at 
,
perpendicular to 
.
be the intersections of 
with 
,
be the projections of 
onto line 
be the second intersections of 
with the incircle 
be the intersection of 
and 
.
bisects segment 
.
be the incentre of 
such that the incircle touches the sides 
and 
meets 
and 
respectively. 
and 
meet the incircle at 
and 
intersect at 
Prove that 
bisects 
then prove 
![$P\in\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/f/3/d/f3d763f4197792a2956273485eb1e94700a43044.png)
has degree 
having 
.
is the leading coefficient of 
Y by Amir Hossein, mathlogician, AlastorMoody, crazyeyemoody907, v4913, HWenslawski, HamstPan38825, Adventure10, Mango247, Rounak_iitr, Tastymooncake2, Funcshun840
Let 
be an acute triangle with circumcircle 
, and let 
be the foot of the altitude from 
to 
. Let 
and 
be the points on with 
and 
. The tangent to at intersects lines 
and 
at 
and 
respectively; the tangent to at intersects lines and at 
and 
respectively. Show that the circumcircles of 
and 
are congruent, and the line through their centers is parallel to the tangent to at .
Ankan Bhattacharya and Evan Chen
be an acute triangle with circumcircle 
, and let 
be the foot of the altitude from 
to 
. Let 
and 
be the points on with 
and 
. The tangent to at intersects lines 
and 
at 
and 
respectively; the tangent to at intersects lines and at 
and 
respectively. Show that the circumcircles of 
and 
are congruent, and the line through their centers is parallel to the tangent to at .Ankan Bhattacharya and Evan Chen
be the center of 
and 
be the midpoints of 
and 
respectively. Refer to the diagram below.![[asy] size(9cm); pair A, B, C, O, M, N, P, Q, E1, F1, E2, F2, O1, O2; A = dir(105); B = dir(190); C = dir(350); O = origin; M = (A + B)/2; N = (A + C)/2; P = point(unitcircle, intersections(unitcircle, M, N)[1]); Q = point(unitcircle, intersections(unitcircle, M, N)[0]); E1 = extension(A, C, P, rotate(90, P) * origin); F1 = extension(A, B, P, rotate(90, P) * origin); E2 = extension(A, C, Q, rotate(90, Q) * origin); F2 = extension(A, B, Q, rotate(90, Q) * origin); O1 = circumcenter(A, E1, F1); O2 = circumcenter(A, E2, F2); fill(O--P--E1--cycle^^O--Q--E2--cycle, lightblue + lightred + gray(0.5)); draw(O--P^^O--Q^^O--M^^O--N, gray(0.5)); draw(E1--F1^^E2--F2); draw(arc(O1, F1, E1)^^arc(O2, F2, E2), dashed); draw(unitcircle); draw(A--B--C--cycle); draw(P--Q); dot(A^^B^^C^^O^^M^^N^^P^^Q^^E1^^F1^^E2^^F2);
label("$A$", A, dir(60)); label("$B$", B, dir(B)); label("$C$", C, dir(C)); label("$O$", O, dir(200)); label("$M$", M, dir(120)); label("$N$", N, dir(60)); label("$P$", P, dir(P)); label("$Q$", Q, dir(Q)); label("$E_1$", E1, dir(P)); label("$F_1$", F1, dir(P)); label("$E_2$", E2, dir(Q)); label("$F_2$", F2, dir(Q)); [/asy]](http://latex.artofproblemsolving.com/0/1/8/018820a874a8c7cc11ecd81ddea63d4ec1a767b9.png)
and 
are cyclic (giving the radical axis is 
)
(giving the congruence of the circles).
,
,
(ii).![\[ \measuredangle F_2E_2O = \measuredangle QE_2O = \measuredangle QNO = \measuredangle MNO = \measuredangle MAO = \measuredangle F_2AO \]](http://latex.artofproblemsolving.com/3/4/d/34d7d37fd890683fe6b1deaa7536bfa9ec54377b.png)
and hence 
is cyclic. Similarly, 
is cyclic.
is the Simson line of 
,
.![\[ E_1P = \sqrt{E_1A \cdot E_1C} = \sqrt{E_2A \cdot E_2C} = E_2P. \]](http://latex.artofproblemsolving.com/0/7/8/0785b4f24c147d7826f9bfed371e1a38b75b0f4a.png)
But also 
.
.
