A curious property of Acute Triangles

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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.

 

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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+Z2X2)/(2YZ)<1. So, YZ ≤ Y2+Z2X2<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 + Z2X2. So YZ ≤ Y2+Z2X2<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, C < π/2 and C > π/2.

Case 1          C = π/2

In this case,   = +   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   < +   while π = X+Y+Z.

Hence, Y > π(π/2-X)/(π-X) and Z <2/2-πX+X2)/(π-X).

Case 3          C > π/2

Likewise it follows that   > +   while π = X+Y+Z.

Hence, Y < π(π/2-X)/(π-X) and Z >2/2-πX+X2)/(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 – YX/2 > 0 and

0 < π/2 – ZX /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 = (1/4)π, (X,Y,Z) as right-angled, and calculate Y using Y = (π/2-X)/(1-X/π), we get = (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 = (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≤(Y2+Z2X2)/(2YZ)<1. So, YZ ≤ Y2+Z2X2<2YZ. It follows that  X2 ≤ Y2 – YZ + Z2Multiplying 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.

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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) = (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.

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