[/quote]
,
,
,
are pairwise distinct real numbers satisfying 
Prove that
Let 
Prove that
chessboard such that no two pawns attack each other?
and 
be two positive integers. Prove that the integer![\[a^2+\left\lceil\frac{4a^2}b\right\rceil\]](http://latex.artofproblemsolving.com/1/4/5/14509b500811314c0c363da994e6de875ba468ba.png)
is not a square. (Here 
denotes the least integer greater than or equal to 
,
be a point on line 
satisfying 
.
be tangent to 
at 
.
on 
and 
on 
analogously, using the excircles opposite 
and 
,
is cyclic, then 
is tangent to the circumcircle of 
.
and 
Prove that
Let 
Prove that
and incenter 
be the midpoint of 
,
are selected on sides 
,
,
.
intersects 
other than 
and 
meet on 
be pairwise different positive real numbers such that![\[a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\]](http://latex.artofproblemsolving.com/8/a/e/8ae0bd1f55eb2fb0b7209b62739a9b98f462ee56.png)
is an integer for every 
Prove that 
such that 
is a perfect square.
Y by Adventure10, Blue_banana4, kiyoras_2001
Let 
be a triangle such that
![\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]](//latex.artofproblemsolving.com/6/9/b/69b1964be5f92eb44a36b0b8604bf473fe27e210.png)
where
and 
denote its semiperimeter and its inradius, respectively. Prove that triangle is similar to a triangle 
whose side lengths are all positive integers with no common divisors and determine these integers.
be a triangle such that![\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]](http://latex.artofproblemsolving.com/6/9/b/69b1964be5f92eb44a36b0b8604bf473fe27e210.png)
where
and 
denote its semiperimeter and its inradius, respectively. Prove that triangle is similar to a triangle 
whose side lengths are all positive integers with no common divisors and determine these integers.This post has been edited 1 time. Last edited by MithsApprentice, Sep 30, 2005, 7:45 PM
,
,
.
,
,
,
.
and simplify to
into sums of squares. We can easily see that 
and after some computations that 
and 
.
we get 
and the sides of the triangle must be 
.
,
condition.
.
.
,
.
.
and 
.![\[\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\frac{(s-a)+(s-b)+(s-c)}{r}=\frac{3s-2s}{r}=\frac{s}{r}. \]](http://latex.artofproblemsolving.com/f/9/5/f95578de3eb6484c5f2432ba170c351f6c7e6732.png)
![\[\left(\left(\cot\frac{A}{2}\right)^2+\left( 2\cot\frac{B}{2}\right)^2+\left( 3\cot\frac{C}{2}\right)^2\right) \left(1+\frac{1}{4}+\frac{1}{9}\right)\]](http://latex.artofproblemsolving.com/7/f/9/7f9d297e6096444232b4d1d1ad079c80029a7ebe.png)
![\[\ge\left(\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\right)^2=\frac{s^2}{r^2}.\]](http://latex.artofproblemsolving.com/1/a/5/1a5ea8d8e7f4656f0dbe6d0dddabce5224906b88.png)
![\[\frac{49}{36}\left(\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left(3\cot\frac{C}{2}\right)^2\right),\]](http://latex.artofproblemsolving.com/3/7/6/37607bc8ac6c81d13d6f800e4e98130559c73973.png)
so our inequality is actually ![\[\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left( 3\cot\frac{C}{2}\right)^2\ge\frac{(6s)^2}{(7r)^2}. \]](http://latex.artofproblemsolving.com/8/b/b/8bb91241fee39d4294149fb3b3e75e195019a7fd.png)
Since 
