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.
, such that 
and 
.
such that 
is prime, 
with 
and ?
be a positive integer. Egor has 
cards with the number “
” written on them, and cards with the number “
” written on them. Egor wants to paint each card red or blue so that no subset of cards of the same color has the sum of the numbers equal to . Find the greatest such that Egor will not be able to paint the cards in such a way.
that has 8097 distinct points with area of a triangle that has 3 points belong to all 
. Prove that there exists a triangle 
that has the area 
contains at least 2025 points that belong to ( each of that 2025 points can be inside the triangle or lie on the edge of triangle )X
where 
. Point 
is the circumcenter of triangle , and 
is the projection of point 
onto line 
. The midpoints of , 
, and 
are 
, 
, and 
, respectively. The line 
intersects 
and 
at points 
and 
, respectively. Prove that is the incenter of triangle 
.
be an acute triangle with orthocenter . Let 
, 
and 
be the feet of the altitudes of triangle opposite to vertices 
, 
, and 
respectively. Let 
and 
be the midpoints of 
and 
, respectively. Let 
be the intersection of lines 
and 
. Prove that is the circumcenter of triangle if and only if is the midpoint of 
.
be the circumcircle of acute triangle . Points 
and 
are on segments 
and 
respectively such that 
. The perpendicular bisectors of 
and 
intersect minor arcs and of at points 
and respectively. Prove that lines 
and 
are either parallel or they are the same line. be a scalene acute triangle with incenter 
and circumcircle 
. 
is the midpoint of small arc 
on and 
is the projection of onto the line passing through the midpoints of and . A circle 
with center 
is internally tangent to at , and touches segment . If the circle with diameter 
meets again at 
, prove that 
bisects 
. be circumscribed about circle 
, and let be the orthocenter of 
. The circle touches line at . The tangent to the circle 
at meets at 
. Let be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
be a square and a point lies on the interior of such that triangle 
is equilateral. Evaluate 
is the foot of the altitude from of triangle . On the lines and points and 
are marked such that the circumcircles of triangles 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles and at and intersect at 
.
. be a scalene triangle such that is its incircle. is tangent to at . A point (
) is located on .
, 
, and 
be the incircles of the triangles 
, 
, and 
, respectively. and is also tangent to .
be a triangle. Triangles 
and 
are constructed outside of triangle such that 
and 
and 
. Segments 
and 
meet at 
. Let be the circumcenter of triangle 
. Prove that 
be a triangle. Triangles and are constructed outside of triangle and and . Segments and meet at . Let be the circumcenter of triangle . Prove that 
then because the triangles 
and 
coincide after rotation with center and angle 
, we have that the angle formed by the segments and 
is equal to . Then we get that 
and 
. Now we take as the origin of the complex plane. Then if 
we have that 
and 
and also 
thus 
. Hence 
which means that 
is perpendicular to 
, as desired. 
with . Since we have so many pairs of equal angles here, probably it would help if we can take advantage of some hidden cyclic quadrilaterals and some spirality.
. It is not hard to see that 
and 
are cyclic. Indeed, the congruence of two triangles 
and 
gives us 
, 
. Let's draw the circles 
and 
. Denote 
respectively the intersections of 
with these two circles.
. Hence, let 's bisect this angle by calling the midpoint of the arc of the circumcircle of 
. Then we have some spiralities here: 
and 
, 
and 
. Thus 
is cyclic and 
are collinear. Similarly, 
is cyclic and 
are collinear.
. This is equivalent to proving 
. Hence, it is sufficient for us to prove that actually 
is isosceles at and 
. The latter is very easy by some angle-chasing. To prve 
, use .
and are congruent so 
. then 
are cyclic.
. But 
so 
through 
meet at 
.
and 
are similar so 
so 
and are similar. Then triangle 
and 
are similar
and 
and 
and 
meets at . we have 
so 
are cyclic.
so 
and are congruent so . are on circle 
. Similarly 
are on circle 
meet 
at 
be the circumcircle of triangle 
meet 
at 
.
so triangles 
are similar. are midpoint of so triangles 
and 
are similar
. Then 
are cyclic
are cyclic we have 
.
are cyclic. (
meets at ) Thus 
is the radical axis of and 
so 
. (because 
and 
)
, i.e. 
.
, the circumcircle 
of 
and 
. Thus, 
the quadrilaterals , are cyclically with the radical axis 
.
the ray 
is the bisector of the angle 
.
. Thus, 
and 
.
