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.
+4 w
such that the base-
integer 
is a perfect square and the base- integer 
is a perfect cube.
positive integers 
such that for every index 
, with 
, 
is a multiple of 
. Determine the greatest possible value of the largest of the numbers.
be a triangle. Let 
and 
be the points in which the median and the angle bisector, respectively, at 
meet the side 
. Let 
and 
be the points in which the perpendicular at to 
meets 
and 
, respectively. And 
the point in which the perpendicular at to meets 
produced.
is perpendicular to .
and 
such that for all 
if 
then 
Find the smallest of sum of all elements of 
such that there exist positive integers 
for which the following holds 
for all 
. where 
is a perfect cube
of a prime number 
and a positive integer 
for which 
is an integer.
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side at , and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
opposite the vertex is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)
such that if 
and 
are distinct rational numbers satisfying 
or 
, then 
? Justify your answer.
be a convex hexagon satisfying 
, 
, 
, and![\[
AB \cdot DE = BC \cdot EF = CD \cdot FA.
\]](http://latex.artofproblemsolving.com/f/9/5/f954fbdd4d7daefddbcdd5c66fb567bac92f85f2.png)
Let 
, 
, and 
be the midpoints of 
, 
, and 
. Prove that the circumcenter of 
, the circumcenter of 
, and the orthocenter of 
are collinear.
be the midpoint of 
, 
the midpoint of 
, and so on. Note that 
passes through 
and 
by trapezoid properties, and so on. The key claim is this:
lie on a circle.
. So 
are cyclic, and so on. Since , , and aren't the same point (otherwise the orthocenter is undefined and the problem is dead anyways), we have that all 
midpoints are cyclic, as desired. 
, 
and 
share a midpoint, and so 
also shares the same circumcenter as the midpoints. Now, apply complex numbers to find that the centroids of 
, 
, and are collinear. Taking a 
homothety at their shared circumcenter, we find that their orthocenters are collinear as well, and we are done!
then the problem still follows from homothety things so yay.
be the orthocenter of , 
be the medial triangle of 
, and 
be the medial triangle of 
. By parallel sides , , 
are collinear. We have 
which is symmetric by the length condition. Thus by symmetry all of 
, 
, , 
, 
, 
have equal power wrt (), so they lie on a circle concentric with 
.
, the center of 
, is the orthocenter of 
. Similarly the center 
of 
is the orthocenter of 
. Use complex numbers with the center of as origin. Compute 
It follows that 
, implying the conclusion.
is the circumcenter of 
. mean?
mean? has coordinates 
, and similar for the other points, then has coordinates![\[(x_A+x_C+x_E-2x_O,y_A+y_C+y_E-2y_O).\]](http://latex.artofproblemsolving.com/8/3/9/83966e2611d596d8900a3341f427e0e837fbf8e3.png)
Equivalently, you can choose an arbitrary origin, and treat every point as the vector from the origin to that point. Then, is the point you get from adding the vector to , the vector to , and the vector to 
, and subtracting two times the vector to .
, 
, 
, 
, 
, 
. In general, observe that 
and (in directed lengths) 
, so 
and 
are homothetic and directly congruent. This implies they are translations of each other.![[asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=red; pen sec2=lightred; pen tri=purple; pen qua=pink+fuchsia; pen fil=invisible; pen sfil=invisible; pen sfil2=invisible; pen tfil=invisible;
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filldraw(X--Y--Z--cycle,tfil,tri+dashed); filldraw(MD--MB--MF--cycle,tfil,tri); filldraw(MA--ME--MC--cycle,tfil,tri); filldraw(circumcircle(Dp,Bp,Fp),sfil2,sec2+dashed); filldraw(circumcircle(Ap,Ep,Cp),sfil2,sec2+dashed); draw(Fp--A,sec); draw(Ep--D,sec); filldraw(Dp--Bp--Fp--cycle,sfil,sec); filldraw(Ap--Ep--Cp--cycle,sfil,sec); filldraw(A--B--C--D--EE--F--cycle,fil,pri);
dot("\(A\)",A,E); dot("\(B\)",B,SE); dot("\(C\)",C,SW); dot("\(D\)",D,W); dot("\(E\)",EE,NW); dot("\(F\)",F,NE); dot("\(D'\)",Dp,Dp); dot("\(B'\)",Bp,W); dot("\(F'\)",Fp,NE); dot("\(A'\)",Ap,N); dot("\(E'\)",Ep,SW); dot("\(C'\)",Cp,SE); dot("\(X\)",X,E); dot("\(Y\)",Y,SE); dot("\(Z\)",Z,W); [/asy]](http://latex.artofproblemsolving.com/8/1/7/8175c98c17679c75afd143731753a62af53561f7.png)
, then the circumcenters of and 
coincide.![\[\operatorname{Pow}(A,(D'B'F'))=AB'\cdot AF'=BC\cdot FE,\]](http://latex.artofproblemsolving.com/8/b/7/8b72171597b6a48fdbf8533a65c6716e6dd09d6b.png)
which is fixed, so 
, 
, 
have equal power with respect to 
. 
and 
be the circumcenters of and 
. I contend that in general, the midpoint of 
coincides with the orthocenter of 
.
and 
be the medial triangles of 
and 
, so their orthocenters are and . It is easy to check that 
, 
, 
are the midpoints of 
, 
, 
.
and 
are translations of each other, and is their vector average, so we conclude the orthocenter of is the midpoint of . This completes the proof.
be the midpoint of 
, 
be the midpoint of , and so on. Once you get that 
is cyclic (which has been proven above so I won't prove it again) and that 
, etc, you can actually directly finish with complex once you realize that the orthocenter of is the midpoint of the orthocenters of 
and 
. Indeed, set up 
to be the unit circle, with values 
; then by standard complex formulas you get that 
, and similar for and .
