[/quote]
,
such that, for every positive integer 
the integer
is a multiple of 
(Note that 
denotes the greatest integer less than or equal to 
For example, 
and 
)
and orthocenter 
.
to the circumcicle of 
in 
.
be the midpoint of 
.
.
and 
Prove that
and 
be positive integers greater than 
.
grid lies a coin with its tail side up. A move consists of the following steps.
square in the grid;
for which it is possible that every coin shows head-side up after a finite number of moves.
and 
.![\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]](http://latex.artofproblemsolving.com/c/3/7/c3708768a84c2bbb1e4fae502a9bb1e92cb13d02.png)
be the incentre of acute-angled triangle 
,
at 
,
respectively. Let line 
intersect the circumcircle of the triangle at 
lies between 
and 
.
so that for every 
.
such that 
.
at 
be the circumcircle of acute triangle 
and 
and 
.
and 
and 
.
are concyclic.
the centre of the circle passing through 
,
are difined similarly. Prove that 
,
,
are concurrent.
> 
,
€ 
be a fixed positive integer. Given a set 
of 
points in the plane such that no three are collinear and no four concyclic, let 
be the number of circles 
that contain 
in their interior, and let ![\[m(S)=a_1+a_2+\cdots + a_n.\]](http://latex.artofproblemsolving.com/4/9/b/49b928b1f20e2d1799a1234c7f25a2bc3d62d0dd.png)
Prove that there exists a positive integer 
depending only on 
such that the points of 
are the vertices of a convex polygon if and only if 
such that 
.
and 
.
Y by farhad.fritl, Davud29_09, ehuseyinyigit, Rounak_iitr, dangerousliri, cubres, MathLuis, Frd_19_Hsnzde, mariairam, Funcshun840
Let 
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let 
and 
be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let 
and 
be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
be the reflection of 
be the reflection of 
,
.
are concyclic and 
are also concyclic.
are concyclic, we have:![\[
\angle HPB' = \angle HPM = \angle MDH = 90^\circ - \angle DAE
\]](http://latex.artofproblemsolving.com/5/0/4/504a19c0d765bfe4ea44cc4074d00bd4f065148a.png)
Similarly,![\[
\angle HQC' = 90^\circ - \angle DAE
\]](http://latex.artofproblemsolving.com/0/b/f/0bfe0fc35c9fc6ee98c355ed36cbafa975c171e7.png)
![\[
\angle HPB' = \angle HC'B' \quad \text{and} \quad \angle HQC' = \angle HB'C'
\]](http://latex.artofproblemsolving.com/3/9/0/390a8b934425590b5ccd85a79e06b838ba1d9e59.png)
So, the points 
lie on a circle.
.
are concyclic. 
and 
be points on 
and 
such that 
.
with transversal 
,
,
are collinear.
,
are collinear.
.
if and only if 
lies on the radical axis of 
and 
.
be the circumcenter of triangle 
.
are concyclic,
and 
)
,
.
,
over 
,
,
.
and 
.![\[\angle MDH=90^\circ-\angle DAE=\angle ONM=\angle OTM=\angle HTM\]](http://latex.artofproblemsolving.com/b/c/f/bcfa3870444512a566434174437a11ebf9eea50b.png)
we have 
,
.
concur. Let 
intersect the 
at 
.
,
.
be the midpoint of 
,
where 
are the orthocenter and circumcenter of 
to be the midpoint of 
,
intersect 
at 
.
intersects the 
-
,
collinear, and by symmetry 
collinear, so we are done.
and 
.
,
be the centroid of 
.
are collinear because 
is an involution and 
.
hence a simple calculation gives 
.