;
,
is the polar of 
by La Hire).![\[ -1 = (PQ;NT) \overset{T}{=} (E_1E_2;N\infty) \]](http://latex.artofproblemsolving.com/3/e/6/3e6eb1446c4e21fab5b48034babd494d5c7ca3a5.png)
with 
the point at infinity. Hence 
and we can proceed as in the previous solution.
is acute is not necessary; it is only present to ensure that 
and 
,
and 
is a right angle, some of the points 
m
,
is a cyclic quadrilateral. So we have 
.
are equal, triangles 
and 
are congruent. Hence their circumcircles are congruent, proving the first part. Also since the radical axis is 
,
intersect 
,
,
,
,
.
,
,
,
.
,
.
and 
are their respective altitudes from 
so their circumcircles are congruent, proving the first part.
be the midpoint of 
the midpoint on 
so 
is cyclic.Also 
is cyclic.After some angle chasing 
( 
is cyclic) , we obtain 
cyclic .Also 
is cyclic . Angle chasing 
.
is circle 
(with 
radius )so after law of sines 
and circle 
is circle 
(with 
radius )so after law of sines 
, then we have 
.
be the centers of 
and 
be the midpoints of 
Easily see that 
are collinear.
we see that 
Moreover, by Law of Sines in 
we see that ![\[\frac{E_2Q}{E_2N}=\frac{\sin\angle E_2NQ}{\sin\angle E_2QN}=\frac{\sin\angle C}{\sin\angle PAQ}=\frac{c}{PQ}.\]](http://latex.artofproblemsolving.com/5/5/d/55d2f79098433987661ed5073b355345d748e13c.png)
By symmetry, we get 
which shows 
Since 
Law of Sines shows that the circumradii of the two triangles are indeed equal.
implies that 
have the same power WRT 
so 
Similarly 
so 
Thus, 
and 
so 
and we are done.
.
are collinear and 
and 
is the Simson Line of 
,
.
.
and 
,
.
and 
.
,![[asy]
pair A = dir(90), B = dir(210), C = dir(330);
pair P = point(unitcircle, intersections(unitcircle, (A+B)/2, (A+C)/2)[1]);
pair E = extension(P, P * (1 + I), A, C);
pair F = extension(P, P * (1 + I), A, B);
draw(unitcircle);
draw(A--B--C--cycle);
draw(A--E^^B--F, gray(0.7));
draw(E--F);
add(pathticks(P--E, 1));
add(pathticks(P--F, 1));
dot(A^^B^^C^^P^^E^^F);
label("$A$", A, dir(60));
label("$B$", B, dir(270));
label("$C$", C, dir(330));
label("$P$", P, dir(P));
label("$E$", E, dir(P));
label("$F$", F, dir(P));
[/asy]](http://latex.artofproblemsolving.com/2/8/a/28af40ac937496f243234a7d1648d19dfc8e6ead.png)
![[asy]
pair A = dir(90), B = dir(210), C = dir(330);
pair P = point(unitcircle, intersections(unitcircle, (A+B)/2, (A+C)/2)[1]);
pair E = extension(P, P * (1 + I), A, C);
pair F = extension(P, P * (1 + I), A, B);
pair Ap = 2*P - A;
draw(unitcircle);
draw(A--B--C--cycle);
draw(A--E^^B--F, gray(0.7));
draw(E--F);
draw(A--Ap, dotted);
draw(B--Ap, dotted);
add(pathticks(P--E, 1));
add(pathticks(P--F, 1));
add(pathticks(P--A, 2));
add(pathticks(P--Ap, 2));
dot(A^^B^^C^^P^^E^^F^^Ap);
label("$A$", A, dir(60));
label("$B$", B, dir(270));
label("$C$", C, dir(330));
label("$P$", P, dir(P));
label("$E$", E, dir(P));
label("$F$", F, dir(P));
[/asy]](http://latex.artofproblemsolving.com/5/5/1/55186abca6bbe4f5d7b9fb78a4c4947e38a353e4.png)
,
.![[asy]
unitsize(100);
pen n_purple = rgb(0.7,0.4,1),
n_blue = rgb(0,0.6,1),
n_green = rgb(0,0.4,0),
n_orange = rgb(1,0.4,0.1),
n_red = rgb(1,0.2,0.4);
pair A, B, C, M, P, D, E, F;
A = dir(90);
B = dir(190); C = reflect(A, origin) * B;
M = (B + C)/2;
P = (-sqrt(1 - (1 + M.y)^2/4), (1 + M.y)/2);
D = extension(P, A, B, C);
E = extension(A, C, P, rotate(90, P) * origin);
F = extension(A, B, P, rotate(90, P) * origin);
draw(B--F^^A--E^^D--B, gray(0.5));
draw(P--B^^P--C, gray(0.7));
draw(P--A^^P--M^^P--D);
draw(E--F);
draw(unitcircle);
draw(A--B--C--cycle);
dot(A^^B^^C^^M^^P^^D^^E^^F);
label("$A$", A, dir(60));
label("$B$", B, dir(130));
label("$C$", C, dir(350));
label("$M$", M, dir(270));
label("$P$", P, P);
label("$D$", D, dir(270));
label("$E$", E, P);
label("$F$", F, P);
[/asy]](http://latex.artofproblemsolving.com/b/1/8/b18e94a7f6138ae8998e112a67ea2f845cadc289.png)
.