(which is the midpoint of the orthocenters of and 
), you just need to check that 
. Since lies on 
, this is equivalent to 
After a not terrible amount of computation, this condition actually nicely reduces to 
which is equivalent to the vectors 
being parallel to 
after you plug in 
, etc. However, this is obvious from the definitions of 
.
and 
, so you're done.
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, 
, 
denote the midpoints of , 
, , and let 
, 
, 
denote the midpoints of , , 
. Let be the circumcenter of . First, we collect some immediate observations.
, , , all lie on the mid-segment of and 
. Similarly, sets 
and 
are colinear.
, so 
. Similarly, 
and 
., , , , , are all concyclic with center .
so , , , are concyclic. However, from the observation above, the center must be , so by repeating this with other two sides, we get the claim.
, 
, are colinear, but these three triangles have the same circumcenter . Thus, it suffices to show that the three centroids are colinear (by Euler line). The easiest way is to use vectors. We have
so summing up gives
done.
be a convex hexagon satisfying , , , andLet , , and be the midpoints of , , and . Prove that the circumcenter of , the circumcenter of , and the orthocenter of are collinear.
. Took me ~5 hours to find the following solution, but the amount of satisfaction is immeasurable.
to be the midpoint of segment .
are concyclic, and we call this circle 
.
and 
are collinear. are collinear. Note that since 
and are the midpoints of segments 
and 
. Then 
. Similarly, we get 
. This is enough to prove that are collinear. Similarly, we could prove that are collinear.
are concyclic.![\[ \frac{AB}{BC} = \frac{FE}{ED} \]](http://latex.artofproblemsolving.com/5/1/d/51d8546943754f6f1121fe2591de16f5bb8d236e.png)
Since 
, we have 
. These two conditions force 
. Similarly 
and 
.
maps 
and 
maps 
. Furthermore, from the previous part, we get that 
, therefore![\[ \measuredangle M_{BF} M_{DF} M_{AC} \equiv \measuredangle M_{BF} M_{DF} Z = \measuredangle BDC = \measuredangle FEA = \measuredangle ZM_{EC} M_{AC} \equiv \measuredangle M_{BF} M_{EC} M_{AC} \]](http://latex.artofproblemsolving.com/1/1/6/116db32a2125c53cce373c85afcf5e556ad414c5.png)
implying the desired concyclic condition.
and are concentric.
. Notice that the homothety proves 
and the homothety 
proves 
, which implies the desired equality. Now, this implies that the midpoint of is the same as midpoint of 
. Furthermore, since 
lies on , then their perpendicular bisector intersects at , center of , which by definition must be also circumcenter of as well.
are the same point. To prove that 
are collinear. We can take homothety 
and by a famous theorem on Euler line states that 
lies on the Euler line of the triangle with the ratio 
. Therefore, it suffices to prove that![\[ G(\triangle ACE), G(\triangle BDF), G(\triangle XYZ) \]](http://latex.artofproblemsolving.com/7/6/0/760e71880a9280ac2fac8fbbd9f055b3ae87cfe8.png)
are collinear.
be six points in plane, no three of which are collinear. Let 
be the midpoints of 
. Then the centroid of 
and are collinear.
. Therefore, we get![\[ \frac{2}{3} (X + Y + Z) = \frac{1}{3} (A + C + E) + \frac{1}{3} (B + D + F) \]](http://latex.artofproblemsolving.com/1/0/5/1051d1d84fdcb8b1a3865d0427359a7d4ae40349.png)
Therefore, 
is the midpoint of segment formed by 
and 
, and we are done.
Solved with some help from DrMath:
be the midpoints of 
respectively; obviously common midlines imply that 
and 
From the length and parallelity conditions, we get 
and 
so by SAS similarity 
hence 
Since 
and 
this implies that 
and by similar logic we have 
so 
is cyclic and symmetrically 
as well. We can further observe that by homothety 
so 
and share a common perpendicular bisector, which implies that and are concentric.
be the respective orthocenters of 
then setting the origin as the common center of and 
we have![\[\vec{O_1}=\vec{B_1}+\vec{D_1}+\vec{F_1}=\vec{A}+\vec{C}+\vec{E}\]](http://latex.artofproblemsolving.com/3/2/2/322bcbdd38042978d1fe6144caf08bae6c6442d1.png)
![\[\vec{O_2}=\vec{X}+\vec{Y}+\vec{Z}=\frac{1}{2}\left(\vec{A}+\vec{B}+\vec{C}+\vec{D}+\vec{E}+\vec{F}\right)\]](http://latex.artofproblemsolving.com/5/9/d/59dd55e5902101764cda37d25a96f73b8adbb2d9.png)
![\[\vec{O_3}=\vec{A_1}+\vec{C_1}+\vec{E_1}=\vec{B}+\vec{D}+\vec{F}.\]](http://latex.artofproblemsolving.com/5/3/8/5380442eb6d6df1be3f16bb06c153466ce41ab6b.png)
Clearly must then be the midpoint of 
but 
are the circumcenters of 
respectively, so we're done. ![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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and similarly for others. Let 
and 
be the circumcenter and circumradius of and . Let 
, and define and 
similarly.