.
and 
be the circumcenters of 
and 
.
and 
.![\[\frac{t-h}{o_b-o_c}=\frac{\frac{2a+4b+4c}{5}}{\frac{(b-c)(a+2b+2c)}{2(b+c)}}=\frac{4}{5}.\frac{b+c}{b-c}\in i\mathbb{R}\]](http://latex.artofproblemsolving.com/2/c/1/2c1fd2fdcf39091bcee5568b4a7972cb947ac03c.png)
As desired
and 
be the reflections of 
has midpoint 
has midpoint 
has midpoint 
is cyclic, which implies the result by Reim's theorem.
is cyclic. Thus![\[\measuredangle HPD' = \measuredangle HPM = \measuredangle HDM = \measuredangle HDA = \measuredangle HE'A = \measuredangle HE'D'\]](http://latex.artofproblemsolving.com/d/8/4/d843c1980c4c2fa6db037ad43f7c6b8805ab4636.png)
implying that 
is cyclic. Similarly, 
is cyclic, and we're done.
be 
intersect 
at 
.
and 
concurrent. Suppose 
Hence, 
are concyclic. And since, 
(
are collinear.
.
be 
.
be orthocenter of 
be midpoint of 
,
is a parallelogram. By doing some more dilation, you will obtain the ratio 
.
intersect 
,
So, 
Hence, we are done ![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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draw((-6.145299148418266,2.1563307052402014)--(-1.407818016280837,2.3107795394165724), linewidth(0.8));
draw((-1.407818016280837,2.3107795394165724)--(-4.315243008383876,4.92491669295181), linewidth(0.8));
draw((-4.440690889711499,3.5663651714587346)--(-3.651110701021928,3.592106643821463), linewidth(0.8));
draw((-4.099769237970145,3.848372064702441)--(-4.232088162837603,2.3742693285864003), linewidth(0.8) + linetype("4 4") + wwqqcc);
draw((-4.066689506753282,4.216897748731452)--(-4.099769237970145,3.848372064702441), linewidth(0.8) + linetype("4 4") + wwqqcc);
draw((-4.066689506753282,4.216897748731452)--(-4.045900795366713,3.579235907640099), linewidth(0.8) + blue);
draw(circle((-4.052310876932325,3.7758553294343264), 0.44127673926345784), linewidth(0.8) + linetype("4 4"));
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dot((-6.145299148418266,2.1563307052402014),dotstyle);
label("$B$", (-6.093024723651059,2.269940052388348), NE * labelscalefactor);
dot((-4.566138771039123,2.2078136499656584),dotstyle);
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dot((-2.98697839365998,2.2592965946911154),dotstyle);
label("$E$", (-2.935122992584353,2.3752034434239038), NE * labelscalefactor);
dot((-1.407818016280837,2.3107795394165724),dotstyle);
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dot((-4.099769237970145,3.848372064702441),linewidth(4pt) + dotstyle);
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dot((-3.7765585823495513,2.233555122328387),linewidth(4pt) + dotstyle);
label("$K$", (-3.7304463915196715,2.328419714074768), NE * labelscalefactor);
dot((-4.145004800767755,3.344423994110871),linewidth(4pt) + dotstyle);
label("$T$", (-4.093020293975479,3.4395332861167454), NE * labelscalefactor);
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/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/d/f/3/df3afb24373444eeb653ce51d117f934f4aaaf68.png)
a point such that 
is a parallelogram and also let 
and 
,
reflections of 
respectively then from parallelogram's spam and midbases checking and projective whatever you like you can easly check that 
are colinear on a line parallel to 
and 
,
is the image of the NPC of 
and thus 
is cyclic, now to finish just notice that if you let 
and in fact 
is diameter so 
are colinear and by PoP 
and thus 
which gives that 
is tangent to 
then since 
we have 
cyclic but then 
and therefore 
are reflections over 
and thus 
and thus 
is a parallelogram which gives that 
and thus 
is cyclic and by Reim's it means 
cyclic and repeating the same process for 
cyclic and by Reim's this gives 
cyclic as desired thus we are done 
.
.
concur, and the conclusion follows.
ratio.
,
,
and 
.
.
Applying Menelaus' Theorem in 
we get that 
,
as well. Then, by Ceva, 
,
,
,
be the center of 
.
are collinear.
and 
are concyclic, we have that 
,
are concyclic. Since 
,
is the bisector of 
-- but so is 
.
are collinear.
,
.
,
is a parallelogram and 
.
