = 1) with the center at the origin C. The circle intersects the x-axis in points D and E.

2. Erect normals to the x-axis at the points D and E. These normals intersect the lines CA and CQ at points D' and E', respectively. This gives us DD' = yA/xA and EE' = -yQ/xQ (remember, xQ < 0).

3. Add these two lengths: Drop a normal to the y-axis from the point E'. Denote the heel of the normal H. Connect the points D and H. Draw a line through the point D' parallel to DH. The line intersects the y-axis at point H'. This gives us CH' = yA/xA - yQ/xQ.

4. Multiply this length by xA: Drop a normal from the point A to the x-axis, denote A' the heel of this normal, CA' = xA. Connect points D and H'. Draw a line parallel to DH' through the point A'. The line intersects the y-axis at point K. This gives as CK = xA (yA/xA - yQ/xQ).

5. Divide the line segment CK in half. This gives as CL = xA/2 (yA/xA - yQ/xQ).

6. Find the square root of this distance: The unit circle intersects the negative portion of the y-axis opposite to the point L at point F. Find the center of the line segment LF - point M. Construct a circle with center M and radius ML = MF. The circle intrsects the y-axis at point N. This gives us v = CN = {xA/2 (yA/xA - yQ/xQ)}.

7. Divide the length xA by the length xQ: Copy the length of the line segment xA = CA' to the negative portion of the y-axis (the 1st quadrant is already too messy), i.e., CR = CA'. Connect points Q' and R. Draw a line through the point E parallel to Q'R. The line intersects the y-axis at the point R'. This gives us CR' = -xA/xQ (remember, xQ < 0).

8. Divide the line segment CR' in four. This gives as u/2 = CT = -xA/(4xQ).

9. The distance TN is the hypotenuse of the right DNTC with sides u/2 = CT and v = CN, i.e., TN = (u2/4 + v2).

10. Find the roots of our quadratic equation by substracting the length u/2 from the hypotenuse or adding the length u/2 to the hypotenuse: Draw a circle with center at the point T and radius TN. The circle intersects the y-axis at points P and P'. CP = TN - CT = (u2/4 + v2) - u/2 and -CP' = -(TN + CT) = -{(u2/4 + v2) + u/2} are the roots of our quadratic equation.

11. Only the first root is acceptable, yP = CP. Connect the points Q and P and extend the line through the DABC. QP is the desired line.

If the point Q is inside the triangle, the analysis and construction are the same as if the point is outside, with one exeption. We cannot trivially determine which two sides of the triangle the area bisector intersects, so that we have to try all 3 pairs. Aparently, when the point Q is close to one side, there is not much difference from the case when the point Q is outside close to the same side. However, when the point Q is deep inside the triangle, there might be 3 different area bisectors through a given point. I will analyze this in the next post.

Attachments
https://www.cut-the-knot.org/htdocs/dcforum/User_files/3f171aa13433dd89.gif
https://www.cut-the-knot.org/htdocs/dcforum/User_files/3f171ae93521d184.gif

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golland
guest
Jul-13-03, 09:54 AM (EST)
 
4. "RE: Cut Triangle in Half"
In response to message #2
 
   Hello Bractals and Vladimir.

I put this problem on the site sometime in October of 2002.

The solution in case the point is on the side of triangle is trivial.

There must be 3 solution lines when the point is "deep" inside the triangle.

When the point is outside or on the perimeter of the triangle where is one solution.

So there must be a locus of points with 2 solutions.


And the construction has to include an angle bisectors.


Bractal, if you use the Sketchpad you can move around vertices

A,B,C and the given point P and see if the arears of the resulting

halves remain equal ( i doubt it is the case for the solution you

provided ).

Anyway, the proof still remains to be seen.


Golland.


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Bractals
guest
Jul-14-03, 10:28 AM (EST)
 
6. "RE: Cut Triangle in Half"
In response to message #4
 
   Hi golland,

I did move the vertices around in Sketchpad and the area was always cut in half. My analysis for the problem awaits information about an html glich and why I cannot upload a gif file ( see message #4 ).

Bractals


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Bractals
Member since Jun-9-03
Jul-14-03, 06:27 PM (EST)
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7. "RE: Cut Triangle in Half"
In response to message #4
 
   Hi Golland,

In my previous reply I said message #4. It's different now - seems messages are renumbered when a new one is inserted.
With respect to how many lines there are at a point; here is my conjecture:
Consider a deltoid with vertices at the midpoints of the medians.
Any point in the exterior of the deltoid has only one solution.
Any point on the boundary of the deltoid has two solutions; unless it is a vertex, then it has only one.
Any point in the interior has two solutions; unless it is on a median, then it has three.
What do you think of the conjecture?

Bractals


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alexb
Charter Member
1029 posts
Jul-14-03, 06:29 PM (EST)
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8. "RE: Cut Triangle in Half"
In response to message #7
 
   >What do you think of the conjecture?

