AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
on the board. For each procces, the stutent deletes the number x and replaces it with either 
or 
or 
. Is this possible to make the number 
on the board?
and 
.
?
lies inside triangle 
such that 
and 
. Point 
is the midpoint of segment 
. Point 
lies on segment 
with 
. Prove that 
.
the number of positive integer 
satisfying 
is less than or equal to 
where 
is a constant independent of 
be different real numbers. Prove that![\[\sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}=\left\{\begin{array}{ll}
0, & \text { if } n \text { is even; } \\
1, & \text { if } n \text { is odd. }
\end{array}\right.\]](http://latex.artofproblemsolving.com/6/5/2/6525fc6b451ff5a3c83ff2a0541b063a7cf065c3.png)
satisfying
for all real numbers 
and 
.
satisfying
Prove
:![\[
1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n + 1) \mid (4n + 3)(4n + 5) \cdot \ldots \cdot (8n + 3).
\]](http://latex.artofproblemsolving.com/3/1/a/31a0eac3c7ca298c8fa221083dc94e127e705784.png)
![$ a,b \in [0 ,1] . $](http://latex.artofproblemsolving.com/9/1/9/91980d5ca5d97fe4b61797e6fbb378211674d9cd.png)
Prove that
.
.
and square 
are placed side by side (i.e. 
is between 
and 
is between 
,
,
.
![[asy]
size(120); defaultpen(linewidth(0.7)+fontsize(10));
pair D2(real x, real y) {
pair P = (x,y);
dot(P,linewidth(3)); return P;
}
int big = 30, small = 14;
filldraw((0,big)--(big+small,0)--(big,small)--cycle, rgb(0.9,0.5,0.5));
draw(scale(big)*unitsquare); draw(shift(big,0)*scale(small)*unitsquare);
label("$A$",D2(0,big),NW);
label("$B$",D2(0,0),SW);
label("$C$",D2(big,0),SW);
label("$D$",D2(big,big),N);
label("$E$",D2(big+small,0),SE);
label("$F$",D2(big+small,small),NE);
label("$G$",D2(big,small),NE);
[/asy]](http://latex.artofproblemsolving.com/6/a/b/6ab722070e0507922b868deb528a8151affbecc8.png)
by subtracting out the areas of the other triangles from the combined areas of squares ABCD and CEFG.![$[ABCD] = x^2$](http://latex.artofproblemsolving.com/1/0/8/108076e1ae974cf9bcc7bd80d95d46212947f669.png)
![$[CEFG] = 14^2 = 196$](http://latex.artofproblemsolving.com/d/c/4/dc4f65f5b82fc3ee90808c7dbb4660efe5b5db11.png)
.![$[ADG] = \frac {AD \cdot DG}{2} = \frac {(x)(x-14)}{2}$](http://latex.artofproblemsolving.com/1/d/4/1d4b11f857bcc965e9baf1c4f9becf8cafd0fca7.png)
.![$[ABE] = \frac {AB \cdot BE}{2} = \frac {(x)(x + 14)}{2}$](http://latex.artofproblemsolving.com/4/e/7/4e704be2960e98658bd5047f585c6d25b425be54.png)
![$[EFG] = \frac {14\cdot14}{2} = 98$](http://latex.artofproblemsolving.com/6/0/e/60e6a4fac3b6748ad6e64c3da96aebcf51303592.png)
.![$[AEG] = x^2 + 196 - \frac {(x)(x-14)}{2} - \frac {(x)(x+14)}{2} - 98 = \boxed {98}$](http://latex.artofproblemsolving.com/b/0/a/b0a6b1dbe61d70ec8d76316f617251f2b28f091c.png)
regardless of the value of 
is the ![$[ABCD]+[CEFG]-[GFE]-[ADG]-[ABC]-[ACE]$](http://latex.artofproblemsolving.com/7/b/4/7b4a207f4fa989bc42e683025b82b25bc5a1cec2.png)
.
.![$[ABCD]=x^2$](http://latex.artofproblemsolving.com/b/a/9/ba9ba3842e7403383fc0e4b20a339734c1dc479b.png)
, ![$[CEFG]=196$](http://latex.artofproblemsolving.com/0/d/a/0da029371bd3f554891ee8c9834888b97207614d.png)
, ![$[GFE]=\frac{196}{2}=98$](http://latex.artofproblemsolving.com/0/3/c/03cb9ae307c7f24d203e68dc22b74cf217c4614c.png)
, ![$[ADG]=\frac{x(x-14)}{2}=\frac{x^2-14x}{2}$](http://latex.artofproblemsolving.com/4/d/1/4d108273996ad40e0fa2f8f004c7f295a83c2a5b.png)
, ![$[ABC]=\frac{x^2}{2}$](http://latex.artofproblemsolving.com/e/4/f/e4f8019d7b235c2ff443b9924c5d2094b83443d6.png)
and ![$[ACE]=\frac{14x}{2}=7x$](http://latex.artofproblemsolving.com/6/0/c/60cf4fd16568f7c8ffa5e5a7afd16610226248be.png)
which simplifies to 
and 
are parallel regardless of the size of 
to base 
(aka the height of the triangle) is the same as the distance from 
.