,![\[ \frac{\cot\frac{A}{2}}{\cot\frac{B}{2}}=4,\]](http://latex.artofproblemsolving.com/7/3/5/7351538572b3e66781c75c8376df2696937125c6.png)
![\[\frac{\cot\frac{B}{2}}{\cot\frac{C}{2}}=\frac{9}{4}, \]](http://latex.artofproblemsolving.com/0/2/a/02a10b88dd575f25811c32fbecc3389dbd5a717e.png)
and ![\[\frac{\cot\frac{C}{2}}{\cot\frac{A}{2}}=\frac{1}{9}. \]](http://latex.artofproblemsolving.com/2/4/9/2491e1d2143f2b172eaceb823b25dbbbad89efb9.png)
Since we proved that ![\[\frac{-a+b+c}{a-b+c}=4, \]](http://latex.artofproblemsolving.com/1/7/d/17dcc68481a27611c6c2929be3ada1a61f3cc5a7.png)
![\[\frac{a-b+c}{a+b-c}=\frac{9}{4}\]](http://latex.artofproblemsolving.com/c/7/7/c77f3fd47e975502e40205fab076eddcdb187ad1.png)
![\[\frac{a+b-c}{-a+b+c}=\frac{1}{9}.\]](http://latex.artofproblemsolving.com/0/a/3/0a348b3718909c1e1a0f453e320a7a0ec67d1467.png)
This is equivalent to the three equations ![\[5a-5b+3c=0, \]](http://latex.artofproblemsolving.com/f/c/d/fcd4d7fe6d658215212eaf4a1cac39788547fe78.png)
![\[5a+13b-13c=0, \]](http://latex.artofproblemsolving.com/5/0/b/50b4f5897a260d10ce1d8cf51211c472df28047d.png)
and ![\[5a+4b-5c=0. \]](http://latex.artofproblemsolving.com/2/a/2/2a21ba232ba014c1c7badb3aa92f52bdefeb4259.png)
Subtracting the third from the second, we find that 
.![\[5a-5b+\frac{27}{8}b=5a-\frac{13}{8}\implies \frac{40}{13}a=b. \]](http://latex.artofproblemsolving.com/6/6/5/6657bc93daeedf204f4aa39ba288f749b8e564f5.png)
Therefore, our triple 
is ![\[(a,b,c)=\left(a,\frac{40}{13}a,\frac{45}{13}a\right). \]](http://latex.artofproblemsolving.com/a/d/8/ad81f251a66f0107e3b0ec8fb44a1aa00cce984b.png)
Clearly the minimal 
,
is similar to the triangle with side lengths 
,
,
,
in the usual fashion, then the equation reads ![\[ x^2 + 4y^2 + 9z^2 = \left( \frac67(x+y+z) \right)^2. \]](http://latex.artofproblemsolving.com/a/8/f/a8f3e6ccf3ead0a2adf194e89b7d516d82462389.png)
However, by Cauchy-Schwarz, we have![\[ \left( 1 + \tfrac14 + \tfrac19 \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge \left( x+y+z \right)^2 \]](http://latex.artofproblemsolving.com/2/f/c/2fcb17b52ffef803ef54cbaf3af0bab887032c4d.png)
with equality if and only if 
,
.
.
as the intouch lengths from 
respectively.![\[ x^2+4y^2+9z^2 = 36(x+y+z)^2 = (6x+6y+6z)^2 \]](http://latex.artofproblemsolving.com/9/d/3/9d3bf6d8b7b7d74239ea8344141909138b10c3c3.png)
Additionally, Cauchy yields ![\[ (x^2+(2y)^2+(3z)^2) \ge (6x+6y+6z)^2 \]](http://latex.artofproblemsolving.com/a/c/2/ac2eb03edfb9e6ed55d9967bdab96cd0e7eccda0.png)
So equality holds in Cauchy, thus 
.
.
![\[ \begin{cases} x=s-a, \\ y=s-b, \\ z=s-c. \end{cases} \]](http://latex.artofproblemsolving.com/1/2/d/12d7f2cc0c18b2ad99557e3a4844d712acc8c274.png)
Then we have 
proving that ![\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 \ge \left( \dfrac{6s}{7r} \right)^2, \]](http://latex.artofproblemsolving.com/3/f/2/3f252b9e6ae68ee5db033cd7486b462e3919b81c.png)
hence 
.
,
,![\[ x^2+4y^2+9z^2 = \frac{36}{49}(x+y+z)^2. \]](http://latex.artofproblemsolving.com/2/2/a/22a8da2e2e08b510b495ae2880cd690e5e47e89b.png)
We can rewrite this as![\[ (x^2+4y^2+9z^2)(k_1^2+k_2^2+k_3^2) = (k_1x+2k_2y+3k_3z)^2 \]](http://latex.artofproblemsolving.com/1/e/1/1e1cfbe92be6becfd0091a74405458cb546684a9.png)
where 
,
.
.
