Problems & Puzzles: Puzzles

Puzzle 519. Triangles with prime sides Frank Rubin poses the following nice puzzle: Find rationals a/b such that two triangles having prime sides have areas in proportion a/b.

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Contributions came from Fred Schneider, Jan van Delden, Giovanni resta & Anton Vrba

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Fred wrote:

Just wondering, do the triangles have to be non-equilateral? All
equilateral triangle's areas (prime-length are not) are multiples of
sqrt(3) and thus the areas of any two with integral sides have a
rational ratio.

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Jan wrote:

Let's call the sides of the triangle p,q,r; let's assume that 2<=p<=q<=r (and p+q>r).

It's possible to show that the area A of this triangle can be written as: A=sqrt(f(p,q,r))/4 with:
f(p,q,r) = 2.p^2.q^2+2.p^2.r^2+2.q^2.r^2-(p^4+q^4+r^4) which can be rewritten as:
f(p,q,r) = (p+q-r).(p+q+r).(r+p-q).(r+q-p)

It's easy to verify that f doesn't depend on the order of the sides p,q,r.

1) Equilateral triangle
If p=q=r we have f(p,p,p)=3.p^4 hence A=sqrt(3)/4.p^2, not a natural number (not in N).
2) Isosceles triangle
If p=q, r>p we have f(p,p,r)=(4.p^2-r^2).r^2
If q=r, p<q we have f(p,r,r)=(4.r^2-p^2).p^2 (similar result)

In "both" cases the first term should be a square, say x^2, for A to be in N.
The term 4.p^2=0 mod 4, but x^2+r^2 is never 0 mod 4 since x^2 is always 0/1 mod 4 and r^2=1 mod 4. (r>2 thus odd).
In the second case the same argument can be used, except in the case where p=2, since then p^2=0 mod 4 is possible.
But then: f(2,r,r)=16.(r^2-1) and r^2-1 can't be a square (r>2).

Note: p,q,r being prime is not used untill now. Only that r is odd>1.

3) Scalene triangle
p<q<r
Anyone with a prove there can't be a solution with f a square?

Although there's (probably) no solution with A a natural number, there are lot's of solutions where the fraction a/b of two such A's are in Q.
For instance using two different equilateral triangles with sides p[1] and p[2] gives a/b=A[1]/A[2]=p[1]^2/p[2]^2.

In general: find two triangles where the decomposition of f shows the same factors with odd multiplicity.
For instance:

Common factor 2 with odd multiplicity: f(2,3,3)=2^7 and f(2,17,17)=2^9.3^2 and f(2,577,577)=2^11.3^2.17^2; etc., giving 2.3 and 4.3.17 and 2.17 and their reciprocals. [Does f(2,g(n),g(n)) with {g(n)=2*g(n-1)^2-1 g(1)=3}, always have one factor 2 with odd multiplicity?]
Common factor 3 with odd multiplicity: f(3,5,7)=3^3.5^2 and f(7,13,19)=3.5^2.13^2 and f(31,37,43)=3.5^2.7^2.37^2; etc., giving 3/13 and 3/(7.37) and 13/(7.37) and their reciprocals.
Common factors 5 and 7 with odd multiplicity: f(3,11,13);f(5,13,17);f(13,13,19);f(17,17,29);f(23,23,29);f(31,47,47); etc.

When p>2, a single odd common factor P should always have P=3 mod 4. Since f=3 mod 4 with odd p,q,r there is at least one factor P=3 mod 4 with odd multiplicity. It can also have factors P=1 mod 4 with odd multiplicity. But never only one factor P=1 mod 4 with odd multiplicity because then f=1 mod 4.

The other situation f(2,r,r)=0 mod 4, so nothing can be derived directly from this. The first r with a single factor P=1 mod 4 with odd multiplicity, r=73 and P=37. These seem to be rare.

In short, the only 'concrete' solutions are:

For a triangle with sides (p,q,r) the area A = sqrt(f(p,q,r))/4. With f(p,q,r)=(p+q-r).(p+q+r).(r+p-q).(r+q-p)

1) Probably no integer area for any triangle. If these triangles exist, they must be scalene (2<p<q<r).
2) The fraction a/b=p^2/q^2 is always possible, using two equilateral triangles with sides p and q.
3) Using g(n)=2*g(n-1)^2-1, g(1)=3 the triangles with sides (2,g(n),g(n)) implicate a fraction in Q. [g(n)=prime for n=1..4, maybe more with n>7].
4) Other types of fractions are possible, but I'm not sure whether each fraction a/b in Q can (or can't) be achieved, it requires matching factors with odd multiplicity of f(p,q,r).
 

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Giovanni wrote:

It is quite simple to find triangles having prime sides whose areas
are in a rational proportion a/b, but I did not find
a general law to tell which values a/b are possible.

Instead, I concentrate on a subproblem, where a/b = 1,
i.e. I searched for triangles which have the same area.

I listed the triangles according to their longest side and
I noticed the first occurrence of a pair, triple, etc.
of prime triangles with the same area. These are the
results (from 2 to 9 triangles with the same area):

2: (11,11,3) (7,7,5)

3: (127,67,61) (113,107,11) (101,97,11)

4: (269,239,53) (191,127,97) (163,157,71) (131,113,101)

5: (787,691,113) (479,373,149) (419,347,137) (349,337,131) (311,263,167)

6: (653,619,37) (521,463,61) (479,467,23) (389,229,163) (197,179,53) (149,139,67)

7: (3527,3049,479) (2819,2789,47) (1361,1181,197) (1109,577,547) (1031,1013,101) (673,401,337) (439,347,293)

8: (7607,7349,293) (5779,5717,191) (5431,3967,1481) (2749,2503,467) (2647,1847,929) (2549,2207,557) (2143,1201,1151) (1259,1063,1013)

9: (14431,14153,619) (9137,8893,911) (7669,7187,1171) (6299,3889,2903) (6287,4211,2593) (6143,5347,1601) (5407,4327,1979) (5351,3677,2473) (5147,4259,1933)

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Anton wrote:

The ratio of areas of two equilateral triangles is always a rational number and thus these form trivial solutions to the puzzle. Now if we add the constraint that no side of one triangle may be same as any other side of any other triangle, then a quick search using small primes finds many solutions making me wonder if we witness Guy's law of small numbers doing its work. An interesting solution to the problem are the following two sets of triangles, both sets have the same integer area proportions.

set 1: A[3, 29, 31] : A[23, 67, 71] = 1 : 23
set 2: A[5, 29, 31] : A[23, 137, 139] = 1 : 23

note the relation 137 = 2 x 67 + 3 and 139 = 2 x 71 - 3

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