Squares packed in triangluar boxes

[bonanova]

We place a square inside a triangle so that a side of the square remains flush with one of the sides of the triangle, but is free to slide along it. As the square is enlarged, eventually two corners of the square will touch the other two sides of the triangle, making further enlargement of the square impossible. Every triangle is thus associated with three "largest" squares. In general the areas of these squares are distinct.

Calabi

I have a particular triangle whose three largest squares are congruent.

What can we say with certainty about this triangle?

Edited February 28, 2015 by bonanova
Re-worded for clarity and accuracy

[gavinksong]

It is equilateral.

[bonanova]

Actually, not.

[gavinksong]

Whaaaaaaat? 

 

First of all, I'd like to note that the construction of the square as you've described it is not possible for all sides of any triangle (see obtuse triangles). But if we assume we are working with a triangle for which it is, it must be acute.

 

After we've constructed one such square, it will divide the triangle into four parts: the square itself, the "wings" on either side of the square, and the "hat" on top of the square. The area of the entire triangle can then be calculated as (for H: triangle height, B: triangle base, S: square side length):

 

A = S² + S (B - S) + SH (S / B)

 

The first term is the area of the square. The second term is the area of the wings, since the wings are both right triangles with a common height S and combined base (B-S). The third term is the area of the hat, since it is similar to the larger triangle with scale factor (S/B). We also know that

 

H = S + H (S/B)

 

In words, the height of the triangle comprises the heights of the square plus hat. Combining the above two equations:

 

B + H = A / S + S

 

For the triangle in question, the area A and the square side S are constant, so the right side of the equation is constant.

 

B + H = K

 

Assuming A and S are constant, H is completely dependent on the value of B, so we have an invertible function ƒ such that

 

ƒ(B) = K

 

for which there is only one unique solution. Since this applies to all three sides of the triangle, I concluded that the triangle must be equilateral.

 

Where did I go wrong?  

 

Edit -- typo

Edited February 28, 2015 by gavinksong

[bonanova]

Perhaps the OP was not worded as well as it could have been. I don't know it it matters, but re-read it now.

[gavinksong]

Looking at it again, ƒ(B) = B + S / (1 - S/B) is not an invertible function... 

I graphed it, and it seems that it has up to two solutions - which would suggest that the triangle is at least isosceles. That is, it can't be scalene since there can't be three distinct base lengths that yield the same triangle area for the same square.

That answers the question somewhat (I think), although it isn't a very satisfactory way of stumbling upon it.

 

I must ponder some more.  

Edited February 28, 2015 by gavinksong

[bonanova]

Yes, of my triangle, it can be said with certainty that it is not scalene.
What more specifics can be given?

[bonanova]

Looking at it again, ƒ(B) = B + S / (1 - S/B) is not an invertible function... 

I graphed it, and it seems that it has up to two solutions - which would suggest that the triangle is at least isosceles. That is, it can't be scalene since there can't be three distinct base lengths that yield the same triangle area for the same square.

That answers the question somewhat (I think), although it isn't a very satisfactory way of stumbling upon it.

 

I must ponder some more.  

 

Did you get the second solution yet?

[gavinksong]

Did you get the second solution yet?

 

No. It is too hard. 

 

I reread the OP, and I noticed that the rewording nullifies my assumption that the triangle must be acute. In other words, if you hadn't confirmed that the triangle wasn't scalene, I would no longer be sure that it was. I would have to introduce new equations to confirm it, but it would make my reasoning even more complicated. I am failing to think of an elegant way to think about this problem.

 

Perhaps you could give us a hint.

[BMAD]

the sums of the square areas is greater than the triangle areas??

[plasmid]

I got the same answer as gavinksong, with a little different approach.

If you draw a square of side length L, you have the line L parallel to the base (L

_horiz in the figure) with length L_horiz = base * (height - L_vert) / height

L = base * (height - L) / height

L = base * (1 - L/height)

L = base - base*L / height

L + base*L / height = base

L * (1 + [base/height]) = base

L * (height+base) / height = base

L * (height + base) = base * height

L = (base * height) / (base + height)

L = 2*TriangleArea / (base + height)

 

Since the squares are all congruent for each side, and since the triangle's area is fixed regardless of which side you're treating as the base, we can conclude that 1 / (base + height) is identical for each of the sides, so (base + height) is identical for each of the sides.

 

Base + height would of course be the same for each side if you're dealing with an equilateral triangle. If your constraints are that the area (base*height)/2 must stay constant and (base + height) must be the same for each side, then there might be another solution where X is the base and Y is the height for one side and Y is the base and X is the height for another side. Those are the only two possible arrangements of values if their sum and product must remain constant, so the triangle would have to be isosceles.

 

Suppose you have an isosceles triangle with side lengths {A, A, B}. The height from line B would be A * sin(angle AB) and the height from line A would be B * sin(angle AB).

So (base + height) for line A would be (A + B*sin(angle AB)) = (A+B) - B*(1-sin(angle AB))

and (base + height) for line B would be (B + A*sin(angle AB)) = (A+B) - A*(1-sin(angle AB))

Those two terms can only be equal if the sin() term is one (which would not be compatible with an isosceles triangle) or if A=B. So the only case of an isosceles triangle that works is an equilateral triangle.

Edit: I guess I ought to attach that figure I was talking about ^.^

Edited March 6, 2015 by plasmid

[bonanova]

the sums of the square areas is greater than the triangle areas??

 

Well, yes. But that can be true of scalene triangles as well.

 

What can we say with certainty about my hypothetical triangle,

[knowing only that its three largest (internal) squares have equal area]

that distinguishes it from any triangle that does not meet that criterion?

 

One possible answer, of the type sought, is that the triangle is equilateral.

But it turns out that my triangle is not equilateral. 