Then the equation is![\[x^2+4y^2+9z^2 = \frac{36}{49}(x+y+z)^2\]](http://latex.artofproblemsolving.com/1/e/a/1ea86b5f6e4bd82e9cf84ae1b9e33e04a4ddc004.png)
But by Cauchy-Schwarz, we know that![\[\left( 1 + \frac{1}{4} + \frac{1}{9} \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge (x+y+z)^2 .\]](http://latex.artofproblemsolving.com/5/c/8/5c82defe2edc31f2e09aa89ffc0e1461b5e29ba3.png)
The equality case occurs when 
If we scale this up by 
we have 
Finally, using the definitions of 
tells us our triangle is one with side lengths 
to the parts of a triangle to derive the similar triangle T?
But:
so that this holds. Since all sides were scaled up by an equal amount, it remains similar.
![\[s=r[\cot A/2+\cot B/2+\cot C/2].\]](http://latex.artofproblemsolving.com/f/d/6/fd66d715c8b6e7538c4acc59f6599985abf552f1.png)
Let 
so we have![\[x^2+4y^2+9z^2=36(x+y+z)^2/49.\]](http://latex.artofproblemsolving.com/6/0/a/60a3f37e1d1e7a3b1856c06cab615224f5e99840.png)
Rearranging, we have![\[13x^2+160y^2+405z^2=72[xy+yz+zx].\]](http://latex.artofproblemsolving.com/c/b/f/cbf20b201dd754be0af82bab3cad49ced5fb14bc.png)
Rewrite this equation as follows:![\[(2x-18z)^2+(3x-12y)^2+(9z-4y)^2=0.\]](http://latex.artofproblemsolving.com/4/0/6/40641d332f595aa817fde8f6fb2c4a934c7ffd5e.png)
Hence we have 
.![\[\cot A/2+\cot B/2:\cot B/2+\cot C/2:\cot C/2+\cot A/2=c:a:b,\]](http://latex.artofproblemsolving.com/2/d/3/2d38a721ef4e04b8cb21c406670f2ba6cfc4ddda.png)
we get 
,
,
,
where 
.
which simplifies to 
However, observe the following application of Cauchy-Schwarz: 
so we know that for equality to hold we need 
,
is 
.
,
which by the CS equality case means that 
and consequently we have that 
is similar to the triangle with side-lengths 
if 
.![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
be the area of triangle ![$$\frac{s}{r}=\frac{[ABC]^2}{sr^3}=\frac{s-a}{r}\cdot \frac{s-b}{r}\cdot \frac{s-c}{r}=\tan\left(90-\frac{A}{2}\right)\tan\left(90-\frac{B}{2}\right)\tan\left(90-\frac{C}{2}\right)=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2},$$](http://latex.artofproblemsolving.com/1/5/f/15f70c5bf8b5dddfe3ead688554596b7d8fdc0e8.png)
and also that 
Clearly equality is achieved, thus 
for some number 
,
.
,
.
,
,
,
By Cauchy-Schwarz Inequality we get that 
Since we know that these two expressions are equal we can consider the equality case: 
.![\[(\frac{x}{r})^2 + (2\cdot \frac{y}{r})^2 + (3\cdot \frac{z}{r})^2 = (\frac67 \cdot \frac{x+y+z}{r})^2\]](http://latex.artofproblemsolving.com/1/1/c/11c0220719dd3a4a8eb69b809a76ce32ba9ec4a7.png)
But by Cauchy,![\[x^2 + (2y)^2 + (3z)^2\cdot (1 + \frac12^2 + \frac13^2) =\lVert \begin{bmatrix} x \\ 2y \\ 3z\end{bmatrix} \rVert^2\cdot \lVert \begin{bmatrix} 1\\ \frac12 \\ \frac13\end{bmatrix} \rVert^2 \geq (x+y+z)^2\]](http://latex.artofproblemsolving.com/1/e/d/1ed8be2eaf6b1ebf451055d10d945d9333d8040d.png)
Note that 
,
.
,
,
and we're done. 
be the contact points of the incircle on 
Let 
The equation becomes 
which rearranges to 
In fact, we can complete the square to get ![\[(2u-18w)^2+(3u-12v)^2+(9w-4v)^2=0\]](http://latex.artofproblemsolving.com/9/1/d/91d584a7945622424de0da4b9aa3add2c25aeaeb.png)
Therefore, we conclude 
Our triangle is 
,
But 
so equality must hold everywhere. Thus the side lengths must be in the ratio 
![\begin{align*}
&[(6r)^2+(3r)^2+(2r)^2]\left[\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left(3\cot\frac{C}{2}\right)^2\right]\\
\ge&\left(6r\cot\frac{A}{2}+6r\cot\frac{B}{2}+6r\cot\frac{C}{2}\right)^2\\
=&~(6s)^2
\end{align*}](http://latex.artofproblemsolving.com/9/5/f/95face314b5265d8e7a2679090fe536c3473b59b.png)
with equality at 
,
.
