Suppose we start with an acute angled triangle where the angle sizes are denoted by X, Y, and Z (assume radians unless otherwise stated). Without loss of generality, assume that the angle sizes X, Y, and Z are such that 0 < X ≤ Y ≤ Z < π/2.
Clearly X < π/2 and so 2X < π = X+Y+Z. It follows that X < Y+Z.
A similar argument shows that Y < X+Z and again that Z < X+Y. So X, Y and Z (the angle sizes of the original acute-angled triangle) can form the lengths of the sides of a new triangle (X,Y,Z), as above, with corresponding internal angles of sizes A, B and C (say), A corresponds to (is opposite) X , B to Y and C to Z. That A + B + C = π is obvious, and it seems clear that A ≤ B ≤ C but this requires some proof.
Proof
First, cos(A) = (Y² + Z² – X²)/(2YZ) and since 0 < X ≤ Y ≤ Z then cos(A)>0.
Similarly, cos(B) = (X² + Z² – Y²)/(2XZ) and so cos(B)>0 also following from 0 < X ≤ Y ≤ Z.
Hence, A and B are acute angles, i.e. < π/2.
Now cos(A) – cos(B) = (Y – X)(X + Y + Z)(X + Y – Z)/(2XYZ) ≥ 0 and therefore cos(A) ≥ cos(B).
Since cos() is a decreasing function over (0, π/2) it follows that A ≤ B.
If C ≥ π/2 then A ≤ B < π/2 ≤ C. If C < π/2 then we can use the fact that
cos(B) – cos(C) = (Z – Y)(X + Y + Z)(Z + Y – X)/(2XYZ) ≥ 0 and so cos(B) ≥ cos(C).
Again since cos() is a decreasing function over (0, π/2) it follows that B ≤ C.
Hence we have A ≤ B ≤ C.
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It is easily checked that 0 < A ≤ π/3, 0 < B < π/2 and π/3 ≤ C < π.
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 since in those cases Z ≥ X + Y and so (X,Y,Z) would not exist. The operation to create (X,Y,Z) is a bit like turning the acute triangle ‘inside out’.
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 (integer) 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.
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.
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The largest angle in triangle (X,Y,Z) is C and so there are 3 possibilities for that angle,
i.e. C = π/2, π/3 ≤ C < π/2 and π/2 < C < π.
Case 1 C = π/2
In this case, Z² = X² + Y² while π = X+Y+Z.
So, (Z + Y)(Z – Y) = X² and Y + Z = (π – X).
It follows that Z – Y = X² / (π – X).
So, Z = 1/2 [(π – X) + X² / (π – X)]
and Y = 1/2 [(π – X) – X² / (π – X)]
Hence if (X,Y,Z) is right angled (at C), 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 π/3 ≤ C < π/2
It follows that 1/2 ≥ cos(C) > 0 and so 1/2 ≥ (X² + Y² – Z²)/2XY > 0.
Hence XY ≥ X² + Y² – Z² > 0 and so Z² ≥ X² + Y² – XY > Z² – XY. Multiplying by X + Y and
using the fact that X + Y = π – Z we get πZ² ≥ X³ + Y³ + Z³ > πZ² + XY(Z – π)
Case 3 π/2 < C < π
So -1 < cos(C) < 0 and therefore -1 < X² + Y² – Z²)/(2XY) < 0.
It follows easily that Z² < (X + Y)² < Z² + 2XY.
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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, both Y and Z -> π/2.
[This last fact is obvious but it follows from the inequalities X > π/2 – Y ≥ X/2 > 0 and
0 < π/2 – Z ≤ X /2.]
Also as X -> π/3, Y -> π/3 and Z -> π/3.
[Again this follows from the inequalities 0 ≤ (Z – π/3) ≤ 2(π/3 – X) and 0 ≤ |π/3 – Y| ≤ 3(π/3 – X).]
As another example, if we take X = (1/4)π, (X,Y,Z) as right-angled, and calculate Y
using Y = (π/2 – X)/(1 – X/π), we get Y = (1/3)π. Then using 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 – scaled).
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If we extend/generalise this method to X = (p/q)π, where p and q are positive integers and
p/q < 1/3, 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.
Similarly, if d is a positive integer and X = (2d-1)π/(6d -1), so X < π/3, and (X,Y,Z) is right-angled, then it follows that
((2d – 1)8d)² + ((2d + 1)(6d – 1))² = (20d² – 4d +1)² .
Likewise if X = 2dπ/(6d+1), so again X < π/3, and (X,Y,Z) is right-angled, then it follows that
(4d(4d+1))² + ((2d+1)(6d+1))² = (20d² + 8d +1)².
It is interesting that all odd primes greater than 3 take the form 6d+1 or 6d-1 for some positive integer d and those terms appear in these identities multiplied by 2d+1.
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Further inequalities (assuming (x,y,z) is acute angled).
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 < YZ + X². Multiplying these last inequalities by Y + Z gives
(Y + Z) X2 ≤ Y³ + Z³ < (Y + Z)(YZ + X² ) and also using Y+Z = π – X we get
πX² ≤ X³ + Y³ + Z³ < πX² + YZ(Y + Z).
Using C ≥ π/3 (since C is the largest angle in triangle A, B, C) it follows similarly to the 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.
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If the original acute-angled triangle with sides (x, y, z), assuming that 0 < x ≤ y ≤ z < √(x² + y²), 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.
Proof
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 √(x² + y²).
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. So, X = A, Y = B and Z = C. 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.
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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.
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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.
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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 ).
Proof
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.
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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.
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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.
Proof
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.
curiousmathsknols 