AoPS Academy
, with 
, is inscribed in the circle 
. Let 
be a variable point on the arc 
that does not contain 
, and let 
and 
denote the incenters of triangles 
and 
, respectively. varies, the circumcircle of triangle 
passes through a fixed point.
![\[
x^2 + 2|x| - 6x + 15 = 0.
\]](http://latex.artofproblemsolving.com/1/1/d/11dbf7a605e8c49bb730eb0bd3737019b277fa41.png)
(Note: the modulus of a complex number 
is 
.)
and 
.
and 
.
.
are the 256 roots (not necessarily distinct) of the equation 
, evaluate 
.
such that 
, 
, 
, and 
are real numbers that satisfy the system of equations
and 
be positive real numbers such that 
and 
. What is the value of 
?
and satisfy ![\[ a^2+b^2+c^2 +abc = 4 . \]](http://latex.artofproblemsolving.com/d/3/a/d3ac01295b077425b5eaae556faa4cd678bce32a.png)
Show that ![\[ 0 \le ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/2/0/9/209b3fcb56c01c335bbcb8d5fb781e28e49a0798.png)
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}
\frac{9}{a+b+c} \geq 1
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 1
if
,
,
?
in terms of 
.
.
to z because we already know the value of 
.
(in the equation 
)
.
.
and cross multiply.
.
.
![\[
\frac{ab + bc + ca}{a^2 + b^2 + c^2} + \frac{1}{6} \left( \frac{(a - b)^2}{a^2 + b^2} + \frac{(b - c)^2}{b^2 + c^2} + \frac{(c - a)^2}{c^2 + a^2} \right) \leq 1.
\]](http://latex.artofproblemsolving.com/d/8/b/d8b3869addff6715d86b73083c77395d64ab1995.png)
,
for the same reason.
and 
.
.
.
,
,
as it is also equal to 
and 
.
but once again substitute 
with 
with 
.
,
.
and 
Prove that
a^2+b^2 \geq 2abb^2+c^2 \geq 2bca^2+c^2 \geq 2aca^2+b^2+c^2 \geq ab+bc+acab+bc+ac+abc \leq 4ab+bc+ac-abc \leq 4-2abc
2abc \ge 2 abc \ge 1 \sqrt[3]{abc} \ge 1
\displaystyle \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \ge 1 3 \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c} 3abc \ge ab+bc+ca
a^2+b^2+c^2 \ge ab+bc+ca3abc \ge a^2+b^2+c^2
a^2+b^2 \geq 2abb^2+c^2 \geq 2bca^2+c^2 \geq 2aca^2+b^2+c^2 \geq ab+bc+acab+bc+ac+abc \leq 4ab+bc+ac-abc \leq 4-2abcabc \ge 1. However, by applying AM-GM to a^2 + b^2 + c^2 + abc = 4,
\displaystyle{{a^2 + b^2 + c^2 + abc}\over{4}} \ge \sqrt[4]{a^3b^3c^3}
abc \le 1.
x, y \leq 1(1+x)(1-y)c \leq 1(1+x)^2+(1-y)^2+c^2+(1+x)(1-y)(c)=4
x^2+y^2+c^2-xyc \geq 0 becausexy+yc+xc-xyc=xy+yc+xc(1-y) \geq 0(2+c)(1+x-y) \leq 4
such that 
,
.![\[ U = \left\{ (a,b,c) \mid a,b,c > 0 \text{ and } a^2+b^2+c^2 < 1000 \right\}. \]](http://latex.artofproblemsolving.com/1/d/4/1d4f5748608b93f806d00a36ade8b939c3af8366.png)
This is an intersection of open sets, so it is open. Its closure is ![\[ \overline U = \left\{ (a,b,c) \mid a,b,c \ge 0 \text{ and } a^2+b^2+c^2 \le 1000 \right\}. \]](http://latex.artofproblemsolving.com/2/0/8/208109ca33a19160d3a04e9d8e00b5e97bf7ef40.png)
Hence the constraint set ![\[ \overline S = \left\{ \mathbf x \in \overline U : g(\mathbf x) = 4 \right\} \]](http://latex.artofproblemsolving.com/1/6/0/1600e8cd8aa5219c1b520dc15c03d94547005f3c.png)
is compact, where 
.![\[ f(a,b,c) = a^2+b^2+c^2+ab+bc+ca. \]](http://latex.artofproblemsolving.com/b/b/0/bb04a96cd90630a62afe19639303a5842a150e20.png)
It's equivalent to show that 
subject to 
.
