be a quadrilateral inscribed in a circle. Show that the centroids of triangles 
lie on one circle.
Prove that
Equality holds when 
or 
,![\[
\frac{(a^2 + bc)^2}{b + c} + \frac{(b^2 + ca)^2}{c + a} + \frac{(c^2 + ab)^2}{a + b} \geq \frac{2abc(a + b + c)^2}{ab + bc + ca}.
\]](http://latex.artofproblemsolving.com/e/7/0/e70d68c260be93ad221f67dbaa3adfd2b121a312.png)
be the circumcircle of the triangle 
.
is tangent to the sides 
and 
,
.
intersecting the interior of triangle 
.
.
.
of size 
with 
such that![\[A^2+B^2=AB+BA=2I\]](http://latex.artofproblemsolving.com/c/3/4/c34e5f7ff2e0a180d2553fec45f278370ac9f8e6.png)
![$f : [0,1] \longrightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/9/a/4/9a4b7997c35990f839742d29866122791ade8b7c.png)
is differentiable with 
.
for all ![$x \in [0,1]$](http://latex.artofproblemsolving.com/d/3/4/d3423a3200d0ad7660fc4265d59d5598a53d551d.png)
,
for all 
.
is differentiable and 
for all 
.
,
.
for which there exists positive integers 
,
,
satisfying![\[
2a^n+3b^n=4c^n.
\]](http://latex.artofproblemsolving.com/0/2/9/0295405b01494edc05ce9867b2ec48a4e069bf26.png)
matrices 
and 
with integer entries, that satisfy the following conditions:
;
;
.
denotes the 
points are chosen at random, independently, and distributed uniformly in the unit disc 
.
be the convex hull of the chosen points. Which probability is larger: that 
be the 
be the set of 
.
is constitued of 2 elements to be determined.
for which there exist a differentiable function 
and exist a sequence 
of real positive numbers, convergent to 
such that![\[
g'(x) = \lim_{n \to \infty} \frac{f(x + a_n) - f(x)}{a_n}, \forall x \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/5/c/1/5c1966d819f6562ec5b2e6e9979729cd944ab781.png)
is continuous on 
,
with 
![$f : [a, b] \to \mathbb{R}$](http://latex.artofproblemsolving.com/4/b/a/4baf2d51eb4d8376d7a95126b727c893b588c70d.png)
satisfies the condition of the mean value theorem and function ![$g : [p, q] \to \mathbb{R}$](http://latex.artofproblemsolving.com/6/2/c/62c861380387bf2de75bd22027dfb1e7ba486736.png)
has the property 
on the interval ![$[a, b],$](http://latex.artofproblemsolving.com/2/5/5/255c58a4522df6b7519d2ccd3abac472060b117a.png)
but 
does not satisfy the condition of the mean value theorem on the interval ![$[p, q].$](http://latex.artofproblemsolving.com/e/b/d/ebd6eb93b7a86b867e34fae52d2fb3e22a962036.png)
Investigate whether there is a value of 
such that 
Prove your answe
Y by Adventure10, Mango247, and 2 other users
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.
![[asy]
unitsize(0.5 cm);
draw((0,0)--(1,0));
draw((0,1)--(1,1));
draw((2,1)--(3,1));
draw((0,2)--(3,2));
draw((0,3)--(3,3));
draw((0,0)--(0,3));
draw((1,0)--(1,3));
draw((2,1)--(2,3));
draw((3,1)--(3,3));
[/asy]](//latex.artofproblemsolving.com/e/f/4/ef4cdb21baa20a696547d85c9c5773637ad93506.png)
Determine all
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.
![[asy]
unitsize(0.5 cm);
draw((0,0)--(1,0));
draw((0,1)--(1,1));
draw((2,1)--(3,1));
draw((0,2)--(3,2));
draw((0,3)--(3,3));
draw((0,0)--(0,3));
draw((1,0)--(1,3));
draw((2,1)--(2,3));
draw((3,1)--(3,3));
[/asy]](http://latex.artofproblemsolving.com/e/f/4/ef4cdb21baa20a696547d85c9c5773637ad93506.png)
Determine all
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.
This post has been edited 3 times. Last edited by djmathman, Aug 1, 2015, 2:55 AM
Reason: No point in having link to broken image if asymptote is already there
Reason: No point in having link to broken image if asymptote is already there
can be tiled with 
rectangles then at least one of 
must be even: we start from a corner of the board and place a 
there. Then we continue to the right, placing 
to the right of 
(so the numbers in the squares will be 
)
'
(so we write 
doesn't satisfy this.
in the squares of the board like in my first post. In each 
,
,
)
or 
.
