Revival?
by agbdmrbirdyface, May 25, 2017, 4:51 PM
I think it's time to revive this blog.
2002 ISL G4 wrote:
Circles 
and 
intersect at points 
and 
. Distinct points 
and 
(not at or ) are selected on . The lines 
and 
meet again at 
and 
respectively, and the lines 
and 
meet at 
. Prove that, as and vary, the circumcenters of triangles 
all lie on one fixed circle.
and 
intersect at points 
and 
. Distinct points 
and 
(not at or ) are selected on . The lines 
and 
meet again at 
and 
respectively, and the lines 
and 
meet at 
. Prove that, as and vary, the circumcenters of triangles 
all lie on one fixed circle.![[asy]
size(250);
pair a1, a2, b1, b2, p, q, c, x, y;
path s1 = unitcircle;
draw(s1, blue);
path s2 = circle((2,0), 1.25);
draw(s2, blue);
pair[] ips = intersectionpoints(s1, s2);
p = ips[0];
q = ips[1];
a1 = dir(130);
b1 = dir(190);
x = 4*p - 3*a1;
pair[] aips = intersectionpoints(p--x, s2);
a2 = aips[1];
y = 4*p - 3*b1;
pair[] bips = intersectionpoints(p--y, s2);
b2 = bips[1];
c= extension(a1, b1, a2, b2);
draw(c--b1--b2--c--a2--a1, red);
draw(circumcircle(c, a1, a2), orange);
draw(circumcircle(c, b1, b2), linetype("8 8"));
dot(p);
dot(q);
dot(a1);
dot(b1);
dot(a2);
dot(b2);
dot(c);
label("$P$", p, dir(91)*1.5);
label("$Q$", q, S*1.5);
label("$A_1$", a1, dir(150));
label("$A_2$", a2, dir(320));
label("$B_1$", b1, dir(220));
label("$B_2$", b2, dir(40));
label("$C$", c, dir(100));
[/asy]](http://latex.artofproblemsolving.com/9/3/2/9322d314427bfc8e30359221baa988cc0df8f3cc.png)
,
,
,
,
,
concur at a point 
.
is a complete quadrilateral, regardless of the choice of 
.
passes through 
,
and 
.
be the circumcenter of an acute-angled triangle 
,
be the projections of 
respectively. Prove that the incenter of 
lies on the Simson line of 
be the reflections of 
and define 
similarly, so it suffices to show the incenter of 
lies on 
.
.
is on 
with 
,
not containing 
,
.
be the desired orthocenter and 
be the orthocenter of 
.
are inversely congruent, i.e. 
share a common perpendicular bisector 
,
correspond in this reflection, so it suffices to show that 
,
.
and let 
be the reflection of 
similarly.
is the midpoint of arc 
that 
and similarly for 
.
and similarly for 
.
so that square roots work out nicely (or more like I'm too lazy to write out the other cases 
are 
,
.
,
be a positive integer. Two players Alice and Bob take turns to write numbers from the set 
on a blackboard. Alice begins the game by writing 
on her first move. If a player has written 
on a certain move, the adversary is then allowed to write either 
or 
(provided the number does not exceed 
is of type A or type B.
whose type is different from the type of 2004.
and 
are of the same type as 
.
will always have winning strategies for 
and 
,
,
.
,
,
.
,
,
.
,
,
.
its first digit is 
and all other digits are even, as otherwise the process can be continued until we reach the first odd digit (reading from the right), making 
in base four is 
that is type B is 
,
and 
intersecting at points 
and 
,
be a line through the center of 
and 
and let 
be a line through the center of 
and 
.
and 
.
