Why is this angle always less than 49ドル°$?

Question

I was drawing this configuration in GeoGebra, repeating it dozens of times, always considering any triangle $ABC$ with the centroid $G$, while maintaining the 25ドル°$ angle.
An observation then occurred to me:
The angle $x$, in this figure, is always strictly less than 49ドル°$!

How is this possible? Is there any proof of this observation?

triangle

• dfnu: Since you like special configurations (math.stackexchange.com/questions/5110093/…) I also propose a case where the two angles $\angle ACG$ and $\angle CBG$ are equal and at the same time $\angle CBG$ is maximized. It turns out that the only solution is when $ABC$ is right angled and has two perpendicular medians.
• Jamil Sanjakdar: Thanks..very nice.

Answer 1

Angle $x$ is in fact strictly less than 49ドル^\circ$, but juuust barely. The maximal $x$ is 48ドル.995050\ldots^\circ$.

Here's a quick-and-dirty justification:

Defining $g:=|GC|$, $b:=|CA|$, $\theta=\angle GCA$ (for generality), we can coordinatize thusly: $$G = (0,0) \qquad C = (-g,0) \qquad A = C + b,円(\cos\theta,\sin\theta) \qquad B = -(A+C)$$ Then, with $x:=\angle CBG$, we have $$\begin{align} \cos x &= \frac{(C-B)\cdot(G-B)}{|BC||BG|} \\[4pt] \quad\to\quad \sin^2x &= 1-\frac{(\;(C-B)\cdot(G-B)\;)^2}{(\;(C-B)\cdot(C-B)\;)\;(\;(G-B)\cdot(G-B)\;)} \\[4pt] &=\frac{b^2 g^2 \sin^2\theta}{(b^2 + 9 g^2 - 6 b g \cos\theta) (b^2 + 4 g^2 - 4 b g \cos\theta)} \tag{$\star$} \end{align}$$ For fixed $\theta$ and $g$, we have that $\sin^2x$ is a function of $b$. We can find its maximum by considering its $b$-respecting derivative (calculated with the assistance of Mathematica) vanishes: $$\frac{2 b g^2\sin\theta \; (6 g^2-b^2)\;(b^2 + 6 g^2 - 5 b g \cos\theta)}{(b^2 + 9 g^2 - 6 b g \cos\theta)^2 \;(b^2 + 4 g^2 - 4 b g \cos\theta)^2} = 0$$ (The denominator is strictly positive, so we don't have to worry about where the derivative is undefined.)

Casually ignoring all factors but 6ドルg^2-b^2$ as degenerate cases, we conclude that $\sin^2x$ is maximized when $b=\sqrt6,円g$. (Interestingly, this is independent of $\theta$. If there's any justice, a fact this nice should have an elegant geometric proof.) As it happens, the resulting relation can be written (for non-negative $x$) $$\sin^2 x \leq \frac{\sin^2\theta}{\left(5 - 2\sqrt6\cos\theta\right)^2} \quad\to\quad x \;\leq\; \arcsin \frac{\sin\theta}{5 - 2\sqrt6\cos\theta}$$ When $\theta=25^\circ$, the right-hand side is 48ドル.995050498065\ldots^\circ$. $\square$

• Jamil Sanjakdar: ..Excellent... Very nice...[+1] .
• Blue: @JamilSanjakdar: Thanks! ... Given the simplicity of the final relation, I suspect there's a clean geometric path to it.
• dfnu: There is justice in this world (that of maths of course... In a more general setting, I wouldn't be so sure).
• Blue: @dfnu: Good catch. I should've left $x$ wrapped in sine, and then made the specific deduction for $\theta=25^\circ$.
• Ataulfo: @Blue: Thank you very much.

Answer 2

A complete geometric proof

Assume, WLOG, that $\overline{BG} = 2$. Starting from this segment, produce it to $M$, so that $GM \cong \frac12 BG$. Clearly $M$ is the midpoint of $AC$. Since we want a fixed angle $\measuredangle MCG$, we must take $C$ on a circlular arc $\gamma$ with center on the axis of segment $GM$.