,
.
and 
are concyclic.
.
,
,
.
,
.
and 
as well. Thus the circumcenter of 
and courtesy of it satisfying 
,
,
,
.
so we are done.
and such that 
lies on the radical axis of circles 
.
,
and line 
.
and 
denote the midpoint of 
.
and 
.
is the midpoint of 
.
and 
and 
must concur at a point, which turns out to be 
and 
.
,
and 
.
and set![\[A=\frac{S_{ABC}(a^2S_A+b^2S_B+c^2S_C)}{S_{AB}+S_{BC}+S_{CA}}.\]](http://latex.artofproblemsolving.com/2/e/4/2e49d77bb462d1592ffb5a8ffdf62d4cbb2242db.png)
This is equal to 
after some manipulation, and indeed it turns out that the radax will require 
.
lies on the radical axes of the 2 circles.Let 
are collinear.
and 
.Note that 
and 
.
,
which implies 
as well. A bit of manipulations give that if 
.
.
bisects 
and also 
.
are concyclic and 
bisects 
.
are collinear.
.
we get that 
,
,
and 
,
and 
intersect 
at 
,
Suffices to show 
.
Subbing coordinates of 
Dear reader, the author of this post had used 
in a previous edit and was bashing his head against a wall when the expressions won't cancel, please send help.
This is clearly symmetric in 
and 
)
and 
by 
and 
,
and 
.
and 
intersect at 
.
we get
Similarly, 
which, combined, are enough to show that 
and 
,
so 
all unit vectors, we have (by Menelaus or areal coordinates), 
.
and 
(since it is the midpoint of 
so 
are collinear (with 
)
so 
is cyclic as required.
.
collinear. Now in complex, 
,
,
.
,
is 
note the obvious scalar multiple.
.![\[
f(X) = pow_{(DNE)}(X) - pow_{(DMH)}(X).
\]](http://latex.artofproblemsolving.com/5/9/2/5928825dbc465635cd491979cb1d06b7d68abea2.png)
We are done if 
.
at 
,
be its angles, and 
be its sides, as usual. Clearly 
.
(and similarly for 
to be the second intersection of 
.
.![\[
MU = MP+PU = \frac{e}{2}\cos \delta + PH\tan\alpha.
\]](http://latex.artofproblemsolving.com/c/6/5/c651a2d8e8b40b9599ab34a7b356fc60c5320ff3.png)
As 
is symmetric in our reference, we don't even need to compute it (even though it is doable). So, now,![\[
f(M) = MN\cdot MU = \frac{ae}{4}\cos\delta + \frac{a}{2}PH\tan\alpha.
\]](http://latex.artofproblemsolving.com/a/e/1/ae1246aed3a5ae3361244d38a5432d42e2dc8ad0.png)
And similarly, if 
is the second intersection of line ![\[
f(N) = -MN\cdot NV = -\frac{ad}{4}\cos\epsilon - \frac{a}{2}PH\tan\alpha.
\]](http://latex.artofproblemsolving.com/e/a/6/ea6cd224c1a36cf1aaeebf1bc1e29026049766cb.png)
Now we claim that 
.
,
by using Manelaus' on 
.![\[
0 = 5f(k) = 2f(M)+2f(N)+f(A) = \frac{ae}{2}\cos\delta + aPH\tan\alpha -\frac{ad}{2}\cos\epsilon - aPH\tan\alpha + \frac{d^2 - e^2}{2}
\]](http://latex.artofproblemsolving.com/8/6/5/865eea8e3c0cee6d30d49c4c76797c5f3a896870.png)
But everything cancels nicely and the equality reduces to![\[
ae\cos\delta-ad\cos\epsilon + d^2 - e^2 = 0.
\]](http://latex.artofproblemsolving.com/5/8/3/5834feaded925ad1f3ca8b6dac74874bb1f58a42.png)
This follows directly from using cosine law to plug 
and 
.
and 
and 
meet again at 
.
are collinear.
,
.
