euclidean-geometrygeometryplane-geometrytrianglestrigonometry
I sent this problem to Presh Talwalkar who suggested me to send it to this site.
I tried many things but was not able to find the correct solution.
I made various segments trying to get an equilateral triangle similar to the Russian triangle problem, but no success.
I also tried to flip the triangle UFO over side NO but again no success.
I tried to find like triangles, but not enough.
Could you please give me a hint?
Thanks,
R. de Souza
Let $a$, $b$, $c$ be the lengths of the sides opposite $A$, $B$, $C$, and let points $D$, $E$, $F$ lie on the sides opposite $A$, $B$, $C$. (The latter is slightly different than in the original question). Let $r = |DE| = |EF| = |FD|$ be the length of the sides of the equilateral triangle; let $s = |AF| = |BD| = |CE|$ be the lengths of the congruent sub-segments of the sides.
The Law of Cosines for each $r$-side in $\triangle AEF$, $\triangle BFD$, $\triangle CDE$ gives
$$\begin{align} r^2 = s^2 + (b-s)^2 - 2 s (b-s) \cos A \\ r^2 = s^2 + (c-s)^2 - 2 s (c-s) \cos B \\ r^2 = s^2 + (a-s)^2 - 2 s (a-s) \cos C \end{align}$$
Setting the right-hand sides equal and subtracting $2s^2$ yields: $$b^2-2bs-2s(b-s)\cos A = c^2-2cs-2s(c-s)\cos B = a^2 - 2 a s - 2 s(a-s) \cos C$$
Using the Law of Cosines (in $\triangle ABC$) to re-write each of $\cos A$, $\cos B$, $\cos C$, and then simplifying a bit, gives
$$\begin{align} &\phantom{=}\;a\left( s^2\left(-a^2+b^2+c^2\right)+ s b\left(a^2-\left(b+c\right)^2\right)+b^3 c\right) \\ &= b\left( s^2\left(\phantom{-}a^2-b^2+c^2\right)+ s c\left(b^2-\left(c+a\right)^2\right)+c^3 a\right) \\ &= c\left( s^2\left(\phantom{-}a^2+b^2-c^2\right)+ s a\left(c^2-\left(a+b\right)^2\right)+a^3 b\right) \end{align}$$
We can break the triple equality into a system of two quadratic equations in the parameter $s$. Eliminating $s$ gives a polynomial equation, $p(a,b,c)=0$; the factors of $p$ provide these equations: $$\begin{align} abc (a+b+c) &= 0 &(1) \\ a^2 b\left(a-b\right)+b^2c\left(b-c\right)+c^2 a\left(c-a\right) &= 0 & (2) \\ a b\left(a-b\right)\left(a\left(b-c\right)\left(a^2-b^2\right)-b^2c^2\right) \\ +b c\left(b-c\right)\left(b\left(c-a\right)\left(b^2-c^2\right)-c^2a^2\right) \\ +c a\left(c-a\right)\left(c\left(a-b\right)\left(c^2-a^2\right)-a^2b^2\right) &= 0 & (3) \end{align}$$
Solutions in each case correspond to ostensible solutions to the problem.
We can ignore (1), which has only trivial solutions. I believe (2)'s only non-trivial solutions (with the Triangle Inequality in play) require $a=b=c$, but I need to double-check this.
As for (3), clearly $a=b=c$ works; and, indeed, if (say) $a=b$, then $a=b=c$. So any non-equilateral solutions would need to be strictly scalene; it turns out that (3) admits scalene solutions.
Below is Mathematica's plot of (3), with $c=1$; the horizontal axis is $a$, and the vertical is $b$. The lines are $a+b=c$ and $b+c=a$ and $c+a=b$, which bound the region of feasibility according to the Triangle Inequality. The point $(1,1)$ seems to have been missed by the implicit plotter, but lots of other points appear in blue. (I believe there are even more. Zooms show an island near the $a+b=c$ border, for instance.)
Here's the figure for $a=8$, $b=8.58667$, $c=1$:
Equilateral $\triangle DEF$ extends beyond the confines of $\triangle ABC$. This may be why @Brian was having difficulty ruling-out non-equilateral $\triangle ABC$s.
I suspect equation (3) can be expressed in a way that's more illuminating; it's not clear to me, for instance, whether every solution to (3) corresponds to an "external" triangle. There's probably a whole other approach to this problem that makes everything obvious, but this is the best I have at the moment.
This Rust program gives the answer 205689153 in about a minute and a half. It’s based on the Euler characteristic formula $V - E + F = 1$ for a connected plane graph with $V$ vertices, $E$ edges, and $F$ faces. But there doesn’t seem to be a nice formula to find $V$ and $E$ without lots of computation, because in some cases, multiple pairs of segments concur at the same intersection point. So we just list all the intersections and count up the duplicates.
use std::collections::hash_map::HashMap;
fn det(a: (i32, i32), b: (i32, i32), c: (i32, i32)) -> i32 {
(b.0 - a.0) * (c.1 - a.1) - (b.1 - a.1) * (c.0 - a.0)
}
fn gcd(mut x: i32, mut y: i32) -> i32 {
while y != 0 {
let z = x % y;
x = y;
y = z;
}
x
}
fn reduce(n: i32, d: i32) -> (i32, i32) {
let g = gcd(n, d);
(n / g, d / g)
}
fn main() {
for &n in &[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100] {
let sides = [
(1..n).map(|i| (i, 0)).collect::<Vec<_>>(),
(1..n).map(|i| (n - i, i)).collect::<Vec<_>>(),
(1..n).map(|i| (0, n - i)).collect::<Vec<_>>(),
];
let segments = (0..)
