[/quote]
,
,
,
.
,
,
,
be the point on 
.
and 
,
,
,
are collinear.
.
-
at 
,
and 
.
on lines 
and 
such that quadrilateral 
is a rhombus.
be non-negative real numbers, such that 
Prove that![\[52 + 17(\sqrt a + \sqrt b + \sqrt c + \sqrt d)^2\geqslant 9(ab+ bc + ca + da + db + dc)^2.\]](http://latex.artofproblemsolving.com/6/e/0/6e06719b43ee8df511295af3581b487ddfb17f5c.png)
and 
prove that![\[\sqrt a + \sqrt b + \sqrt c \geqslant ab+bc+ca.\]](http://latex.artofproblemsolving.com/c/e/9/ce9dc64acc63c6bcce137e963ca8f86ab75018bf.png)
[/quote]
and 
and there is no other palindrome number between them. Determine the possible values of 
.
to 
.
coins, the sum of denominations of which is 
.
at which we can confidently say that Vlad is able to pay 
,
be the largest prime number that divides 
.
,
.
,
,
are all less than 
.
.
be a cyclic quadrilateral. Let 
which is outside the circumscribed circumference of 
.
if and only if 
.
.
of real numbers which satisfy the equality![\[ b_1^2 + b_2^2 + \cdots + b_k^2 = 4\left(a_1^2 + a_2^2 + \cdots +a_k^2\right)\]](http://latex.artofproblemsolving.com/1/f/8/1f8225dd21e6249fff12f0c072daf9355b3ac927.png)
where 
for 
.
.
with 
such that 
for every 
.
and 
and 
.
and 
are perpendicular.
,
![\[\frac{3(a^2 + b^2 + c^2 + d^2)}{ab + bc + ca + da + db + dc} + \frac{256abcd}{(a + b + c + d)^4} \geqslant 3.\]](http://latex.artofproblemsolving.com/0/5/d/05dc58b0b3bbc1347ed7fa2a837b8f197b2c17b1.png)
Y by Davi-8191, tenplusten, anantmudgal09, OlympusHero, Adventure10, HWenslawski, Jc426, math_comb01, Mango247, Rounak_iitr, and 3 other users
In a convex quadrilateral 
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](//latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.Attachments:
This post has been edited 2 times. Last edited by djmathman, Aug 1, 2015, 2:53 AM
Reason: formatting
Reason: formatting
,
,
,
be the orthogonal projections of the point P on the lines AB, BC, CD, DA, and let 
,
,
,
be the reflections of P in these lines, or, equivalently, the reflections of P in the points 
is the line BD. Hence, the points 
, we have 
.
and 
,
.
.
.
and 
,
has the segment AP as diameter. Therefore, by the Extended Law of Sines, we have
,
.
.
has remained the same, so 
is the perpendicular bisector of 
are the symmetrics of 
respectively wrt 
is an isosceles trapezoid which has 
to 
and 
respectively on the circumcircle of 
and 
since they subtend equal arcs. The circle also gives us 
by equal arcs 
and 
.
since they are vertical, and 
by subtraction. The equal angles and two adjacent equal sides force 
and 
at 
and 
.
and triangles 
and 
are similar in ratio 
.
by subtraction, and triangles 
and 
are also similar in ratio 
and 
are similar in ratio 
,
.
is 
.
is cyclic, its reflection 
is also cyclic and the original quadrilateral 
. ]
to 
.
,
and by an analogous argument that 
.
so that by definition, 
.
,
be the associated intersection point for quadrilateral 
.
is cyclic, so you agree that 
.
,
,
while 
.
.
and 
and 
are both isosceles trapezoids and therefore 
(ii) Assume 
be a point such that 
,
which implies 
are isogonal conjugates with respect to 
,
bisects 
.
be the circumcenter of 
is cyclic. Since 
,
and 
or 
is cyclic. However, the latter is absurd because 
and 
,
on 
to the corresponding point 
.
is cyclic, then define 
.
such that 
and 
,
so one direction is completed
.
.
both belong to the 
,
.
and thus 
is also cylcic. Hence 
is cyclic.
denote the reflection of 
and 
be 
.
,
,
are cyclic with diameters 
,
respectively.
and 
Now, because 
.
,
.
In addition, we have 
Thus, 
is cyclic if and only if 
.
It follows that 
is cyclic, as this is the harder portion.
and 
and 
.
). In the diagram below, I've extended PC and marked the equal angles red, in case it is confusing what it means for 
and 
to be isogonal in this context.
and 
,
and 
excenters, respectively. Now comes the key claim:
and 
are both cyclic.
is cyclic, as the others are analogous. Observe that 
is the angle bisector of 
,
,
!
on 
,
.
bisects 
.
is the external angle bisector of 
,
.