.
and similarly for others. Cyclicly multiplying all of them shows that![\[\frac{AE}{BD}\cdot\frac{EC}{FB}\cdot\frac{CA}{DF}=1\Longrightarrow AC\cdot CE\cdot EA=BD\cdot DF\cdot FB.\]](http://latex.artofproblemsolving.com/4/5/3/453ebb7a48135b7a90edd80e9d4b8a6c1d0d2706.png)
Since the area of a triangle with side lengths 
and circumradius 
is 
, to show the claim it suffices to show that and have the same area. This follows from the parallelisms.![\[[ACE]=[B_1AC]+[A_1CE]+[C_1EA]+[A_1B_1C_1]=[B_1DF]+[A_1FB]+[C_1BD]+[A_1B_1C_1]=[BDF]. \blacksquare\]](http://latex.artofproblemsolving.com/4/2/b/42b3384ef60e979de478f1cc28365a94b6fa943c.png)
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draw(O1--A,blue);
draw(O2--D,red);
draw(O1--foot(O1,B,C), dotted);
draw(O2--foot(O2,B,C),dotted);
draw(H--foot(H,B,C),dotted);
draw(O1--O2);
draw(anglemark(M1,O1,A,5));
draw(anglemark(D,O2,N1,5));
draw(rightanglemark(B,foot(O1,B,C),O1,4));
draw(rightanglemark(B,foot(O2,B,C),O2,4));
draw(X--Y--Z--cycle);
dot("$X$",X);
dot("$Y$",Y,dir(270));
dot("$Z$",Z,dir(180));
dot("$A_1$",A1,dir(120));
dot("$B_1$",B1,dir(30));
dot("$C_1$",C1);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$O_1$",O1,dir(270), blue);
dot("$O_2$",O2,dir(180), red);
dot("$H$",H,dir(180));
dot("$M$",M1,dir(350));
dot("$N$",N1,dir(60));
[/asy]](http://latex.artofproblemsolving.com/2/d/b/2db9e037009851e53701fab5503192f289775fea.png)
.
which is true, since 
. be the midpoint of segment 
; we claim that is the orthocenter of . Since 
, it suffices to show that 
. Let and be points on , such that 
and 
are perpendicular to . Then by claim 2, we have![\[\measuredangle MO_1A=90^\circ-\angle (O_1A,BC)=90^\circ-\angle(BC,O_2D)=\measuredangle DO_2N.\]](http://latex.artofproblemsolving.com/8/8/8/888f26452fc77d0fd8f817bb23f4235650f7d811.png)
Therefore, by the law of sines,![\[AM=\frac{O_1A}{\sin\angle O_1MA}\cdot\sin\angle MO_1A=\frac{O_2D}{\sin\angle O_2ND}\cdot\sin\angle NO_2D = DN.\]](http://latex.artofproblemsolving.com/4/a/7/4a7b3b5c869754c9bcd8e2debc293003c6023ce9.png)
Therefore, is also the midpoint of segment 
. Consequently, 
is parallel to and hence perpendicular to 
as desired.
so that 
is a parallelogram. Define 
similarly. Note that 
and 
Similarly, 
thus 
all have equal power wrt 
So the center of is the center of 
call this 
Similarly, the center of is the center of 
call this 
are the orthocenters of the medial triangles of 
and 
It's not hard to see 
are parallelograms, so is the midpoint of both 
and 
and 
is (since 
is a parallelogram). Using the same argument on 
as well yields that the midpoint of the orthocenters of and 
which is the midpoint of 
is the orthocenter of as desired. 
, , and be the midpoints of , 
, and , , and be the midpoints of , , and . Also, let be the orthocenter of . Note that we can use parallel sides to see that , , and are collinear. Thus we have ![\[ \text{Pow}(M_3,(XYZ)) = M_3Z \cdot M_3X = \frac 14 AB \cdot DE \]](http://latex.artofproblemsolving.com/1/b/d/1bd918887213fe712aced60bebe2ea59a3323eed.png)
by midlines. Applying this argument cyclically, and noting the condition 
, , , , , , all lie on a circle concentric with . of is the orthocenter of , and likewise the circumcenter of is the orthocenter of . is the origin. Then we have ![\[ o_1 = m_1+m_2+m_3 = (c+e)/2+(a+e)/2+(a+c)/2=a+c+e \]](http://latex.artofproblemsolving.com/4/2/1/42199c85d802eee7e511d3679fb71eb9ba85f3ce.png)
![\[ o_2 = n_1+n_2+n_3 = (b+d)/2+(d+f)/2+(b+f)/2=b+d+f \]](http://latex.artofproblemsolving.com/9/7/5/975b3e334f6a9bfc3cfa35a93e0c31f19a512212.png)
![\[ h = x+y+z = (a+d)/2+(b+e)/2+(c+f)/2=(a+b+c+d+e+f)/2.\]](http://latex.artofproblemsolving.com/6/f/6/6f6e6322c158659c3526734020106906ffa2bec8.png)
Note that from the above we have 
, so is the midpoint of segment . In particular, , , and are collinear, as required.
, 
, 
, then there exists an involution swapping 
. If we define points and similar to , then the involution also swaps 
.
be and define other points similarly. Let's also define 
in the usual fashion. There are immediate benefits: we can get rid of "circumcenter of " and replace with "orthocenter of 
", something which is much easier to prove in foresight due to nice homotheties. If you don't find this step, you might end up proving something about radical axes. It's not clean.