.
bits of an 
on the 
,
,
rectangles that can be covered without gaps and without overlaps with hooks such that
where row wise you name the cell 
and column wise you name the cells 
and name the cells of the board correspondingly as 
.
"
(grobber talks about "... form 
...). But do you noticed you are asking a question on a 2004 or 2005 post? Never too late ... !
rectangles (with 
)
rectangles (with 
and 
)
or 
rectangle can be tiled. So if 
)
.
is not one of the above rectangles, and that it can be tiled (we will prove a contradiction). Note that the hole of each hook must be covered by some other hook, and it's easy to see we can pair up the hooks like this. This forms 
.
is not of the above rectangles, then we have that
,
,
or 
,
snapes:
'
snapes:
snapes:
snapes:
snapes:
snapes:
snapes:
snapes:
,
,
squares 
squares 
squares 
rectangle will cover 3 squares of each color.
,
snape will cover 3 squares 
,
is even.
is even.
,
,
rectangles. The top left and bottom left corners can't both be covered by vertical rectangles. If they are both covered by horizontal rectangles, then the 3rd square of the left-most column can't be covered. So, without loss of generality the top left corner is covered by a horizontal rectangle, and the bottom left is covered by a vertical one. Then it's easy to see that that horizontal rectangle can't be united with any other rectangle. Contradiction!
rectangles, with one shifted one square. Every hook must pair with some other hook in this way, so the number of hooks is even. Thus, the number of squares is divisible by 
.
.
rectangles and 
rectangles with 
work. It's easy to see that 
don't work.
and 
.
,
,
.
pattern 
Notice that both of the 
.
and 
works, since this can be tiled with 
and 
also works since 
and 
are possible. These can be placed one above the other to give 
rectangle can be shown to be un-tilable using some small casework for corner pieces.
.
.
![[asy]
size(6cm,0);
draw((0,0)--(3,0)--(3,3)--(2,3)--(2,1)--(1,1)--(1,2)--(0,2)--(0,0));
draw((1,2)--(1,4)--(4,4)--(4,2)--(3,2));
draw((6,0)--(6,3)--(10,3)--(10,0)--(6,0));
draw((7,0)--(7,2)--(8,2)--(8,1)--(9,1)--(9,3));
[/asy]](http://latex.artofproblemsolving.com/d/5/b/d5b2b844117fd210b1b40280888170ad3d437695.png)
.
and 
.
forces that anyway. WLOG, assume 
.
or 
.
for some nonnegative integers 
and it's each to check that those don't work.
rectangles, as well as all 
rectangles, where 
.
.
and 
,
.
;
in order, and for 
let 
denote the number of type A rectangles whose leftmost cells lie in a column whose label is congruent to 
.
similarly. Now,
,
.
,
.
.
,
extra 
.
.
and 
,
is odd. Therefore there are an odd number of type C shapes. Also, since 
,
rectangle must be divisible by 
,
,
.
is easily checked as impossible. For other 
for nonnegative integers 
,
such that that the bottom left corner is the following square.![[asy]
size(5cm); int N = 4; for (int i = 0; i <= N; ++i) { draw((0,i)--(N,i)); draw((i,0)--(i,N)); }
pen r = red+white+white;
fill(shift(0,0)*unitsquare, r); fill(shift(1,0)*unitsquare, r); fill(shift(1,1)*unitsquare, r); fill(shift(2,1)*unitsquare, r); fill(shift(2,2)*unitsquare, r); fill(shift(3,2)*unitsquare, r); fill(shift(3,3)*unitsquare, r); fill(shift(0,3)*unitsquare, r); [/asy]](http://latex.artofproblemsolving.com/4/2/3/423d816cabfb8802842b71d87ed34516bad553f5.png)
![[asy]
size(5cm); int N = 4; for (int i = 0; i <= N; ++i) { draw((0,i)--(N,i)); draw((i,0)--(i,N)); }
pen r = red+white+white;
fill(shift(0,0)*unitsquare, r); fill(shift(3,0)*unitsquare, r); fill(shift(1,1)*unitsquare, r); fill(shift(2,1)*unitsquare, r); [/asy]](http://latex.artofproblemsolving.com/8/4/5/845b56dd9539fe0e3a9cb8f6f29ca06682a065ff.png)
,
with its subscript as its residue and 
for vertical. Define 
similarly.
pieces and the coloring places an even number of red squares (consider each 
separately), this implies that there are an odd number of pieces in 
have an odd number of pieces as well.
,
.
and 
rectangles.
is not possible. The latter immediately follows from caseworking on a corner.
is odd, that means the total number of 
structure, so I wanted my coloring to loop mod 
and 
.
.
from this would result in a standard checkerboard coloring that wouldn't do anything, so I decided to work in 
.
colorings work for this problem; the one I picked just happened to be the first one I chose.