Angle $\angle GBC$ is maximized if we choose $C$ so that $BC$ is tangent to the circlular arc $\gamma$ (see figure below, where, additionaly, a non optimal solution $A'BC'$ is shown).

enter image description here

Since $\angle BCG \cong \angle BMC$ (Inscribed Angle Theorem), by AAA criterion we have $BCG \sim BCM$. Thus, recalling that $\overline{BG} = 2$ and $\overline{BM} = 3$, we get $$\frac{\overline{BM}}{\overline{BC}}=\frac{\overline{BC}}{\overline{BG}},$$ i.e. $$\overline{BC}^2 = 6.$$ The same similarity now gives us $$\frac{\overline{CM}}{\overline{CG}} = \frac3{\sqrt 6}.$$ Since $AC \cong 2 CM$ we immediatly get $$\boxed{\overline{AC} = \sqrt 6\ \overline{CG}},\tag{1}\label{1}$$ as per Blue's answer.

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For the sake of completeness: without using \eqref{1} explicitely, we can take advantage directly of tangency condition discussed above, and express $x=\measuredangle GBC$ in terms of $\theta = \measuredangle GCA$ as follows.

First note that $2ドル \measuredangle GCB + \theta +x = 180^\circ,$$ using the above similarity, and the interior angles of $BMC$. Hence $$\measuredangle GCB = 90^\circ-\frac{\theta +x}2.$$ Now we get an equation by expressing $\overline{GC}$ via sine rule in both $GMC$ and $BCG$, whence $$\cos^2 \frac{\theta +x}2 = 2\sin x \sin \theta,$$ or, equivalently, $5ドル \sin x \sin \theta-\cos x \cos \theta =1.$$ With the change of variable $\tan\frac{x}2\mapsto t$ we get to $$(1−\cos a)t^2−(10\sin a) t+1+\cos a=0,$$ which has the "nice" solution $$t = \frac{(5\pm 2\sqrt 6)\sin \theta}{1-\cos \theta}.$$ Of course, for each $\theta\in(0,180^\circ)$ we only have one solution in $x$. So which sign should we take?

By geometric constraints, $$x < 180^\circ-\theta,$$ and this constraint is satisfied only if we use the minus sign in the above expression. Therefore, the correct solution, for the entire range of $\theta \in (0,180^\circ)$ is

$$\boxed{x_{max} = 2\arctan\left[\frac{(5- 2\sqrt 6)\sin \theta}{1-\cos \theta}\right]}.$$

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Later edit

A long comment to the comment below by Blue: as he pointed out the relation above can be written as $$\tan \frac{x}2\tan\frac{\theta}2 = K,\tag{3}\label{3}$$ for some constant $K$ that only depends on the ratio $$R = \sqrt{\frac{\overline{BG}}{\overline{BM}}}=\frac{\overline{CM}}{\overline{CG}}.$$ ($x$ here represents again the maximum value of $\measuredangle GBC$, given $\measuredangle ACG=\theta$.

enter image description here

To see this, consider the figure above, where we added the projection $H$ of $C$ on $BM$. Recall that $\measuredangle HCM = \frac{\theta+x}2$. Therefore we can write \begin{eqnarray} R &=& \frac{\overline{CM}}{\overline{CG}}\\ &=&\frac{\overline{CM}}{\overline{CH}}\cdot \frac{\overline{CH}}{\overline{CG}}\\ &=& \frac{\cos\left(\frac{x}2-\frac{\theta}2\right)}{\cos\left(\frac{x}2+\frac{\theta}2\right)}. \end{eqnarray} Note that this expression does not change if $H$ (unlike what is shown in the figure) lies between $G$ and $M$. Using addition formulas and dividing numerator and denominator by $\cos \frac{x}2 \cos\frac{\theta}2$ we obtain $$\frac{1+\tan\frac{x}2\tan\frac{\theta}2}{1-\tan\frac{x}2\tan\frac{\theta}2}=R,$$ i.e. $$K=\tan\frac{x}2\tan\frac{\theta}2 = \frac{R-1}{R+1}.$$ In the case proposed by OP, $R =\sqrt{\frac{3}{2}}$, which yields, as expected, $K = 5-2\sqrt 6$.