,
lies on the boundary, that means one of the components is zero (since 
is clearly impossible). WLOG 
,
for 
,
,
.![\[ \nabla f = \left<2a+b+c, 2b+c+a, 2c+a+b \right> \]](http://latex.artofproblemsolving.com/2/f/1/2f1641cf05573314b51e84dd484ce7ea77820a9c.png)
and ![\[ \nabla g = \left<2a+bc, 2b+ca, 2c+ab\right>. \]](http://latex.artofproblemsolving.com/8/8/9/8894849f0b47c6f236d637bcf3b60d98577a893f.png)
Of course, 
everywhere, so introducing our multiplier yields ![\[ \left<2a+b+c,a+2b+c,a+b+2c\right> = \lambda \left<2a+bc,2b+ca,2c+ab\right>. \]](http://latex.artofproblemsolving.com/3/2/0/32066df167b8e8026f5b37a7142b9bd5fdd9fbd1.png)
Note that 
since 
.
from 
implies that ![\[ (a-b)(\left[ 2\lambda - 1 \right] - \lambda c) = 0. \]](http://latex.artofproblemsolving.com/3/5/c/35c5d53b45364c038d6381f1e58094e85436c39a.png)
We can derive similar equations for the others. Hence, we have three cases.
,
,
.
,
.
(of course our conditions force 
)![\[ 2a+2c = a+b+2c = \lambda (2c+ab)
= \frac{1}{2-a} (2c+a^2) \]](http://latex.artofproblemsolving.com/2/f/5/2f5bd28dd016a11506f79087ae947b82c063cb9a.png)
which implies ![\[ 4a+4c-2a^2-2ac = 2c+a^2 \]](http://latex.artofproblemsolving.com/2/5/d/25de76451946469fa8b64a2761d4ff2b737d0a3d.png)
meaning (with the additional note that 
)![\[ c = \frac{3a^2-4a}{2-2a}. \]](http://latex.artofproblemsolving.com/a/7/c/a7c3aa790b033d583e45895f3c7873e5154bb18c.png)
Note that at this point, 
forces 
.
now gives ![\[ \left( 3a^2-4a \right)^2 + a^2 \left( 3a^2-4a \right)\left( 2-2a \right) + \left( 2a^2-4 \right)\left( 2-2a \right)^2 = 0. \]](http://latex.artofproblemsolving.com/4/4/a/44a7145a619354c46158819a02556091be59f9b2.png)
Before expanding this, it is prudent to see if it has any rational roots. A quick inspection finds that 
is such a root (precisely, 
)
The only root 
is 
.![\[ c = \frac{a(3a-4)}{2-2a} = \frac{1}{8} \left( 7 - \sqrt{17} \right) \]](http://latex.artofproblemsolving.com/5/c/e/5cedfe5a14c0dc14ac2ad8d94c925db853d6f89a.png)
and ![\[ f(a,b,c)
= 3a^2+2ac+c^2
= \frac{1}{32} \left( 121 + 17\sqrt{17} \right)
. \]](http://latex.artofproblemsolving.com/c/9/0/c90708fee647b986cc0fa70de085fb892b6c71a8.png)
This is the last critical point, so we're done once we check this is less than 
.
;![\[ \frac{1}{32} \left( 121 + 17\sqrt{17} \right) \approx 5.97165. \]](http://latex.artofproblemsolving.com/3/4/1/341d40f0bc1f71d45c9667e28a94c264e9b764c1.png)
This completes the solution.
contradiction.
and satisfy ![\[ a^2+b^2+c^2 = 3. \]](http://latex.artofproblemsolving.com/0/b/d/0bd4f8fac6a40a1c5e80d9ec6248b2000a9c185e.png)
Show that ![\[ ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/4/a/3/4a3a07f268e8b8466e2523d60f68e3b69add1452.png)
and 
.
and:
for 
.
for a triangle 
.![\[ab+bc+ca\ge 3\sqrt[3]{a^2b^2c^2}\ge abc\]](http://latex.artofproblemsolving.com/2/3/9/239e232909ba58470014a9c1e61e0ed72d30b59b.png)
because 
,![\[3\ge \sum_{\mbox{cyc}}(\cos A+\cos B)^2.\]](http://latex.artofproblemsolving.com/4/1/e/41e879befc0392770f447ed1a603f6a528284066.png)
Note since triangle 
for some nonnegative 
,![\[3\ge \sum_{\mbox{cyc}}\left(\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}\right)^2 = 2\sum_{\mbox{cyc}}\frac{x^2}{(x+y)(x+z)}+2\sum_{\mbox{cyc}}\frac{yz}{(y+z)\sqrt{(x+z)(x+y)}}.\]](http://latex.artofproblemsolving.com/1/2/d/12ddbe56c59931fbb6116cbd8b87c807c733a519.png)
By rearranging, the problem amounts to checking![\[6xyz+3\sum_{\mbox{sym}}x^2y\ge 2\sum_{\mbox{cyc}}x^2(y+z)+2\sum_{\mbox{cyc}}yz\sqrt{(x+z)(x+y)}.\]](http://latex.artofproblemsolving.com/e/f/1/ef123180374f03331d46f539b07d85e31d2ef5dc.png)
By cancelling, it is equivalent to show![\[6xyz+\sum_{\mbox{sym}}x^2y\ge 2\sum_{\mbox{cyc}}yz\sqrt{(x+z)(x+y)}.\]](http://latex.artofproblemsolving.com/8/f/5/8f588f9cba0cfae70f6bd57a3c0527beede5f1e5.png)
But by AM-GM, the latter expression is at most![\[\sum_{\mbox{cyc}}(2xyz+y^2z+z^2y),\]](http://latex.artofproblemsolving.com/e/d/c/edca09990ebd2bf8b09c0ab85ad5917b4f4cef26.png)
so done.
