First, define triangle space T to be the set {(x,y,z): x,y,z>0 and x<y+z, y<x+z, z<x+y}. Define (x,y,z)=(a,b,c) iff x=a, y=b, and z=c, where (x,y,z) and (a,b,c) are members of T.

If (x,y,z) and (a,b,c) T then it is easy to show that if + is defined in the obvious way, i.e. (x,y,z) + (a,b,c) = (x+a,y+b,z+c) then (x,y,z) + (a,b,c) T. + is commutative and associative.

Clearly if k>0, and k(x,y,z) is defined by (kx,ky,kz) then k(x,y,z) T.

A type of multiplication * can also be defined, e.g.

(x,y,z)*(a,b,c) = (xa+yc+zb, yb+xc+za, zc+xb+ya)

This * operation is closed on T, commutative, distributive over addition, but not associative.

If we define the perimeter function p:T—>R ^{+} in the obvious way, i.e. p((a,b,c))=a+b+c, then if t1, t2, and t3T , p(t1+t2) = p(t1)+p(t2), p(kt1) = kp(t1), and p(t1*t2) = p(t1)*p(t2)

Even although (t1*t2)*t3 ≠ t1*(t2*t3) the perimeter function makes

p((t1*t2)*t3)=p(t1*(t2*t3)).

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If (x,y,z) T is an acute angled triangle with corresponding angles **X**,**Y**,**Z**, then it is easy to show that (x^{2},y^{2},z^{2}) and (**X**,**Y**,**Z**) also belong to T. This is not true for obtuse or right triangles. For a proof that (**X**,**Y**,**Z**) belongs to T if (x,y,z) T is an acute angled triangle, see

‘A curious property of Acute Triangles’.

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If (x,y,z) T then (x ^{1/n} , y ^{1/n} ,z ^{1/n} ) T for any integer n>=2 and each of these triangles is also acute. Clearly, since x ^{1/n} -> 1 for any x>0 as n -> ∞, (x ^{1/n} , y ^{1/n} , z ^{1/n} ) -> (1,1,1)

as n -> ∞.

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Now, suppose (a,b,c) T with a<b<c. Suppose also that (a,b,c)+(c,a,b) is right angled. Then (a,b,c)+(c,a,b) =(a+c,b+a,c+b), where b+a<c+a<c+b.

If (a+c,b+a,c+b) is right-angled then (c+b) ^{2} = (b+a) ^{2} + (a+c) ^{2} from which it follows that

c = a(a+b)/(b-a). Since c>b then it follows that a(a+b)>b(b-a) and so (a+b) ^{2} > 2b ^{2.}

Hence a>(√ 2 – 1)b, and c<a+b means that a<b/2.

Using the above we can easily generate some Heronian triangles.

Choose a value for b such as 17. Then (√ 2-1) x 17<a<17/2 or 7.04163056<a<8.5.

Take a=8. Then c=200/9 or 22 2/9. Hence, if b=17 and a=8,

(a+c, b+a, c+b) = (30 2/9, 25, 39 2/9) or 1/9(272, 225, 353) which is Heronian.

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Now suppose (a,b,c) is scalene i.e.a<b<c for example. Can (a,b,c) be written as a sum of 2 isosceles triangles? Suppose so. Then, assume that (a,b,c)=(x,x,y)+(p,q,q). It follows that a=x+p, b=x+q, and c=y+q. From these equations then q-p=b-a>0 and y-x=c-b>0.

In (x,x,y) then x<y<2x, or 0<y-x<x. From (p,q,q) we get no additional information. So, given (a,b,c) with a<b<c, choose x so that c-b=y-x<x<a. Then choose y=x+(c-b). p then follows from p=a-x and q=p+(b-a).

Now suppose we apply this to (5,12,13) then y-x=1 and q-p=7. So, 1<x<5. If x=3 then y=4. Hence, p=2 and q=9. So, (5,12,13)=(3,3,4)+(2,9,9). Clearly this is not the only way this can be done. E.g. (5,12,13)=(2,2,3)+(3,10,10)=(4,4,5)+(1,8,8) and these are just integer side lengths.

**Note:**the 2 isosceles triangles mentioned here are fundamentally different since (x,x,y), x<y, can be acute, right or obtuse, while (p,q,q), p<q, can only be acute. Perhaps (x,x,y) should be called type i1 and (p,q,q) type i2. Hence, all triangles are either e(equilateral), i1, i2 or s(scalene) = i1+i2.

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If we start with 2 positive numbers x and y, and assume that x<y<2x, we can generate 2 isosceles triangles from these immediately, i.e. (x,y,y) and (x,x,y). If we add these together we get … (x,y,y) + (x,x,y) = (2x,x+y,2y). Clearly, 2x<x+y<2y so (2x,x+y,2y) is scalene. Right angled?? If so then (2x) ^{2} + (x+y) ^{2} = (2y) ^{2} .

From that it follows that 3y ^{2} – 2xy -5x ^{2} = 0.

Solving this equation for y in terms of x we get y = 5x/3 or -x. So, y= 5x/3. Hence (2x,x+y,2y) = (2x,8x/3,10x/3) = 2x/3(3,4,5) or a scaled version of a (3,4,5) triangle.