.
is cyclic, which immediately yields the desired collinearity. 
intersects 
at point 
Hence, 
,
.
gives that 
are collinear, and similarly, 
are collinear. Power of point at 
which gives the result.
be circumcenter and centroid of 
, 
.
.
and 
implies that 
so 
and 
implies that 
.
it is easy to see by menelos that 
and 
collinear.
.
,
,
,
.
and 
.
,
.
.
,
.
.
.
is the radical axis of the two circles, but since 
.
the lengths of sides 
.
.
.
.
is the midpoint of 
Let 
.![$[ABX]=[BXD]=[XDE]=[XEC] \implies \frac{AX}{XB}=\frac{[AXB]}{[BXC]}=\frac{1}{3}$](http://latex.artofproblemsolving.com/3/6/1/36100f35eb68f8b519dcf7d7cf9cfa8c749d2d80.png)
Similarly, 
(Van Aubel’s Theorem, easily provable by areas). Also note that since 
is the midpoint of 
meet 
bisects 
.
.
is the midpoint of 
,
,
and 
and 
.
and 
,
are concyclic.
and 
.
are concurrent at 
are collinear because
By letting reflections of 
we see that 
are collinear and we get that 
as 
and 
are parallelograms. Since 
we get that 
intersects 
.
so 
,
so that 
,
and 
.
and by applying menelaus on 
we get that 
,
and we are done!
s.t.
by Menelaus' Thm. we kan get 
collinear. Similarly 
collinear.
cut 
respectively.
by a corollary of Ptolemy's Thm. we get
Similarly
So
,
and
So
That is 
and we are done.
meet at 
meet 
at 
and 
.
.
bisects 
by Menelaus on 
wrt 
and 
wrt 
we have 
so 
.
is midpoint of 
so 
is isosceles and since 
.
.
.
which is well-known that 
so 
cyclic. From cyclic we get that
so we get that 
collinear. Let 
.
so we get that 
,
and 
we get that 
collinear and 
.
.
.
,
and 
are collinear, so from Desarguess theorem we get that triangles 
and 
are perspective as desired. So we are done!
and 
are paralellograms 
,
,
,
)
and 
are collinear triples. Notice that 
and 
are also parallelograms since 
and 
.
and 
are also collinear.
.
are isogonal with respect to 
.
follows from reflecting 
we have![\[\measuredangle MPQ = \measuredangle YPQ = \measuredangle YXQ = \measuredangle YXN = \measuredangle MNX = \measuredangle MNQ\]](http://latex.artofproblemsolving.com/4/4/0/44025ed2e5b6f70b67a1cd32766a361a062ee60e.png)
are concyclic.
![[asy]
/* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */
pair A = (4.86240,33.99728); pair D = (-1.88462,-8.34117); pair E = (32.20988,-8.34117); pair B = (-35.97913,-8.34117); pair C = (66.30439,-8.34117); pair M = (1.48889,12.82805); pair N = (18.53614,12.82805); pair X = (8.98249,17.06190); pair T = (15.16262,-8.34117);
import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps);
draw(A--D, linewidth(0.6)); draw(D--T, linewidth(0.6)+orange); draw(T--E, linewidth(0.6)+orange); draw(E--A, linewidth(0.6)); draw(D--B, linewidth(0.6)+red); draw(E--C, linewidth(0.6)+red); draw(A--B, linewidth(0.6)); draw(A--C, linewidth(0.6)); draw(B--X, linewidth(0.6)+blue); draw(X--C, linewidth(0.6)+blue); draw(A--T, linewidth(0.6));
dot("$A$", A, NW); dot("$D$", D, NW); dot("$E$", E, NE); dot("$B$", B, NW); dot("$C$", C, NE); dot("$M$", M, NW); dot("$N$", N, NE); dot("$X$", X, NW); dot("$T$", T, NE);
[/asy]](http://latex.artofproblemsolving.com/1/a/0/1a026efbc78f4ac291d71a12175ac3365a0d83a7.png)
.
with 
as the transversal, we get,![\[
\frac{AM}{MD}\cdot \frac{DB}{BT}\cdot \frac{TX}{XA}=-1
\]](http://latex.artofproblemsolving.com/1/d/0/1d061c602321e9045e072d4f39e5c2549eaea969.png)
which gives 
.