as![\[0\le -a((b-1)(c-1)-1)+bc.\]](http://latex.artofproblemsolving.com/a/2/0/a20c7a6d8f25b0ebec703c1beb543cc523330afa.png)
This is clearly true since 
must be nonpositive.
and 
have the same sign. Then 
.
,
.
![\[c+ab \ge ac+bc+ab-abc.\]](http://latex.artofproblemsolving.com/6/d/c/6dc05a262e3fb7c92e8b3c0c44863a4726d44adb.png)
This means to prove the upper bound it suffices to prove 
.
.
contradiction. Hence, 
so wlog any two of 
or 
and 
so we have ![\[ab+bc+ca-abc=a(b+c)+bc(1-a)\geq 0\]](http://latex.artofproblemsolving.com/c/a/8/ca8cc0695484c910f6dac3780919825cb603e0c9.png)
So lower bound is proved.So now we need to prove upper bound so ![\[a^2+b^2+c^2+abc=4\]](http://latex.artofproblemsolving.com/0/5/0/050bd59f3f1c19aa5159b3b1901a4bb2a47dcdc4.png)
and we know that 
by AM-GM so now put this up![\[a^2+2bc\leq 4\implies bc(2+a)\leq 4-a^2=(2+a)(2-a)\\\implies bc\leq 2-a\]](http://latex.artofproblemsolving.com/5/6/f/56fa77a104934e960055cc4a0fdfd8beaf2fa609.png)
now lets see what we have until now,
for both the cases
![\[ab+bc+ca-abc\leq ab+2-a+ca-abc=2-a(1-b-c+bc)=2-a(1-b)(1-c)\leq 2\]](http://latex.artofproblemsolving.com/a/1/a/a1a63f4a952705cc03b54b7720def2d27e7d5751.png)
So we are done now !
Equality when 
Q.E.D.
then 
We have to prove 
Equality when 
Q.E.D.
follows from Erdos-Mordell from the circumcenter of ![\[abc+2-ab-bc-ca=\sum_{\text{cyc}}bc\cdot\frac{(y-z)^2}{2yz}\]](http://latex.artofproblemsolving.com/e/f/e/efe79336f7e51119827c4f0113e52c8e16c466fe.png)
which is nonnegative so we are done.
is at most 
with positive coordinates such that 
,
is given by the set of all points 
.
subject to the constraint 
.
achieves a global maximum at some critical point in the constraint set 
.
.
(either directly or if 
)
given the condition 
,![\begin{align*}
\nabla f &= [2a+b+c, 2b+c+a, 2c+a+b] \\
\nabla g &= [2a+bc, 2b+ca, 2c+ab]
\end{align*}](http://latex.artofproblemsolving.com/0/5/2/052124a799b7ae72539848a4b0b1f8c4b21854bf.png)
and observe that the equation 
implies that 
or 
and cyclic permutations. In particular, either 
.
so 
.![\[ab+bc+ca-abc = a^2+2a\left(2-a^2\right)-a^2\left(2-a^2\right) - 2 = \left(a-\sqrt 2\right)(a-1)^2 \leq 0\]](http://latex.artofproblemsolving.com/f/e/c/fec792717357d82f90bd5f6001ad2a16c6ef9985.png)
which is clearly true as 
.
Hence
or
Q.E.D.
and satisfy![\[ a^2 + b^2 + c^2 + abc = 4 .
\]](http://latex.artofproblemsolving.com/0/7/b/07b9ff1d86f856cf428bf6dcff37ae4507907777.png)
Show that![\[ ab + bc + ca - abc \leq 2.
\]](http://latex.artofproblemsolving.com/e/4/1/e414e5bf9bc548b454f6f1e9d77bb4091810c2a6.png)
we have:
works,
,
,
,
