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Question Number 118811 by ajfour last updated on 19/Oct/20

Commented by ajfour last updated on 19/Oct/20

What is the side length of the equal  sided hexagon inscribed within the  triangle ABC as shown?

$${What}\:{is}\:{the}\:{side}\:{length}\:{of}\:{the}\:{equal} \\ $$$${sided}\:{hexagon}\:{inscribed}\:{within}\:{the} \\ $$$${triangle}\:{ABC}\:{as}\:{shown}? \\ $$

Commented by I want to learn more last updated on 20/Oct/20

MrW, please come and solve.

$$\mathrm{MrW},\:\mathrm{please}\:\mathrm{come}\:\mathrm{and}\:\mathrm{solve}.\: \\ $$

Commented by mr W last updated on 21/Oct/20

ajfour sir:  i think there is no unique solution.

$${ajfour}\:{sir}: \\ $$$${i}\:{think}\:{there}\:{is}\:{no}\:{unique}\:{solution}. \\ $$

Commented by ajfour last updated on 21/Oct/20

fine sir, how about extreme values?

$${fine}\:{sir},\:{how}\:{about}\:{extreme}\:{values}? \\ $$

Answered by 1549442205PVT last updated on 20/Oct/20

This problem has infinite many of  solutions in case the triangle  ABC   is equilateral since we can construct  an infinite number of hexagons with  equal sides and side length is different

$$\mathrm{This}\:\mathrm{problem}\:\mathrm{has}\:\mathrm{infinite}\:\mathrm{many}\:\mathrm{of} \\ $$$$\mathrm{solutions}\:\mathrm{in}\:\mathrm{case}\:\mathrm{the}\:\mathrm{triangle}\:\:\mathrm{ABC}\: \\ $$$$\mathrm{is}\:\mathrm{equilateral}\:\mathrm{since}\:\mathrm{we}\:\mathrm{can}\:\mathrm{construct} \\ $$$$\mathrm{an}\:\mathrm{infinite}\:\mathrm{number}\:\mathrm{of}\:\mathrm{hexagons}\:\mathrm{with} \\ $$$$\mathrm{equal}\:\mathrm{sides}\:\mathrm{and}\:\mathrm{side}\:\mathrm{length}\:\mathrm{is}\:\mathrm{different} \\ $$

Commented by ajfour last updated on 20/Oct/20

no, you dont follow;  given that sides  of triangle are AB=c, BC=a, CA=b ;  find side s of the hexagon in terms  of a, b, c.

$${no},\:{you}\:{dont}\:{follow};\:\:{given}\:{that}\:{sides} \\ $$$${of}\:{triangle}\:{are}\:{AB}={c},\:{BC}={a},\:{CA}={b}\:; \\ $$$${find}\:{side}\:\boldsymbol{{s}}\:{of}\:{the}\:{hexagon}\:{in}\:{terms} \\ $$$${of}\:{a},\:{b},\:{c}. \\ $$

Commented by 1549442205PVT last updated on 20/Oct/20

Commented by 1549442205PVT last updated on 20/Oct/20

Put AD=m,AE=n,CM=p,CH=q,ME=t  BF=r,BG=s,BC=a,AC=b,AB=c  From the cosine theorem we have :   { ((t^2 =m^2 +n^2 −2mncosA(1))),((t^2 =p^2 +q^2 −2pqcosC(2))),((t^2 =r^2 +s^2 −2rscosB(3))),((n+t+p=b(4))),((q+t+r=a(5))),((m+t+s=c(6))) :}  Thus,we have the system of 6 equations  with 6 unknowns where a,b,c,cosA,  cosB,cosC are parameters which   are independent together and now  we need  have to solve it.

$$\mathrm{Put}\:\mathrm{AD}=\mathrm{m},\mathrm{AE}=\mathrm{n},\mathrm{CM}=\mathrm{p},\mathrm{CH}=\mathrm{q},\mathrm{ME}=\mathrm{t} \\ $$$$\mathrm{BF}=\mathrm{r},\mathrm{BG}=\mathrm{s},\mathrm{BC}=\mathrm{a},\mathrm{AC}=\mathrm{b},\mathrm{AB}=\mathrm{c} \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{cosine}\:\mathrm{theorem}\:\mathrm{we}\:\mathrm{have}\:: \\ $$$$\begin{cases}{\mathrm{t}^{\mathrm{2}} =\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} −\mathrm{2mncosA}\left(\mathrm{1}\right)}\\{\mathrm{t}^{\mathrm{2}} =\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} −\mathrm{2pqcosC}\left(\mathrm{2}\right)}\\{\mathrm{t}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} −\mathrm{2rscosB}\left(\mathrm{3}\right)}\\{\mathrm{n}+\mathrm{t}+\mathrm{p}=\mathrm{b}\left(\mathrm{4}\right)}\\{\mathrm{q}+\mathrm{t}+\mathrm{r}=\mathrm{a}\left(\mathrm{5}\right)}\\{\mathrm{m}+\mathrm{t}+\mathrm{s}=\mathrm{c}\left(\mathrm{6}\right)}\end{cases} \\ $$$$\mathrm{Thus},\mathrm{we}\:\mathrm{have}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{6}\:\mathrm{equations} \\ $$$$\mathrm{with}\:\mathrm{6}\:\mathrm{unknowns}\:\mathrm{where}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{cosA}, \\ $$$$\mathrm{cosB},\mathrm{cosC}\:\mathrm{are}\:\mathrm{parameters}\:\mathrm{which}\: \\ $$$$\mathrm{are}\:\mathrm{independent}\:\mathrm{together}\:\mathrm{and}\:\mathrm{now} \\ $$$$\mathrm{we}\:\mathrm{need}\:\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}. \\ $$$$ \\ $$

Commented by I want to learn more last updated on 20/Oct/20

Sir, 1549442205PVT  which app you use to draw.

$$\mathrm{Sir},\:\mathrm{1549442205PVT}\:\:\mathrm{which}\:\mathrm{app}\:\mathrm{you}\:\mathrm{use}\:\mathrm{to}\:\mathrm{draw}. \\ $$

Commented by 1549442205PVT last updated on 21/Oct/20

Geogebra,Drawing of tinkutara done this

$$\mathrm{Geogebra},\mathrm{Drawing}\:\mathrm{of}\:\mathrm{tinkutara}\:\mathrm{done}\:\mathrm{this} \\ $$

Commented by mr W last updated on 21/Oct/20

6 equations, but 7 unknowns,  does it mean that there is no unique  solution?

$$\mathrm{6}\:{equations},\:{but}\:\mathrm{7}\:{unknowns}, \\ $$$${does}\:{it}\:{mean}\:{that}\:{there}\:{is}\:{no}\:{unique} \\ $$$${solution}? \\ $$

Commented by 1549442205PVT last updated on 25/Oct/20

Yes,Sir is right.I  also have such thinkings

$$\mathrm{Yes},\mathrm{Sir}\:\mathrm{is}\:\mathrm{right}.\mathrm{I}\:\:\mathrm{also}\:\mathrm{have}\:\mathrm{such}\:\mathrm{thinkings} \\ $$

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