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.
the points 
and 
are the feet of the altitudes through 
and 
respectively. The incenters of the triangles 
and 
are 
and 
respectively; the circumcenters of the triangles 
and 
are 
and 
respectively. Prove that 
and 
are parallel.
with 
prime and ![\[a^p+b^p=p!.\]](http://latex.artofproblemsolving.com/f/e/b/feb27894b1f6497559eeb094cfe92256bc26688e.png)
such that 
. Show that 
. When will equality hold?
be a cyclic quadrilateral with circumcenter 
, such that 
is not a diameter of its circumcircle. The lines 
and 
intersect at point 
, so that 
lies between 
and , and 
lies between and . Suppose triangle 
is acute and let 
be its orthocenter. The points 
and on the lines and , respectively, are such that 
and 
. The line through , perpendicular to , intersects at 
, and the line through , perpendicular to , intersects at 
. Prove that the points , , are collinear.
which satisfy the condition![\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]](http://latex.artofproblemsolving.com/4/b/d/4bddce124cf0da82cb8bb6c6ae2325239354f27c.png)
Find all possible values of 
be a positive integer and 
be a positive integer coprime to 
. Let 
, and for 
, define
Find, in terms of and 
, the greatest positive integer 
for which there exists an index 
such that 
is divisible by 
.
table are initially white. Alice and Bob play a game. First Alice paints of the fields in red. Then Bob chooses 
rows and columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of such that Alice can win the game no matter how Bob plays.
was meant in both places. Thanks, I'll fix it.
of course: as you say any large number suffices. The point is to capture the constraint set in some compact set, for which the boundary (
in this case) is a non-issue.
. Hence, we focus on the right side.![\[ U = \left\{ (a,b,c) \mid a,b,c > 0 \text{ and } a^2+b^2+c^2 < 1000 \right\}. \]](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 
. Excellent.![\[ f(a,b,c) = a^2+b^2+c^2+ab+bc+ca. \]](http://latex.artofproblemsolving.com/b/b/0/bb04a96cd90630a62afe19639303a5842a150e20.png)
It's equivalent to show that 
subject to 
. Over 
, it must achieve a global maximum. Now we consider two cases.
lies on the boundary, that means one of the components is zero (since is clearly impossible). WLOG 
, then we wish to show 
for 
, which is trivial.
, we may use the method of Lagrange Multipliers. Consider a local maximum 
. Compute ![\[ \nabla f = \left<2a+b+c, 2b+c+a, 2c+a+b \right> \]](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 
. Subtracting 
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.
, then 
, and this satisfies 
., 
, 
are pairwise distinct, then we derive 
, contradiction.
. That means (of course our conditions force 
). Now ![\[ 2a+2c = a+b+2c = \lambda (2c+ab)
= \frac{1}{2-a} (2c+a^2) \]](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 
) we have ![\[ c = \frac{3a^2-4a}{2-2a}. \]](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, 
). Now, we can expand and try to factor:
The only root in the interval 
is 
. You can guess what comes next -- write![\[ c = \frac{a(3a-4)}{2-2a} = \frac{1}{8} \left( 7 - \sqrt{17} \right) \]](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 
. This follows from the inequality 
; in fact, we actually have ![\[ \frac{1}{32} \left( 121 + 17\sqrt{17} \right) \approx 5.97165. \]](http://latex.artofproblemsolving.com/3/4/1/341d40f0bc1f71d45c9667e28a94c264e9b764c1.png)
This completes the solution.
contradiction.
and satisfy ![\[ a^2+b^2+c^2 +abc = 4 . \]](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)
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 
. Now, obviously 
and:
for , so:
. And we will use this in second equation:
for a triangle . Observe that![\[ab+bc+ca\ge 3\sqrt[3]{a^2b^2c^2}\ge abc\]](http://latex.artofproblemsolving.com/2/3/9/239e232909ba58470014a9c1e61e0ed72d30b59b.png)
because 
, which shows the first part of the inequality. For the latter part, rewrite it as showing![\[3\ge \sum_{\mbox{cyc}}(\cos A+\cos B)^2.\]](http://latex.artofproblemsolving.com/4/1/e/41e879befc0392770f447ed1a603f6a528284066.png)
Note since triangle is not obtuse, we can let 
for some nonnegative 
, so the problem amounts to showing![\[3\ge \sum_{\mbox{cyc}}\left(\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}\right)^2 = 2\sum_{\mbox{cyc}}\frac{x^2}{(x+y)(x+z)}+2\sum_{\mbox{cyc}}\frac{yz}{(y+z)\sqrt{(x+z)(x+y)}}.\]](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 
. Multiplying by and subtracting 
, we get 
. Adding 
![\[c+ab \ge ac+bc+ab-abc.\]](http://latex.artofproblemsolving.com/6/d/c/6dc05a262e3fb7c92e8b3c0c44863a4726d44adb.png)
This means to prove the upper bound it suffices to prove 
. Assume FTSOC that 
. Then,
contradiction. Hence, and we are done.
so wlog any two of 
can be 
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.
for 
acute. Then the stronger lower bound 
follows from Erdos-Mordell from the circumcenter of to its tangential triangle. has side lengths we can write in terms of by Law of Cosines.![\[abc+2-ab-bc-ca=\sum_{\text{cyc}}bc\cdot\frac{(y-z)^2}{2yz}\]](http://latex.artofproblemsolving.com/e/f/e/efe79336f7e51119827c4f0113e52c8e16c466fe.png)
which is nonnegative so we are done.
is at most 
. For the right side, let denote the open set of all points 
with positive coordinates such that 
, such that its closure 
is given by the set of all points with nonnegative coordinates such that 
.
subject to the constraint 
. By LM, 
achieves a global maximum at some critical point in the constraint set 
.
. Then 
(either directly or if 
), and thus it suffices to show that 
given the condition 
, which is clear.![\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 or 
. In this case, the constraint implies 
so 
. along . It remains to verify that ![\[ab+bc+ca-abc = a^2+2a\left(2-a^2\right)-a^2\left(2-a^2\right) - 2 = \left(a-\sqrt 2\right)(a-1)^2 \leq 0\]](http://latex.artofproblemsolving.com/f/e/c/fec792717357d82f90bd5f6001ad2a16c6ef9985.png)
which is clearly true as 
. This completes the proof.
such that 
.
such that 
,
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:
