Let $G$ be the centroid of triangle $ABC$. Find the biggest $\alpha$ such that there exists a triangle for which there are at least three angles among $\angle GAB, \angle GAC, \angle GBA, \angle GBC, \angle GCA, \angle GCB$ which are $\geq \alpha$.
Problem
Source: IZhO 2024, P6
Tags: geometry
Complete_quadrilateral
11.01.2024 05:48
A little modified version of my solution from contest (I moved $B'$ on $AC$ to get that those $2$ angles are equal even though it's not necessary)
WLOG $a\geq b\geq c$. By chasing some lengths using sine theorems we can get that the biggest angle is $\angle GAB$, and that the second and third are $\angle GAC$ and $\angle GBA$ in some order. Let $B'$ be the midpoint of $AC$. We have that $AB'\geq BB'\sin \angle B'AB\geq BB' \sin \alpha$ and that $GB'\geq AB' \sin \angle GAB'\geq AB' \sin \alpha$, so $GB'\geq BB' \sin^2\alpha$, and since $GB'=\frac{BB'}{3}$we have $\sin^2\alpha\leq \frac{1}{3}$, so $\alpha\leq\arcsin \frac{1}{\sqrt{3}}$. This is achieved for triangle in which $\angle A=\frac{\pi}{2}$ and $\angle C=\arcsin \frac{1}{\sqrt{3}}$.
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Complete_quadrilateral
11.01.2024 05:56
Is there some more inspiring solution? I hope that there is but the final answer doesn't give me much hope for that.
Assassino9931
11.01.2024 06:28
@above congratulations on not being misled by the difficulty on the problem due to its position! This problem was proposed, as one would expect, by Nairi Sedrakyan from Armenia. I hoped that this problem from Bulgaria 1997 https://artofproblemsolving.com/community/c6h601002p3567951 would be useful for the meme, but at least it seems similarly flavoured.
gvole
11.01.2024 10:54
You can smooth to get $\alpha = \angle GBA=\angle GAC$, then calculate that this gives $a^2+c^2=2b^2$. Plugging into formula for $\angle GAC$, you get $\sin^2\alpha=\frac{\sin^2\angle BAC}3$. Set $\angle BAC = 90^{\circ}$ and you're done...
Assassino9931
17.01.2024 21:32
Presenting a simplified version of the crazy superhero cosmic solution of 3BUL12 (Angel Hristov) from the competition. A downside is that it is easy to mess up a lot of details when rushing on white paper and hence barely get half of the points even after a literally 2.5 hour coordination! At the end he was very proud and fortunately not affected from the materialistic side by the medal cut-offs. We thank all people involved in correcting the details in the solution.
Without loss of generality let $a \leq b \leq c$, where as usual $a = BC$, $b = CA$, $c = AB$. Denote $\angle 1 = \angle GAC$, $\angle 2 = \angle GAB$, $\angle 3 = \angle GBA$, $\angle 4 = \angle GBC$, $\angle 5 = \angle BCG$, $\angle 6 = \angle GCA$. The parallelogram $ACBC_0$, particularly the triangle $CBC_0$, shows that $\angle 5 \geq \angle 6$. Analogously we get $\angle 1 \geq \angle 2$ and $\angle 3 \geq \angle 4$. Since $\angle ACB$ is largest in $ABC$, we obtain that angles $1$, $2$, $3$, $4$, $6$ are acute. Now since $GB \leq GA$ (follows directly by the median formula) and $S_{AGC} = S_{BGC}$, we obtain $\frac{\sin \angle 1}{\sin \angle 4} = \frac{a}{b} \cdot \frac{GB}{GA} \leq 1$, thus $\angle 1 \leq \angle 4$. Similarly, $\angle 3 \leq \angle 6$. From $GB \geq GC$ (again via the median formula) it follows that $\angle 5 \geq \angle 4$. Therefore the three largest angles are $4$, $5$ and $6$, with $5$ being at least as large as $4$ and $6$.
Fix $A$ and $C$ and a value $\alpha_1$ of the parameter $\alpha$. All introduced points afterwards are treated to be in the same halfplane $H$ with respect ot $AC$. Let $M$ be the midpoint of $AC$. Let $\omega_1$ be the circle with center $C$ and radius $CA$, let $\omega_2$ be the circle with center $A$ and radius $AC$, let $C' = AC \cap \omega_1$ be the antipode of $A$ and let $Z \in \omega_1$ be such that $\angle AC'Z = \alpha_1$. Let $\omega_3$ be the circle through $C$ and $M$ such that the arc $\widehat{CM}$ has measure $2\alpha_1$ - equivalently, any inscribed angle in $H$ with middle vertex on $\omega_3$ has measure $\alpha_1$. Then $\angle 4 = \angle CBM \geq \alpha_1$ if and only if $B$ is in the interior of $\omega_3$ or on $\omega_3$ itself. On the other hand, if $N$ is the midpoint of $AB$, then $\angle 6 = \angle ACN = \angle AC'B$ by the midsegment $CN$ in triangle $AC'B$, so $\angle 6 \geq \alpha_1$ if and only if $B$ lies in the sector of $\omega_1$ enclosed by the chord $C'Z$ and the arc $\widehat{C'Z}$. (Note in particular that $C'Z$ and $\omega_3$ should have at least one common point for $B$ to exist.) Moreover, $AB \geq AC \geq BC$ holds if and only if $B$ lies inside or on the boundary of $\omega_1$ and outside or on the boundary of $\omega_2$.
Let $\alpha_0$ be the value of $\alpha$ such that $C'Z$ touches $\omega_3$ - we claim that $\alpha_1 \leq \alpha_0$ for any value $\alpha_1$ of the parameter $\alpha$. Suppose the contrary, i.e. that there is a value $\alpha_1$ of the parameter $\alpha$ which strictly exceeds $\alpha_0$. Then the portion of $\omega_3$ in the halfplane we are working at shrinks (since the measure of the inscribed angles is now larger), while $Z$ gets closer to $C'$ in counterclockwise direction (since $\angle AC'Z$ is now larger) - hence the intersection of $C'Z$ and $\omega_3$ which we require from Step $2$ is no longer possible, contradiction.
Now it remains to compute $\alpha_0$, which is also the value of $\angle 4$ and $\angle 6$, and to really observe that $AB \geq AC \geq BC$ so that $\angle 5$ is indeed at least as large, by Step $1$. The tangency implies $\angle C'BC = \angle BMC$, so in triangle $BMC'$ we get $\angle CB'M = 90^{\circ}$ and hence $\angle C'BM = \angle BMC = 90^{\circ} - \alpha_0$. Hence from triangle $BMC$ we obtain $\angle BCM = 90^{\circ}$. Now if $AM = MC = x$, then $CC' = 2x$, so $BC = \sqrt{CC' \cdot CM} = x\sqrt{2}$ and $AB > AC > BC$, as well as $BM = \sqrt{CM^2 + B'C^2} = x\sqrt{3}$ and in conclusion $\alpha_0 = \arcsin\left(\frac{\sqrt{3}}{3}\right)$.