coincides with and is the midline of trapezoid 
and cyclically. Now we can begin. I claim that 
lie on the common nine-point circle of and . Notice by homothety that![\[YA_1\cdot YD_1=\frac{1}{2}BC\cdot \frac{1}{2}EF\]](http://latex.artofproblemsolving.com/0/b/9/0b9db6ea8b8c51cf2ff1b313686707800343632e.png)
which is fixed, which should mean that 
are cyclic. We get similar results for two other cyclic quadrilaterals. But if the circles are different, then we get a contradiction since the radical axes and cyclically do not concur.![\[YA_1=\frac{1}{2}BC=D_1Z\]](http://latex.artofproblemsolving.com/0/5/b/05b4c93ffa694a39def75ce3e143e4808c5f279e.png)
which means that and have a shared perpendicular bisector. As a result, the circumcenter of coincides with the circumcenter of . We should also realize, though, that![\[G_{\triangle A_1C_1E_1}=\frac{\overrightarrow{A}+\overrightarrow{C}+\overrightarrow{E}}{3}\]](http://latex.artofproblemsolving.com/f/4/c/f4c6aba8be4dc6f0ff4c6ad62bec7f472fe4dd58.png)
![\[G_{\triangle B_1D_1F_1}=\frac{\overrightarrow{B}+\overrightarrow{D}+\overrightarrow{F}}{3}\]](http://latex.artofproblemsolving.com/e/0/2/e029ce20392e3a8e15de9b46bfe657320b1f9944.png)
![\[G_{\triangle XYZ}=\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}+\overrightarrow{D}+\overrightarrow{E}+\overrightarrow{F}}{6}\]](http://latex.artofproblemsolving.com/e/3/0/e30adcc0bbd3dbd28117d739f47d0729f2165401.png)
hence the midpoint of 
and 
is just 
.
).
, 
, and 
inside the hexagon such that 
, 
, and 
are all parallelograms, making the hexagon look like the Google Drive logo. The length condition is equivalent to ![\[CD^{\prime}\cdot CF^{\prime} = AD^{\prime}\cdot AB^{\prime} = EF^{\prime}\cdot EB^{\prime}\]](http://latex.artofproblemsolving.com/d/b/7/db73a9a2c324720575e5f2cd3f67a72a75723a1a.png)
in other words, 
. Thus, the circumcenter of coincides with that of 
.
, 
, and 
inside the hexagon such that 
, 
, and 
are all parallelograms. Then rewrite the length condition as ![\[FC^{\prime}\cdot FA^{\prime}=DE^{\prime}\cdot DA^{\prime}=BE^{\prime}\cdot BC^{\prime}\]](http://latex.artofproblemsolving.com/9/7/2/97280767b87ef995d5388e5d61cc0b898b670113.png)
so 
, hence the circumcenters of and 
are the same point., the circumcenter of , and the orthocenter of are collinear. We start with the following claim: and are translations of each other.
The triangle formed by 
has sides that are parallel to the vectors ![\[\left[(d+f-e)-(f+b-a), (f+b-a)-(b+d-c), (b+d-c)-(d+f-e)\right]=\left[d-e-b+a, f-a-d+c, b-c-f+e\right]\]](http://latex.artofproblemsolving.com/a/6/2/a62b94ce1f7dd93a4af3df518d6a37558dfa6020.png)
while the triangle formed by 
has sides that are parallel to the vectors ![\[\left[(e+a-f)-(a+c-b), (a+c-b)-(c+e-d), (c+e-d)-(e+a-f)\right]=\left[e-f-c+b, a-b-e+d, c-d-a+f\right],\]](http://latex.artofproblemsolving.com/4/4/e/44e3cd3e6801042d7cb87902de17f1d8fac0998f.png)
which are the same vectors as above, but in a different order. The claim is proven. ![\[a-a^{\prime}=a-(d+f-e)=a-d-f+e=(a-f+e)-d=b^{\prime}-d,\]](http://latex.artofproblemsolving.com/0/e/f/0ef1564351005b8bc86f8b0516a1931f5ef7387a.png)
so 
is a parallelogram and by a similar logic so are 
and 
. Then, the midpoints of , , and become the midpoints , , and of 
, 
, and 
. We see that is the vector average for the medial triangles of , . But the orthocenter of a medial triangle is the circumcenter of the original triangle (use the Euler line and a homothety of ratio 
at centroid), so the orthocenter of lies halfway between the circumcenters of , . Q.E.D.
, , 
be the midpoints of , , . Then 
by midlines on 
and 
, so 
is cyclic, and similarly 
and 
. It is easy to check by radical axis that 
is cyclic. Also we get that , and have equal power wrt 
so the circumcenter of lies on the nine-point center of 
(since 
is the nine-point circle).
is the circumcenter of 
since the altitude of to is the perpendicular bisector of . So it suffices to show that the orthocenters of , 
and 
are collinear. However since these three triangles share a circumcenter it suffices to show that the three centroids of the triangles are collinear by homothety at the circumcenter along the Euler line of the three triangles.
, 
and 
which is the midpoint of the previous two centroids, done.
be the unique points inside the hexagon such that 
, 
, 
are parallelograms. Likewise, let 
be the unique points inside the hexagon such that 
, 
, 
are parallelograms. Note that and 
are congruent and are simply translations of each other. Denote and to be the circumcenters of and respectively, and let be the orthocenter of . has the same cirumcenter as , and has the same circumcenter as 
Likewise, also has the same power with respect to 
as and , so they all lie the same distance away from , which clearly implies the claim. Likewise, we can do the same thing for . 
be the triangle such that is its medial triangle. Notice that is now the circumcenter of .
is the midpoint of , and et cetera for 
and 
.
, 
, and 
are all parallelograms. On top of this, we have that 
, 
, 
are all paralellograms. Therefore, is the midpoint of 
and , etc. Representing this in vector form, we have 
Because is the medial triangle of we also have 
and et cetera for and . From these two equations we get 
and et cetera for and , so the claim is proven. and are translations of each other, must also be a translation that is halfway in between these two triangles. This implies their circumcenters are linear, and in fact must be the midpoint of and , so we are done. 
