Suppose we start with an acute angled triangle where the sides are of length x, y, and z, where 0<x≤y≤z. Let the corresponding angle sizes be denoted by X, Y, and Z (assume radians).
It follows that the angle sizes X, Y, and Z are such that 0<X≤Y≤Z<π/2.
Then X < π/2 since the triangle is acute-angled and so 2X < π = X+Y+Z. It follows then that X<Y+Z.
Similarly, it is easy to show that Y<X+Z and Z<X+Y. So X, Y and Z (the angle sizes of the original acute-angled triangle) can also be the lengths of the sides of a new triangle ABC (say), as above, with corresponding internal angles of sizes A, B and C. It is easily checked that 0<A≤π/3, 0<B<π/2 and π/3≤C.
It should be fairly obvious that this only works if the original triangle is acute-angled, i.e. it does not work if the original triangle is right or obtuse.
Notice that for an acute triangle with degree angle sizes 45o , 60o and 75o , the angle sizes also form a Pythagorean triple, i.e. 45² + 60² = 75². Basically, the numbers 45o , 60o and 75o are just a common multiple of (3, 4, 5). Any others?
The only other degree measure acute-angled triangles where this property holds are those where the angles are 30o , 72o and 78o (5, 12, 13) and 18o , 80o and 82o (9, 40,41).
The angle sizes of the original acute angled triangle can easily be shown to have a number of interesting and unexpected properties, such as …
3YZ/2π ≤ π/2 – X < 2YZ/π, [this follows from 0<A≤π/3]
XZ/π < π/2 – Y < 2XZ/π, [follows from 0<B<π/2] and
0 < π/2 – Z ≤ 3XY/2π [follows from π/3≤C and Z<π/2].
Proof that 3YZ/2π ≤ π/2 –X< 2YZ/π.
Assume that (x,y,z) is acute-angled as above and that its corresponding angles are X,Y, and Z.
Let the triangle (X,Y,Z) formed by (side lengths) X, Y and Z have corresponding angle sizes A, B and C as above.
Since we have assumed that 0<x≤y≤z, it follows that 0<X≤Y≤Z and so, since Z<π/2, A, B and C exist and 0<A≤B≤C.
We also have, π = X+Y+Z and π = A+B+C. It must also be true that 0<A≤π/3 since A is the smallest angle in the triangle A, B, C.
From 0<A≤π/3 we must have that 1/2≤cos(A)<1.
Hence 1/2≤(Y2+Z2–X2)/(2YZ)<1. So, YZ ≤ Y2+Z2–X2<2YZ.
Now π = X+Y+Z and so π – X = Y+Z.
Hence (squaring both sides of π – X = Y+Z) we get π2 – 2πX+X2 = Y2+2YZ+Z2 or
π2– 2πX – 2YZ = Y2 + Z2 – X2. So YZ ≤ Y2+Z2–X2<2YZ becomes
YZ ≤ π2– 2πX – 2YZ < 2YZ or 3YZ ≤π2– 2πX < 4YZ.
Finally, dividing by 2π gives 3YZ/2π ≤ π/2 –X< 2YZ/π. ♦
Note: If the triangle (X,Y,Z) is also acute angled, then the third set of inequalities above,
i.e. 0 < π/2 –Z ≤ 3XY/2π becomes XY/π < π/2 –Z ≤ 3XY/2π.
The largest angle in triangle (X,Y,Z) is C and so there are 3 possibilities for that angle, i.e. C = π/2, C < π/2 and C > π/2.
Case 1 C = π/2
In this case, Z² = X² + Y² while π = X+Y+Z. If we eliminate Z from these 2 equations we get
Y = π(π/2-X)/(π-X) and from π = X+Y+Z we get Z= (π2/2-πX+X2)/(π-X).
Hence if (X,Y,Z) is right angled Y and Z are effectively determined/parametrised by X.
For example if X = π/6 then Y = 2π/5 and Z= 13π/30.
It is easily checked that π = π/6 +2π/5 +13π/30 and that …
(13π/30)2 = (π/6)2 + (2π/5)2. (Note: (X, Y, Z) is just a (5,12,13) triangle).
Case 2 C < π/2
It follows that Z² < X² + Y² while π = X+Y+Z.
Hence, Y > π(π/2-X)/(π-X) and Z < (π2/2-πX+X2)/(π-X).
Case 3 C > π/2
Likewise it follows that Z² > X² + Y² while π = X+Y+Z.
Hence, Y < π(π/2-X)/(π-X) and Z > (π2/2-πX+X2)/(π-X).
If (X,Y,Z) is right angled and isosceles then X = π/(2+√2), Y= π/(2+√2), and Z= π/(1+√2).
Now X lies in the range 0<X≤π/3 and as X-> 0, Y and Z-> π/2.
However, as X->π/3, Y-> π/4 and Z->5π/12, which is just a (3,4,5) triangle scaled by π/12.