,
.
and 
.
,
is an isoceles trapezoid, so 
.
.
,
is an isoceles trapezoid as well. This means that 
and 
share a perpendicular bisector.
and 
.
to meet the circumcircle of 
respectively.
so it follows that 
and 
which implies that 
is an isosceles trapezoid with legs 
we have 
so 
It follows that 
is an isosceles trapezoid with legs 
so 
is parallel with 
we have 
so 
Since 
so 
as desired. 
such that the fourth point is not outside the circumcircle of those three points.
then 
and let 
be the second intersections of the lines 
with 
(a property we proved in the necessity that it must be isosceles), but since 
intersects that circle at only two points, either 
is an isosceles triangle. We proceed when 
because the other case is analogous.
Since P lies on the perpendicular bisector of 
,
,
which implies that 
It follows that P lies on the perpendicular bisector of 
,
is an isosceles trapezoid (perpendicular bisectors are parallel). Therefore, since 
by SSS congruence, we have 
Hence, 
,
.
share a perpendicular bisector, and the result follows.
.
such that 
and 
similarly. The triangles 
and 
are congruent as 
,
.
,
is cyclic, and similarly 
is cyclic. It follows that all the points 
are concyclic, as needed.
and 
,
and 
are isogonal in 
.
and 
are isosceles trapezoids (with 
)
are collinear, as are 
,
which lies on the perpendicular bisector of 
by symmetry.
be points and 
such that 
are isogonal in 
,
etc. The isogonality condition is equivalent to
Upon expansion and simplification this means that any 
)
,
,
and 
.
,
.
is placed at 
,
so 
but 
.
are valid positions for 
, then construct 
is cyclic. Let 
intersect again a
.
and 
the midpoint of 
are collinear.
.
which finishes.
.
means orange, 
means green and 
and 
are like the diagram that is (hopefully) attached.
,![\[ \measuredangle PAC = \measuredangle ACP \iff \measuredangle BAC + \measuredangle CBD + \measuredangle DCA + \measuredangle ADB = 0. \qquad (\spadesuit). \]](http://latex.artofproblemsolving.com/7/e/5/7e57e142c8f1fa0c11c8588dfc18cc3f5af78fa0.png)
A careless reader would be forgiven for thinking that 
implies the problem or at least one direction, but it turns out the situation is more subtle. The second part analyzes the angle conditions more carefully and provides a complete proof.
.
.
,![[asy]
import geometry;
size(9cm); pair A = (-0.50099,1.69422); pair B = (-1.8,-0.6); pair C = (0.09182,-1.79600); pair D = (1.55802,-0.60384); pair P = (-1.04359,-0.19363);
draw(circumcircle(A,B,C), gray+dashed); filldraw(A--B--C--D--cycle, invisible, black); draw(B--D); draw(A--C);
draw(C--P--A, brown); draw(B--P--D, gray);
markangle(radius=15, n=2, P, A, C, red, StickIntervalMarker(1,1, red)); markangle(radius=15, n=2, A, C, P, red, StickIntervalMarker(1,1, red)); markangle(radius=10, n=3, A, P, B, deepgreen, StickIntervalMarker(1,2, deepgreen)); markangle(radius=10, n=3, C, P, D, deepgreen, StickIntervalMarker(1,1, deepgreen)); markangle(radius=12, n=1, B, A, C, blue); markangle(radius=12, n=1, C, B, D, blue); markangle(radius=12, n=1, D, C, A, blue); markangle(radius=12, n=1, A, D, B, blue);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$P$", P, dir(45));
[/asy]](http://latex.artofproblemsolving.com/0/e/c/0eca105f1f7fddffcbb253a96d5434767c875c18.png)
and 
in the claim (colored green with three rings) so that their only dependence on 
and 
(colored red with two rings), as follows: 
Since the claim says 
,
in terms of only 
These four latter angles are colored blue with one ring in the figure. This proves 
.![\[ \measuredangle BAC + \measuredangle CBD + \measuredangle DCA + \measuredangle ADB = 0 \]](http://latex.artofproblemsolving.com/b/8/6/b86d4b46e9417c6e66aa186a00663048a68a1057.png)
is equivalent to 
.
meets line 
.
meets line 
,
.
(that is, 
)
if and only if 
is not degenerate) and the second lemma turns our angle condition into 
and hence 
be the point for which 
is cyclic and harmonic; then 
.
