For reference:The angle B of a triangle ABC measures 60°. The AN and CM medians are plotted. The radius of the circle inscribed in the MBNG quadrilateral (G is centroid(barycenter) of ABC) measures $\sqrt3$ . Calculate AC.
My progress
$\triangle BED: \\sen30 = \frac{\sqrt3}{BD}\therefore BD = 2\sqrt3\\ cos 30 = \frac{BE}{BD}\rightarrow \frac{\sqrt3}{2}=\frac{BE}{2\sqrt3}\therefore BE = 3\implies BN = 3+\sqrt3\\ \triangle BNG:cos 30 = \frac{BN}{BG}\rightarrow \frac{\sqrt3}{2}=\frac{3+\sqrt3}{BG}\rightarrow BG = \frac{6+2\sqrt3}{\sqrt3}=2\sqrt3+2\\ BG = \frac{2BP}{3}\rightarrow BP = 3\sqrt3 + 3\\ \triangle BPC: tg30 = \frac{PC}{BP}\rightarrow \frac{\sqrt3}{3} = \frac{PC}{3\sqrt3+3} \implies \boxed{ PC = 3+\sqrt3}\\ \therefore \boxed{\color{red}AC = 2(3+\sqrt3) = 6+2\sqrt3}$
My question...only the equilateral triangle meets the conditions? Why if the quadrilateral is indescribable $\measuredangle MGN=120^o$
$\endgroup$ 93 Answers
$\begingroup$Given $ \small MBNG$ is a tangential quadrilateral, it is easy to see that,
$ \small BN + MG = BM + GN$
[How? $ \small BE + EN = BF + NH, MK + KG = MF + GH$. In fact there is a theorem called Pitot Theorem that states the same.]
So if $ \small AN = d, CM = e$, we have $ \small MG = \frac{e}{3}, \small GN = \frac{d}{3}$ and we rewrite $\small BN + MG = BM + GN$ as,
$\frac{c}{2} + \frac{d}{3} = \frac{a}{2} + \frac{e}{3}$
$ d + \frac{3c}{2} = e + \frac{3a}{2} \tag1$
Now given the circle is also the incircle of $\triangle BCM$ and $\triangle ABN$, AND the area of both $\triangle ABN$ and $\triangle BCM$ are same, which is half the area of $\triangle ABC$, we conclude they must have the same perimeter.
$c + d + \frac{a}{2} = a + e + \frac{c}{2}$
$d + \frac{c}{2} = e + \frac{a}{2} \tag2$
Subtracting $(2)$ from $(1)$, $c = a$.
As $a = c ~$ and angle between the sides is $60^\circ$, $\triangle ABC$ must be an equilateral triangle.
So we easily find that,
$ \small AC = BC = 2 BN = 2 (BE + EN) = 2 (3 + \sqrt3)$
$\endgroup$ $\begingroup$Since $BNGM$ has an inscribed circle, it must follow pitot theorem i.e. the opposite sides must sum to the same value. Let $x,y,d$ be the lengths of $BM, BN, BG$ respectively.
By law of cosines, we have$$MG=\sqrt{x^2+d^2-xd\sqrt{3}}$$$$NG=\sqrt{y^2+d^2-yd\sqrt{3}}$$By pitot theorem,$$x+\sqrt{y^2+d^2-yd\sqrt{3}}=y+\sqrt{x^2+d^2-xd\sqrt{3}}$$$$x-\sqrt{x^2+d^2-xd\sqrt{3}}=y-\sqrt{y^2+d^2-yd\sqrt{3}}$$Consider the function $f(x)=x-\sqrt{x^2+a^2-xa\sqrt{3}}$ for some real parameter $a$. We have that$$f'(x)=1-\frac{2x-a\sqrt{3}}{2\sqrt{x^2+a^2-xa\sqrt{3}}}$$If $u=\frac{x}{a}$, this is equivalent to$$1-\frac{2u-\sqrt{3}}{2\sqrt{u^2-u\sqrt{3}+1}}$$$$=1-\sqrt{\frac{4u^2-4u\sqrt{3}u+3}{4u^2-4u\sqrt{3}+4}}$$$$=1-\sqrt{1-\frac{1}{4u^2-4u\sqrt{3}+4}}$$Since $u\in\mathbb{R}$, we have $4u^2-4u\sqrt{3}+4\in [1,\infty)$, which implies$$\frac{1}{4u^2-4u\sqrt{3}+4}\in (0,1]$$$$\implies 1-\frac{1}{4u^2-4u\sqrt{3}+4}\in [0,1)$$$$\implies \sqrt{1-\frac{1}{4u^2-4u\sqrt{3}+4}}\in [0,1)$$$$\implies 1-\sqrt{1-\frac{1}{4u^2-4u\sqrt{3}+4}}\in (0,1]$$Hence, $f'(x)$ is positive$~\forall x\in\mathbb{R}$, which means $f(x)$ is monotonically increasing.
This means that the only solution to$$f(x)=f(y)$$is when $x=y$. So $BM=BN$. From there it follows that the original triangle is equilateral. The rest of the answer follows.
$\endgroup$ 0 $\begingroup$Since $$BM+GN=BN+MG,$$ in the standard notation we obtain:$$\frac{c}{2}+\frac{1}{6}\sqrt{2b^2+2c^2-a^2}=\frac{a}{2}+\frac{1}{6}\sqrt{2a^2+2b^2-c^2}$$ or$$3(a-c)+\sqrt{2a^2+2b^2-c^2}-\sqrt{2b^2+2c^2-a^2}=0$$ or$$(a-c)\left(1+\frac{a+c}{\sqrt{2a^2+2b^2-c^2}+\sqrt{2b^2+2c^2-a^2}}\right)=0,$$ which gives $a=c$ and our triangle is an equilateral triangle.
Thus, $$AC=2BN=2(EN+BE)=2(\sqrt3+3).$$
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