This can’t be the case though because we have assumed that x≤y≤z and so X≤Y≤Z.
Y>X while X is near 0 but Y decreases as X increases towards π/3 until Y=X= π(2-√2)/2. At that point (X,Y,Z) is right angled at Z, and Z= π(√2-1). If (X,Y,Z) is right-angled, and π(2-√2)/2<X≤π/3, it follows that Y<X, but this is not possible since we have assumed that X≤Y.
If π(2-√2)/2<X≤π/3 it is straightforward to show that (X,Y,Z) is also acute angled.
First π = X + Y + Z and X≤Y≤Z mean that Z ≤ π – 2X and since π(2-√2)/2<X it follows that Z<π(√2-1)<π/2. Hence the result.
Put another way, if you start with an acute-angled triangle where the smallest angle lies roughly from 52.7 degrees to 60 degrees and you generate another triangle using these angle sizes as the lengths of the sides, the generated triangle must also be acute-angled.
If however, 0<X≤π(2-√2)/2 then it is easy to show that the generated triangle can be acute, right or obtuse. Note: (2-√2)/2 ≈ 0.292893219.
As another example, if we take X = (1/4)π, and calculate Y using Y = (π/2-X)/(1-X/π), we get Y = (1/3)π, and then use π = X+Y+Z, we get Z = (5/12)π and of course …
((5/12)π)² = ((1/4)π)² + ((1/3)π)². (Note: (X, Y, Z) is just a (3,4,5) triangle).
If we extend/generalise this method to X = (p/q)π, where p and q are positive integers and
p/q < (2-√2)/2, and calculate Y and Z as above, then we arrive at the (familiar?) identity …
((q-p)² + p²)² = ((q-p)² – p²)² + (2p(q-p))² but having used little discernible number theory along the way.
From 0<A≤π/3 we must have that 1/2≤cos(A)<1.
Hence 1/2≤(Y2+Z2–X2)/(2YZ)<1. So, YZ ≤ Y2+Z2–X2<2YZ. It follows that X2 ≤ Y2 – YZ + Z2 . Multiplying this last inequality by Y + Z gives (Y + Z) X2 ≤ Y³ + Z³ and so using π – X = Y+Z we get X³ + Y³ + Z³ ≥ πX² .
Using C ≥ π/3 (since C is the largest angle in triangle A, B, C) it follows as above that X³ + Y³ + Z³ ≤ πZ² .
So, for any acute angled triangle with angles X, Y and Z where X≤Y≤Z it is true that
πX² ≤ X³ + Y³ + Z³ ≤ πZ².
If the original acute-angled triangle with sides (x, y, z), assuming that x≤y≤z, and its ‘inside-out’ (X,Y,Z) are similar then …
x/X= y/Y= z/Z. But X+Y+Z = π. So, it follows easily that …
X= xπ/(x + y + z)
Y= yπ/(x + y + z) and
Z= zπ/(x + y + z)
Obviously an equilateral triangle is similar to its ‘inside-out’. I strongly suspect that this is the only acute angled triangle with this property.
Suppose we start with an acute-angled triangle where the sides are of length x, y, and z. Assume without loss of generality that 0 <x≤y≤z.
Let the corresponding angle sizes be denoted by X,Y, and Z (assume radians).
Suppose also that (x, y, z) is similar to (X,Y,Z). Then (x, y, z) is equiangular to (X,Y,Z). By equiangular I mean that corresponding angles are equal. Also, corresponding sides are in the same ratio, so
x/X = y/Y = z/Z. Applying the sine rule to (X,Y,Z) it follows that X/sin(X) = Y/sin(Y) = Z/sin(Z).
The function f: (0,π/2) -> R defined by f(x) = x/sin(x), where x ∈ (0,π/2) is 1 – 1
and so if f(a) = f(b), a, b ∈ (0,π/2), then a = b.
It follows then that X/sin(X) = Y/sin(Y) = Z/sin(Z) implies that X = Y = Z, since X, Y and Z∈ (0,π/2), and so
naturally x = y = z, i.e. (x, y, z) is equilateral.
So if an acute angled triangle is similar to its ‘inside-out’ it can only be equilateral.
Note: If (x,y,z) is an acute-angled triangle as above, with corresponding angles X, Y and Z, then (X,Y,Z) forms a triangle where X + Y + Z = π, or what might be called a π-triangle.
If (a,b,c) represents any triangle, then πa/(a+b+c), πb/(a+b+c), and πc/(a+b+c) can represent the angles of an acute angled triangle.
Clearly πa/(a+b+c) + πb/(a+b+c) + πc/(a+b+c) = π and if (say) πa/(a+b+c)≥π/2 then a≥ b + c which contradicts the assumption that (a,b,c) forms a triangle.
Finally, a connection with Fermat’s Last Theorem.