You may want to stop by https://mathforum.org and search on of their geometry forums. Sometime ago there was a discussion with John Conway on that deltoid.


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Vladimir
Member since Jun-22-03
Jul-18-03, 04:48 PM (EST)
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18. "RE: Cut Triangle in Half"
In response to message #7
 
   LAST EDITED ON Jul-18-03 AT 08:23 PM (EST)
 
Hi, Bractals

Your conjecture is not entirely correct:
Any point inside of the deltoid (on a median or not) has 3 different area bisectors. In addition, the deltoid boundaries are not circular arcs, but segments of hyperbolas, the extended triangle sides being their asymptotes.

Draw a line from a point Q inside the DABC such that it cuts the triangle area in half.

Put the origin of the xy coordinate system into the triangle vertex C and the y-axis along the side a (i.e., through the vertex B). If the triangle ABC is labeled counterclockwise, the y-coordinate of the vertex B is positive. If the triangle DABC is labeled clockwise, the y-coordinate of the vertex B is negative. In that case, reflect both the DABC and the point Q in the x-axis, reducing the problem to the former case. Select the unit length a = CB = 1.

To simplify the problem, we transform the general DABC to an isosceles right angle DA'B'C' by the following linear transformation (see the attached drawing):

u = x/xA
v = y - yA/xA x

Subsequently, we refrain from using the dashes for DA'B'C' and re-label it'simply as DABC.

Consider the area bisectors intersecting the line segments BMa at point P and AMb at point S as they slide on the triangle sides. Their equations are

u/uS + v/vP = 1

Since the lines are area bisectors, we have

uSvP = 1/2 or uS = 1/(2vP)

Substituting this to the bisector line equation

2vPu + v/vP = 1 or 2vP2u + v - vP = 0

This is a family of lines dependent on the parameter vP. The equation of their envelope (if it exists) is given by

F(u, v, vP) = 2vP2u + v - vP = 0
F(u, v, vP)/vP = 4vPu - 1 = 0

Eliminating the parameter vP we get the envelope equation

v = 1/(8u)

So the envelope is a segment of hyperbola with asymptotes along the sides a (v = 0) and b (u = 0) of the DABC. The equations of the medians ma, mb and their midpoints Da, Db are:

ma: 2v + u = 1, Da = (1/2, 1/4)
mb: v + 2u = 1, Db = (1/4, 1/2)

The hyperbola intersects (actually, touches - after all it is an envelope) the medians at their midpoints: Combining v = 1/(8u) and 2v + u = 1 (ma) and solving for u, v, we get u = 1/2, v = 1/4, or Da. Combining v = 1/(8u) and v + 2u = 1 (mb) and solving for u, v, we get u = 1/4, v = 1/2, or Db.

If we select a point (u0, v0) on the hyperbolic segment, then v0 = 1/(8u0). Substituting this to the general equation of the bisector

2vPu0 + v0/vP = 1
2vPu0 + 1/(8u0vP) = 1
16vP2u02 - 8u0vP + 1 = 0
(4vPu0 - 1)2 = 0
vP = 1/4u0

We have a double root - the only solution. Therefore, there is only one area bisector through the point (u0, v0) intersecting the DABC on the line segments BMa and AMb. If we select a point (u1, v1) above the hyperbolic segment, then v1 > 1/(8u1). Substituting this to the general equation of the bisector just like for the point (u0, v0) we get

(4vPu1 - 1)2 < 0

Since this is impossible, there is no area bisector through the point (u1, v1) intersecting the DABC on the line segments BMa and AMb. If we select a point (u2, v2) below the hyperbolic segment, then v2 < 1/(8u2). Denote w2 = 1 - 8u2v2 > 0. Substituting this to the general equation of the bisector just like for the point (u0, v0) we get

(4vPu2 - 1)2 = w2 > 0
vP = (1 w) / (4u2)

Taking into account the limits 1/2 vP 1 separately for each root

1/2 (1 w) / (4u2) 1
2u2 - 1 w 4u2 - 1

In order to square an inequality, we have to make sure that the smaller side is positive (or zero). Therefore, we can square the left inequality for the - sign root (after multiplcation by -1) and the right inequality for the + sign root. For the - sign root we get

w 2u2 - 1
w2 = 1 - 8u2v2 (2u2 - 1)2
u2 + 2v2 1

For the + sign root we get

w 4u2 - 1
w2 = 1 - 8u2v2 (4u2 - 1)2
2u2 + v2 1

Therefore, the point (u2, v2) must be above the median ma for the - sign root to be acceptable and above the median mb for the + sign root to be acceptable. Any point inside the DAGMb and any point inside the DBGMa is below the hyperbolic envelope, above one median, and below the other. Therefore, only one root is acceptable and there is only one area bisector through the point (u2, v2) intersecting the DABC on the line segments BMa and AMb. For any point below the hyperbolic segment DaDb and above both medians ma, mb, there are two such area bisectors.