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
![[asy] // Generated By AsyPadv2.0
import olympiad;
import markers;
import math;
import graph;
// change the unit size to fit your needs
unitsize(1.0cm);
// colored pens
pen c000000 = rgb("000000");
// dependency level 0
/* You can change the coordinates of these points of dependency level 0.
The drawing will retain the same relationships and qualities.
Please be aware that as a result of this some of the image may be clipped off. */
pair A = (7.01, 5.90); dot(A, c000000); label("$A$", A, dir(74.05460409907846));
pair B = (3.29, 5.89); dot(B, c000000); label("$B$", B, dir(123.02386755579636));
pair C = (1.86, 3.11); dot(C, c000000); label("$C$", C, dir(175.60129464500335));
pair D = (4.21, 0.99); dot(D, c000000); label("$D$", D, dir(306.4962760793981));
// dependency level 1
// Do not change anything below, unless you are experienced in Asymptote.
path segA_B = A--B; draw(segA_B, c000000);
path segB_C = B--C; draw(segB_C, c000000);
path segC_D = C--D; draw(segC_D, c000000);
pair X = (A + D)/2; dot(X, c000000); label("$X$", X, dir(315.0));
// dependency level 2
path parD_segA_B = (D-100.0*dir(segA_B))--(D+100.0*dir(segA_B));
path parA_segC_D = (A-100.0*dir(segC_D))--(A+100.0*dir(segC_D));
// dependency level 3
pair E = relpoint(parD_segA_B, 0.48919865743013735); dot(E, c000000); label("$E$", E, dir(269.9735171788647));
// dependency level 4
path parE_segB_C = (E-100.0*dir(segB_C))--(E+100.0*dir(segB_C));
pair Y = (B + E)/2; dot(Y, c000000); label("$Y$", Y, dir(279.8074719848136));
// dependency level 5
pair F = intersectionpoint(parA_segC_D, parE_segB_C); dot(F, c000000); label("$F$", F, dir(19.436760465123882));
path lineA_Y = (A-100.0*unit(Y-A))--(Y+100.0*unit(Y-A));
// dependency level 6
path segE_F = E--F; draw(segE_F, c000000);
path segF_A = F--A; draw(segF_A, c000000);
pair Z = (F + C)/2; dot(Z, c000000); label("$Z$", Z, dir(315.0));
pair Yp = intersectionpoint(parD_segA_B, lineA_Y); dot(Yp, c000000); label("$Y'$", Yp, dir(217.44717441236094));
// dependency level 7
path lineA_Z = (A-100.0*unit(Z-A))--(Z+100.0*unit(Z-A));
path segE_Yp = E--Yp; draw(segE_Yp, c000000);
path pbYp_D = ((Yp+D)/2-100.0*unit(((D-Yp).y, -(D-Yp).x)))--((Yp+D)/2+100.0*unit(((D-Yp).y, -(D-Yp).x)));
// dependency level 8
pair Zp = intersectionpoint(segC_D, lineA_Z); dot(Zp, c000000); label("$Z'$", Zp, dir(89.73160074040601));
// dependency level 9
path ccZp_Yp_D = circumcircle(Zp, Yp, D);
pair ccZp_Yp_Dcenter = circumcenter(Zp, Yp, D); real ccZp_Yp_Drad = circumradius(Zp, Yp, D); draw(ccZp_Yp_D, c000000);
path segZp_Yp = Zp--Yp; draw(segZp_Yp, c000000);
path pbZp_Yp = ((Zp+Yp)/2-100.0*unit(((Yp-Zp).y, -(Yp-Zp).x)))--((Zp+Yp)/2+100.0*unit(((Yp-Zp).y, -(Yp-Zp).x)));
// dependency level 10
pair Op = intersectionpoint(pbYp_D, pbZp_Yp); dot(Op, c000000); label("$O'$", Op, dir(315.0));
// dependency level 11
pair O = (Op + A)/2; dot(O, c000000); label("$O$", O, dir(315.0));
// clip the drawing view
clip((0.0, 0.0)--(0.0, 7.0)--(10.0, 7.0)--(10.0, 0.0)--cycle); [/asy]](http://latex.artofproblemsolving.com/b/7/e/b7e067d182462a672a8f310c63b596d74b131bfd.png)
and 
such that 
and 
are parallelograms. Let 
be the circumcenter of 
.
with center 
.![\[ED \cdot EY' = ED \cdot AB = CD \cdot AF = CD \cdot CZ',\]](http://latex.artofproblemsolving.com/9/7/4/9743e9a17eaebaefc7ea4e43d1e3bc73834dbece.png)
so 
,
,
are collinear. The third is just the midpoint of the first two, so we are done.
,
and 
as well. (A habit when working with 
,
.
and 
,
and 
are not necessarily collinear. But after cleaning up my diagram a bit with just these points: I notice that 
-- WOW! That's where the length condition was useful after all.
and 
and the intersections are 
and from 
.
to be the center 
:
)
and circumcenter 
.
,
are collinear.
collinear with the order 
with length ratio 
.
,
are collinear.
,
lies at infinity, this will imply that the centers of the three circles are collinear, as desired. Equivalently, we will show that 
are concurrent, where 
is the radical axis of 
,
is the radical axis of 
,
is the line at infinity.
.
.
,
be positive numbers such that 
.
and 
,![\[a^2x(1-y-z) = b^2y(1-x-z) = c^2z(1-x-y) = L\]](http://latex.artofproblemsolving.com/3/f/4/3f4f83229d963ad6cefdeb800bc8c620b5f9cc69.png)
for some constant 
,![\[a^2 = Tyz(1-x-z)(1-x-y),\]](http://latex.artofproblemsolving.com/b/1/7/b17c01eabfb25ce5f9e149bf3b8ae744f671f7d5.png)
![\[b^2 = Txz(1-x-y)(1-y-z),\]](http://latex.artofproblemsolving.com/8/9/0/890312d2c16cc96f65b896b693e1eef7b655f7a5.png)
and ![\[c^2 = Txy(1-x-z)(1-y-z).\]](http://latex.artofproblemsolving.com/1/1/7/117b359017bd81ad48c9d08ebcda98d1f8fa6556.png)
for some constants 
.
for some constants 
,
is of the form 
for some constants 
.
