[superprismatic]

Consider the triangle with vertices at

(0,0), (a,0), and (b,c) with 0 < b < a.

Consider a point (x,y) inside this

triangle. Shoot perpendiculars from

(x,y) to each side of the triangle.

This divides the triangle into 3

quadrilaterals. Find (x,y) such that

these quadrilaterals all have the same

area.

[Guest]

I think I got it!

x=a/2

y=c/3

Because each quadrilateral has two right angles (created by the perpendiculars) you can cut them in half into 6 equal right triangles. Meaning thier areas all equal a*c/12. I used that to solve for y and then substituted the answer for y and solved for x. Let me know if I am wrong in any way. If I am not mistaken (x,y) would also create equal aread triangles if lines where drawn to the corners of the original triangle and it is the balance point for the triangle.

a*c/12= a*y/4

c/12=y/4

c/3=y

a*c/12=x*y/2

a*c/12=x*c/6

a/12=x/6

a/2=x

[Guest]

I think I got it!

x=a/2

y=c/3

Because each quadrilateral has two right angles (created by the perpendiculars) you can cut them in half into 6 equal right triangles. Meaning thier areas all equal a*c/12. I used that to solve for y and then substituted the answer for y and solved for x. Let me know if I am wrong in any way. If I am not mistaken (x,y) would also create equal aread triangles if lines where drawn to the corners of the original triangle and it is the balance point for the triangle.

a*c/12= a*y/4

c/12=y/4

c/3=y

a*c/12=x*y/2

a*c/12=x*c/6

a/12=x/6

a/2=x

I may be wrong but my geometry is very rusty...

Assuming that the quadrilaterals can be cut in equal halves by line joing the circumcenter (where perpendiculars meet) to the side indirectly assumes that the circumcenter and the incenter (where angle bisectors meet) are the same points. This would be a very specific case: an equilateral triangle only! Which is why you got x = a/2 and y = c/3.

Now it remains to be seen whether the 3 quadrilaterals can have equal areas if and only if the triangle is an equilateral triangle!

Edited August 28, 2009 by DeeGee

[Guest]

For starters, Redsnuff’s answer does not work (in my view,) since I am not convinced that the triangles he constructs have to be equal in area. As a simple counter-example to his solution, take a = 6, b = 4, c = 3. The point (3,1) does not work.

Secondly, I think that for any triangle for which all angles are acute, and many others, there must be a solution to Superprismatic’s problem.

Take a random point inside the triangle. One of the quadrilaterals will be smallest. Move the point away from its vertex until it has an equal area to the largest of the other quadrilaterals. There will now still be a smallest quadrilateral. Repeat the procedure ad infinitum. A point will be found to equilibrize the situation.

[Guest]

I can't find a way too solve this without it getting rather messy. I'm sure there is a simple way to solve it.

[superprismatic]

Consider the triangle with vertices at

(0,0), (a,0), and (b,c) with 0 < b < a.

Consider a point (x,y) inside this

triangle. Shoot perpendiculars from

(x,y) to each side of the triangle.

This divides the triangle into 3

quadrilaterals. Find (x,y) such that

these quadrilaterals all have the same

area.

Sorry, I misstated the problem. In order to guarantee

being able to shoot perpendiculars, the triangle

needs to be acute. So, the problem should read:

Consider an acute triangle with

vertices at (0,0), (a,0), and (b,c)

with 0 < b < a. Consider a point

(x,y) inside this triangle. Shoot

perpendiculars from (x,y) to each

side of the triangle. This divides

the triangle into 3 quadrilaterals.

Find (x,y) such that these

quadrilaterals all have the same area.

Edited August 28, 2009 by superprismatic

[bonanova]

Take the sides in order given.

Three distances from (x, y) to the sides are d₁, d₂, d₃.

Distance from point (x₀, y₀) to line containing points (x₁, y₁) and (x₂, y₂) is

This gives d₁, d₂ and d₃.

Three points of intersection are p₁, p₂, p₃.

Point of intersection (x₀, y₀) of perpendicular

from point (x, y) to line containing points (x₁, y₁) and (x₂, y₂) is

x₀ = [(y-y₁) + mx₁ +x/m]/[m+1/m]

y₀ = y₁ + m(x₀-x)

where m = (y₂-y₁)/(x₂-x₁)

Apply from (x, y) to the three sides of the triangle to get p₁, p₂ and p₃.