.
,
,
.
But by Cauchy,
so we have an equality. Therefore, 
,
.
By Cauchy-Schwarz Inequality,
Equality holds if and only if 
similarly.
.
Similarly, 
Substituting these into the original equation gives us 
But observe that 
so equality must hold in the step with Titu's lemma, i. e. 
.
.
.
.
,
be its ![\[x = AE = AF, y = BF = BD, z = CD = CE\text{.}\]](http://latex.artofproblemsolving.com/b/a/c/bac97dbd6bc4ea40025f54bb596b013c97f89a6f.png)
Therefore we have
which is the equality case of Cauchy-Schwarz. Therefore
Now we have
Therefore 
.
with equality iff 
so 
,
,
,
.
at 
,
such that 
,
,
.
Substituting to the given equation yields:
But, by Cauchy Schwarz, note that
but the equality case of this resulting inequality is equivalent to the given equation. Therefore, 
.
.
.
and define similarly for others. Note that by considering the segments that the intouch points partition the sides of 
Cauchy-Schwarz yields![\[ \bigg[x^2 + (2y)^2 + (3z)^2 \bigg] \left[1 + \frac{1}{2^2} + \frac{1}{3^2} \right] \ge (x + y + z)^2 = \frac{s^2}{r^2} \]](http://latex.artofproblemsolving.com/9/f/f/9ffa730bd817ee8d084d8eed243f735386574652.png)
![\[ \iff x^2 + (2y)^2 + (3z)^2 \ge \left( \frac{6s}{7r} \right)^2. \]](http://latex.artofproblemsolving.com/c/8/0/c809061f8f627f13fdd20e9638de0451907e423f.png)
The condition of the problem implies equality, so 
,
,
.
so our answer is 
.
(and similarly for 
)
Thus, we must be in the equality case of Cauchy, or 
Therefore, the ratio of the side lengths must be 
.
terms, we apply Cauchy-Schwarz in the following manner: ![\[\left(1 + \frac14 + \frac19\right)\left(\left(\cot \frac A2\right)^2 + \left(2\cot \frac B2\right)^2 + \left(3\cot \frac C2\right)^2\right) \geq \left(\cot \frac A2 + \cot \frac B2 + \cot \frac C2\right)^2.\]](http://latex.artofproblemsolving.com/0/6/0/06079be514ffa117c310515bf7446498a860decf.png)
This seems to suggest that 
,
gives ![\[\left(\frac67\left(\cot \frac A2 + \cot \frac B2 + \cot \frac C2\right)\right)^2\]](http://latex.artofproblemsolving.com/1/e/c/1ec2619634307141fd9c303c729327519703279e.png)
on the RHS, which would imply equality in the above Cauchy application. This is in fact true, since ![\[r \cot \frac A2 + r \cot \frac B2 + r \cot \frac C2 = (s - a) + (s - b) + (s - c) = s\]](http://latex.artofproblemsolving.com/5/6/c/56c0661a07355fa200588d880be5a9327b7ee6de.png)
by dropping the perpendicular from the incenter to each of the sides.
,
yields ![\[b + c - a = 4(c + a - b) = 9(a + b - c).\]](http://latex.artofproblemsolving.com/2/b/6/2b60295d908a63b5f6394d48e75be5a8ba2e7dc4.png)
Let 
denote the common value. Scaling and adding these quantities pairwise, we obtain 
,
respectively. Then our condition is equivalent to 
which is in fact the equality case of Cauchy-Schwartz. Thus, for some constant 
we have that 
Therefore, the three side lengths are 
![\[\left(\left(\cot\frac A2\right)^2+\left(2\cot\frac B2\right)^2+\left(3\cot\frac C2\right)^2\right)\left(\left(\frac67\right)^2+\left(\frac37\right)^2+\left(\frac27\right)^2\right)\ge\left(\frac{6s}{7r}\right)^2\]](http://latex.artofproblemsolving.com/a/b/d/abddc3b0d88c545cf29a1c7be066049dfbe579b6.png)
BY CAUCHY . EQUALITY IS WHEN 
SO![\[\boxed{a:b:c=13:40:45}\]](http://latex.artofproblemsolving.com/6/f/6/6f686df12e94fbb328119c63e31d958af0d74818.png)
SUSUS.