, 
to the sides 
, 
respectively :
. I used the simple relations 
and 
.
and intersect 
and 
at 
and 
, respectively. Since quadrilaterals 
and 
are similar, it follows that 
and hence that and 
are parallel. Now let 
and 
be the circles centered at and with radii 
and 
, respectively. Let the second intersection of and be . By the inscribed angle theorem, it follows that 
which implies that 
. Now note that since is the center of spiral similarity taking 
to 
, it follows that and are cyclic which implies that 
and hence that 
. Therefore 
is cyclic and since 
bisects 
and 
, it follows that , and are collinear. Hence is the radical axis of and , which implies that is perpendicular to and therefore perpendicular to since is parallel to .
be the circumcenter of 
.
be the midpoint of 
, respectively .
.
are concyclic ,
. ... 
,
,
. ... 
and we get 
,
and 
.
if only if 
(It's well known)
and 
and 
By power of point in 
we get : if only if 
since 
(By power of point in 
we get: 
we get: 
we get: which it is true ... 
, 
and 
, where 
. It remains to verify 
. To ease the amount of calculation we also set to be the origin. Also, we let 
.Then we get ![\[ \frac{o}{\overline{o}} = \frac{p-q}{\overline{p}-\overline{q}} \iff \frac{\frac{b-cz^2}{1-z^2}}{\frac{\frac{1}{b}-\frac{1}{cz^2}}{1-\frac{1}{z^2}}} = \frac{\frac{b}{z}-cz}{\frac{z}{b}-\frac{1}{cz}} \iff \dots \]](http://latex.artofproblemsolving.com/a/a/c/aacd03db49aa05bac448f9aed1096e0a4c37ac26.png)
That above is how the author of the article presented the solution; I didn't formulate it like him.
. Why is that true?
as the origin, 
, and 
. By standard formulas, we have 
and 
. and are both cyclic. Observe 
, with spiral center at , so the standard Yufei spiral similarity configuration tells us 
and 
. Now angle chase to get 
.
. Now we want 
as pure imaginary, that is![\[ \frac{(b-e^{i 2\theta})}{(1-e^{i 2\theta})(e^{-i \theta}b-e^{i \theta})} = \frac{-(\overline{b}-\frac{1}{e^{i 2\theta}})}{(1-\frac{1}{e^{i 2\theta}})(e^{i \theta} \overline{b}-\frac{1}{e^{i \theta}})} \]](http://latex.artofproblemsolving.com/a/0/8/a0822fcfad7251f745c78f54fb0461a687a3bc3c.png)
After some simplifying, one can see that this is clearly true, so 
and we are done.
outside such that 
. By considering the rotation sending 
to (which are congruent by SAS) we see that the angle between lines 
and is equal to 
. Thus , , , and lie on a common circle, so is the circumcenter of 
. We will subsequently dispose of point .
and 
the projections of onto and onto respectively, and let be the midpoint of 
. I claim that 
. To prove this, remark that by simple angle chasing 
, so 
is directly similar to 
. By the duality of spiral similarity, 
with similarity ratio 
. Analogously, 
with the same similarity ratio 
. Since is taken to 
and 
under these spiral similarities, we must have .
denote the measure of angle 
, and remark that ![\[\angle D_1MD_2 = 180^\circ - \angle BMD_1 - \angle CMD_2 = 180^\circ - \angle BOC = 2\psi.\]](http://latex.artofproblemsolving.com/9/1/c/91cf19b961c36095c55c18a8e763de858cb8bb7b.png)
It follows that, upon letting the perpendicular from to 
intersect at , we have 
, so quadrilateral 
is cyclic. Thus the angle between 
and is equal to 
, which implies 
. But 
by similarity, and so we obtain the desired perpendicularity. 
. By Spiral Similarity, 
so 
. Thus 
. Now for the key step, construct point outside 
such that 
. Then 
thus,
Furthermore, 
so
Hence, using directed angle, we find
as desired.
and 
are inscribed; next, meet and 
at , and and 
at . Since 
, 
is inscribed. So, if 
is the radius of 
, ![\[
PO^2-PA^2 = PR\cdot PC + \rho^2 - PA^2 = PA\cdot PX +\rho^2 - PA^2 = PA\cdot AX + \rho^2 = \mathcal P(A,(PQXY)),
\]](http://latex.artofproblemsolving.com/4/e/4/4e464ba39da2a0edbd77dc9ca1f2fd750fe70389.png)
where 
denotes power. Since the last expression is the same for 
, this means 
, done!