,
,
.![\[\begin{vmatrix}u_2-u_{ace} & u_2 - u_{bdf} & 1 \\ v_2-v_{ace} & v_2 - v_{bdf} & 1 \\ w_2-w_{ace} & w_2 - w_{bdf} & 1 \end{vmatrix} = 0.\]](http://latex.artofproblemsolving.com/f/2/3/f23eb7bb14cdacc7238fc798a753f8463008bb8b.png)
,
to the equation 
.
and 
.![\[u_{ace} = \frac{-xy(1-y)(1-z)a^2+xyz(1-z)b^2+x(1-x)(1-y)(1-z)c^2}{xyz+(1-x)(1-y)(1-z)}\]](http://latex.artofproblemsolving.com/1/6/c/16ccc02305047e2db4439519c2b6190164036a49.png)
and similarly for 
and 
.
etc. into the previous equation for 
.
(this ![\[u_{ace}A = -xy(1-y)(1-z)Tyz(1-x-z)(1-x-y) + xyz(1-z)Txz(1-x-y)(1-y-z)+x(1-x)(1-y)(1-z)Txy(1-x-z)(1-y-z)\]](http://latex.artofproblemsolving.com/8/2/c/82c084f0693eaa7de5e1651f0274d24b482444c2.png)
![\[=-xy(1-y)(1-z)Tyz(1-x-z)(1-x-y) + xyz(1-z)Txz(1-x-y)(1-y-z)+x(1-x)(1-y)(1-z)Txy(1-x-z)(1-y-z) + x(xyz)Txy(1-x-z)(1-y-z) - x(xyz)Txy(1-x-z)(1-y-z)\]](http://latex.artofproblemsolving.com/c/c/9/cc9caa6a5f069cae12c8d0fc59b1b709dbd97068.png)
![\[=A(xTxy(1-x-z)(1-y-z))-xy(1-y)(1-z)Tyz(1-x-z)(1-x-y) + xyz(1-z)Txz(1-x-y)(1-y-z)- x(xyz)Txy(1-x-z)(1-y-z)\]](http://latex.artofproblemsolving.com/9/6/6/9666ac2fa83f414a71b59edaf1e5a785cba27a66.png)
![\[= AB -xy(1-y)(1-z)Tyz(1-x-z)(1-x-y) + xyz(1-z)Txz(1-x-y)(1-y-z)- x(xyz)Txy(1-x-z)(1-y-z)\]](http://latex.artofproblemsolving.com/a/8/2/a8275b9a58f0450315a74fe272108dea57fb03cf.png)
where 
.![\[u_{ace}A = AB + xyzT(-y(1-y)(1-z)(1-x-z)(1-x-y) + xz(1-z)(1-x-y)(1-y-z)-x^2y(1-x-z)(1-y-z)) \]](http://latex.artofproblemsolving.com/e/0/a/e0a18315c4f42e3364a6ec75e819af7f197412bc.png)
![\[= AB + xyzT(-y(1-x-z)( (1-y)(1-z)(1-x-y) +x^2(1-y-z) )+ xz(1-z)(1-x-y)(1-y-z)) \]](http://latex.artofproblemsolving.com/6/b/3/6b35e6fd6381cf975da20c39c09dae901376b51d.png)
![\[= AB + xyzT(-y(1-x-z)( (1-y)(1-z)(1-x)-y(1-y)(1-z) +x^2(1-y-z) )+ xz(1-z)(1-x-y)(1-y-z)) \]](http://latex.artofproblemsolving.com/c/5/2/c52f99e5ed787287153ed78a6229760ff5a4e327.png)
![\[= AB + xyzT(-y(1-x-z)( A-xyz-y(1-y)(1-z) +x^2(1-y-z) )+ xz(1-z)(1-x-y)(1-y-z)) \]](http://latex.artofproblemsolving.com/3/4/7/347179fc3e4c9e3a3b3bbb0316525cf03561f306.png)
![\[= AB + AC + xyzT(-y(1-x-z)(-xyz-y(1-y)(1-z) +x^2(1-y-z) )+ xz(1-z)(1-x-y)(1-y-z)), \]](http://latex.artofproblemsolving.com/2/f/1/2f119c627f762e1eb074052332d4b738cbdf2c16.png)
where 
.![\[u_{ace}A = AB+AC+xyzT(-y((1-x)-z)(-xyz-y(1-y)(1-z) +x^2(1-y-z) )+ xz(1-z)((1-x)-y)((1-y)-z))\]](http://latex.artofproblemsolving.com/b/9/5/b95d896358f54f6577c65fd87f4df4fd1883355f.png)
