Question

A carom board is a square board with a symmetrical design as shown in the diagram
below. E, F, G and ,H are the midpoints of AD,BA,CB and DC respectively And wants to pocket his last coin which is at E. Taking the laws of reflection to be applicable, he strikes from the point M, so that the striker reflects at F, then at G, then at H, and finally hits the coin at E. Choosing D to be the origin (right direction is +ve X-axis and upward direction is +ve Y−axis)

1: M does not lie on the line segment EF
2:The coordinate of point of M is (-60,10)
3:The coordinate of point of M is (-40,10)
4:The slope of MF is -1
5:FG and GH are perpendicular line segments
6:Equation of GH will be X+Y=150.
7:Segment HE intersects the inner square at(-10,40)

Answer 1

By law of reflection ∠i=∠r  

or  tani=tanr  (from triangle OAB and BDC)

x

2−x

​  

=  

2

4

​  

 

Or  4−2x=x  

⇒3x=2⇒x=  

3

2

​  

 

We get  BD=2−  

3

2

​  

=  

3

4

​  

 

(i) : Thus displacement of queen from front edge to hole    

BC

=  

3

2  

i

^

 

​  

+2  

j

^

​  

 

Magnitude of displacement  ∣  

OB

∣=  

(  

3

2

​  

)  

2

+2  

2

 

​  

=  

3

2

​  

 

10

​  

ft  

(ii) : Thus displacement of queen from center to front edge    

BC

=  

3

4  

i

^

 

​  

−4  

j

^

​  

 

Magnitude of displacement  ∣  

BC

∣=  

(  

3

4

​  

)  

2

+4  

2

 

​  

=  

3

4

​  

 

10

​  

ft  

(iii) : Thus displacement of queen from center to hole    

OC

=2  

i

^

−2  

j

^

​  

 

Magnitude of displacement  ∣  

OC

∣=  

(2)  

2

+2  

2

 

​  

=2  

2

​  

ft