Parallel lines and circles!

Geometry Level 4

Three circles each of radius 1cm touch each other externally and they lie between two parallel lines.

Then what is the minimum possible distance (in cm) between the lines?

Answer correct to three decimal places!


The answer is 3.732.

Prateek Mehra by Prateek Mehra , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

This is the diagram for Ajay Nandugopal's solution. I have the same solution.
The minimum distance between parallel lines is = r + 3 2 ∗ ( r + r ) + r = 1 + 3 + 1 = 3.732 \text{The minimum distance between parallel lines is }\\ =r + \dfrac{ \sqrt3} 2 *(r +r)+r=1 + \sqrt3+1 =3.732

0 Helpful 0 Interesting 0 Brilliant 0 Confused

As per Pythagoras theorem the hieght of the equilateral triangle should be √5 and hence the answer should be 4.236

Biraj Kundu - 5 years, 9 months ago

Log in to reply

Not 5 b u t 3 \sqrt5~~but~~\sqrt3 .

Niranjan Khanderia - 5 years, 9 months ago

Sir I guess you have a typo in the last line of your solution. It should be (3)^(1/2) instead of ((3^(1/2)/2)

Sathvik Acharya - 4 years ago

Log in to reply

Thank you. I have corrected. Sorry for delay.

Niranjan Khanderia - 3 years, 11 months ago
Ajay Nandugopal
Jan 24, 2015

by drawing the fig, we can notice that the line joinin the center forms an equilateral triangle, therefore the dist btw them is = r + (3^1/2)/2 *(r+r) +r= 2 +3^1/2=3.732

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...