Lines from (x, y) to the vertices define six right triangles

whose areas are t₁, t₂, t₃, t₄, t₅ and t₆.

t₁ = .5 x d₃ x dist [p₃, (0, 0)] - where dist is distance between the two points.

t₂ = .5 x d₁ x dist [p₁, (0, 0)]

t₃ = .5 x d₁ x dist [p₁, (a, 0)]

t₄ = .5 x d₂ x dist [p₂, (a, 0)]

t₅ = .5 x d₂ x dist [p₂, (b, c)]

t₆ = .5 x d₃ x dist [p₃, (b, c)]

The areas of the three quadrilaterals are

A₁ = t₁+t₂

A₂ = t₃+t₄

A₃ = t₅+t₆.

Algebraic:

Plug all the formulas together from the first spoiler, and set A₁ = A₂ = ac/6 [= 1/3 of the triangle area.]

Solve for x and y.

I'll leave that for someone else: Tag team, Bushindo?

Numeric:

For an initial x and y

Calculate A₁ and A₂.

Iterate.

I did the numeric approach for a specific case of a=12, b=9, c=8.

Triangle area is .5 x 12 x 8 = 48

Solution:

x = 6.4815, y = 2.4815 gives A₁ = A₂ = 16.

Edit: As an aside, this disproves redsnuff's

His solution would be x=6 [slightly too small] and y=2.6667 [slightly large].

Edited August 31, 2009 by bonanova
Add reference in 2nd spoiler to redsnuff's post

[Guest]

I am working a different approach to that of Bonanova, one which does not require iteration. I hope that we will find eventual agreement, though.

[bonanova]

I gave the equations to calculate of A1(x,y) and A2(x,y).

Equating them to A/3 lets you solve for x,y.

Algebraically, you get a closed form expression.

Numerically, you get specific values.

In this problem, algebraic is a bit clumsy; but the equations are there.

Are there other approaches?

Perhaps the equal area condition is stationary point that presents a simpler equation to solve.

Don't know, but I'm interested to see what you come up with.

I've missed the Aha! solution often enough.

[bushindo]

I gave the equations to calculate of A1(x,y) and A2(x,y).

Equating them to A/3 lets you solve for x,y.

Algebraically, you get a closed form expression.

Numerically, you get specific values.

In this problem, algebraic is a bit clumsy; but the equations are there.

Are there other approaches?

Perhaps the equal area condition is stationary point that presents a simpler equation to solve.

Don't know, but I'm interested to see what you come up with.

I've missed the Aha! solution often enough.

Alright, tag team. Here's an idea that may reduce the computational load.

Here's an image of a triangle.

Let the triangle on the right be the original triangle, and let the blue point be (x,y) such that the 3 quadrilaterals created are equal. Let a be the angle in degrees that the triangle makes at the origin. Note that a is not the length of the side, non-bold a is the length of a side (bad choice of notation, I know, but I didn't think of that when I drew the graphs). Note that solving for the area of the two lower pink triangles on the right hand side is easy. The main idea here is to rotate the entire triangle so that another side is flat along the x axis. Then we can solve the another set of pink triangles easily.

Recall that the rotational matrix that rotates vectors a degrees counterclockwise is

cos( [b]a[/b] )    -sin( [b]a[/b] )

sin( [b]a[/b] )     cos( [b]a[/b] )

We rotate the entire triangle by (180 - a) degrees counterclockwise so that another side is now flat along the x axis. The new coordinate for the blue center point is

(x*cos( 180 - a ) - y sin( 180 - a ), x*sin( 180 - a ) + y*cos( 180 - a ) )

From here we can compute two new sub-triangles easily (denoted by pink triangles on left hand side). To do the last rotation, we'll need to do a translation first, and then a rotation. But the idea remains the same. Of course, once we have expressions for the areas of the 6 sub-triangles, we can solve for x and y.

So, this seems to be less cumbersome by a smidge. Tag team, someone else?

Edited August 31, 2009 by bushindo

[Guest]

I attach my approach, which very probably is equivalent to that of Bonanova's and Bushindo's (indubitably correct.) I may have committed some trivial inaccuracies, but the method speaks for itself. Would Superprismatic please come out of hiding and spread his pearls of wisdom on the problem. I certainly cannot find an "aha" approach, as BN calls it.

P.S. I have just found one error, though it does not nullify the approach.

Brainden Question.doc

Edited August 31, 2009 by jerbil

[Guest]

I have just amended the silly errors occurring in my last post.

Brainden Question.doc

[bushindo]

That is a great approach. Definitely more elegant than my proposed approach. Well done.

[bonanova]

Solve three quadratic equations.

Your two equations have the form

.

1. a₁x² + b₁xy + c₁y² + d₁ = 0.
2. a₂x² + b₂xy + c₂y² + d₂ = 0.
   .
   Where a_i, b_i, c_i, d_i are expressions involving only a, b, c.
   Recast these in the standard quadratic form for x:
   .
3. x² + B₁x + C₁ = 0.
4. x² + B₂x + C₂ = 0.

.

where B_i = b_iy/a_i and C_i = [c_iy² + d_i]/a_i

Solve 3 and 4 to get expressions for x in terms of a, b, c, and y.

Equate them to get a quadratic for y in terms of a, b, c only.

Plug y back into either expression for x and you have x, y in terms of a, b, c.

Probably will give four values of x, y - only one of which will be inside the triangle.

Not all that messy:

[1] the process is laid out - just solve 3 quadratics

[2] the final expressions probably simplify.

Agree with Bushindo - nice approach.

Now if I can just figure out why the new site permanently and unalterably makes half my posts bold...

Rookie?

site admin comment: Well spotted, I have removed the default bold font for span in spoiler. Thanks.

Edited August 31, 2009 by rookie1ja
site admin added comment/answer

[Guest]

Thanks BN for your description as to how to complete the problem most easily. I will definitely do this myself when I have the time, whether you beat me to it or not.

Looking at Superprismatic's original post, I note the two conditions

(a) The point to be found must be internal to the triangle,

(b) 0 < b < a.

However, the problem clearly has a solution for some values of c when the triangle is right angled, i.e. b = a. By extension, there must be solutions for some values for b when b > a, so that one angle is obtuse. Inevitably I ask myself how obtuse can I get! Seriously though, to answer this I think an explicit formula needs to be found.

[bonanova]

Thanks BN for your description as to how to complete the problem most easily. I will definitely do this myself when I have the time, whether you beat me to it or not.

Looking at Superprismatic's original post, I note the two conditions

(a) The point to be found must be internal to the triangle,

(b) 0 < b < a.

However, the problem clearly has a solution for some values of c when the triangle is right angled, i.e. b = a. By extension, there must be solutions for some values for b when b > a, so that one angle is obtuse. Inevitably I ask myself how obtuse can I get! Seriously though, to answer this I think an explicit formula needs to be found.

x = [-b ⁺/_- sqrt(b² - 4ac)] / 2a etc... is an explicit formula, no?

It's double valued, which gives you two y values and then two x values for each of them.

But it seems impossible for more than one of the four combinations to lie within the triangle [for any a, b, c.]

p.s. You will beat me to it.

p.p.s. to OP: Doesn't 0 < b < a guarantee an acute triangle?

p.p.p.s. To site admin. It was the span tag. Thanks for the fix.

site admin: You are welcome

Edited September 1, 2009 by bonanova
site admin added comment

[superprismatic]

x = [-b ⁺/_- sqrt(b² - 4ac)] / 2a etc... is an explicit formula, no?

It's double valued, which gives you 2 y values and then two x values for each of them.

But it seems impossible for more than one of the four combinations to lie within the triangle [for any a, b, c.]

p.s. You will beat me to it.

p.p.s. to OP: Doesn't 0 < b < a guarantee an acute triangle?

p.p.p.s. To site admin. It was the span tag. Thanks for the fix.

site admin: You are welcome

To guarantee an acute triangle, I think that one must have 0 < b < a as well as c > sqrt(b*[a-b]). If (b,c) is too close to the x-axis the triangle will be obtuse.

[bonanova]

To guarantee an acute triangle, I think that one must have 0 < b < a as well as c > sqrt(b*[a-b]). If (b,c) is too close to the x-axis the triangle will be obtuse.

Yeah ... didn't think of the short case.

The perpendiculars fall outside two of the bases.

[Guest]

Attached are my most recent thoughts on the matter. It includes an analysis of Bonanova's example as a useful adjunct.

Brainden Question.doc