of 
intersect line 
,
inversion), 
.
,
and 
,
and 
.
,
.
and 
.![\[PE^2 = EA \cdot EC = FB \cdot FA = PF^2\]](http://latex.artofproblemsolving.com/7/8/2/782496bc94e5f74f47deadf99ea2499d7d290c59.png)
as desired.
passes through the center of 
,
and 
.
.
then we can easily show that triangles 
and 
are congruent 
and 
)So 
lies on radical axis of 
and 
.
,
(this can be vertified by checking the lengths by the length formula) and 
.
.
or 
.
gives us a quadratic equation 
.
by vieta.
and the line through 
is 
.
.
(if this is true, then law of sines implies the problem). Just use the length formula to prove
, and see that since 
,
.
. To find those two circumcenters, just find the equation of the bisector 
and 
.
, and the circumcenter of 
And thus we don't need circumcenter 
and 
.
,
,
on the unit circle. We see that 
and 
satisfy
,
and 
both satisfy in this denominator, we get
)
are concyclic. Indeed, this is equivalent to
Then 
and 
respectively. Then the perpendicular bisector of 
and the perpendicular bisector of 
These intersect at the circumcenter 
with coordinates 
We compute that 
Note that 
Then their 
-
This is equivalent to the quadratic 
This has solutions 
Let 
for convenience. Then choose 
and 
One can then check that 
so line 
which is perpendicular to 
has equation 
There is no need to compute the equation 
as here and henceforth it will be the same but with each 
replaced by 
Intersecting 
note that by symmetry, the coordinates of 
Then we can easily construct the perpendicular bisector of 
with slope 
with equation 
The equations of the other three perpendicular bisectors follow from symmetry. Then we intersect the perpendicular bisectors of 
and 
:
Plugging this in to the equation for either of our two lines gives 
Then we compute 
This can be simplified further, but we only need to note that all the terms have even degree in 
so replacing all appearances of 
(as is necessary to compute the other circumradius) would not change the value, so the circles are congruent.
as all terms in the numerator not containing 
is 
and the slope of 
so they are perpendicular, as desired.
Note that 
It follows that 
is the Simson line of 
Similarly, 
Therefore, 
have radical axis 
cut by 
This involution must be reflection in 
It follows from symmetry in 
Therefore by the law of sines, 
and 
have equal circumradii, hence congruent circumcircles.
such that 
is a parallelogram. Now we have 
,
,
and we are done.
So, 
are concyclic and similarly 
are concyclic, which implies 
and we are done.
lies on 
and 
the perpendiculars from 
are collinear. Simson line finishes. 
and 
so 
Similarly, 
and so 
Since 
and 
we are done with the first part of the problem.
But the tangent to 
at 
,
is cyclic, and since 
,
is cyclic. Similarly, 
and 
are cyclic. Now ![\[\angle AF_1O = \angle EF_1O = \angle EPO = \angle FPO = \angle FE_1O = \angle AE_1O,\]](http://latex.artofproblemsolving.com/b/c/5/bc5cdc20045e3d557107fd1789a8dd612a362a2b.png)
so 
is cyclic, and similarly 
is cyclic. Furthermore, ![\[\angle OE_1E_2 = \angle OE_1F = \angle OPF = \angle OQF = \angle OE_2F = \angle OE_2E_1,\]](http://latex.artofproblemsolving.com/5/7/5/575da9e604026d7600d625ab87656ee3a637cb74.png)
so 
,
.
implies 
.
be their centers. By the Extended Law of Sines, ![\[2R_1 = \frac{E_1F_1}{\sin \angle E_1AF_1} = \frac{E_2F_2}{\sin \angle E_2AF_2} = 2R_2,\]](http://latex.artofproblemsolving.com/9/9/d/99dc49399bd0fa9beedfaab1b6d3f07a3baad698.png)
so 
,
be the tangent to 
gives 
,
are all collinear.
