A scalene triangle $ABC$ was drawn, and Elmo marked its incenter $I$, Feuerbach point $X$, and Nagel point $N$. Sadly, after taking the abcdEfghijkLMnOpqrstuvwxyz, Elmo lost the triangle $ABC$. Can Elmo use only a ruler and compass to reconstruct the triangle? Proposed by Karn Chutinan
Problem
Source: 2024 RELMO 7
Tags: geometry, incenter, Feuerbach Point, Nagel point, constructive geometry
YaoAOPS
28.06.2024 09:20
Fakesolve.
pieater314159
28.06.2024 10:35
The answer is no. We will show this by showing that, if this were possible, then Elmo would be able to construct a regular $13$-gon. It is well known that this cannot be constructed with a ruler and compass. (We will do a bit more computation than is necessary, partly to avoid some things that feel like handwaving and partly to give what I think are some really amusing complex-number calculations.) Construct a circle $\gamma$ with center $I$ and marked point $P_0$. We associate the plane Elmo can draw in with the complex plane by letting $I=0$ and $P_0=1$. Say that Elmo can construct a number $x\in\mathbb C$ if he can construct the point which corresponds to $x$ under this association. It is well-known that we can "solve quadratics": if we can construct $a$, $b$, and $c$, then we can construct the roots of $ax^2+bx+c$. Construct $u:=\frac{\sqrt{13}-1}2$, one of the roots of $x^2+x-3$.
We compute
\[s_1+s_2=\omega+\omega^3+\omega^4+\omega^9+\omega^{10}+\omega^{12}=\frac12\sum_{n=1}^{12}e^{\frac{2\pi i}{13}n^2}=\frac{\sqrt{13}-1}2=u,\]by the theory of quadratic Gauss sums, and
\[s_1s_2=(\omega+\omega^3+\omega^9)(\omega^4+\omega^{10}+\omega^{12})=3+\omega^2+\omega^5+\omega^6+\omega^7+\omega^8+\omega^{11}=3+(-1-u)=2-u.\]So, $s_1$ and $s_2$ are the roots of $x^2-ux+(2-u)$, as desired.
Let $s_3=1$. Construct $X=\frac{s_2}{s_1}$ and $N=\frac{2s_2^2+2s_1s_3}{s_1s_2-s_3}$. Construct the triangle $ABC$ with incenter $I$, Feuerbach point $X$, and Nagel point $N$. Note that $\gamma$ contains $X$, since $s_2=\overline{s_1}$ so $|X|=1$, and that $\gamma$ is centered at $I$, so the incircle of $\triangle ABC$ is $\gamma$ Construct the tangency points of the sides of $\triangle ABC$ to the incircle $\gamma$. Let $P_1$ be the closest of these points to $P_0$. Construct $P_n=P_1^n$ for $n\in\{0,1,\ldots,12\}$. We claim that the points $\{P_0,\ldots,P_{12}\}$ form a regular $13$-gon. It suffices to show that the tangency points of the sides of $\triangle ABC$ to its incircle are $d:=\omega$, $e:=\omega^3$, and $f:=\omega^9$. Since the ability to reconstruct $\triangle ABC$ implies that it is unique, it suffices to show that the triangle $ABC$ defined by intersecting the tangents to $\gamma$ at $d$, $e$, and $f$ has incenter $I$, Feuerbach point $X$, and Nagel point $N$. Note that $s_1=d+e+f$, $s_2=de+ef+fd$, and $s_3=def$. The fact that the incenter is $I$ follows from the fact that the triangle formed by $\omega$, $\omega^3$, and $\omega^9$ is acute, so $\gamma$ is the incircle of this triangle (and not one of the three excircles).
