In the equilateral triangle below, $CD, AD$, and $BD$ concur at point $D$. The measures of the angles ($a, b, c, d, e,$ and $f$) are also defined as given in the figure below.
Conjecture:
There aren't any $a, b, c, d, e,$ and $f$ such that all are
different integers without using the combinations of the numbers: $6, 54, 42, 18, 48$, and $12$. (to clarify, all six of these numbers can not be used together, but they can be used together in groups)
How can I prove this conjecture mathematically, and what is so special about these six numbers?
My thoughts and work on the problem
I am intentionally omitting the degree signs for convenience.
Using, the Trigonometric Ceva's Theorem, I obtained the following expression:
$$\frac{\sin(a)}{\sin(b)}\cdot\frac{\sin(c)}{\sin(d)}\cdot\frac{\sin(e)}{\sin(f)}= 1$$
which is equivalent to:
$$\begin{aligned} \sin(a) \sin(c) \sin(e) = \sin(b) \sin(d) \sin(f)
\end{aligned} \label{a}\tag{1}$$
We also know that,
$$a+b=c+d=e+f=60$$
To convert the triple sine product to summations, I used the following "trick". Since,
$$\sin(a+c+e)=\sin(b+d+f)$$
and also due to $Eq. (1)$, it is true that:
$$\sin(-a+c+e)+\sin(a-c+e)+\sin(a+c-e)$$
$$=\sin(-b+d+f)+\sin(b-d+f)+\sin(b+d-f)$$
Let's substitute $b=60-a, d=60-c$, and $f=60-e$ and let $x=(c+e-a), y=(a+e-c),$ and $z=(a+c-e)$ to simplify things a bit:
$$\sin(x) +\sin(y) +\sin(z)=\sin(60-x)+\sin(60-y) +\sin(60-z)$$
I expanded the $(60-x)$'s using compound angle formulae:
$$\sqrt{3}/2 (\cos(x) + \cos(y) + \cos(z)) – 3/2 (\sin(x) + \sin(y) + \sin(z)) = 0$$
which simplifies to:
$$\sin(30-x) +\sin(30-y) +\sin(30-z)=0$$
but I feel lost in the algebra and don't know what to do next. I want to get to a point where I can easily plug numbers and show that no other numbers satisfy this equation, other than the combinations of the six numbers given at the beginning.
The trigonometric proof of one of the cases may aid to formulate a proof, so I decided to include it as well:
For $a=6, b=54, c=42, d=18, e=48, f=12$, $Eq. (1)$ is satisfied, so let's prove it.
Proof.
$$ \sin(12°) \sin(18°) \sin(54°)\stackrel{?}{=} \sin(6°)\sin(42°)\sin(48°)$$
Now we use the "trick":
$$ \sin(60°)+ \sin(48°)- \sin(24°)\stackrel{?}{=} \sin(84°)+\sin(12°)+\sin(0°)$$
$$\sin(60°)=\frac{\sqrt{3}}{2} \stackrel{?}{=} \sin(12°)+ \sin(24°) – (\sin(48°) -\sin(84°))$$
$$ = 2 \sin(18°) \cos(6°) + 2 \cos(66°) \sin(18°)$$
$$ = 2 \sin(18°) (\cos(6°) + \cos(66°)$$
$$ = 2 \sin(18°) (2\cos(36°)\cos(30°) )$$
$$1 \stackrel{?}{=} 4 \sin(18°) \cos(36°)$$
$$\frac{1}{2} = \cos(36°) \cos(72°)$$
which is true as it is a well-known identity.
Here is the figure of this triangle (rotated):
Given only 6 and 12 degrees (the red angles) the rest of the angles can be found. For the sake of brevity, I will omit the synthetic proofs. For the interested reader, see this link, as it also shows the beauty of synthetic solutions in these types of problems from the perspective of a geometrician. The combinations of $6, 54, 42, 18, 48$, and $12$ often come up in synthetic geometry problems: I even managed to find one in SE.
@KorayUlusan helped me develop a code to test the conjecture by brute force in Python. Turns out that the conjecture is correct, but I'm looking for a mathematical explanation. Here is the code:
import math
epsilon = 2.22044604925e-14 # for integer overflow
for a in range(60):
b = 60 - a # because a+b=60
for c in range(60):
d = 60 - c
for e in range(60):
f = 60 - e
if a != b and b != c and c != d and d != e and e != f:
try:
rightHandSide = (
math.sin(math.radians(a))
* math.sin(math.radians(c))
* math.sin(math.radians(e))
) / (
math.sin(math.radians(b))
* math.sin(math.radians(d))
* math.sin(math.radians(f))
)
if abs(1 - rightHandSide) < epsilon:
print(f"a is {a}, b is {b}, c is {c}, d is {d}, e is {e}, f is {f}")
except ZeroDivisionError:
pass
The Python output for the code (of course, most of these results are the same triangles, just rotated):
a is 6, b is 54, c is 42, d is 18, e is 48, f is 12
a is 6, b is 54, c is 48, d is 12, e is 42, f is 18
a is 12, b is 48, c is 18, d is 42, e is 54, f is 6
a is 12, b is 48, c is 54, d is 6, e is 18, f is 42
a is 18, b is 42, c is 12, d is 48, e is 54, f is 6
a is 18, b is 42, c is 54, d is 6, e is 12, f is 48
a is 42, b is 18, c is 6, d is 54, e is 48, f is 12
a is 42, b is 18, c is 48, d is 12, e is 6, f is 54
a is 48, b is 12, c is 6, d is 54, e is 42, f is 18
a is 48, b is 12, c is 42, d is 18, e is 6, f is 54
a is 54, b is 6, c is 12, d is 48, e is 18, f is 42
a is 54, b is 6, c is 18, d is 42, e is 12, f is 48
Thanks in advance for any sort of help or contribution.
Best Answer
Not an answer but an extended comment.
First: you have already proved the conjecture. You may think that the exhaustive search is not a nice tool, but you checked all $59\times 59\times 59$ possibilities and proved that there are no other solutions.
Second: there is no mystics about the integer numbers. They appear to be integer because some ancient human being decided that it is nice to divide the circle in 360 parts which were named "degrees". The obtained solutions would remain integer if he/she decided to divide it instead in 60 parts but cancel to be integer if he/she would prefer 100 parts.
Third: the problem can be however formulated as a really challenging one.
The following formulation of the problem is due to @Blue
Namely: find all triples $\{p,q,r\}$ of non-zero rational multiples of $\pi$: $$-\frac\pi6 <p,q,r<\frac\pi6$$ such that $$ \sin\left(\frac{\pi}{6}+p\right) \sin\left(\frac{\pi}{6}+q\right) \sin\left(\frac{\pi}{6}+r\right) = \sin\left(\frac{\pi}{6}-p\right) \sin\left(\frac{\pi}{6}-q\right) \sin\left(\frac{\pi}{6}-r\right)\tag1 $$ which neatly reduces to: $$ \cot p\cot q+\cot q\cot r+\cot r\cot p=−3.\tag2 $$
I have checked all triplets with denominator up to $n=120$ and the result is: there are no others except for $$ \frac\pi{30}\{2,3,-4\}, $$ which you have found. Is the triplet the only one?