[/quote]
,
chessboard contains a rook. One by one, rooks which attack an odd number of others are removed. Find the maximal number of rooks that can be removed. (A rook attacks another rook if they are on the same row or column and there are no other rooks between them.)
intersect at the point 
.
,
is the altitude of the triangle, the point 
is marked. Points 
and 
are the feet of perpendiculars from point 
and 
,
and 
intersect at 
.
.
with respect to 
is greater than 0 if 
and 
.
and since 
?
be real numbers such that 
.
and 
.
Prove that
,
be a point on arc 
of circle 
and 
.
be the reflection of 
and 
,
intersects 
,
intersects 
.
lies on 
Prove that
Let 
Prove that![$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$](http://latex.artofproblemsolving.com/f/5/e/f5e68204507aa9c3e71c28844bbee3a804546b94.png)
Prove that
Let 
Prove that
Let 
Prove that![$$ a+b^2+c\geq \frac{135 \sqrt[3]{90}-2}{2}$$](http://latex.artofproblemsolving.com/8/6/6/866623c65f54d6147a43b1b241c2cba96863b624.png)
such that 
for all 
such that 
.
be a positive integer such that the sum of its positive divisors is at least 
.
distinct prime factors.
![\[{\ \{\frac{n}{10^k}} \} > \frac{n}{10^{10}} \]](http://latex.artofproblemsolving.com/0/c/a/0ca11fa76313b65fd76a3cceee90955960e44ef6.png)
.
Y by Davi-8191, Mathuzb, Blackhawk, mathematicsy, samrocksnature, icematrix2, donotoven, megarnie, jhu08, suvamkonar, Rounak_iitr, Adventure10, Mango247, ItsBesi
Let 
, 
, 
, 
be real numbers such that 
and all zeros 
and 
of the polynomial 
are real. Find the smallest value the product 
can take.
, 
, 
, 
be real numbers such that 
and all zeros 
and 
of the polynomial 
are real. Find the smallest value the product 
can take.This post has been edited 2 times. Last edited by v_Enhance, Jun 14, 2014, 7:45 AM
Reason: descriptive title
Reason: descriptive title
.
is a construction.
and bash some more to get 
,
and 
,
.
.
gives the desired minimum 16. Taking 
shows that the minimum can be attained so we are done.
![\begin{align*}\prod_{j=1}^4(x_j^2+1)&=\prod_{j=1}^4(x_j-i)(x_j+i)=\left[\prod_{j=1}^4(i-x_j)\right]\left[\prod_{j=1}^4(-i-x_j)\right]\\&=(i^4+ai^3+bi^2+ci+d)(i^4-ai^3+bi^2-ci+d)\\&=[(1+d-b)+i(c-a)][(1+d-b)-i(c-a)]\\&=(b-d-1)^2+(c-a)^2.\end{align*}](http://latex.artofproblemsolving.com/7/c/d/7cdfeb76892c111e8f2dbfb9ac1992906f0b02e7.png)
As 
,
.
,
.
real condition.![\[ \prod (x_i^2+1) = P(i)P(-i) = (b-d-1)^2 + (c-a)^2 \ge (5-1)^2 + 0^2 = 16. \]](http://latex.artofproblemsolving.com/7/e/e/7eeae8a7246bfae630bfc9121d88f67eec0f4251.png)
.
)
we have 
we have to find the value of ![$$\boxed{(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)=\prod_{cyc}{}[(\alpha+i)(\alpha-i)]_{min}}$$](http://latex.artofproblemsolving.com/f/4/2/f42116c594fde739a2185bc44be0de4d74b4c55b.png)
Putting the 
and 
in given equation we get since they ask the minimum value then have 
Multiply these we get 
therefore we get 
is the correct answer!!!!!
with equality at 
where 
.
To minimize 
let 
with equality at 
.
.
for 
.![\[(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)=P(i)P(-i)\]](http://latex.artofproblemsolving.com/b/9/a/b9acec2c27d761c7a860283e815d96e8ca657f7c.png)
to get
.
Therefore if we expand our desired product, we obtain ![\[(a - c)^2 + (b - d)(b - d - 2) + 1 \geq 0^2 + 5(3) + 1 = 16. \blacksquare\]](http://latex.artofproblemsolving.com/0/d/5/0d5c1fe3d15f203a2defc50bbabc19da9b39a44c.png)
We therefore claim the minimum is 
,
Putting in 
,
Clearly 
,
.