be center of 
From radical axes 
is the orthocenter of 
But since 
goes through the orthocenter and is perpendicular to 
satisfying ![\[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]](http://latex.artofproblemsolving.com/2/6/e/26edbd87f49eb68ec9b662b602db712e5151dc9c.png)
![\[a_1b_1 + a_2b_2 + \ldots + a_n b_n \le \sqrt{a_1^2+a_2^2+ \ldots + a_n^2} \cdot \sqrt{b_1^2+b_2^2+ \ldots + b_n^2}.\]](http://latex.artofproblemsolving.com/5/4/f/54fd6110e36d0f1a2fc5ba503c59c41f0d2b88cd.png)
WLOG let 
so that ![\[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x+xyz} = \sqrt{(x-1+1)(1+yz)}\]](http://latex.artofproblemsolving.com/5/2/a/52a2b979127ca95766c2404e3380546c317ae1c8.png)
, and from Cauchy, ![\[\sqrt{(x-1+1)(1+yz)} \ge \sqrt{x-1}+\sqrt{yz}=\sqrt{x-1}+\sqrt{(y-1 +1)(1+ z-1)} \ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]](http://latex.artofproblemsolving.com/d/7/a/d7a084b72655f0eafc291733da4e2758f05ceeea.png)
In order for the original equation to be true, equality must be present in both inequalities, giving us ![\[x-1 = \frac{1}{yz}\]](http://latex.artofproblemsolving.com/d/3/b/d3b3cd562d28b9398742a479b5d56c84561d394d.png)
![\[y-1 = \frac{1}{z-1},\]](http://latex.artofproblemsolving.com/1/3/c/13c1c6cfa9e0f7b8e2ae8062a216e36ce14153b5.png)
which gives us the triplet ![\[(x, y, z) = \left(\frac{y^2+y-1}{y^2}, y, \frac{y}{y-1} \right)\]](http://latex.artofproblemsolving.com/3/5/c/35c03ce49e20f2eef7a81031dea41728542052d9.png)
for 
![\[y \ge x \Rightarrow y \ge \frac{y^2+y-1}{y^2} \Rightarrow y^3 -y^2-y+1 = (y+1)(y-1)^2 \ge 0\]](http://latex.artofproblemsolving.com/7/4/6/7468b2a9d10d21dd1938e77aabb8cbb1fcf2e174.png)
![\[z \ge x \Rightarrow \frac{y}{y-1} \ge \frac{y^2+y-1}{y^2} \Rightarrow y^3 \ge y^3-2y+1 \Rightarrow 2y \ge 1.\]](http://latex.artofproblemsolving.com/c/a/5/ca5f502f389874abbe54c7fe899c78db5a177375.png)
Also, 
as otherwise that would make 
undefined. Specifically, the original condition breaks down to ![\[\sqrt{1+zx} = \sqrt{x-1}+\sqrt{z-1}\]](http://latex.artofproblemsolving.com/6/a/5/6a50a0a4977c1de1a5330dff7a5dbbf70e6feaed.png)
but as we previously had gotten ![\[\sqrt{1+zx} > \sqrt{zx} = \sqrt{(z-1+1)(1+x-1)} \ge \sqrt{z-1}+\sqrt{x-1},\]](http://latex.artofproblemsolving.com/9/d/d/9dd08f065e62a2add79fe682b9542c2bf5aaa52a.png)
contradiction. Finally, we check that ![\[\frac{y}{y-1} = 1 + \frac{1}{y-1} > 1\]](http://latex.artofproblemsolving.com/5/8/2/582df949d003d20ef57befe9a550e354f748063b.png)
![\[\frac{y^2+y-1}{y^2} = 1 + \frac{y-1}{y^2} > 1.\]](http://latex.artofproblemsolving.com/f/a/1/fa123f6d78ac187d316df4f5b2a9630b3a708415.png)
Therefore, we conclude 
for 
generates our solution set.
