AoPS Academy
be points on the sides 
respectively of a triangle 
. Suppose that 
and 
meet at 
and meet at 
, and 
and meet at 
, such that 
, and 
. Compute the ratio of the area of 
to the area of 
.
and 
intersect at different points 
and 
. The straight lines tangents to that pass through and intersect at 
. Let 
be a point on that is out of . The 
line intersects at 
again, the 
line intersects again to in 
and the line 
intersects again to the circumference in 
. Prove that the 
line passes through the midpoint of the 
segment.
are positive real numbers such that 
, prove that ![\[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]](http://latex.artofproblemsolving.com/0/b/e/0bedd9ed8362ac5202845f8a4d0d24519423e9b4.png)
with 
, points 
are such that 
and 
and 
, and 
and 
, as shown on the picture. Suppose that 
is a right angle. Prove that the points 
are collinear.
. Prove that![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](http://latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality. be an acute angled triangle.And altitudes 
and 
intersects at point 
.Let 
be a point on ray such that 
.Circumcircle of 
intersects line 
at and so prove that 
,![\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]](http://latex.artofproblemsolving.com/4/3/1/431b0ed375e037ee3d19e48a2bb4099f12f87784.png)
Proposed by David Stoner
such that![\[ x f(x + xy) = x f(x) + f \left( x^2 \right) f(y) \quad \forall x,y \in \mathbb{R}.\]](http://latex.artofproblemsolving.com/b/0/f/b0f1a1d3b9de5d23f1b5a2e9be3b88d8bd575883.png)
is not divisible by 
for any positive integer 
.
, s.t :
as 
. Then we may show that 
through angle chasing, redefine as 
. Then let 
, note that 
is cyclic and tangent to 
, likewise we may angle chase that 
is cyclic so 
and thus 
is tangent to 
, thus 
, and thus 
. Finally, the result is another painful angle chase.
lies on 
, where 
is the radius of by trig bash
and then angle/arc bash the required angle equality.
. Let the circumcircle of triangle 
intersects segments and 
again at 
and 
, respectively. Since 
it follows that 
is tangent to 
. lies on 
has equal length to the diameter of . Notice that 
Since 
, right triangles 
and 
are similar. Then 
Which is equal to the diameter of . Notice that the power of with respect to is equal to 
which is equal to 
. Denote the radius of as 
. The power can also be written as 
so 
Then clearly 
, finishing the claim.![[asy]
import geometry;size(12cm);pair A=(0,0);pair B=(4,0);pair D=(0,-3);pair C=D+rotate(220)*(A-D);pair E=C+.17(B-C);pair G=intersectionpoint(line(D,E),line(A,B));pair tac=circumcenter(G,E,B);line cat=line(B,B+rotate(90)*(tac-B));pair F=intersectionpoint(cat, line(E,D));pair O=circumcenter(C,D,E);pair P=intersectionpoint(circle(A,D,G),circle(C,E,D));circle w=circle(B,D,E);point[] X=intersectionpoints(w,line(A,D));point[] Y=intersectionpoints(w,line(B,F));pair X=X[1];pair Y=Y[0];pair K=circumcenter(A,B,C);pair Z=intersectionpoint(line(B,D),line(X,Y));pair Q=intersectionpoint(line(A,C),line(B,P));filldraw(A--B--D--cycle,white+palegreen);filldraw(B--X--Y--cycle,white+palegreen);filldraw(Z--X--B--cycle, palegreen);draw(A--B--C--D--cycle);draw(F--G);draw(circle(C,D,E), dotted);draw(P--F);draw(circle(A,D,G),dashed+purple);draw(w,blue);draw(B--F);draw(circle(G,E,B),red);draw(G--A);draw(A--C,grey ); draw(B--P,grey );draw(circle(A,B,C),dashed+orange);perpendicularmark(line(A,D),line(A,B));draw(A--D,StickIntervalMarker(1,1));draw(D--C,StickIntervalMarker(1,1));markangle(B,Q,A,radius=.4cm);markangle(B,Q,A,radius=.5cm);markangle(F,P,D,radius=.4cm);markangle(F,P,D,radius=.5cm);draw(G--P--D,grey);dot("A", A,dir(60));dot("B", B,dir(20));dot("C", C,dir(-30));dot("D", D,2*dir(180));dot("E",E,1.5*dir(100));dot("F",F,dir(-20));dot("G",G,dir(160));dot("P",P,dir(180));dot("O",O,dir(250));dot("X",X,dir(160));dot("Y",Y);dot("Q",Q,dir(-25));
[/asy]](http://latex.artofproblemsolving.com/8/0/4/804e42ee6ebbbf626198747fd3b000247b300d3c.png)
and 
are cyclic. 
Then to finish the problem 
, while allowing 
to vary.
whose diagonals meet at (see the figure). Assume, 
. The circumcenter of the trapezoid lies on the perpendicular bisectors of both 
and 
. Call this circumcircle as 
. moves along the line 
(in the part, which is inside the circle ). Let the projection of on 
be . The circle 
that is centred at with radius 
meets the diagonal 
at . This construction gives us exactly the configuration described in the original problem. is defined as the intersection of 
and . Then 
and are tangent to each other at . This follows from the fact about harmonic points. Note that 
is harmonic, so is the Apollonian circle of and , i.e. locus of such that 
. This implies ![\[ \angle PED = \angle PFD + \angle PTD = \angle PYP' + \angle DPP', \]](http://latex.artofproblemsolving.com/5/6/9/569945d95679cdba4a56b4ae6bcb01aac393369e.png)
so the circles are tangent to each other at . meets again at . A simple calculation shows that ![\[ \frac{DC'}{DA} = \frac{r}{R}\cdot \frac{BT}{BE}, \]](http://latex.artofproblemsolving.com/1/5/6/15653c722f6562042d20edd7cc571d888a962e80.png)
where and are radii of and . On the other hand, we have 
and ![\[ \frac{r}{R} = \frac{PE}{PT}. \]](http://latex.artofproblemsolving.com/d/e/3/de3ac739b36ecaabd016adeb8df0be788642b546.png)
Now, noting that is the Apollonian circle of and , we get 
. This yields 
. coincides with the point given in the problem.
, where the points 
are chosen on the sides, we further obtain that lies on 
.
, where 
.
be the center of 
. Then 
. Thus, it suffices to show that 
. Indeed, we have ![\[ \angle AQB + \angle PED = \angle APB + \angle PBC + \angle PFX +\angle PTD = \]](http://latex.artofproblemsolving.com/c/f/e/cfeed8fd236d80c2d763025ea0bd675058cff6a4.png)
![\[ = \angle ATP' + \angle BYP' + \angle PYH' + \angle PYP' + \angle P'TD \]](http://latex.artofproblemsolving.com/0/5/1/0516d8e5d54d21da857e6309f5a45c6d50bd130a.png)
![\[ = ( \angle ATP' + \angle P'TD) + (\angle BYP'+ \angle P'YP + \angle PYH') \]](http://latex.artofproblemsolving.com/9/0/e/90edfe5448f8203e6b4ca9a1c8feec81e9690003.png)
![\[ = \angle ATD + \angle BYH' = \angle ATD + \angle BHH' = 90^{\circ}. \]](http://latex.artofproblemsolving.com/c/9/9/c99dc83026a4dd48b99fe59ce1a42add7e517ab9.png)
This finishes the proof.
,such that 
P
.
be a point on side 
such that 
.
,
satisfying 
.
and 
.
Let 
so that 
tangent 
and 
tangent to 
,
,
