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Question Number 209637 by mr W last updated on 17/Jul/24

Commented by mr W last updated on 17/Jul/24

find minimum of AB+BC+CD=?

findminimumofAB+BC+CD=?

Commented by Frix last updated on 17/Jul/24

I get 2(√7)

Iget27

Commented by mr W last updated on 17/Jul/24

that′s correct sir!  your method?

thatscorrectsir!yourmethod?

Commented by Frix last updated on 17/Jul/24

I will post it later.

Iwillpostitlater.

Commented by Frix last updated on 19/Jul/24

I “fold” it  A= ((2),(0) )  B∈y=xtan 20°  C∈y=xtan 40°  D∈y=xtan 60° ∧ ∣OD∣=6 ⇒ D= ((3),((3(√3))) )  The shortest path is  ∣AD∣=2(√7)

IfolditA=(20)By=xtan20°Cy=xtan40°Dy=xtan60°OD∣=6D=(333)TheshortestpathisAD∣=27

Commented by mr W last updated on 19/Jul/24

thanks sir!

thankssir!

Commented by Frix last updated on 19/Jul/24

But this method doesn′t work for n points  with (n−1)α≥180°

Butthismethoddoesntworkfornpointswith(n1)α180°

Commented by mr W last updated on 19/Jul/24

for (n−1)α>180° there exists no  minimum at all.

for(n1)α>180°thereexistsnominimumatall.

Answered by mr W last updated on 19/Jul/24

Commented by mr W last updated on 19/Jul/24

let′s treat OA, OD as mirrors.  (AB+BC+CD)_(min) =A′D′  =(√(2^2 +6^2 −2×2×6 cos (3×20°)))  =2(√7)

letstreatOA,ODasmirrors.(AB+BC+CD)min=AD=22+622×2×6cos(3×20°)=27

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