.
which further implies that 
,
.![[asy]
/* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (4.86240,33.99728); pair D = (-1.88462,-8.34117); pair E = (32.20988,-8.34117); pair M = (1.48889,12.82805); pair N = (18.53614,12.82805); pair H = (4.86240,-3.98309); pair X = (8.98249,17.06190); pair O = (10.01251,22.32315); pair T = (15.16262,-8.34117); pair G = (6.77954,5.80944);
import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps);
draw(A--D, linewidth(0.6)); draw(D--E, linewidth(0.6)); draw(E--A, linewidth(0.6)); draw(circle((-4.36993,2.90829), 11.52073), linewidth(0.6) + blue); draw(circle((19.24485,-1.71489), 14.56020), linewidth(0.6) + blue); draw(H--O, linewidth(0.6) + linetype("4 4") + red); draw(A--T, linewidth(0.6));
dot("$A$", A, NW); dot("$D$", D, dir(270)); dot("$E$", E, SE); dot("$M$", M, SW); dot("$N$", N, NE); dot("$H$", H, NW); dot("$X$", X, NW); dot("$O$", O, NW); dot("$T$", T, NE); dot("$G$", G, NW);
[/asy]](http://latex.artofproblemsolving.com/6/0/2/6025205ee2b0fd216152c0d91233619cc4684d3f.png)
is cyclic, it suffices to show that 
.
.![\[ \measuredangle NGM=\measuredangle HGM+\measuredangle NGH =\measuredangle HDM+\measuredangle NEH =2(90^{\circ}-\measuredangle DAE) =\measuredangle NOM \]](http://latex.artofproblemsolving.com/b/9/7/b970773afceb417cd143e3cac4fcf1b89a02f71e.png)
which implies that 
is cyclic. Furthermore,![\[ \measuredangle NGO=\measuredangle NMO =90^{\circ}-\measuredangle MAN =\measuredangle AEH=\measuredangle NEH=\measuredangle NGH \]](http://latex.artofproblemsolving.com/6/3/c/63c613c6fb7f71ff1dfdd23355790d560e54079c.png)
which gives us that 
are collinear.
.
denote the affix of the orthocenter of 
sends 
.
.![\[ x=\frac{3a+2t}{5}=\frac{3a+d+e}{5} =\frac{3a+(2n-a)+(2m-a)}{5}=\frac{a+2n+2m}{5} .\]](http://latex.artofproblemsolving.com/d/9/0/d9079a370dbd71fc876fa169194f94a5438dddcb.png)
![\[ \frac{h}{x}=5\cdot \frac{2m+2n+a}{a+2n+2m} =5\in\mathbb{R} \]](http://latex.artofproblemsolving.com/4/5/1/451af57ead3cf7633097ba034093434bede057e6.png)
which is clearly true and our claim is proved. 
are collinear and we are done.
,
,
.
