5-Con triangles

Two triangles are said to be 5-Con or almost congruent if they are not congruent triangles but they are similar triangles and share two side lengths (of non-corresponding sides). These 5-Con triangles are important to properly understand the solution of triangles. Indeed, knowing three angles and two sides is not enough to determine a triangle up to congruence. A triangle is said to be 5-Con capable if there is another triangle which is almost congruent to it.

The 5-Con triangles have been discussed by Pawley, and later by Jones and Peterson. They are also mentioned by Martin Gardner in his book Mathematical Circus. An earlier reference is the following exercise: Explain how two triangles can have five parts (sides, angles) of one triangle congruent to five parts of the other triangle, but not be congruent triangles.

Examples

There are infinitely many 5-Con triangles.

• The smallest 5-Con triangles with integral sides have side lengths (8;12;18) and (12;18;27). This is an example with obtuse triangles.
• Acute 5-Con triangles are for example (1000;1100;1210) and (1100;1210;1331).
• Right 5-Con triangles are for example $(1;m;m^{2})$ and $(m;m^{2};m^{3})$ with $m={\sqrt {\frac {1+{\sqrt {5}}}{2}}}\sim 1.3$. The right 5-Con triangles are exactly those obtained from scaling this pair. There is no example of right 5-Con triangles with integral sides.
• There are no 5-Con triangles that are equilateral, and none that are isosceles.

Results

1. Consider 5-Con triangles with side lengths $(a;b;c)$ and $(ma;mb;mc)$ where $m$ is the scaling factor, which we may suppose to be greater than $1$. We may also suppose $a\geq b\geq c$. Then we must have $b=ma$ and $c=mb$. The two triples of side lengths are then of the form$$a(1;m;m^{2})\qquad \mathrm {and} \qquad a(m;m^{2};m^{3}).$$Conversely, for any $a>0$ and $1<m<{\frac {1+{\sqrt {5}}}{2}}$, such triples are the side lengths for 5-Con triangles (Supposing w.l.o.g. $a=1$, the greatest number in the first triple is $m^{2}$ and we only need to ensure $m^{2}<1+m$; the second triple is obtained from the first by scaling with $m$. So we have two triangles: They are clearly similar and exactly two of the three side lengths coincide.). Some references work with $m^{-1}<1$ instead, which leads to the inequalities ${\frac {{\sqrt {5}}-1}{2}}<m^{-1}<1$.
2. Any 5-Con capable triangle has different side lengths and the middle one is the geometric mean of the other two. The ratio between the largest and the middle side length is then equal to that between the middle and the smallest side length. We can use this ratio and its inverse for scaling and obtaining two 5-Con triangles.
3. To study the possible shapes of 5-Con triangles we may restrict to studying the triangle with side lengths $$(1;m;m^{2})\qquad \mathrm {where} \qquad 1<m<{\frac {1+{\sqrt {5}}}{2}}.$$The greatest angle is a strictly increasing function of $m$ and varies from 60° to 180° (the limit cases are excluded). The right triangle corresponds to the value $m={\sqrt {\frac {1+{\sqrt {5}}}{2}}}$.
4. Having two 5-Con triangles with integral sides amounts (in the above notation) to taking any rational number $1<m<{\frac {1+{\sqrt {5}}}{2}}$ and then choosing $a>0$ in such a way that $am^{3}$is an integer. The four involved side lenghts $(a;am;am^{2};am^{3})$ are of the form $z(x^{3};x^{2}y;xy^{2};y^{3})$where $x,y,z$ are positive integers and $x,y$ are coprime. One is of course only interested in primitive 4-tuples, i.e. those where the entries do not have any common prime factor (in other words, $z=1$).

Further remarks

Defining almost congruent triangles gives a binary relation on the set of triangles. This relation is clearly not reflexive (it is irreflexive) and it is symmetric. It is not transitive: As a counterexample, consider the three triangles with side lengths (8;12;18), (12;18;27), and (18;27;40.5).

There are infinite sequences of triangles such that any two subsequent terms are 5-Con triangles. It is easy to costruct such a sequence from any 5-Con capable triangle: To get an ascending (respectively, descending) sequence, keep the two greatest (respectively, smallest) side lengths and choose a third greater (respectively, smaller) side length as to obtain a similar triangle. One may easily arrange the triangles in such a sequence in a neat way, for example by forming a spiral or by reflecting one ray inside a given angle , see Figures TODO.

One generalization is considering 7-Con quadrilaterals: One example given in is displayed in Figure TODO. In loc.cit. the author introduces also "(2n-1)-Con n-gons under a similarity".