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### RE: Message sent from website

This letter and its reply may be of general interest. The reply is given in italics.

-----Original Message-----
From: Tamaghna Kumar Banerjee
Sent: Saturday, February 19, 2005 1:59 AM
Subject: Message sent from website

query: I am a 12pass student & about to write the IITJEE 2005.I want to know ,how can I join others in cracking new problems here (I am a new user of 123 & I see all the problems already solved).

The most recent problem is not solved. You can type your solution in .doc file and email us at mail@123iitjee.com along with your name, class, school/college and place.

I also want to know whether the pattern of IITJEE 05 mains (10 ques. of 2 marks & 10 of 4 marks) going to change or not.

JEE is known for surprises, hence nothing can be said for sure. Be prepared for anything.

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$