![\[=AB+AC+xyzT(y((1-x)-z)(xyz+y(1-y)(1-z) -x^2(1-y-z) )+ xz(1-z)((1-x)(1-y) - y(1-y) - z(1-x)+yz))\]](http://latex.artofproblemsolving.com/1/e/8/1e88e37c71f0e486c064818904509d22679799ab.png)
![\[=AB+AC+xyzT(y(1-x)y(1-y)(1-z) + y(1-x)(xyz-x^2(1-y-z))-yz(xyz+y(1-y)(1-z) -x^2(1-y-z) )+ xz(1-z)(1-x)(1-y) + xz(1-z)( - y(1-y) - z(1-x)+yz))\]](http://latex.artofproblemsolving.com/d/3/f/d3f25504bf9462d5699378f49bfab6611c5a6a4e.png)
![\[= AB+AC+xyzT((y^2+xz)(1-x)(1-y)(1-z) + y(1-x)(xyz-x^2(1-y-z))-yz(xyz+y(1-y)(1-z) -x^2(1-y-z) )+ xz(1-z)( - y(1-y) - z(1-x)+yz))\]](http://latex.artofproblemsolving.com/b/e/4/be4d217985151f3e1a07baece04f23595040a503.png)
![\[= AB+AC+xyzT((y^2+xz)(A-xyz) + y(1-x)(xyz-x^2(1-y-z))-yz(xyz+y(1-y)(1-z) -x^2(1-y-z) )+ xz(1-z)( - y(1-y) - z(1-x)+yz))\]](http://latex.artofproblemsolving.com/2/8/a/28aeb058a928fdddfb3a7e111af0fade0594a5d6.png)
![\[= AB+AC+AD + xyzT(-(y^2+xz)xyz + y(1-x)(xyz-x^2(1-y-z))-yz(xyz+y(1-y)(1-z) -x^2(1-y-z) )+ xz(1-z)( - y(1-y) - z(1-x)+yz)),\]](http://latex.artofproblemsolving.com/f/1/a/f1a225d256269ff20d36a43ee10f148c2271389f.png)
where 
.![\[u_{ace}A = AB+AC+AD+xyzT(-xy^3z-x^2yz^2 + y(1-x)(xyz-x^2+x^2y+x^2z)-yz(y-y^2-yz+y^2z +xyz-x^2+x^2y+x^2z )+ xz(1-z)( - y+y^2 - z+xz+yz))\]](http://latex.artofproblemsolving.com/3/8/d/38df008ea2855b36aa382b71ac9dd63a4016436e.png)
![\[= AB+AC+AD+xyzT(-xy^3z-x^2yz^2 + (1-x)(xy^2z-x^2y+x^2y^2+x^2yz)-(y^2z-y^3z-y^2z^2+y^3z^2 +xy^2z^2-x^2yz+x^2y^2z+x^2yz^2 )+ (1-z)( - xyz+xy^2z - xz^2+x^2z^2+xyz^2))\]](http://latex.artofproblemsolving.com/e/2/2/e22c5b07f03da3598ca544d01ac8e435ae1697d6.png)
![\[= AB+AC+AD+xyzT(-xy^3z-x^2yz^2 + xy^2z-x^2y+x^2y^2+x^2yz - x^2y^2z+x^3y-x^3y^2-x^3yz -y^2z+y^3z+y^2z^2-y^3z^2 -xy^2z^2+x^2yz-x^2y^2z-x^2yz^2 - xyz+xy^2z - xz^2+x^2z^2+xyz^2 + xyz^2 - xy^2z^2 + xz^3 - x^2z^3 - xyz^3)\]](http://latex.artofproblemsolving.com/0/0/2/002e14613519a5f58b88f06e2c60c4a364277cb9.png)
![\[= AB+AC+AD+xyzT(-x^3yz-xy^3z-xyz^3-2x^2y^2z-2x^2yz^2-2xy^2z^2+2x^2yz+2xy^2z+2xyz^2-xyz-y^2z+y^3z+y^2z^2-y^3z^2-x^2y+x^3y+x^2y^2-x^3y^2-z^2x+z^3x+z^2x^2-z^3x^2)\]](http://latex.artofproblemsolving.com/2/a/9/2a999f1912bfb1185cf33a26d44311473b8130c5.png)
![\[=AB+AC+AD+AE,\]](http://latex.artofproblemsolving.com/4/7/1/4711cefa1fc83b41c01414b643cdf0e773240e75.png)
where 
.![\[u_{ace} = B+C+D+E = Tx^2y(1-x-z)(1-y-z)-Txy^2z(1-x-z)+Txyz(y^2+xz)-Txyz(xyz+y^2z+x^2y+z^2x)\]](http://latex.artofproblemsolving.com/1/a/9/1a9e8cc7424b31eab0cf857105b7da1d35428073.png)
![\[=xc^2+Txyz(-y+y^2+xy+xz+yz)-Txyz(xyz+y^2z+x^2y+z^2x).\]](http://latex.artofproblemsolving.com/5/d/f/5dfbafeae63394f5d5978b38ed3d46975a76a30a.png)
.
