Menu Close

Question-23131




Question Number 23131 by ajfour last updated on 26/Oct/17
Commented by ajfour last updated on 26/Oct/17
Q.23122    (B) pentagonal
$${Q}.\mathrm{23122}\:\:\:\:\left({B}\right)\:{pentagonal} \\ $$
Commented by math solver last updated on 26/Oct/17
 i think answer can also be circular   region ...  in q. it is said points P_i  are on   circumference  of circle of R=1.  so let′s make  a circle of radius   say r=0.25 so all the points in  circular region will definitely be   closer to O than P_i .
$$\:\mathrm{i}\:\mathrm{think}\:\mathrm{answer}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\:\mathrm{circular}\: \\ $$$$\mathrm{region}\:… \\ $$$$\mathrm{in}\:\mathrm{q}.\:\mathrm{it}\:\mathrm{is}\:\mathrm{said}\:\mathrm{points}\:\mathrm{P}_{\mathrm{i}} \:\mathrm{are}\:\mathrm{on}\: \\ $$$$\mathrm{circumference}\:\:\mathrm{of}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{R}=\mathrm{1}. \\ $$$$\mathrm{so}\:\mathrm{let}'\mathrm{s}\:\mathrm{make}\:\:\mathrm{a}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\: \\ $$$$\mathrm{say}\:\mathrm{r}=\mathrm{0}.\mathrm{25}\:\mathrm{so}\:\mathrm{all}\:\mathrm{the}\:\mathrm{points}\:\mathrm{in} \\ $$$$\mathrm{circular}\:\mathrm{region}\:\mathrm{will}\:\mathrm{definitely}\:\mathrm{be}\: \\ $$$$\mathrm{closer}\:\mathrm{to}\:\mathrm{O}\:\mathrm{than}\:\mathrm{P}_{\mathrm{i}} .\: \\ $$
Commented by math solver last updated on 26/Oct/17
sir plz do tell where i m wrong
$$\mathrm{sir}\:\mathrm{plz}\:\mathrm{do}\:\mathrm{tell}\:\mathrm{where}\:\mathrm{i}\:\mathrm{m}\:\mathrm{wrong} \\ $$
Commented by ajfour last updated on 26/Oct/17
the boundary of the area will be  the points that are equidistant  from the centre O and the  nearest point among P_1 ,P_2 ,P_3 ,P_4 ,  and P_5 .
$${the}\:{boundary}\:{of}\:{the}\:{area}\:{will}\:{be} \\ $$$${the}\:{points}\:{that}\:{are}\:{equidistant} \\ $$$${from}\:{the}\:{centre}\:{O}\:{and}\:{the} \\ $$$${nearest}\:{point}\:{among}\:{P}_{\mathrm{1}} ,{P}_{\mathrm{2}} ,{P}_{\mathrm{3}} ,{P}_{\mathrm{4}} , \\ $$$${and}\:{P}_{\mathrm{5}} . \\ $$
Commented by math solver last updated on 26/Oct/17
? what you mean?  can′t be answer circular region
$$?\:\mathrm{what}\:\mathrm{you}\:\mathrm{mean}? \\ $$$$\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{answer}\:\mathrm{circular}\:\mathrm{region} \\ $$
Commented by ajfour last updated on 26/Oct/17
consider pentagon divided into  5 triangles with common vertex  O. In any one triangle the  boundary of the required region  is the set of points equidistant  from O and as well as one of  the other two vertices that is  nearer to tbe boundary locus point.  Not circular. It is the region  witbin the pentagon formed by  the perpendicular bisectors of  OP_1 , OP_2 , ..., OP_5  .
$${consider}\:{pentagon}\:{divided}\:{into} \\ $$$$\mathrm{5}\:{triangles}\:{with}\:{common}\:{vertex} \\ $$$${O}.\:{In}\:{any}\:{one}\:{triangle}\:{the} \\ $$$${boundary}\:{of}\:{the}\:{required}\:{region} \\ $$$${is}\:{the}\:{set}\:{of}\:{points}\:{equidistant} \\ $$$${from}\:{O}\:{and}\:{as}\:{well}\:{as}\:{one}\:{of} \\ $$$${the}\:{other}\:{two}\:{vertices}\:{that}\:{is} \\ $$$${nearer}\:{to}\:{tbe}\:{boundary}\:{locus}\:{point}. \\ $$$${Not}\:{circular}.\:{It}\:{is}\:{the}\:{region} \\ $$$${witbin}\:{the}\:{pentagon}\:{formed}\:{by} \\ $$$${the}\:{perpendicular}\:{bisectors}\:{of} \\ $$$${OP}_{\mathrm{1}} ,\:{OP}_{\mathrm{2}} ,\:…,\:{OP}_{\mathrm{5}} \:. \\ $$
Commented by math solver last updated on 26/Oct/17
thank you sir !
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:! \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *