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is inscribed in a circle of radius 2 with 
, and 
is a real number satisfying the equation 
, where 
, 
, 
. Find all possible values of .
satisfy the equation, We substitute 
Note that 
Now 
Now note that 
so since 
is decreasing for acute angles, 
Similarly 
so the inequality follows.
Thus 
Also, we need 
so 
Solving this gives 
in some order, so it follows that 
We substitute back to get 
I think?
. Now, note that ![\[ a^2+c^2 \le b^2 \le 4b \]](http://latex.artofproblemsolving.com/1/f/b/1fb990043ead259d4da62c1ee910295410c306c2.png)
since 
(the diameter of its circumcircle). Then, ![\begin{align*} 0 &= x^4 + ax^3 + bx^2 + cx + 1 \\ &= x^2 \left[ \left( x + \frac{1}{2} a \right)^2 + \left( \frac1x + \frac{1}{2} c \right)^2 + \left( b - \frac{a^2+c^2}{4} \right) \right] \\ &\ge 0+0+0 = 0. \end{align*}](http://latex.artofproblemsolving.com/9/0/6/906ca1500959a1104e84801556a41d70456aefe5.png)
In order for equality to hold, we must have 
, 
, and 
. This gives us 
, 
, 
. Solving for 
implies ![\[ \{a,c\} = \left\{ \sqrt6 \pm \sqrt 2 \right\}. \]](http://latex.artofproblemsolving.com/0/4/3/04386e54f28c097a1f59885da2c7e51f167ba811.png)
This gives the values claimed above; by taking 
, 
, 
as deduced here, we find they work too.
how many points? anyway here is a (correct) solution
we have
Equality for 
and 
which gives the desired solutions.
. We complete the square:
However since 
and 
, 
, so 
. Then by the Trivial Inequality all terms of the above expression are zero. In particular, 
, and since 
(as 
contradicts 
), ![\[0=x+\frac a2=\frac c2x+1\implies x=-\frac a2=-\frac2c.\]](http://latex.artofproblemsolving.com/5/c/7/5c72aea8b86929af46ef1dd90c624657d1c93471.png)
Thus we have 
and . Solving, 
, and the desired result follows.
. Completing the square twice, note
and 
lies inside a circle with diameter 4, 
. Hence all three terms are nonnegative and![\[x + \frac{a}{2} = 1 + \frac{c}{2}x = 4b - a^2 - c^2 = 0.\]](http://latex.artofproblemsolving.com/f/8/a/f8a6dd676cde3df7a3a81a9dc503f8561294cdf9.png)
In particular and 
; also 
. Hence and![\[\{a, c\} = \{\sqrt{6} + \sqrt{2}, \sqrt{6} - \sqrt{2}\} \implies \boxed{x = -\frac{\sqrt{6} \pm \sqrt{2}}{2}}\]](http://latex.artofproblemsolving.com/0/6/5/065c172676f108651b9fedb6d280be2b062fa732.png)
and 
. Now,
From this, we can conclude that 
. We can manipulate this to get:
So
From this, we can conclude that 
and 
. Also, note that this is the equality, so 
. Plugging this into 
and applying the Trivial Inequality yields 
.
, or 
. Applying the quadratic formula, we obtain
This simplifies to 
It is easy to verify that these both work.
turned into 
. Then, I played around with other cases and noticed I never got a negative result, so I tried to show that is was always 
with completing the square and showing that remaining term (which nicely has a degree of 
) was always positive
doesn't have any positive solution. So, we will replace with 
where is positive.
[because 
]
as 
and 
and 
or 
. 
or 
. Thus 
. But this is the value of .
or 
. Consider a solution . Observe that![\[0=x^4+ax^3+bx^2+cx+1 = (x+a/2)^2x^2+(cx/2+1)^2 + (b-c^2/4-a^2/4)x^2\ge\]](http://latex.artofproblemsolving.com/d/e/7/de729cce6f5b196e2e048e826b30f263a9cf87cb.png)
![\[(b-c^2/4-a^2/4)x^2.\]](http://latex.artofproblemsolving.com/d/b/a/dba0c4964180a31936640666a871f131706e3fd5.png)
As , this implies![\[b \le c^2/4+a^2/4 \le b^2/4.\]](http://latex.artofproblemsolving.com/2/9/b/29bec5a89ca64df05d3879cdffc34b39b2540acb.png)
Rearranging yields 
. It is clear that , so we get . Remark that equality holds, so 
. Note 
, so by the previous rearrangement we have![\[0=(x+a/2)^2x^2+(cx/2+1)^2.\]](http://latex.artofproblemsolving.com/5/a/0/5a0df21624a643e84f51d42db93240bb1f514809.png)
By the Trivial Inequality, this forces 
, so 
. Rearranging yields![\[ac=4.\]](http://latex.artofproblemsolving.com/b/a/4/ba45d118b15ec021e95b9552118b78c3817d5203.png)
Thus, we can write![\[(a+c)^2=a^2+c^2+2ac = 16+8=24.\]](http://latex.artofproblemsolving.com/2/7/a/27a5291788bf91f45daa39bfa40939bb9f81cd3d.png)
Then, we get that 
are roots to![\[0=x^2-(a+c)x+ac = x^2-2\sqrt{6}x+4.\]](http://latex.artofproblemsolving.com/c/d/e/cde21b4834158a05006da444045744bf6d8da272.png)
By the quadratic formula,![\[a = \frac{2\sqrt{6}\pm \sqrt{24-16}}{2} = \sqrt{6}\pm \sqrt{2}.\]](http://latex.artofproblemsolving.com/3/0/1/301abeda4f69b46ab0f5d6cc77bcd54481427150.png)
Finally, we conclude![\[x = \frac{-\sqrt{6}\pm \sqrt{2}}{2}.\]](http://latex.artofproblemsolving.com/7/a/7/7a71132727f575d1150e59f2d521fa9e12b4c1d7.png)
Equality can be attained by taking![\[(a,b,c) = \left(\sqrt{6}+\sqrt{2},4,\sqrt{6}-\sqrt{2}\right),\left(\sqrt{6}-\sqrt{2},4,\sqrt{6}+\sqrt{2}\right).\]](http://latex.artofproblemsolving.com/e/a/3/ea330281804ad99143410f5b64f445ec72ebae9e.png)
in this case. However, I didn't see any convenient way to force that (in fact, that isn't even the equality case). Then, I noticed that I wanted to use 
, so I tried to get 
to materialize. After a bit work, I managed to get the to show up by completing the square, then the rest of the proof was mostly abusing the equality case and manipulating to solve for 
.
are positive, is negative. Let 
So we have 
now seeing the 
and 
that gives us motivation to factor it as 
Claim: 
but we have 
since 
and 
we get that 
also From 
we get that 
Solving the equation gives us 
unless messing around somehow revealed this.
, so let 
. So 
. Divide everything by . So we get ![\[y^2-ay+b-\frac{c}{y}+\frac{1}{y^2}=0\]](http://latex.artofproblemsolving.com/7/8/8/788f70700df96a10f2833b783680bdce0afb84c9.png)
.![\[\left(y-\frac{a}{2}\right)^2+\left(\frac{1}{y}-\frac{c}{2}\right)^2=y^2-ay+\frac{a^2}{4}+\frac{c^2}{4}-\frac{c}{y}+\frac{1}{y^2}=\frac{a^2}{4}+\frac{c^2}{4}-b\ge 0,\]](http://latex.artofproblemsolving.com/d/0/3/d030806dcb85893b34ef807c5c1a07d0e73d7cdb.png)
as desired .
.
. Since , 
, which implies 
. 
.
and , so and .
, so 
and 
. This gives . and . So 
. Thus, 
.
or 
.
, which work.
, 
(the Pythagorean inequality), and 
, since the side of the triangle cannot be longer than the diameter of the circumcircle. Then:
so:
Equality holds everywhere we mentioned an inequailty, so and 
and 
. The latter two equations rearrange to and .
into the second equation, we have , and so 
. Then 
. Note that both of these values produce valid triangles subject to the problem conditions and, as a result, 
.
![\[ x^4 + ax^3 + bx^2 + cx + 1 = x^2\left(x + \frac{a}{2}\right)^2 + \left(\frac{c}{2}x + 1\right)^2 + \left(b - \frac{a^2}{4} - \frac{c^2}{4}\right)x^2.\]](http://latex.artofproblemsolving.com/b/9/5/b9580e3ef073db7702aadf11456ae8b9aa0ece96.png)
Now, we will show that 
. Notice that it is equivalent to show![\[ b\ge \frac{1}{4}(b^2 + 2ac\cos B)\iff 4\sin B\ge \frac{1}{4}(16\sin^2 B + 2ac\cos B)\]](http://latex.artofproblemsolving.com/7/6/4/76444d06a2080e163bec8de8cd462697e1aab148.png)
![\[ \iff \sin B\ge \sin^2B + \frac{1}{8}ac\cos B.\]](http://latex.artofproblemsolving.com/1/5/0/1501365da4b554be89d40af9d47a02bd9d246283.png)
Noting that 
and 
, we have that the above inequality is true, so 
is as well. Additionally, note that equality holds exactly when 
. Thus we have the equation![\[ x^2\left(x + \frac{a}{2}\right)^2 + \left(\frac{c}{2}x + 1\right)^2 + \left(b - \frac{a^2}{4} - \frac{c^2}{4}\right)x^2 = 0.\]](http://latex.artofproblemsolving.com/8/8/4/884742f56b27d48812d3e87decd2dd699fe54bbb.png)
Note that each individual term on the LHS is at least zero, so for equality to hold, each individual term must be exactly zero. This implies 
, 
, so . We also get 
. It follows that is a diameter of the circumcircle, so 
, and consequently . Hence, 
and 
, both of which are positive. Then,![\[ x = -\frac{a}{2} = \boxed{-\frac{1}{2}(\sqrt{6}\pm\sqrt{2})}.\]](http://latex.artofproblemsolving.com/3/0/f/30fbeac202ee7c60d6943e841acfbe1a08d3fcf9.png)
Constructions for both values of are 
, and it is easy to see that these satisfy the triangle inequality as well as the Pythagorean inequality 
, so we are done.
is negative, so set 
, now our equation is:
we can divide by 
since :
Claim: 
and , thus:
and the claim follows 
, , and 
only.
and cyclic by the extended law of sines.![\[ \sin(A)\cos(C) + \cos(A)\sin(C) \ge \sin^2 A + \sin^2 C. \]](http://latex.artofproblemsolving.com/4/7/4/474d7c9b0eae5131cd5b7cdc0c97b43d71e77b07.png)
Divide by 
:![\[ \tan A + \tan C \ge \tan A \frac{\sin A}{\cos C} + \tan C \frac{\sin C}{\cos A}.\]](http://latex.artofproblemsolving.com/4/1/c/41cba0fd3444d74626fcd9d151e401d93725aeef.png)
If 
, then clearly since 
, we must have 
, and also since 
, it follows that 
. Similarly, 
.
. 
The first term is 
if 
or 
. The second term is only if 
. Clearly is ruled out, so the only case left is 
. Since for equality to hold we require 
, it thus follows that 
, or 
. Straightforward computations give the answer of 
.
where we use Cauchy Schwarz on line two, the fact that 
on line three, AM-GM on line four and on line six. For equality to hold, we must have and 
(for the AM-GM equality). Also, must be negative for equality to hold on line 
. This implies that is 
, which can easily be checked to work.
![$$[ABC]=\frac{abc}{4R}=\frac{abc}{8}=\frac{ab\sin \angle BCA}{2}$$](http://latex.artofproblemsolving.com/4/2/8/428a30eec6252fa0a1ea9168e871c83289c5f19c.png)
so 
.
and 
.
.
,
.
and by our Lemma, we have 
Now we get the equality case so 
,
and so everything is an equality. In particular, 
.
is true, as then you can use the trivial inequality to finish. All you need to do then is prove it and then you're done.