 

So the answer takes the form, it is EITHER equilateral OR it is ___________________.

[gavinksong]

 

I got the same answer as gavinksong, with a little different approach.

If you draw a square of side length L, you have the line L parallel to the base (L

_horiz in the figure) with length L_horiz = base * (height - L_vert) / height

triangle square.jpg

L = base * (height - L) / height

L = base * (1 - L/height)

L = base - base*L / height

L + base*L / height = base

L * (1 + [base/height]) = base

L * (height+base) / height = base

L * (height + base) = base * height

L = (base * height) / (base + height)

L = 2*TriangleArea / (base + height)

 

Since the squares are all congruent for each side, and since the triangle's area is fixed regardless of which side you're treating as the base, we can conclude that 1 / (base + height) is identical for each of the sides, so (base + height) is identical for each of the sides.

 

Base + height would of course be the same for each side if you're dealing with an equilateral triangle. If your constraints are that the area (base*height)/2 must stay constant and (base + height) must be the same for each side, then there might be another solution where X is the base and Y is the height for one side and Y is the base and X is the height for another side. Those are the only two possible arrangements of values if their sum and product must remain constant, so the triangle would have to be isosceles.

 

Suppose you have an isosceles triangle with side lengths {A, A, B}. The height from line B would be A * sin(angle AB) and the height from line A would be B * sin(angle AB).

So (base + height) for line A would be (A + B*sin(angle AB)) = (A+B) - B*(1-sin(angle AB))

and (base + height) for line B would be (B + A*sin(angle AB)) = (A+B) - A*(1-sin(angle AB))

Those two terms can only be equal if the sin() term is one (which would not be compatible with an isosceles triangle) or if A=B. So the only case of an isosceles triangle that works is an equilateral triangle.

Edit: I guess I ought to attach that figure I was talking about ^.^

 

 

But I thought the triangle is not equilateral. 

And yet your reasoning for it seems solid enough.

 

We actually used the same assumptions, and one of them is that the triangle is acute. If it is obtuse, then the diagram changes entirely.

 

Edit -- that must be the OR that bonanova is talking about.

Edited March 6, 2015 by gavinksong

[gavinksong]

The triangle is either equilateral or obtuse with base = B = S / (1 - 1/√2) for some S?

 

Edit -- Sorry, I kept messing up.

Edited March 6, 2015 by gavinksong

[bonanova]

 

Did you get the second solution yet?

 

No. It is too hard. 

 

I reread the OP, and I noticed that the rewording nullifies my assumption that the triangle must be acute. In other words, if you hadn't confirmed that the triangle wasn't scalene, I would no longer be sure that it was. I would have to introduce new equations to confirm it, but it would make my reasoning even more complicated. I am failing to think of an elegant way to think about this problem.

 

Perhaps you could give us a hint.

 

Mathematically, or more simply with a careful sketch, we can

observe that for acute triangles, the square that sits on the

shortest side has the largest area. That tells us all we need

to know about acute triangle solutions. We can also observe

(or calculate) that for significantly obtuse triangles, the square

that sits on the longest side (opposite the obtuse angle) has

the largest area.

 

These two observations [1] show there is a second solution and

[2] specify the nature of it.

 

All we need now is an angle or a ratio of sides.

[bonanova]

But I thought the triangle is not equilateral. 

 

It cannot be said with certainty that my triangle is equilateral.

[bonanova]

The triangle is either equilateral or obtuse with base = B = S / (1 - 1/√2) for some S?

 

Edit -- Sorry, I kept messing up.

 

Yes to the first part, not sure about your formula.

[plasmid]

Consider the isosceles obtuse triangle in the figure.

Call the acute angles theta.

Call the square's side length L.

In the first figure, we have

H = L/2 tan theta

In the second, we have

H+L = L (sqrt 2) cos(45 - theta)

H = L [(sqrt 2) cos(45 - theta) – 1]

Substituting, we get

L [(sqrt 2) cos(45 - theta) – 1] = L/2 tan theta

(sqrt 2) cos(45 - theta) – 1 = 1/2 tan theta

Resorting to a computer at that point, theta comes out to be near 40 degrees.

But I wonder: could a scalene triangle work if it's a right triangle?

The observation that the square's size will be different for the smallest side versus the other sides for acute and obtuse triangles might not be applicable for a right triangle: the squares for the two sides that are involved in the right angle clearly must be the same.

[bonanova]

Nice solve. There are cubic equations for sin, cos, or tan of the acute angle. But if God had wanted to impose on us the agony of solving cubic equations, he would not have invented computers.

 

Your triangle is the unique alternative to the equilateral triangle. It is true that for acute triangles the squares relate to the length of their respective sides. The case of a right scalene triangle is a degenerate case in which the two (coalesced) squares on the two (unequal) legs have, necessarily, the same area. But they are both larger than the square that rests on the hypotenuse. Not until the triangle becomes (slightly) obtuse (apex slightly larger than 110^o), and of course isosceles, does the square opposite the apex increase to equal the other two.

 

What is surprising to me is that this triangle was apparently unknown (unpublished, at least) until less than 20 years ago. A fact that makes this, I think, an interesting puzzle question.

 

Calabi's triangle and his analysis of it.

[bonanova]

I had forgotten that plasmid published these same figures, that I just made.

FWIW I'll post them.

 

I computed the areas as a function of h, and found they were equal when h = .813.

This gave theta as 39.132^o and the apex angle as 101.736^o.

 

 

[bonanova]

Here is a plot of the fractional area of the two squares.

Fraction of triangle occupied by the two squares as a function of theta.

 

Data are shown only for oblique triangles: Theta <= 45 degrees.

The formula for slanted squares in acute triangles is different.

 

Note the slanted square occupies exactly 1/2 of a right isosceles triangle.