If we let n be a whole number, n>=2, and assume that the (usual) equation,
z n = x n + y n , has a non-trivial solution in positive whole numbers x, y and z, then we can also assume without loss of generality that 0 < x < y < z. It is easy to show that, for such a non trivial solution, z < x + y, and so it follows that x, y, and z can form the sides of a triangle. In the case n=2, that triangle is right-angled, but in every other case, i.e. n>=3, it is straightforward to show that the triangle (x, y, z) is acute-angled.
Proof – that (x, y, z) is acute-angled if n>=3.
If n>=3, it is straightforward to show that z 2 < x 2 + y 2 if a non-trivial solution x, y, z to Fermat’s equation exists.
Let the corresponding angle sizes in triangle (x, y, z) be denoted by X,Y, and Z.
Then cos(Z) = (x2+ y2– z2)/(2xy), and so cos(Z)>0, i.e. Z(largest angle) is acute, so the triangle is acute angled.
For a non-trivial solution to the Fermat equation,
z = cos(Y)x + cos(X)y and also z = (x/z)n-1x + (y/z)n-1y,
where 0<cos(Y)<1 and 0<(x/z)n-1<1 etc.
Is it true then that cos(Y)= (x/z)n-1and cos(X)=(y/z)n-1??
In the case n=2 this is certainly true.
If either of these equations, cos(Y)= (x/z)n-1or cos(X)=(y/z)n-1, is true, then it follows that
z n-2 = (y n – x n )/(y 2 – x 2 ).
If we assume that z n-2 >= (y n – x n )/(y 2 – x 2 ) it is easy to show that a contradiction arises. Hence it is certainly true that z n-2 < (y n – x n )/(y 2 – x 2 ).
If z n = x n + y n then z i < x i + y i is easily established for all i in the range 1, 2, … n-1.
Hence, z n-2 < x n-2 +y n-2 .
So, z n-2 >= (y n – x n )/(y 2 – x 2 ) implies that (x n-2 + y n-2 )(y 2 – x 2 ) > y n – x n .
Multiplying out this last inequality and simplifying gives x n-4 >y n-4 which is not possible if n>=4.
If n=3, we start with z>= (y 3 – x 3 )/(y 2 – x 2 ). Squaring both sides of this inequality, and using the fact that x 2 + y 2 > z 2 , gives after some rearrangement …
(y 4 – x 4 )(y 2 – x 2 ) > (y 3 – x 3 ) 2 . If this is multiplied out and simplified it leads to …
0 > x 2 – 2xy + y 2 or 0 > (x – y) 2 which again is not possible.
Hence the result is proved.
If n=3 this inequality becomes z<(y3-x3)/(y2-x2) or z<(y2+xy+x2)/(x+y). This last inequality leads to z-y<x2/(x+y) and z-x<y2/(x+y). It is straightforward to show that these inequalities are also true for all n>3.
If a non-trivial solution, (x,y,z), n>2, to the Fermat equation exists, then, as has been shown above, (x,y,z) is an acute angled triangle. An interesting construction can be drawn inside this triangle, which I would show here, if this system would only allow me to display it. Basically, a smaller (and similar) triangle (x(x+y-z)/z, y(x+y-z)/z,(x+y-z)) can be drawn inside (x,y,z).
Clearly since zn = xn + yn then (x+y-z)n = (x(x+y-z)/z)n + (y(x+y-z)/z)n. x+y-z is obviously an integer and it is easily shown that 0<x+y-z<z, 0<x(x+y-z)/z<x and 0<y(x+y-z)/z<y. If it can be shown that x(x+y)/z and y(x+y)/z are also integer then the method of Infinite Descent could be invoked.
In the paragraphs immediately above, the terms x+y-z and xy/z appear. I wondered if there was some connection between these terms independent of Fermat’s equation. It turns out that there is, i.e. x+y-z<xy/z if z≠0 and (z-x)(z-y)>0.
Assume that z≠0 and (z-x)(z-y)>0. Consider xy/z-(x+y-z).
xy/z-(x+y-z) = (xy-z(x+y)+z2)/z = (x-z)(y-z)/z = (z-x)(z-y)/z>0. Hence the result is true.
So if we had (say) 0<x<y<z then (using explicit * for multiplication and / for division) we have x+y-z<x*y/z. Clearly the LHS and RHS of < have the same structure but just different operators inserted.
There are other similar inequalities e.g. if 0<x<y<z, then x-y+z>xz/y. If we consider the expression x-y+z-xz/y then this is equal to …
-(1/y)(y2 – y(x+z) + xz) = -(1/y)(y-x)(y-z) = (1/y)(y-x)(z-y)>0. Hence this result is also true.
Again if 0<x<y<z, then -x+y+z<yz/x.
Consider -x+y+z-yz/x = (-1/x)(x2 -x(y+z) +yz).
Then,-x+y+z-yz/x = (-1/x)(x-y)(x-z) =(-1/x)(y-x)(z-x)<0. Hence the result.