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An even later edit

Since the problem seems to attract many, I wish to add another insight on the role played by equation \eqref{3}. Consider the figure below, where we added $R$ (the center of circle $\gamma$), $Q$ (the midpoint of $GM$), and $P$ (the intersection between $\gamma$ and the perpendicular bisector of $GM$), so that $CP$ bisects $\angle MCG$. Since we know that $\measuredangle MCH = \frac{\theta + x}2$, we have that $$\measuredangle KCH = \measuredangle KPQ = \frac{x}2,$$ and that $CP \perp BN$, where $BN$ is the bisector of $\angle GBC$. As a result, triangle $BKC$ is isosceles and $\overline{BK} = \overline{BC} = \sqrt 6$, independently of $\theta$.

enter image description here

Therefore, for every $\theta$, $\overline{KQ} = \frac{5}2-\sqrt 6$. We have, then \begin{eqnarray} \frac{5}2-\sqrt 6 &=& \overline{KQ}\\ &=& \overline{PQ}\ \tan\frac{x}2\\ &=&\left(\overline{PT}-\overline{QT}\right)\ \tan\frac{x}2\\ &=&\left(\overline{GT}-\overline{QT}\right)\ \tan\frac{x}2\\ &=&\frac12 \left(\frac1{\sin\theta}-\frac{\cos\theta}{\sin\theta}\right)\ \tan\frac{x}2\\ &=& \frac1{2}\tan\frac{\theta}2 \tan\frac{x}2 , \end{eqnarray} yielding again \eqref{3}.

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Yet another edit

I think one of the most beautiful configurations is the one in which $x$ and $\theta$ are equal.

With the constraint 0ドル< x < 180^\circ$, from \eqref{3} we obtain, in this case, $$\tan\frac{x}{2} = \sqrt{5-2\sqrt 6} = \sqrt3-\sqrt2,$$ i.e. $$\tan x = \frac{1}{\sqrt2}.$$

enter image description here

This is a right-angled triangle with sides proportional to $\overline{BC} = 1$, $\overline{AC} = \sqrt 2$, and $\overline{AB} = \sqrt3$, as depicted in the figure above, where the red markers show the angle whose measure corresponds to $x = \theta = \arctan\frac1{\sqrt2}$.

This is, by the way, the only right-angled triangle having two perpendicular medians. In fact, if we only know this fact and set $\overline{BC} = 1$, by Thales Theorem we have that the projections of the sides $BG$ and $CG$ on $BC$ have measures $\frac23$ and $\frac13$ respectively. By the Geometric Mean Theorem applied to $BCG$, we get that the distance of $G$ from $BC$ is equal to $\sqrt{\frac13\cdot \frac{2}3} =\frac{\sqrt 2}{3}$. We can now apply Thales Theorem "horizontally" and obtain $\overline{AC} = 3\cdot \frac{\sqrt2}3 = \sqrt2$, as desired.

• Blue: Very nice. Hooray for justice! :)
• Blue: It's interesting boxed $x_{max}$ relation can be written $$\tan\frac12\theta\;\tan\frac12x_{max} \;=\; 5 - 2\sqrt6 = \frac{1}{5+2\sqrt{6}}$$ especially given that, with $Z$ the midpoint of $GM,ドル $|BZ|:|BC| = 5:2\sqrt{6}$. I wonder if the relation is "obvious" in light of some general property. ... I've found a tan-half-angle formula for the angle made by opposite sides of a cyclic quad (of which $\triangle CGM$ can be viewed as a limiting case, with two vertices coinciding at $C$), but applying it here isn't particularly elegant.
• dfnu: @Blue Nice! I also noted that, but couldn't come up with much, so far. Note that in general you have $\tan\frac{x}{2}\tan\frac{\theta}{2} = \mbox{constant}$ where the constant depends on the ratio $\overline{BG}:\overline{BM},ドル with $\overline{BC} = \sqrt{\overline{BG}\cdot\overline{BM}}$.
• dfnu: @Blue I analyzed the situation under a slightly different perspective, that emphasizes perhaps a little bit more the role played by the product of the tangents.
• dfnu: @Blue: Another interesting insight is this. Draw bisector of $\angle GCM,ドル and let $K$ be its intersection with $BC$. Let also $Q$ be the midpoint of $GM$. This bisector forms with $CH$ (or with the perpendicular bisetor of $GM$) an angle whose measure is $\frac{x}2$. The relation $$\overline{KQ}=\frac12 \tan\frac{x}2\tan\frac{\theta}2 = \frac{5}2-\sqrt 6 $$ implies that point $K$ remains fixed, with $\overline{BK} = \sqrt 6,ドル for all possible configurations. Note in particular that $C \to K,ドル as $\theta \to 180^\circ$ (or as $x\to 0$).