.![[asy]
/* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-2.5,1.5); pair D = (-3.42948,-1.49748); pair E = (0.,-1.5); pair B = (-6.85897,-1.49496); pair C = (3.42948,-1.50251); pair H = (-2.50163,-0.72426); pair M = (-2.96474,0.00125); pair N = (-1.25,0.); pair T = (-2.18589,0.30050); pair P = (-4.01792,-0.40338); pair Q = (-0.59976,-0.20878);
size(11cm); pen yqqqyq = rgb(0.50196,0.,0.50196); pen zzttqq = rgb(0.6,0.2,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(A--D--E--cycle, linewidth(0.6) + zzttqq);
draw(circle((-3.29251,-0.71852), 0.79090), linewidth(0.6) + yqqqyq); draw(circle((-1.31746,-1.32705), 1.32876), linewidth(0.6) + yqqqyq); draw(A--D, linewidth(0.6) + zzttqq); draw(D--E, linewidth(0.6) + zzttqq); draw(E--A, linewidth(0.6) + zzttqq); draw(B--D, linewidth(0.6) + gray); draw(C--E, linewidth(0.6) + gray); draw(B--T, linewidth(0.6) + gray); draw(C--T, linewidth(0.6) + gray);
dot("$A$", A, dir((2.269, 5.764))); dot("$D$", D, dir((-7.595, -25.118))); dot("$E$", E, dir((4.622, -19.609))); dot("$B$", B, dir((-5.791, -19.529))); dot("$C$", C, dir((-5.942, -20.526))); dot("$H$", H, dir((5.937, -8.391))); dot("$M$", M, dir((-19.603, 16.026))); dot("$N$", N, dir((2.278, 4.468))); dot("$T$", T, dir((2.402, 4.794))); dot("$P$", P, dir((-10.671, 6.253))); dot("$Q$", Q, dir((2.095, 4.902)));
[/asy]](http://latex.artofproblemsolving.com/3/f/f/3ff5b176c14418534eb55af7e294d9a8b7800bf7.png)
.
and 
.![\[ \det \begin{bmatrix} 3 & 1 & 1 \\ 1 & 1 & 0 \\ 0 & 2 & -1 \\ \end{bmatrix} \]](http://latex.artofproblemsolving.com/8/2/1/8218d1eb4abbcbd02829da5d78fae19e8b85d1b9.png)
so ![\[ 0 = -a^2yz - b^2zx - c^2xy + (x+y+z)\left( \frac{c^2}{2} x + w z \right) \]](http://latex.artofproblemsolving.com/b/8/a/b8a17ae5c0607256a1ab6e511bc423391fff6e20.png)
for some constant 
(by plugging in ![\begin{align*} \frac{c^2}{2} S_B S_C + w \cdot S_A S_B &= S_A S_B S_C \cdot \frac{a^2S_A + b^2S_B + c^2S_C}{S_B S_C + S_C S_A S_A S_B} = S_A S_B S_C \cdot \frac{8[ABC]^2}{4[ABC]^2} \\ \implies \frac{c^2}{2} S_C + S_A \cdot w &= 2 S_A S_C \\ \implies S_A \cdot w &= S_C \left(b^2 + \frac{1}{2} c^2-a^2 \right). \end{align*}](http://latex.artofproblemsolving.com/b/0/1/b011c6b3cb44a50c18b5076a53a0af4b441a52c4.png)
In other words, ![\[ 0 = -a^2yz - b^2zx - c^2xy + (x+y+z)\left( \frac{c^2}{2} x + \frac{S_C(b^2+c^2/2-a^2)}{S_A} z \right). \]](http://latex.artofproblemsolving.com/4/4/7/4471a70eb08577c507bc7eff1acb6d8e3fec38da.png)
Similarly, ![\[ 0 = -a^2yz - b^2zx - c^2xy + (x+y+z)\left( \frac{b^2}{2} x + \frac{S_B(c^2+b^2/2-a^2)}{S_A} y \right). \]](http://latex.artofproblemsolving.com/2/7/9/27974a042223311b604c91c0a2519e5da22997ff.png)
To check ![\[ \frac{c^2}{2} \cdot 3 + \frac{S_C (b^2+c^2/2-a^2)}{S_A} = \frac{b^2}{2} \cdot 3 + \frac{S_B (c^2+b^2/2-a^2)}{S_A}. \]](http://latex.artofproblemsolving.com/3/3/9/339bacf1d3a68f526b9486068d0323e986f579fc.png)
Using a common denominator for the left-hand side, we get 
Since this is symmetric in 
and the problem is solved.
.
both intersect the 
,
and 
is 
while the perpendicular bisector of 
is 
Intersecting for the centre of 
and 
The centre of 
,
is 
so 
.
to meet the circles again at 
,
.
,
(antipode of 
.
analogously. Then as 
(in a directed sense) and 
,
is 
be on the line 
,
.
be the reflections of 
and 
intersect on the altitude of 
,
.
where the middle equality comes from rotating the pencil by 
,
.
,
where 
is the altitude from 