By cyclic permutation of the vertices A, B, and C, we can find the hyperbolic segments DbDc and DcDa and all other results obtained for the vertex C at the origin. For any point inside the triangle pair DBGMc and DCGMb, there is only one area bisector through this point intersecting the DABC on the line segments CMb and BMc. For any point bounded by the hyperbolic segment DbDc and the medians mb, mc there are two such area bisectors. Likewise, for any point inside the triangle pair DCGMa and DAGMc, there is only one area bisector through this point intersecting the DABC on the line segments AMc and CMa. For any point bounded by the hyperbolic segment DcDa and the medians mc, ma there are two such area bisectors. Adding up the 3 families of area bisectors, for any point inside the DABC and outside of the hyperbolic "deltoid" DaDbDc, there is only one area bisector through this point. For any point inside the hyperbolic deltoid, there are 3 different area bisectors through this point. For any point on the hyperbolic deltoid boundary and not coincident with one of the deltoid vertices Da, Db, Dc, there are 2 different area bisectors through this point. At the vertices, 2 different hyperbolic segments touch the median (and therefore touch each other). The two area bisectors on the deltoid boundary, each from a different family, fuse into the single median. Therefore, there is only one area bisector through each vertex of the deltoid.

Attachments
https://www.cut-the-knot.org/htdocs/dcforum/User_files/3f185050569e7500.gif
https://www.cut-the-knot.org/htdocs/dcforum/User_files/3f18506256b0d683.gif

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Bractals
Member since Jun-9-03
Jul-23-03, 09:13 PM (EST)
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19. "RE: Cut Triangle in Half"
In response to message #18
 
   Hi Vladimir,

I agree with your conclusions on the deltoid. I think I needed that week's vacation.

With respect to a question I put to Golland: Do we lose any generality if we only look at an equilateral triangle?

Here are my thoughts. For any three non-collinear points in Euclidean 3-space it'seems that we can find a plane (and if one plane, then an infinite number) such that the projections of the points on the plane would be the vertices of a equilateral triangle. Therefore, given a non-equilateral triangle and point we project the problem to an equilateral triangle and a point. Solve that problem and reverse the projection to solve the original problem (since the ratios of areas are invariant under a projection). What do you think?

Bractals


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Vladimir
Member since Jun-22-03
Jul-24-03, 06:41 AM (EST)
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20. "RE: Cut Triangle in Half"
In response to message #19
 
   Hi Bractals

You are correct - we do not lose any generality by looking only at an equilateral triangle, or an isosceles right angle triangle which I used in my analysis (because it'simplifies the analysis, for one family of bisectors at least, even more than the equilateral triangle). You can imagine a projection in 3D space, but you have to be careful what kind of projection you use - it must be the parallel projection. Other projections of a plane into another plane (perspective projection) do not have to preserve area ratios.

But I do not think you have to go into the 3D space. You can just perform a linear transformation using any 2x2 matrix with non-zero determinant and apply it to every 2D vector in a plane - namely the 3 triangle vertices and the point through which the area bisector should pass:

x' = a11x + a12y
y' = a21x + a22y

(I cannot post a matrix equation)

D = a11a22 - a12a21 0

Such transformation preserves the area ratios. If two shapes are the same before such transformation, they are also the same after such transformation. You can divide the area of any shape before the transformation into extremaly little squares, the area ratio of two different shapes being the ratio of the number of their little squares. Each little square is transformed into the same little parallelogram, but of course the number of little parallelograms in each transformed shape equals to the number of little squares in each shape before the transformation.

This transformation can be performed using the straightedge and compass. For the triangle itself, it is trivial, because you choose the transformation to get the desired triangle shape and position. So you just draw the desired triangle. You only have to use the straightedge and compass construction to transform the point through which the area bisector must pass at the start and transform back the point at which the area bisector intersects the triangle (or another arbitrary point of the bisector) at the end. I tried that and it only complicated the construction. However, a suitable transformation makes the analysis much simpler.

Regards, Vladimir


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Bractals
Member since Jun-9-03
Jul-14-03, 09:24 PM (EST)
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9. "RE: Cut Triangle in Half"
In response to message #0
 
   Hi Alex,

At the start of this topic I clicked on the "Subscribe to this topic" link and I got the following error message. I got the same message when I went to Conferences > User menu > Topic Subscription.

Bractals

SCRIPT ERROR!!!
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There was an error in processing your request.
Following is the error message:
System Message: Permission denied

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Please notify the administrator of this site.
Thank you.


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alexb
Charter Member
1029 posts
Jul-14-03, 09:26 PM (EST)
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10. "RE: Cut Triangle in Half"
In response to message #9
 
   Thank you for bringing this to my attention. I hope that the problem has been resolved. Please try subscribing to a topic now.


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Bractals
Member since Jun-9-03
Jul-14-03, 11:43 PM (EST)
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11. "RE: Cut Triangle in Half"
In response to message #10
 
   Thanks Alex it works now.

Bractals


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