Answer 3

For a geometric approach, let circle with radius $BC$ intersect medians $CD$, $BF$ extended at $H$, $K$. Extend $CA$ to $J$, and let $JK$ and $HC$ extended meet at $L$.

angle B < 50 degrees Since $\angle ACG$ is exterior to triangle $CJL$, then$$\angle ACG=25^o=\angle JLC+\angle CJL$$and$$\angle CJL<25^o$$But$$\angle CJL=\frac{1}{2}\angle CBK$$Therefore$$\angle CBK<50^o$$

• David K: Interesting, but the other solution gives a tighter bound on the angle.

Answer 4

Fix cevian $CGF$. Point $A$ must vary along a line $\ell$ that forms a 25ドル^\circ$ angle with $CG$ at $C$. This uniquely determines $B$ by reflection about $F$, so $B$ varies along a line $\ell'$ parallel to $\ell$ which passes through $C'$ (the reflection of $C$ about $F$).

Therefore, $\angle CBG$ is maximized when $(CBG)$ is tangent to $\ell$' (see here). Let $\alpha=\angle GCB$. It follows that $\angle GBC'=\alpha$, so $\alpha=\frac{155^\circ-x}{2}$. By Ratio Lemma on $\triangle CBC'$ with cevian $BG$, then on $\triangle ACB$ with cevian $CF$, we have $$\frac{1}{2}=\frac{\sin x}{\sin\alpha}\frac{BC}{BC'}=\frac{\sin x}{\sin\alpha}\frac{BC}{AC}=\frac{\sin x}{\sin\alpha}\frac{\sin 25^\circ}{\sin\alpha}$$ Note that $f(x)=\frac{\sin x\sin 25^\circ}{\sin^2\left(\frac{155^\circ-x}{2}\right)}$ is increasing when $x<\frac{\pi}{2}$ since the numerator is increasing and the denominator is decreasing. We evaluate $f(49^\circ)\approx 0.50007$. Thus, the solution to $f(x)=\frac{1}{2}$ must satisfy $x<49^\circ$. enter image description here

• Jamil Sanjakdar: Nice method . Thanks for your answer [+1].

Answer 5

Here is a geometric solution that explicitly finds the optimal value of $x$.

The setup is the same as my other answer—we know that $A$ varies along some line $\ell$, and by reflection across $F$, point $B$ varies along some line $\ell'$. The optimal angle occurs when $(GCF)$ is tangent to line $\ell'$.

By angle chasing, $\angle CDG=\angle DBC'=\angle C'CB=\angle CC'A$, so $C'ADG$ is cyclic (call its circumcircle $\omega$). Power of a Point from $C$ to $\omega$ gives $$CD\cdot CA=CG\cdot CC'\implies 2CD^2=3CG^2,$$ so $\frac{CD}{CG}=\frac{\sqrt{3}}{\sqrt{2}}$. Multiplying by 2ドル$ gives the $\sqrt{6}:1$ ratio found in @Blue's answer, which is independent of $\theta$.

Now for the formula for $x$. More angle chasing gives $\angle DFG=\angle AC'C=\angle HDG$ (black angles) and $\angle FDG=x=\angle DHG$ (red angles), so $\triangle HDG\sim\triangle DFG$. Thus, \begin{equation} \frac{HG}{DG}=\frac{DG}{FG}\implies \frac{2DG^2}{4FG^2}=\frac{HG}{2FG}\implies \left(\frac{\sqrt{2}DG}{CG}\right)^2=\frac{HG}{CG}.\tag{*} \end{equation}

Law of Cosines on $\triangle CDG$ gives $DG=5-2\sqrt{6}\cos\theta$ when $CG$ is normalized to $\sqrt{2}$. The LHS of (*) becomes 5ドル-2\sqrt{6}\cos\theta$. Law of Sines on $\triangle HCG$ yields $\frac{HG}{CG}=\frac{\sin \theta}{\sin x}$. Using these relations, we rewrite the lengths in (*) to angles: $$\sin x=\frac{\sin\theta}{5-2\sqrt{6}\cos\theta}$$ When $\theta=25^\circ$, the optimal value of $x$ is $\approx 48.995$, which is less than 49ドル^\circ$.

enter image description here

• Jamil Sanjakdar: Excellent...[+1]

Answer 6

I give here a distinct solution of the precedent ones.

