A Few Problems
by djmathman, Sep 26, 2014, 4:11 PM
MOSP Homework 2013 wrote:
Let 
be a positive integer. Given non-overlapping circular discs on a rectangular piece of paper, prove that one can cut the piece of paper into convex polygonal pieces each of which contains exactly one disc.
be a positive integer. Given non-overlapping circular discs on a rectangular piece of paper, prove that one can cut the piece of paper into convex polygonal pieces each of which contains exactly one disc.
be the 
to denote the radical center of circles 
,
,
.
and 
,
,
is any integer between 
and 
or 
.
and of 
)MOSP Homework 2013 wrote:
Let 
, 
be real numbers such that 
. Determine the maximum value of
![\[(a_{n+1}+a_{n+2}+\cdots+a_{2n})-(a_1+a_2+\cdots+a_n).\]](//latex.artofproblemsolving.com/6/d/8/6d86f12525865cbc74c9729105905015ea503d46.png)
, 
be real numbers such that 
. Determine the maximum value of![\[(a_{n+1}+a_{n+2}+\cdots+a_{2n})-(a_1+a_2+\cdots+a_n).\]](http://latex.artofproblemsolving.com/6/d/8/6d86f12525865cbc74c9729105905015ea503d46.png)
be the expression we wish to maximize. Define 
to be the sequence of real numbers such that 
for all 
and 
otherwise. The condition can be rewritten as 
.
case gives 
.![\[2\sum_{i=1}^{n-1}i^2+n^2=\dfrac{(n-1)n(2n-1)}3+n^2=\dfrac{n(2n^2+1)}3.\]](http://latex.artofproblemsolving.com/f/6/7/f672f703e71f17e49ccf13a1db9f0bf42ca42793.png)
![\begin{align*}\left[\sum_{j=1}^{n-1}jb_j+nb_n+\sum_{j=n+1}^{2n-1}(2n-j)b_j\right]^2&\leq\left[\sum_{i=1}^{2n-1}b_i^2\right]\left[\sum_{i=1}^{n-1}i^2+n^2+\sum_{i=n+1}^{2n-1}(2n-i)^2\right]\\&=\dfrac{n(2n^2+1)}3\\\implies S&\leq\boxed{\sqrt{\dfrac{n(2n^2+1)}3}}.\end{align*}](http://latex.artofproblemsolving.com/5/6/5/565021f592b8046cf07e43a80866e602757827f9.png)
This is attainable by considering the equality condition of C-S.
case, when Cauchy-Schwarz gave me![\[\sum_{i=1}^{3}(a_{i+1}-a_i)^2\geq\left[\sum_{i=1}^{3}(a_{i+1}-a_i)\right]^2=(a_{4}-a_1)^2.\]](http://latex.artofproblemsolving.com/f/2/5/f2578743ebec65e8f783faff4463fa3cfec9022b.png)
Ultimate Elimination Tourney G2 wrote:
Let 
be a point in the interior of triangle 
such that 
is isosceles and its base angles 
and 
measures the same as angle 
. Reflect about 
and 
to get 
and 
respectively. Suppose that the perpendicular bisector of 
intersects 
at 
. Also, let 
intersect at 
and 
intersect at 
. Prove that 
are collinear.
be a point in the interior of triangle 
such that 
is isosceles and its base angles 
and 
measures the same as angle 
. Reflect about 
and 
to get 
and 
respectively. Suppose that the perpendicular bisector of 
intersects 
at 
. Also, let 
intersect at 
and 
intersect at 
. Prove that 
are collinear.
.
be the projections of 
respectively. Let 
be the intersection of 
with 
and let 
and 
be the midpoints of 
and 
respectively. The homothety centered at 
sends 
to 
respectively so it suffices to show 
,
,
are all cyclic quads from the right angles, so from the givens 
and 
.
.
.
,
.
,
.
be 
,
respectively. From Ratio Lemma we get![\[\dfrac{B_1M}{MC_1}=\dfrac{A_1B_1\sin\angle B_1A_1D}{AC_1\sin\angle DA_1C_1}=\dfrac{A_1B_1\sin\angle DCB_1}{A_1C_1\sin\angle C_1AD}.\]](http://latex.artofproblemsolving.com/a/1/7/a175f489c872b18849c58e3332f3de32af3fbd82.png)
gives 
and similarly 
.![\[\dfrac{B_1M}{MC_1}=\dfrac{A_1B_1}{A_1C_1}\left(\dfrac{DY}{DZ}\right)\left(\dfrac{\sin\angle B}{\sin\angle C}\right)=\dfrac{DY}{DZ}\left(\dfrac{A_1B_1/\sin\angle C}{A_1C_1/\sin\angle B}\right).\]](http://latex.artofproblemsolving.com/f/f/8/ff8dca0fa1c69320a614e868fbb6b659b7bb659e.png)
and 
have the same circumradius, so do 
and 
.![\[\dfrac{MB_1}{MC_1}=\dfrac{DY}{DZ}=\dfrac{B_1N}{C_1P}.\]](http://latex.artofproblemsolving.com/9/3/7/93725589f6b980441a6c941df6bb510f5418f5ca.png)
by SAS similarity as desired.This post has been edited 5 times. Last edited by djmathman, Mar 5, 2015, 3:43 AM
Reason: align tags
Reason: align tags
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