Problems from the Bus
by djmathman, May 23, 2013, 10:56 PM
And no, I'm not making that up. This probably needs a bit of explanation.
Today, all of the Physics students were invited to a "Physics Day" trip to Six Flags (since Physics is not a required science and is taught in 12th grade normally if one does not double up). Being a student in AP Calculus, the invitation was extended to me, and so I went.
And there was a chance of rain.
So to prepare for any long indoor sessions from rain I "created" a "Practice Olympiad" of three questions to work on. I solved two of the problems on the bus, and it turns out that no rain affected the amusement park until we were about to leave.
Here are the two questions and their solutions:
Eh, this is wrong - will correct later
Trivial
Today, all of the Physics students were invited to a "Physics Day" trip to Six Flags (since Physics is not a required science and is taught in 12th grade normally if one does not double up). Being a student in AP Calculus, the invitation was extended to me, and so I went.
And there was a chance of rain.
So to prepare for any long indoor sessions from rain I "created" a "Practice Olympiad" of three questions to work on. I solved two of the problems on the bus, and it turns out that no rain affected the amusement park until we were about to leave.
Here are the two questions and their solutions:
All-Russian MO 2003 Grade 9 #6 wrote:
Let
be positive numbers with the sum
. Prove the inequality
![\[\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}.\]](//latex.artofproblemsolving.com/c/6/f/c6fce19452f41bc112788355e1679c5b277e8e07.png)


![\[\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}.\]](http://latex.artofproblemsolving.com/c/6/f/c6fce19452f41bc112788355e1679c5b277e8e07.png)
Eh, this is wrong - will correct later
Note that the inequality in question is equivalent to proving that
By the Engel form of Cauchy-Schwartz, we have

![\[\sum\left(\dfrac1{1-a}-\dfrac2{1+a}\right)=\sum \dfrac{3a-1}{1-a^2}\geq 0.\]](http://latex.artofproblemsolving.com/d/a/6/da66303f6824670117724bc407c809f6da74731f.png)
\[\begin{align*}\sum\dfrac{3a-1}{1-a^2}&=\sum\dfrac{(3a-1)^2}{(3a-1)(1-a^2)}\geq\dfrac{\left[\sum(3a-1)\right]^2}{\sum (3a-1)(1-a^2)}\\&=\dfrac{(3\sum a - 3)^2}{\sum (3a-1)(1-a^2)}=\dfrac{(3-3)^2}{\sum (3a-1)(1-a^2)}=0,\end{align*}\]as desired.

Sharygin 2009 #14 wrote:
Given triangle
of area 1. Let
be the perpendicular from
to the bisector of angle
. Determine the area of triangle
.





Trivial
For simplicity, let
and
. Note that from the fact that
we have
so
. Next, note that
, so
. This means that


![$[ABC]=1$](http://latex.artofproblemsolving.com/4/a/0/4a08c4c1431ba612a7f5f04de67b275518a068b2.png)




\[\begin{align*}[AMC]&=\dfrac12b(MC)\sin\dfrac C2=\dfrac12ab\cos\dfrac C2\sin\dfrac C2\\&=\dfrac{\cos\dfrac C2\sin\dfrac C2}{\sin C}=\dfrac{\cos\dfrac C2\sin\dfrac C2}{2\sin\dfrac C2\cos\dfrac C2}=\boxed{\dfrac12}.\end{align*}\]
This post has been edited 2 times. Last edited by djmathman, Sep 5, 2013, 1:33 AM