in some order, and hence 
are 
in some order. Plugging in 
to the equation 
yields ![\[-a^2(x+y)(x+z)-b^2(2-2x-y-z)(x+y)-c^2(2-2x-y-z)(x+z)+(2)(u_2(2-2x-y-z)+v_2(x+z)+w_2(x+y)) = 0,\]](http://latex.artofproblemsolving.com/f/d/a/fda34623da79efa5c7212663cd1458d1b89c569c.png)
or ![\[u_22(2-2x-y-z)+v_22(x+z)+w_22(x+y) = a^2(x+y)(x+z) + b^2(2-2x-y-z)(x+y) + c^2(2-2x-y-z)(x+z).\]](http://latex.artofproblemsolving.com/6/e/1/6e1f49ba52feb632d622de9e5c79c955b3b781f6.png)
Similarly, ![\[u_22(y+z) + v_22(2-x-2y-z) + w_22(x+y)=a^2(2-x-2y-z)(x+y) + b^2(y+z)(x+y) + c^2(y+z)(2-x-2y-z).\]](http://latex.artofproblemsolving.com/2/b/c/2bc9e4b87b043b4a39613342628ba2855cf71e8f.png)
Subtracting the second equation from the first equation yields ![\[u_22(2-2x-2y-2z)-v_22(2-2x-2y-2z)=a^2(x+y)(-2+2x+2y+2z)+b^2(x+y)(2-2x-2y-2z)+c^2((2-2x-2y-2z)(x-y)+(y+z)(x+z)-(x+z)(y+z)).\]](http://latex.artofproblemsolving.com/3/7/3/373b33a45cdfff85924a89c16e0806062b53ec0f.png)
Dividing by 
yields ![\[2u_2-2v_2=-a^2(x+y)+b^2(x+y)+c^2(x-y).\]](http://latex.artofproblemsolving.com/5/5/1/55156cc79e260dd759246e687cf74501d854c0f8.png)
Similarly, ![\[2v_2-2w_2 = a^2(y-z)-b^2(y+z)+c^2(y+z)\]](http://latex.artofproblemsolving.com/6/8/7/6874260f93c83fb26c1c9b6c558ba1d53afb658a.png)
and ![\[2w_2-2u_2 = a^2(x+z)+b^2(z-x)-c^2(x+z).\]](http://latex.artofproblemsolving.com/0/f/6/0f697015923d79220f77ed2f20d00ea317fbf127.png)
This means 
and 
.![\[2u_2(2-2x-y-z) + (2u_2+a^2(x+y)-b^2(x+y)+c^2(y-x))(x+z)+(2u_2+a^2(x+z)+b^2(z-x)-c^2(x+z))(x+y) = a^2(x+y)(x+z) + b^2(2-2x-y-z)(x+y) + c^2(2-2x-y-z)(x+z).\]](http://latex.artofproblemsolving.com/1/1/9/119e2e09e737976791a490eb5ad98246aa4077ff.png)
This becomes ![\[4u_2 + 2a^2(x+y)(x+z)+b^2(-(x+y)(x+z)+(z-x)(x+y)) + c^2((y-x)(x+z)-(x+z)(x+y)) = a^2(x+y)(x+z) + b^2(2-2x-y-z)(x+y) + c^2(2-2x-y-z)(x+z),\]](http://latex.artofproblemsolving.com/a/f/4/af41e6c8116e6e41c468a7579973f8ee925559ac.png)
then ![\[4u_2 + 2a^2(x+y)(x+z) + b^2(x+y)(-2x) + c^2(x+z)(-2x) = a^2(x+y)(x+z) + b^2(x+y)(2-2x-y-z) + c^2(x+z)(2-2x-y-z)\]](http://latex.artofproblemsolving.com/1/c/f/1cf2452ff08affed2defdf1cc40daa2bb17e47a1.png)
implying ![\[4u_2 = -a^2(x+y)(x+z) + b^2(x+y)(2-y-z) + c^2(x+z)(2-y-z).\]](http://latex.artofproblemsolving.com/7/6/b/76ba266975ecb61b951c58c70da9b8cef38b2ef6.png)
![\[4u_2 - 4u_{ace} = -a^2(x+y)(x+z)+b^2(x+y)(2-y-z)+c^2(x+z)(2-y-z)-4xc^2+4Txyz(y-y^2-xy-xz-yz)+4Txyz(xyz+y^2z+x^2y+z^2x).\]](http://latex.artofproblemsolving.com/2/0/8/2085b82582013154a59de8925c2f432fbed2315f.png)
Now, at each step, we will drop terms that do not affect the determinant. We can drop the 
,![\[2b^2(x+y)+2c^2(x+z)-4xc^2+4Txyz(y-y^2)\]](http://latex.artofproblemsolving.com/a/e/f/aef11908de965e30c582905ececdc66428385eca.png)
![\[= 2b^2(x+y) + 2c^2(-x+z) + 4Txyz(y-y^2).\]](http://latex.artofproblemsolving.com/b/3/7/b378e6ab3f9dc42c0f1b847fad2973522a6986ee.png)
Dividing by 2, we have ![\[b^2(x+y)+c^2(-x+z)+2Txyz(y-y^2)\]](http://latex.artofproblemsolving.com/7/4/0/7401a6119aeb3b042d63e186256142049814c0bf.png)
![\[=(x+y)Txz(1-x-y)(1-y-z) + (-x+z)Txy(1-x-z)(1-y-z) + 2Txyz(y-y^2)\]](http://latex.artofproblemsolving.com/1/3/a/13adf59b28b8cfb954620a8931149b5cc8982a9b.png)
![\[=Tx(1-y-z)((x+y)z(1-x-y)+(-x+z)y(1-x-z))+2Txyz(y-y^2).\]](http://latex.artofproblemsolving.com/2/e/9/2e942899f059fd29d553068654edc0e248730f7b.png)
We can drop some terms (precisely, we drop 
)![\[T(x^2(-y+z)+(y-z)(x)(x+y)(x+z)+(-y+z)(x)(y+z)(x^2+yz)).\]](http://latex.artofproblemsolving.com/e/2/5/e25bac5c32a0cfabd3517b1050f51c36407e2b20.png)
(I did this part of the calculation using Chinese Bashing Notation, and I don't want to write up all the triangles of numbers here.) This is seen to flip sign when 
are swapped, meaning that the first two columns of the determinant sum to the zero vector, as desired. 
,
,
.
and cyclic variations, which quickly forces the hexagon to be centrally symmetric, which was too simple for my liking.
,
instead? This suggests trying ![\[ AB \cdot DE = BC \cdot EF = CD \cdot FA. \]](http://latex.artofproblemsolving.com/d/5/0/d50ac073f87d13a92c19a405178fd577395cb886.png)
and defining 
,
,
similarly, the orthocenters of 
,