(the corresponding Apollonian circle only meets line 
and 
,
.![\[p=\frac{-abc\overline{q}^2+(ab+bc+ca)\overline{q}-a-b-c+q}{q\overline{q}-1}\]](http://latex.artofproblemsolving.com/8/b/c/8bc439b8cf9ac0eee00e9373e153f7d6533c6056.png)
![\[p+q+abc\overline{p}\overline{q}=a+b+c.\]](http://latex.artofproblemsolving.com/9/5/b/95b1c22e456f315e9e9034d6f227186436c7af22.png)
This entire bash was performed from start to finish on paper, so I will only give a summary here of what happened. After expanding the ![\[\frac{1}{bc}(b+c-q)(b+c-bc\overline{q}-p)-b\overline{p}-c\overline{p}+bc\overline{p}\overline{q}=1-\frac pa-a\overline{p}.\]](http://latex.artofproblemsolving.com/4/6/d/46dcb22d266fd73c4b0d4fdb086a716d229da9e0.png)
The 
can be manipulated using our second well-known formula, and similarly for the 
on the LHS. Finally, replace the remaining appearances of 
using the first well-known formula. We end up with the sad-looking equation![\[(a+b+c-q)\left(b+c-bc\overline{q}+\frac{bc}{a}\right)(q\overline{q}-1)=(b+c)\left(-abc\overline{q}^2+ab\overline{q}+bc\overline{q}+ca\overline{q}-a-b-c+q-\frac1a q^2+\left(1+\frac ba+\frac ca\right)q-\left(b+c+\frac{bc}{a}\right)+\overline{q}bc\right).\]](http://latex.artofproblemsolving.com/e/6/6/e667069428756a4ed27d6f86a26c79ea5c3fbedb.png)
Thankfully for us, our goal is simply to show that![\[bcq\overline{q}-(b+c)q+(b^2+bc+c^2)-bc(b+c)\overline{q}=0\]](http://latex.artofproblemsolving.com/9/a/2/9a2bc1c53cfe74ec2b6859dbc0c4112efc86947e.png)
(this is the condition for 
)![\[(bcq\overline{q}-(b+c)q+(b^2+bc+c^2)-bc(b+c)\overline{q})(q-a)\left(\overline{q}-\frac1a\right),\]](http://latex.artofproblemsolving.com/9/e/d/9ed89b0198357c0c9ebca7a53f8deae6811ad723.png)
which can be checked by verifying the coefficients of 
since it's equal to its conjugate. (None of these coefficients end up being more than a short line long.) ![[asy]
import geometry;
size(10cm);
pair A = (8,5.75);
pair B = (10,5);
pair D = (0,0);
path circ = circumcircle(A,B,D);
pair C = intersectionpoints(line((0,0),(1,0)),circ)[0];
pair P = intersectionpoint(
reflect(bisector(line(C,D),line(D,A)))*line(B,D),
reflect(perpendicular(B,bisector(line(A,B),line(B,C))))*line(B,D)
);
// if
pair I_ = incenter(B,P,D);
pair E_ = intersectionpoint(line(P,I_),line(A,D));
pair F = intersectionpoint(line(C,P),line(B,E_));
draw(B--F--P,blue); draw(P--I_,blue);
draw(incircle(B,P,D),blue); draw(I_--E_,blue);
draw(circumcircle(A,B,P),dotted); draw(circumcircle(F,E_,P),dotted);
// only if
pair M = (A+C)/2;
pair B_ = reflect(line(M,P))*B;
pair D_ = reflect(line(M,P))*D;
draw(P--M,red); draw(P--B_--D_--P,red); draw(B_--C--D_,red); draw(B--B_,red); draw(D--D_,red);
//
markangle(2,A,P,E_,radius=10);
markangle(2,E_,P,F,radius=10);
markangle(2,P,A,C,radius=10);
markangle(2,A,C,P,radius=10);
draw(B--D); draw(A--C); draw(B--P--D); draw(A--P--C);
draw(circumcircle(A,B,C),dotted);
draw(A--B--C--D--A);
dot("$A$",A,(0,1));
dot("$B$",B,(0.5,1));
dot("$C$",C,(1,-0.5));
dot("$D$",D,(-1,-0.5));
dot("$E$",E_,(-0.5,1),blue);
dot("$F$",F,(-0.5,1),blue);
dot("$P$",P,(0.5,-1.5));
dot("$M$",M,(-0.25,-2.5),red);
dot("$I$",I_,(0.25,1.25),blue);
dot("$B'$",B_,(0.5,1),red);
dot("$D'$",D_,(-0.5,-1),red);
[/asy]](http://latex.artofproblemsolving.com/5/d/f/5df8b1c682d9c50f9e024df443f6dff58ed23918.png)
be the incenter of 
and 
.
and so is 
.
is cyclic since 
.
and 
are similar (because of 
and 
)
.
and let 
.
and 
are cyclic quadrilaterals.
,
.
and 
,
is isosceles and hence 
.
.
by symmetry. Thus, 
and 
.
gives us 
)
is cyclic. By the symmetry, 
are both isosceles trapezoids, and thus both are cyclic. Note that 
