# [Easy]Inside a rectangle

Arjun Shastry

Ranch Hand

Posts: 1903

1

posted 13 years ago

Recently appeared in one regional newspaper.

Draw a rectangle ABCD with length 'x' and width 'y'.From point A ,draw an angle bisector line AE(bisecting angle BAD) ,intersecting BC at E.From point E,draw a line perpendicular to AE,intersecting CD/AD at F.Continue this procedure until you meet in one of the corners of the rectangle.Question is:

1)Is this procedure infinite(i.e.unable to meet at one of the corners) or finite? For what ratios of x/y this procedure is finite?

2)How many of such lines AE,EF.. can be drawn before meeting one of the corners?Answer in terms of x and y.

[ October 27, 2003: Message edited by: Capablanca Kepler ]

Draw a rectangle ABCD with length 'x' and width 'y'.From point A ,draw an angle bisector line AE(bisecting angle BAD) ,intersecting BC at E.From point E,draw a line perpendicular to AE,intersecting CD/AD at F.Continue this procedure until you meet in one of the corners of the rectangle.Question is:

1)Is this procedure infinite(i.e.unable to meet at one of the corners) or finite? For what ratios of x/y this procedure is finite?

2)How many of such lines AE,EF.. can be drawn before meeting one of the corners?Answer in terms of x and y.

[ October 27, 2003: Message edited by: Capablanca Kepler ]

MH

Arjun Shastry

Ranch Hand

Posts: 1903

1

posted 13 years ago

Ok, Here is what I did:

Total horizontal distance traversed = Total vertical distance traversed = L.C.M. of(x,y).

If x=y,there will be only one line.If x and y are natural numbers(and y is not a multiple of x) ,then total number of lines before meeting one of the corners is x+y-1.If y = Nx,then total number of lines will be N.

Total horizontal distance traversed = Total vertical distance traversed = L.C.M. of(x,y).

If x=y,there will be only one line.If x and y are natural numbers(and y is not a multiple of x) ,then total number of lines before meeting one of the corners is x+y-1.If y = Nx,then total number of lines will be N.

MH

Jim Yingst

Wanderer

Sheriff

Sheriff

Posts: 18671

posted 13 years ago

If x = y = 1, then shouldn't the number of walls before hitting a corner be 0? And if x = 30, y = 40, shouldn't the number of walls be 5?

**If x and y are natural numbers(and y is not a multiple of x) ,then total number of lines before meeting one of the corners is x+y-1.**

If x = y = 1, then shouldn't the number of walls before hitting a corner be 0? And if x = 30, y = 40, shouldn't the number of walls be 5?

"I'm not back." - Bill Harding, *Twister*

Arjun Shastry

Ranch Hand

Posts: 1903

1

posted 13 years ago

Thanks for corrcetion,I think then we have to make 30/40 -->3/4 so that number of lines(of type AE,EF etc) for the rectagles of sizes 3/4 or 15/20 or 30/40 etc is 3+4-1 = 6.So 6 lines can be drawn before reaching the corner of the rectagle.If rectangle is 30/41 then I think number of lines will be 30+41-1 = 70.

[ October 29, 2003: Message edited by: Capablanca Kepler ]

[ October 29, 2003: Message edited by: Capablanca Kepler ]

MH