.zip(&sides)
.flat_map(|(i, side0)| {
sides[i + 1..].iter().flat_map(move |side1| {
side0
.iter()
.flat_map(move |&a| side1.iter().map(move |&b| (a, b)))
})
})
.collect::<Vec<_>>();
let mut regions = 1 + segments.len() as i64;
let mut intersections = HashMap::new();
for (i, &(a, b)) in (0..).zip(&segments) {
for &(c, d) in &segments[i + 1..] {
let p = det(c, d, a);
let q = det(c, d, b);
if p * q < 0 && det(a, b, c) * det(a, b, d) < 0 {
if *intersections
.entry((
reduce(a.0 * q - b.0 * p, q - p),
reduce(a.1 * q - b.1 * p, q - p),
))
.or_insert(i)
== i
{
regions += 1;
}
}
}
}
println!("{} {}", n, regions);
}
}
Output:
1 1
2 4
3 27
4 130
5 385
6 1044
7 2005
8 4060
9 6831
10 11272
100 205689153
Here are the results when dividing each side into $n$ parts for all $1 \le n \le 120$:
1, 4, 27, 130, 385, 1044, 2005, 4060, 6831, 11272, 16819, 26436, 35737, 52147, 69984, 92080, 117952, 157770, 193465, 249219, 302670, 368506, 443026, 546462, 635125, 757978, 890133, 1041775, 1191442, 1407324, 1581058, 1837417, 2085096, 2365657, 2670429, 3018822, 3328351, 3771595, 4213602, 4694337, 5140756, 5769306, 6279934, 6987991, 7661637, 8355580, 9122179, 10077408, 10860478, 11882437, 12859392, 13960045, 15028393, 16394970, 17583472, 18980292, 20342943, 21871402, 23445913, 25385163, 26876233, 28911262, 30947106, 32961190, 35048842, 37459587, 39569107, 42324415, 44890158, 47731083, 50294455, 53649654, 56360842, 59879101, 63420084, 66857380, 70408212, 74445273, 78040573, 82622160, 86647137, 91124683, 95665744, 101133132, 105569497, 110811364, 116310795, 122023012, 127352503, 134068833, 139524337, 146093875, 152642448, 159496621, 166630228, 174340821, 180991705, 189418792, 197333184, 205689153, 213416806, 223144743, 231395536, 241509546, 251118018, 260392267, 270368527, 282027867, 291604741, 303685314, 314632365, 326674581, 337687342, 351301695, 363291763, 376664530, 390047007, 403508989, 417603979, 433264083
Best Answer
Consider a regular 36-gon $A_1A_2\ldots A_{36}$ inscribed in a circle of radius $R$. Inscribed angle over any side is $5^\circ$. We can see our configuration as it is shown on the picture.
It suffices to prove that $UF$ is parallel to the diagonal $A_{13}A_{34}=EA_{34}$; then we have $\angle NFU=\angle NEA_{34}=25^\circ$, so $\angle UFO= \angle NFO-\angle NFU=100^\circ-25^\circ=75^\circ$.
To prove $UF\parallel A_{34}E$, it is enough to prove $\frac{NU}{UT}=\frac{NF}{FE}$ ($T$ is as on the picture). For we can use the following two formulae:
The length of a chord of a circle with inscribed angle $\alpha$ is $2R\sin\alpha$.
If $E$ is on a side $BC$ of a $\triangle ABC$, then $\frac{BE}{EC}= \frac{AB\sin\angle BAE}{AC\sin\angle CAE}$.
Now, from $\triangle OEN$ we have: $$\frac{NF}{FE}= \frac{ON\sin\angle NOF}{OE\sin\angle EOF}=\frac{2R\sin 40^\circ\sin 20^\circ}{2R\sin 60^\circ\sin 60^\circ}.$$ From $\triangle NET$ we have: $$\frac{NU}{UT}= \frac{EN\sin\angle NEU}{ET\sin\angle TEU}= \frac{2R\sin 80^\circ\sin 5^\circ}{ET\sin 20^\circ}.$$ By the law of sines on $\triangle NET$, $\frac{ET}{NE}=\frac{\sin 60^\circ}{\sin 95^\circ}$, so $ET= NE\ \frac{\sin 60^\circ}{\sin 95^\circ}= 2R\sin 80^\circ\frac{\sin 60^\circ}{\sin 95^\circ}$ and thus $$\frac{NU}{UT}= \frac{2R\sin 80^\circ\sin 5^\circ}{2R\sin 80^\circ\frac{\sin 60^\circ}{\sin 95^\circ}\sin 20^\circ}= \frac{2R\sin 95^\circ\sin 5^\circ}{2R\sin 60^\circ\sin 20^\circ}.$$
So, for $\frac{NU}{UT}=\frac{NF}{FE}$ it is enough to check: $\sin 40^\circ\sin 20^\circ\sin 20^\circ= \sin 95^\circ\sin 5^\circ\sin 60^\circ$.
We have: $$\sin 95^\circ\sin 5^\circ\sin 60^\circ=\frac{1}{2}(\cos 90^\circ-\cos 100^\circ)\sin 60^\circ= \frac{1}{2}\cos 80^\circ\sin 60^\circ= \frac{1}{4}(\sin 140^\circ-\sin 20^\circ)= \frac{1}{4}(\sin 40^\circ-\sin 20^\circ),$$ and: $$\sin 40^\circ\sin 20^\circ\sin 20^\circ=\frac{1}{2}(\cos 20^\circ-\cos 60^\circ)\sin 20^\circ= \frac{1}{2}(\cos 20^\circ\sin 20^\circ-\frac{1}{2}\sin 20^\circ)= \frac{1}{2}(\frac{1}{2}\sin 40^\circ-\frac{1}{2}\sin 20^\circ)= \frac{1}{4}(\sin 40^\circ-\sin 20^\circ).$$