Question-118811

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

Commented by ajfour last updated on 19/Oct/20

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

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

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

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

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

Commented by ajfour last updated on 21/Oct/20

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

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

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

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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