[/quote]
,
and 
be positive integers. James has 
marbles with weights 
,
,
and 
for which 
is rational.
of the acute-angled triangle 
is tangent to its side 
at a point 
.
be an altitude of triangle 
be the midpoint of the segment 
is the common point of the circle 
(distinct from 
are tangent to each other at the point 
with the property that 
is divisible by 
.
of positive real numbers,
be an acute triangle with circumcircle 
,
be the foot of the altitude from 
to 
.
and 
be the points on 
and 
.
and 
at 
and 
respectively; the tangent to 
and 
respectively. Show that the circumcircles of 
and 
are congruent, and the line through their centers is parallel to the tangent to 
be an integer. Prove that if 
is an integer, then it is divisible by 11.
such that![\[ f(x)f\left(x + 4f(y)\right) = xf\left(x + 3y\right) + f(x)f(y) \]](http://latex.artofproblemsolving.com/d/b/1/db14b64aa71cb3c8dbf53707f719f5d9a43bd7e4.png)
for any positive real numbers 
.
and 
be circles with centres 
and 
,
.
intersects 
and 
,
lie on 
.
meets 
and meets 
.
.
.
such that 
is divisible by 
and 
is divisible by 
.
natural numbers such that there exist natural numbers 
is a prime number.
can be 
cyclically with period 
.
,
Thus the terms from 
up to 
split into 
sums of 
.
.
,
Hence the entire sum equals 
.
given that
![\[
\frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta}
\]](http://latex.artofproblemsolving.com/e/6/1/e618ab30bf3fcdb952f92d2c0bdb7d6ff8593391.png)
![\[
\sin(2\theta) = 2\sin\theta\cos\theta \Rightarrow \sin^2(2\theta) = 4\sin^2\theta\cos^2\theta
\Rightarrow \sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2(2\theta)
\]](http://latex.artofproblemsolving.com/b/2/d/b2d779116d132d79446c983e8f2473583f518901.png)
![\[
\frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} = \frac{4}{\sin^2(2\theta)}
\]](http://latex.artofproblemsolving.com/1/c/4/1c4fd8393d35c1125b8eebdefec338168c012eed.png)
,![\[
\sin(2\theta) = \frac{\tan(2\theta)}{1 + \tan^2(2\theta)} = \frac{4}{\sqrt{17}}
\]](http://latex.artofproblemsolving.com/a/c/6/ac63c4b53e60424e0df8c3026df1ca0c1febbcda.png)
![\[
\frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = \frac{1}{[\frac{1}{4}][\sin^2(2\theta)]}=\frac{4}{(\frac{4}{\sqrt{17}})^2} = \boxed{\frac{17}{4}}
\]](http://latex.artofproblemsolving.com/5/d/b/5db53b8e0befb44ed07e17933d3fa854a44c7ab9.png)
Y by Rajukian
Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.
Determine all mxn rectangles that can be covered without gaps and without overlaps with hooks such that
- the rectangle is covered without gaps and without overlaps
- no part of a hook covers area outside the rectangle.
Determine all mxn rectangles that can be covered without gaps and without overlaps with hooks such that
- the rectangle is covered without gaps and without overlaps
- no part of a hook covers area outside the rectangle.
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