,
,
,
are all cyclic.
.
to 
.
.
.
.
and 
gives 
.
to 
.
.
is the radical axis of 
and 
.
.
,
.
,
and 
,
respectively. We also let 
is cyclic.![\[\measuredangle OE_1A = \measuredangle OE_1M = \measuredangle OPM = \measuredangle NQO = \measuredangle NF_2O = \measuredangle AF_2O = \measuredangle AE_2O \]](http://latex.artofproblemsolving.com/8/e/2/8e2f362f5b32ce2e33adc84e14098daf028f8147.png)
![\[\implies OE_1 = OE_2
\implies \text{Pow}_{(ABC)}(E_1) = \text{Pow}_{(ABC)}(E_2) \implies E_1P = E_2Q.\]](http://latex.artofproblemsolving.com/4/5/0/4500b7037718e1f55b1596047ce468a128695200.png)
In a similar fashion, we have 
,
implies 
,
and 
are cyclic.
are collinear. Then the feet of the perpendiculars from 
and 
respectively, so 
and 
,
are the circumcenters of 
,
,
means spiral sim taking 
is a rotation, since OP=OQ, so 
,
be where 
,
.![\[\angle BPK = 180 - \angle PBA - \angle PBK = 180 - (\angle C - \alpha) - \angle B = \angle A + \alpha\]](http://latex.artofproblemsolving.com/5/2/d/52d8fdbc18515f3c292c6a5990ebbd5f6163633a.png)
Notice since 
,
is cyclic. Now![\[\angle POK = \angle PF_1K = 180 - \angle F_1PB - \angle F_1BP = 180 - \alpha - (180-\angle C+\alpha) = \angle C - 2\alpha\]](http://latex.artofproblemsolving.com/b/d/c/bdc84a504ed14f551281c7bd8a4bf514d2d15b96.png)
![\[\implies \angle KPO = 180 - (90+\angle B) - (\angle C - 2\alpha) = 2\alpha + \angle A - 90 \implies \angle KF_1O = 2\alpha + \angle A - 90\]](http://latex.artofproblemsolving.com/e/4/4/e44228b08956af18987df1cddff5c4570609a964.png)
Now a quick check yields 
,
.
cyclic. Lastly,![\[\angle AQP = \angle PBA = \angle C - \alpha \implies \angle QAC = \alpha\]](http://latex.artofproblemsolving.com/6/f/c/6fc811d692c4647e93daddc83b15bd41c08c6507.png)
Repeating the answer chase gives 
cyclic. Now the centers of these two circles are perpendicular to 
.
radius of 
,
is cleary cyclic. By right angles, 
is cyclic. Thus ![\[\measuredangle OF_2E_2=\measuredangle OF_2Q=\measuredangle ONQ=\measuredangle ONM=\measuredangle OAM=\measuredangle OAE_2\]](http://latex.artofproblemsolving.com/6/9/5/69534c5ccf581595dc2b1bf486fe4c06ca5ce96a.png)
and the claim is proved. ![\[\angle AF_2O=\angle NF_2O=\angle NQO=\angle PQO=\angle QPO=\angle MPO=\angle ME_1O=\angle AE_1O\]](http://latex.artofproblemsolving.com/7/d/1/7d146e8c33069eafeeeac98d8b1d8de1c99d9f7f.png)
. The other part follows from 
and 
.
.
lie on different sides of 
bisects 
.
,
.
.
and 
,
is cyclic with diameter 
.
is cyclic with diameter 
.
implies that 
.
we then get 
.
.
.
,
,
,
;
.
and 
lie on the perpendicular bisector of 
,
.
,
.
;
similarly. Also let 
respectively. Observe that 
are collinear since they all lie on the perpendicular bisector of 
By inverting this common line about 
are concyclic. In particular, 
are concyclic, so by the Simson line from 
we get that 
and the foot of the altitude from 
The former point lies on both 
and 
,
Similarly, the inverse of the latter point is 
Therefore, 
are concyclic. Similarly, 
are concyclic. Hence 
meaning that the line through their centers is perpendicular to 
This solves the second part of the problem.
and 
we discover that 
to 
In particular, it sends the foot of the altitude from 
Therefore, the spiral similarity described above is in fact a rotation, implying that 
are congruent. This solves the first part of the problem, and we are done.