be a circle and let 
be a line such that 
be a diameter of the circle 
be an arbitrary point on the circle 
be the point of intersection of the lines 
and 
;
the point of intersection of the lines 
and 
intersect the circle 
,
.![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.976002749378333, xmax = 34.86531199295848, ymin = -10.481189671238441, ymax = 12.239223899756556; /* image dimensions */
/* draw figures */
draw(circle((4.02,0.64), 4.671316730858656));
draw((xmin, -1.4436090225563913*xmin + 27.16711976319434)--(xmax, -1.4436090225563913*xmax + 27.16711976319434)); /* line */
draw((0.18,-2.02)--(17.401986181172095,2.0454555016526608));
draw((17.401986181172095,2.0454555016526608)--(5.17527077355485,5.166206959449599));
draw((5.17527077355485,5.166206959449599)--(24.587486816833866,-8.327598047573366));
draw((0.18,-2.02)--(24.587486816833866,-8.327598047573366));
draw((0.18,-2.02)--(13.720717655728016,7.359767959436592));
draw((7.86,3.3)--(8.644181009288122,-0.021928994182223827));
draw((0.8275136207952526,4.050165790484705)--(24.587486816833866,-8.327598047573366));
draw((0.18,-2.02)--(5.17527077355485,5.166206959449599));
/* dots and labels */
dot((7.86,3.3),dotstyle);
label("$B$", (6.338195399361469,3.090290649029001), NE * labelscalefactor);
dot((0.18,-2.02),dotstyle);
label("$A$", (-0.5150645633606531,-2.5813727683961987), NE * labelscalefactor);
dot((13.720717655728016,7.359767959436592),dotstyle);
label("$P$", (13.866653387967544,7.681637225039878), NE * labelscalefactor);
dot((8.644181009288122,-0.021928994182223827),dotstyle);
label("$C$", (8.90394789772049,-1.5685757295702702), NE * labelscalefactor);
dot((17.401986181172095,2.0454555016526608),linewidth(3.pt) + dotstyle);
label("$D$", (17.917841543271265,1.9762139063204798), NE * labelscalefactor);
dot((5.17527077355485,5.166206959449599),linewidth(3.pt) + dotstyle);
label("$E$", (5.257878557947144,5.487243640917033), NE * labelscalefactor);
dot((24.587486816833866,-8.327598047573366),linewidth(3.pt) + dotstyle);
label("$F$", (25.074940617641165,-8.489355494880781), NE * labelscalefactor);
dot((6.090428041178006,-3.547424951721978),linewidth(3.pt) + dotstyle);
label("$G$", (5.865556781242702,-4.6069668460480555), NE * labelscalefactor);
dot((0.8275136207952526,4.050165790484705),linewidth(3.pt) + dotstyle);
label("$G'$", (0.2614131664058928,4.440686700796906), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/0/c/6/0c60f19dac196db75f17aa7938bb3bf472112cf8.png)
We show 
is the reflection of 
We'll proceed with straight angle chasing. It suffices to show ![\[\angle{G'AB}=\angle{G'CB}=\angle{BAG}\]](http://latex.artofproblemsolving.com/1/7/9/179c53ddd54aecfb450ca523aa47cd766a93ae02.png)
![\[90-\angle{DCF}=\angle{BAG} \leftrightarrow 90-\angle{BAG}=\angle{PFA}=\angle{DCF} \leftrightarrow \triangle{DCF} \sim \triangle{DFA}\]](http://latex.artofproblemsolving.com/4/3/a/43af76ca0e56f19dc4afd15233ad15fa2fd2c5ef.png)
![\[\frac{DF}{DC}=\frac{AD}{DF} \Rightarrow AD \cdot DC = DF^2.\]](http://latex.artofproblemsolving.com/7/b/5/7b5aad8b4f78989c99e7d393085642ae73997a5c.png)
But by power of a point, 
so it suffices to show 
which follows from a final angle chase: ![\[\angle{DFE}=\angle{PFE}\stackrel{AEPF \text{cyclic}}{=}\angle{BAE}=\angle{DEB},\]](http://latex.artofproblemsolving.com/c/5/1/c51263bb41ac43d7f258345954acd490af1a6a8f.png)
showing 
is isosceles and we're done.
.
