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 2**X** < 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 45^{o} , 60^{o} and 75^{o} , the angle sizes also form a Pythagorean triple, i.e. 45² + 60² = 75². Basically, the numbers 45^{o} , 60^{o} and 75^{o} 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 30^{o} , 72^{o} and 78^{o }(5, 12, 13) and 18^{o} , 80^{o} and 82^{o} (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 …

3**YZ**/2π ≤ π/2 – **X **< 2**YZ/**π**, **[this follows from 0<**A≤**π/3]

**XZ/**π < π/2 – **Y **< 2**XZ/**π, [follows from 0<**B**<π/2] and

0 < π/2 – **Z ≤ 3****XY/**2π [follows from π/3≤**C** and **Z**<π/2].

**Proof** that 3**YZ**/2π ≤ π/2 –**X**< 2**YZ/**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≤(**Y**^{2}+**Z**^{2}–**X**^{2})/(2**YZ**)<1. So, **YZ ≤ ****Y**^{2}+**Z**^{2}–**X**^{2}<2**YZ**.

Now π = **X**+**Y**+**Z **and so π – **X **= **Y**+**Z**.

Hence (squaring both sides of π – **X **= **Y**+**Z**) we get π^{2} – 2π**X**+**X**^{2} = **Y**^{2}+2**YZ**+**Z**^{2} or

π^{2}– 2π**X** – 2**YZ **= **Y**^{2 }+ **Z**^{2} – **X**^{2}. So **YZ ≤ ****Y**^{2}+**Z**^{2}–**X**^{2}<2**YZ** becomes

**YZ ≤ **π^{2}– 2π**X** – 2**YZ** < 2**YZ** or 3**YZ ≤**π^{2}– 2π**X** < 4**YZ**.

**Finally**, dividing by 2π gives 3**YZ**/2π ≤ π/2 –**X**< 2**YZ/**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 ≤ **3**XY/**2π becomes **XY**/π < π/2 –**Z ≤ **3**XY/**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, **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**= (π^{2}/2-π**X**+**X**^{2})/(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 <** (π^{2}/2-π**X**+**X**^{2})/(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 >** (π^{2}/2-π**X**+**X**^{2})/(π-**X**).

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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).

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** 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**

From 0<**A≤**π/3 we must have that 1/2≤cos(**A**)<1.

Hence 1/2≤(**Y**^{2}+**Z**^{2}–**X**^{2})/(2**YZ**)<1. So, **YZ ≤ ****Y**^{2}+**Z**^{2}–**X**^{2}<2**YZ. **It follows that **X ^{2} ≤ Y^{2} – YZ + Z^{2} . **Multiplying this last inequality by

**Y + Z**gives

**(Y + Z) X**and so using

^{2}≤ Y³ + Z³**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.**

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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.

**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.

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.**

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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**) = (x^{2}+ y^{2}– z^{2})/(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-1}x + (y/z)^{n-1}y,

where 0<cos(**Y**)<1 and 0<(x/z)^{n-1}<1 etc.

Is it true then that cos(**Y**)= (x/z)^{n-1}and 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-1}or 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<(y^{3}-x^{3})/(y^{2}-x^{2}) or z<(y^{2}+xy+x^{2})/(x+y). This last inequality leads to z-y<x^{2}/(x+y) and z-x<y^{2}/(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 z^{n} = x^{n} + y^{n} 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)+z^{2})/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)(y^{2} – 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)(x^{2} -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.