Interestingly, if we try to ‘multiply’ the triangles (x,y,y) and (x,x,y) using the * operation we get (x ^{2} + xy + y ^{2}, 3xy, x ^{2} + xy + y ^{2} ) which is also isosceles but not right-angled.

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**Acute angled triangles have some interesting properties.** If we add in the conditions that such a triangle is also integer, i.e. has integer sides, and is scalene, then only certain lengths of side are possible.

Suppose a scalene, integer, acute angled triangle has sides (x,y,z) with 0<x<y<z.

**What is the ‘smallest’ such triangle?**

Clearly, 1≤x, 2≤y, and 3≤z and also z^{2}<x^{2}+y^{2}.

Then (z – y + y)^{2}<x^{2} + y^{2 }and so (z-y)^{2}+2y(z-y)<x^{2}. From this (last) inequality it follows that x^{2}>5 and so x≥3. x≥3 implies that y≥4 and so from the same inequality it follows that x^{2}>9, or that x≥4.

If x=4, then y≥5 and so, using z^{2}<x^{2}+y^{2}, it follows easily that z-y<x^{2}/(z+y)≤16/11. Hence, z-y≤1. But z-y>=1, and so it must be that z-y=1. From this it follows that 1+2y<16 or y<=7. So y=5, 6, or 7. Hence the ‘smallest’ scalene, integer, acute triangles are (4,5,6), ( 4,6,7) and (4,7,8).

Note that the ‘largest’ of these is very nearly right-angled since 8² = 4² + 7² – 1.

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So, if (x,y,z) is a scalene, integer, acute-angled triangle then, assuming that x<y<z, we can also assume that x>=4. Also, z^{2}<x^{2}+y^{2} since the triangle is acute-angled.

Then (z – y + y)^{2}<x^{2} + y^{2} and so (z-y)^{2}+2y(z-y)<x^{2}.

Now x+1 ≤ y so it follows that (z – y + x + 1)² < x² + (x+1)² .

Hence, z – y + x + 1 ≤ [√(x² + (x+1)²)] and so z – y ≤ [√(x² + (x+1)²)] – (x+1).

(z-y)^{2}+2y(z-y)<x^{2 }in turn implies that y<(x^{2} -(z-y)^{2})/2(z-y).

So, to generate such triangles, first choose **x>=4**.

Then choose **z-y** so that 1 ≤ z – y ≤ [√(x² + (x+1)²)] – (x+1).

Now choose **y** so that x+1 ≤ y < (x^{2}-(z-y)^{2})/2(z-y).

z is of course z-y + y.

Suppose, for example, that x=10, then 1≤z-y≤3, or z-y is 1, 2 or 3.

If z-y=1 then 11≤y≤49. The triangles are therefore (10, 11, 12), (10, 12,13) … (10, 49, 50).

If z-y=2 then 11≤y≤23. The triangles are (10, 11, 13), (10, 12, 14) … (10, 23, 25).

If z-y=3 then 11≤y≤15. The triangles are therefore (10, 11, 14), (10, 12, 15) … (10, 15, 18).

In all such triangles we have z<x+y, so z≤x+y-1 since x, y and z are integers.

Suppose there is a case where z=x+y-1. Then since z^{2}<x^{2}+y^{2} it follows that (x+y-1)² < x² + y²

or, after some simplification, 2(xy – x – y)<-1.

From this point it can be shown easily that x<2, which is not possible since, clearly, x≥4.

Hence, z<x+y-1.

If we repeat this argument with z=x+y-2, then it follows, similarly, that z<x+y-2.

If we try to push this further to z<x+y-3 we get x=4 and z=y+1, which is the case for the ‘smallest’ scalene, integer, acute triangles (4,5,6), ( 4,6,7) and (4,7,8).

**So, for a scalene, integer, acute triangle (x,y,z), where x<y<z, we must have x≥4 and z<x+y-2.**

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If (a,b,c) is an acute angled triangle, a≤b≤c, and (d,e,f) is also acute angled, d≤e≤f, is it true that (a+d,b+e,c+f) is also acute angled?

**Note**: while this question initially seemed to be answered with a ‘yes’ it is in fact ‘no’.

**Proof**

If 0<x<1, then (3, 4, 5-x) and (5, 12, 13-x) are both **acute** angled but their sum

(3,4,5-x) +(5,12,13-x) or (8,16,18-2x) is easily shown to be obtuse if x < 9 -4√5 = 0.05572809.

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However, if (a,b,c) is a triangle, a≤b≤c, but not acute angled, and (d,e,f) is also not acute angled, d≤e≤f, then(a+d,b+e,c+f) is also not acute angled.

Since (a,b,c) and (d,e,f) are not acute angled then a² + b² ≤ c² and d² + e² ≤ f².

We need to prove that (a+d)² + (b+e)² ≤ (c+f)² to show that (a+d, b+e, c+f) is not acute angled. It follows easily that (a+d)² + (b+e)² – (c+f)² = a² + b² – c² + d² + e² – f² +2ad + 2be – 2cf

and so (a+d)² + (b+e)² – (c+f)² ≤ 2ad + 2be – 2cf.

Suppose ad + be – cf >0. Then (ad+be)² >c²f² ≥ (a²+b²)(d²+e²) .

So, 2adbe > a²e² + b²e² or 0 > (ae – bd)² which is clearly not true.

Hence the result is proved.

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