# Some properties of Triangle Space

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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 (x2,y2,z2) 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 z2<x2+y2.

Then (z – y + y)2<x2 + y2 and so (z-y)2+2y(z-y)<x2. From this (last) inequality it follows that x2>5 and so x≥3. x≥3 implies that y≥4 and so from the same inequality it follows that x2>9, or that x≥4.

If x=4, then y≥5 and so, using z2<x2+y2, it follows easily that z-y<x2/(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, z2<x2+y2 since the triangle is acute-angled.

Then (z – y + y)2<x2 + y2 and so (z-y)2+2y(z-y)<x2.

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)<x2 in turn implies that y<(x2 -(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 < (x2-(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 z2<x2+y2 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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