For the Feuerbach point, it suffices to show that $s_2/s_1$ is on the nine-point circle of $\triangle ABC$, since we already know from the previous point and the fact that $|s_2|=|s_3\overline{s_1}|=|s_1|$ that it lies on the incircle. That is, we must show that
\[\left\{\frac{de+ef+fd}{d+e+f},\frac{de}{d+e}+\frac{df}{d+f},\frac{ef}{e+f}+\frac{ed}{e+d},\frac{fd}{f+d}+\frac{fe}{f+e}\right\}\]are concyclic. To do this, we express
\[\frac{de}{d+e}+\frac{df}{d+f}-\frac{de+ef+fd}{d+e+f}=k\frac{ef(d^2-ef)}{(d+e)(d+f)(d+e+f)}\]for some constant $k$. (In fact, $k=1$, but we won't need that.) To show this, note that, when multiplied by $(d+e)(d+f)(d+e+f)$, both sides are homogeneous polynomials of degree $4$ in $d$, $e$, and $f$ which vanish when $e=0$, $f=0$, or $ef=d^2$. The vanishing when $e=0$ or $f=0$ is easy, and the vanishing when $ef=d^2$ is because in this case $\frac{s_2}{s_1}=d$, and the statement is equivalent to "when $AB=AC$, the midpoint of $BC$ and the tangency point of the incircle of $\triangle ABC$ with $BC$ coincide." Now, the right side can be expressed as
\[k\frac{def}{(d+e)(d+f)(e+f)(d+e+f)}\cdot\frac{(e+f)(d^2-ef)}d=k\frac{s_3\left(s_2-\frac{ef}ds_1\right)}{s_1(s_1s_2-s_3)}.\]So, the circle through the three midpoints of $AB$, $BC$, and $CA$ can be parametrized as
\[\left\{\frac{s_2}{s_1}+\frac{ks_3(s_2-us_1)}{s_1(s_1s_2-s_3)}\colon |u|=1.\right\}\](This is a Möbius transformation of the unit circle, and thus parametrizes a circle.) At $u=s_2/s_1$, this equals $s_2/s_1$, and so $s_2/s_1$ lies on this circle, as desired.
For the Nagel point, it suffices to show that $N$, $-d$, and $\frac{2ef}{e+f}$ are collinear. This is because the homothety at $A$ which sends the tangent to $\gamma$ at $d$ to the tangent to $\gamma$ at $-d$ sends the tangency point $D'$ of the $A$-excircle to $-d$, so the Nagel point lies on the line between $-d$ and $\frac{2ef}{e+f}$. So, it we can show that $N$ lies on this line, we have by symmetry that $N$ lies on all three lines defining the Nagel point. We compute
\begin{align*}
N+d
&=\frac{2s_2^2+2s_1s_3+d(s_1s_2-s_3)}{s_1s_2-s_3}\\
&=\frac{(s_2+ef)(2s_2-2ef+ds_1)+2e^2f^2-defs_1+2s_1s_3-ds_3}{(d+e)(d+f)(f+d)}\\
&=\frac{(s_2+ef)(d^2+3de+3df)+2e^2f^2+def(e+f)}{(d+e)(d+f)(f+d)}\\
&=\frac{(s_2+ef)(d^2+3de+3df+ef)}{(d+e)(d+f)(f+d)}.
\end{align*}We have $\frac{2ef}{e+f}+d=\frac{s_2+ef}{e+f}$. So
\[\frac{N+d}{\frac{2ef}{e+f}+d}=\frac{d^2+3de+3df+ef}{d^2+de+df+ef}=1+\frac{2d(e+f)}{(d+e)(d+f)}\in\mathbb R,\]as desired. noteThe somewhat peculiar algebraic manipulation here is inspired by first computing the second difference, and noting that $s_2+ef$ has to be a factor of the numerator of $N+d$ for the ratio to be real. So, given the symmetric form, we do our best to factor out an $s_2+ef$, and collect the remainder into an agreeable form at the end. Of course, unless you happen to know this formula (why would you?), you probably have to intersect some lines to find it, and you might as well just write that up instead. If you're like me and have the bad habit of first doing computations on scratch paper, this write-up method is pretty efficient.
popop614
28.06.2024 18:15
Hi. I wrote this mess :3 Note that we can construct $G$ easily as $2IG = GNa$. Then notice that if we reflect $Na$ over $X$ to $Na'$, from homothety at $G$ we obtain that $HNa' \parallel IX$ since $N_9$ lies on $IX$. Moreover $H$ lies on the feuerbach hyperbola, and hence $H$ is constructable by converse Pascal. Thus $O$ is also constructable, and we obtain the circumcircle by reflecting $H$ across $X$. However we are now trying to intersect a circle with a hyperbola given one point on both; chances are we get an irreducible cubic if we pick rational coordinates for $X$, $I$, $Na$. Set $X = (0,0)$, $I = (1, 4)$, and $Na = (2,2)$. After computation, we must solve the system \[ \begin{cases} \left(x-\frac{7}{4}\right)^{2}+\left(y-1\right)^{2}=\left(\frac{81}{16}+81\right) \\ y = \frac{4}{x}. \end{cases} \]However when we substitute $y = \frac{4}{x}$, we must be able to solve \[ 8(2x + 1)(x^3 - 4x^2 - 80x + 32) = 0. \]One can check that the first factor corresponds to the reflection of $H$ over $X$. The other three must be the vertices of the triangle (this is easily checkable), but the polynomial $x^3 - 4x^2 - 80x + 32$ is irreducible. Bad.