,
works, the answer is in fact 
.
,
,
,
.
,
,
,
.
,
,
,
,![\[(a-c)^2 + (b-d-1)^2\]](http://latex.artofproblemsolving.com/1/b/9/1b9c87b00c5b2551c67e08f5d0d754ccd8ecc513.png)
Since 
.
.
is 0, which occurs when 
.
,
is at least 4. The smallest square of a number that is at least 4 is 
.
.
)
,
.
.
Clearly 
so the RHS of the equation above is 
.
Now, notice that:![\[\prod_{k=1} ^4 (x_k-i)(x_k+i)=\prod_{k=1}^4(x_k-i)\cdot\prod_{k=1}^4(x_k+i).\]](http://latex.artofproblemsolving.com/5/5/c/55c92c786eade005506d5dcc3bc47f5939ed8c4e.png)
The definition of 
where the leading coeffecient is 
since 
we have:![\[P(i)=(i-x_1)\cdots(i-x_4)=(-1)^4(x_1-i)\cdots(x_4-i).\]](http://latex.artofproblemsolving.com/4/4/b/44bb498a520c1c85a9a63625b25e99ae00c46561.png)
This is exactly our first product. Our second product can be found as follows:![\[P(-i)=(-i-x_1)\cdots(-i-x_4)=(-1)^4(x_1+i)\cdots(x_4+i).\]](http://latex.artofproblemsolving.com/c/1/a/c1aec22034e4c74c87716009a66b7d085011a5a2.png)
Hence, what we want to find is 
But substituting in 
gets us 
and 
Hence:![\[P(i)P(-i)=((1-b+d)+(c-a)i)((1-b+d)+(a-c)i)=(1-b+d)^2-(a-c)^2=(-1)^2(b-d-1)^2-(a-c)^2 \ge 4^2 -0^2=16\]](http://latex.artofproblemsolving.com/3/9/e/39e97365c11932145102cc761474b7c7073758d3.png)
where equality holds at 
and 
Hence, the minimum is 
Hence![\[P(x)=x^4+cx^3+(d+5)x^2+cx+d.\]](http://latex.artofproblemsolving.com/d/1/1/d113ecb028a635f3293a60d31b5f3c6ab882a9b3.png)
We set 
to get as much symmetry as possible within our polynomial. This leads to 
However, note that 
and 
so with some wishful thinking this leads us to think about the binomial theorem. We can try 
and realize that those solutions do work.
are equal to 
![\[\prod_{k = 1}^{4} (x_k^2 + 1)\]](http://latex.artofproblemsolving.com/2/a/2/2a21d6fd71b53685481566d913d9ab1eb3378bc5.png)
![\[\prod_{k = 1}^{4} (x_k + i)(x_k - i) = \prod_{k = 1}^{4} (x_k + i) \prod_{k = 1}^{4} (x_k - i)\]](http://latex.artofproblemsolving.com/3/8/6/3868303aae4754081e002369c89105184dbfe94a.png)
,![\[(x + i)^4 + a(x+ i)^3 + b(x+i)^2 + c(x+i) + d\]](http://latex.artofproblemsolving.com/8/c/c/8cccfd684a229f1672f5053ae69d304c8e99825d.png)
and![\[(x - i)^4 + a(x - i)^3 + b(x-i)^2 + c(x-i) + d.\]](http://latex.artofproblemsolving.com/c/c/b/ccb90bc20178119ee132be4ff389f187cdf84702.png)
![\[(1 + ai - b - ci + d)(1 - ai - b + ci + d)\]](http://latex.artofproblemsolving.com/4/e/4/4e4d2106956cf6dedd37c2960c98815a6f0f55f9.png)
![\[= (1 - (b - d))^2 + (a - c)^2\]](http://latex.artofproblemsolving.com/f/6/d/f6dff77901623eaad555be58cf082df39f0381f1.png)
,
as the value of 
and take 
.
and 
.
or not? This only further verifies 
.
We verify that 
.
which finishes.
with the construction 
Plugging in 
we get that 
which simplifies to 
