Question

Length `AC+C C_1+C_1C_2+C_2C_3+C_3C_4+...=?`

Given:
`AC=x`
`/_ACB=90^o`
`/_BAC=60^o`
`C C_1_|_AB`
`C_1C_2_|_BC`
`C_2C_3_|_AB`
`C_3C_4_|_BC`
and forth

Answer 1

The answer is option `C=2x(2+sqrt3)`

Explanation:

The sum is

`S=AC+C C_1+C_1C_2+.........`

`S=x+C C_1+C_1C_2+.........`

In the triangle `Delta ACC_1`

`C C_1=ACsin60=xsqrt3/2`

In the triangle `Delta CC_1C_2`

`C_ 1C_2=CC_1sin60=xsqrt3/2*sqrt3/2=x(sqrt3/2)^2`

Therefore,

`S=x+xsqrt3/2+x(sqrt3/2)^2+.........`

This is a Geometric Progression with common ratio `r=sqrt3/2`and the sum is

`S=x/(1-r)=x/(1-sqrt3/2)` as `|r|<1`

`=(2x)/(2-sqrt3)`

`=(2x)/(2-sqrt3)*(2+sqrt3)/(2+sqrt3)`

`=(2x)((2+sqrt3))/1`

Answer 2

`C: 2x(2+sqrt3)`

Explanation:

Let `AC=a`,
`=> C C_1=a*sin60=a*sqrt3/2`,
`=> C_1C_2=C C_1sin60=a*sqrt3/2*sqrt3/2=a*(sqrt3/2)^2`
`=> C_2C_3=C_1C_2sin60=a*(sqrt3/2)^2*sqrt3/2=a*(sqrt3/2)^3`
let `sin60=r`
`=> S=AC+C C_1+C_1C_2+C_2C_3+ ..= a+ar+ar^2+ar^3+...`
the sum of geometric series :
`S=a+ar+ar^2+ar^3+.... = a/(1-r)`, `0 < r <1`
here `a=x`
`=> S=x/(1-sqrt3/2)=(2x)/(2-sqrt3)=(2x(2+sqrt3))/((2-sqrt3)(2+sqrt3))=(2x(2+sqrt3))/1=2x(2+sqrt3)`