(other than 
intersects sides 
(other than 
,
.
is tangent to the circumcircle of triangle 
.![[asy]
import graph;
size(250);
pair A, B, C, D, M, H, L, Q, P, N, d, m;
A = (0,0);
B = dir(250);
C = 1.5*dir(320);
draw(A--B--C--A, red);
path circum = circumcircle(A, B, C);
draw(circum, blue);
M = midpoint(B--C);
d = bisectorpoint(B, A, C);
D = extension(A, d, B, C);
m = rotate(90, M) * C;
L = extension(A, D, M, m);
draw(M--L, orange);
path circ = circumcircle(A, D, M);
draw(circ, magenta);
pair[] abinters = intersectionpoints(circ, A--B);
Q = abinters[1];
pair[] acinters = intersectionpoints(circ, A--C);
P = acinters[1];
N = midpoint(Q--P);
draw(Q--P);
H = foot(L, D, N);
draw(N--H, orange);
draw(A--L);
draw(circumcircle(N, M, H), magenta);
draw(M--H--L, orange);
draw(N--M);
dot(M);
dot(D);
dot(L);
dot(Q);
dot(P);
dot(N);
dot(H);
label("$A$", A, dir(100));
label("$B$", B, dir(215));
label("$C$", C, dir(345));
label("$D$", D, dir(220));
label("$M$", M, dir(315));
label("$P$", P, dir(20));
label("$Q$", Q, dir(170));
label("$N$", N, dir(65));
label("$L$", L, dir(270));
label("$H$", H, dir(290));
[/asy]](http://latex.artofproblemsolving.com/c/7/f/c7f12ad4de9fe474d68f91883839448a8b3e08d7.png)
and 
.
is cyclic yields 
,
and 
are parallel. This looks like a job for barycentric coordinates!
and 
.
into 
and we get 
.
We solve for 
and 
and we find that 
and 
.
to find 
to find 
We have to find the midpoint of this segment 
We have our point at infinity along 
.
thus implying 
are collinear and so 
and we are done.
be an odd prime. The sequence 
is defined as follows: 
and, for all 
is the least positive integer that does not form an arithmetic sequence of length 
in base 
and reading the result in base 
.
positive integers, there is at least one with a 
digit in its base 
then we simply observe that ![\[\{a, a+d, \ldots, a+(p-1)d \} \]](http://latex.artofproblemsolving.com/f/a/7/fa75bc131f32ded1314f97de8ec11fb9bfe6e010.png)
forms a complete residue class mod 
and thus the set of unit digits for the base 
If the common difference 
then the same argument can be made for the "p's" digit, which must form a complete residue class. In general, if 
we can make the same argument for the 
'
to start the sequence. Now, assume for the sake of contradiction that the statement has been proved up to ![\[X=\overline{a_{k}a_{k-1}\ldots a_1 a_0}\]](http://latex.artofproblemsolving.com/d/4/1/d41706884ce7086f3ac512c88e5cf129f70f3aed.png)
and let the culprit be 
Now consider the numbers ![\[X, X-p^j, X-2p^j, \ldots, X-(p-1)p^j.\]](http://latex.artofproblemsolving.com/2/4/6/2460b0788356abbba12cc93e385887cd8952f24b.png)
If there're multiple culprits, i.e. 
consider the numbers ![\[X, X-(p^{j_1}+p^{j_2} + \ldots + p^{j_m}), \ldots, X-(p-1)(p^{j_1}+p^{j_2} + \ldots + p^{j_m}).\]](http://latex.artofproblemsolving.com/9/e/a/9eaec5fc68a522d921c1d4347868674ff5bd45b4.png)
The motivation is that except 
each of the following numbers in the arithmetic sequence we've constructed is positive and doesn't contain a 
is the p-th term of an arithmetic progression of terms in the sequence, and we're done.
lie on sides 
,
,
,
denote the circumcircles of triangles 
,
,
,
intersects 
,
,
.
to 
Observe that ![\[\angle{MYX}=\angle{MYP}=\angle{MBP}\]](http://latex.artofproblemsolving.com/6/6/7/6672fdc32d97d7d8b7968bd3118d278146da1a8a.png)
![\[\angle{MZX}=\angle{MZA}=\angle{MCP},\]](http://latex.artofproblemsolving.com/9/b/9/9b91dee49ba1d77e13f06987321fca9913c1a81e.png)
and thus 
as desired.
which follows from ![\[\angle{YXM}=\angle{AXM}=\angle{ARM}=\angle{BPM}\]](http://latex.artofproblemsolving.com/0/d/6/0d64245125506129aa02583a30c124e789f2e786.png)
as desired.
May 2017
April 2017