We take the median $CC'$ of the given triangle $\triangle{ABC}$ on the $x$-axis, $C=(0,0),G=(2,0)$ and $C'=(3,0)$. The locus of points $B$ from which angle $\angle{CBG}=49^{\circ}$ is given by the equation $$x^2+y^2-2x-2y\cot(49^{\circ})=0;\space\space y\ge0\hspace2cm(1)$$ This is not other than the locus of inscribed angles of 49ドル^{\circ}$ respect to the chord $\overline{CG}$ in the circle of radius $\sqrt{1+\cot^2(49^{\circ})}$ and center $(1,\cot(49^{\circ})$.

Further we have as given condition the line $y=-\tan(25^{\circ})x$ enter image description here

So the vertex $B$ should be in $(1)$; let $B=(a,b)$ which, because of median $CC'$, gives immediately vertex $A=(6-a,-b)$ which give $\tan(25^{\circ})=\dfrac{b}{6-a}$ which in turn implies that point $B$ is in the line $$y=\tan(25^{\circ})(6-x)\hspace2cm(2)$$ Plotting line $(2)$, it seems to be tangent to the curve $(1)$ in a point, $P$, defined by the intersection of line $(2)$ with its perpendicular $y-\cot(49^{\circ})=\cot(25^{\circ})(x-1)$ passing through the center $(1,\cot(49^{\circ})$ . Calculation gives $P=(1.56007,2.07037)$.

enter image description here

Now if $P=B$ as it seems, then angle $X$ of the problem can be equal to 49ドル^{\circ}$ which would be contrary to the given statement. Putting $P$ in curve $(1)$, we have $$\cot(49^{\circ})=\frac{1.56007^2+2.07037^2-2\times1.566007}{2\times 2.07037}=0.8694364634$$ and $$\cot^{-1}(0.8694364634)=48.9951140793^{\circ}$$ Thus $P\ne B$ which proves that $X\lt49^{\circ}$.

QUESTIONS.-We see that for all angle $X$ such that it is possible the desired configuration for triangle $\triangle{ABC}$, there are just two possibilities, $B_1$ and $B_2$, for the point $B$.

Do the two possible vertices give two different configurations, or is only one valid? Or are two sometimes valid? This can be verified by equating $\overline{BG}=2\overline{GB'}$ and $\overline{BC'}=\overline{C'A}$. For small values of angle $X$, it is not obvious that this holds true, but geometric intuition can be misleading

• Blue: +1. ... If you treat the inscribed 49ドル^\circ$ angle more generally (say, calling it $\theta$), then with a bit of extra work you can solve for the $\theta$ that makes the $B$-line tangent to the circle. More than merely confirming the bound of 49ドル^\circ,ドル this improves it to 48ドル.995\ldots$ as indicated in other answers.
• Ataulfo: @Blue: I assume you're referring to a sort of supremum of the possible values of angle $X$. I wasn't going to publish this answer, after having made a mistake with the construction of the "capable arc." I did so because it was a different method that might interest the O.P. and others drawn to the problem, including yourself as an important member of MSE. But also because I'm still wondering about small values of angle $X$. It wouldn't be a supremum but an infimum (I feel it shouldn't be zero, although perhaps I'm wrong). Thank you very much for your message.
• Jamil Sanjakdar: [+1] .. thank you for your perseverance. I will study the case of the (difficult) case of the minimum of x.
• Ataulfo: @Jamil Sanjakdar: Congratulations. I just suggest this problem thinking in you, who posted the original problem, Remember there are for each (possible) angle exactly two possibilities so you can see if the two $B$ gives solutions; also you have the case of tangency which gives the máximum possible of angle $X$. Also you can explore the inicial data angle distinct of 25ドル$ degrees. Good luck. Regards.