and 
Let meet 
for a second time at 
Let meet 
for a second time at 
is the center of spiral similarity sending 
, we know that is the second intersection point of and 
By Reim's theorem for 
cut by 
it follows that 
Because 
is isoceles, is parallel to the tangent to at 
Therefore, by the converse of Reim's theorem for cut by 
it follows that is cyclic. Thus, there exists 
such that ![\[ r^2 = \text{pow}(A, \odot(PQXY)) = AP \cdot AX = AQ \cdot AY. \]](http://latex.artofproblemsolving.com/3/d/5/3d5774263f7deef82fe95cb3a435d20f3e89eb90.png)
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[/asy]](http://latex.artofproblemsolving.com/7/e/e/7ee16b7b39bddd644abf9b85838155c620752078.png)
Because ![\[ PR \cdot PC = PA \cdot PX = PA(PA + AX) = PA^2 - r^2, \]](http://latex.artofproblemsolving.com/5/d/c/5dc109bb15bd6a8914ec3c7448b6993d1efe259d.png)
we see that 
lies on the radical axis of 
and 
Similarly, lies on the radical axis of and Therefore 
is the spiral similarity that sends to , so it rotates to implying 
. Also, 
are cyclic. be the intersection of 
. Then sends to , but 
, so is the midpoints of arcs and in the circles.
so bisects and , 
concur on 
. Therefore, 
, so (since 
), by (a working SSA similarity), 
. to 
, and to 
. Thus, the angle between and is ![\[\measuredangle (AO, PN)+\measuredangle (PN, PQ)=\measuredangle ABP+\measuredangle NPQ = 90^{\circ}\]](http://latex.artofproblemsolving.com/e/0/0/e001f13d5f3191924dc6e3be41b6c2f1e1d38d9b.png)
. Note that 
thus 
. Therefore 
. Let 
be the reflection of across .
thus 
which means 
. Moreover, 
thus 
. Thus 
. Combining this with 
gives the conclusion.
are cyclic let 
be intersection of this circles with lines and respectively.
is cyclic quadrilateral and with this observations according power of point theorem we have: 
be a point outside and 
. By Jacobi, 
are concurrent at and lies on . Hence, 
. Let 
be the foot of perpendicular from 
to 
respectively. By Jacobi, 
are concurrent.Then Jacobi 
, 
but 
(
). Therefore, 
.
. Note that is the center of spiral similarity 
, so it also maps 
. Thus 
, in particular 
. Hence 
. This eliminates entirely! Indeed, is now located on the perpendicular bisector of with 
.
. Let 
. Then
We know and . Hence 
, so actually
Finally, confirm
is the miquel point of 
. This means that we must have 
and 
are cyclic. Since![\[\angle ARP = \angle ABP = \angle ACQ = \angle ARQ\]](http://latex.artofproblemsolving.com/c/3/f/c3f9f68354ff7339dc00d8df6c62e6e2dd2516c4.png)
this means lies on the angle bisector of 
., so 
. For convenience, let 
. Since lies on the angle bisector, let 
, for some number 
. Now, the equation for 
is![\[-a^{2}yz - b^{2}xz - c^{2}xy + (wz)(x+y+z)\]](http://latex.artofproblemsolving.com/d/5/d/d5db686867c24ef410684bfb5a7071433bea59b5.png)
because passes through 
. Plugging in gives![\[-a^{2}bc - b^{2}tc - c^{2}tb + wc(t+b+c) = 0\Rightarrow w = \frac{a^{2}b + b^{2}t + cbt}{t+b+c}\]](http://latex.artofproblemsolving.com/0/2/5/025e55bd10f8afbed0a1ffd1cf7d2741e354d64e.png)
Now, since 
, if 
, then we can plug this into our equation for to get![\[-b^{2}p(1-p) + \frac{a^{2}b + b^{2}t + cbt}{t+b+c}\cdot (1-p) = 0 \Rightarrow p = \frac{a^{2} + bt + ct}{bt + b^{2} + bc}\]](http://latex.artofproblemsolving.com/4/7/3/473d66445a375350b0d05465be852e6e46d9c271.png)
Therefore, by symmetry,![\[P = \left(\frac{a^{2} + bt + ct}{b^{2}+bt + bc} : 0 : \frac{b^{2} + bc - ct - a^{2}}{bt + b^{2} + bc}\right), Q = \left(\frac{a^{2} + bt + ct}{c^{2} + ct + bc} : \frac{c^{2} + bc -bt - a^{2}}{c^{2} + ct + bc} : 0 \right)\]](http://latex.artofproblemsolving.com/8/d/e/8de027e7eba88a9de6e86616b0e40787f73f0edd.png)
To show that , we can use generalized perpendicularity (Theorem 7.25 of EGMO). Assume that was the origin. We have![\[\overrightarrow{PQ} = (a^{2} + bt + ct)\cdot \frac{c-b}{bc(b+c+t)}\vec{R} + \frac{bt + a^{2} - bc - c^{2}}{c(b+c+t)}\vec{B} + \frac{b^{2} + bc - ct - a^{2}}{c(b+c+t)}\vec{C}\]](http://latex.artofproblemsolving.com/6/3/f/63f95790f4b3ea81bdb4c89c5048b9e788d8d44c.png)
Since is the origin, we have 
. By Theorem 7.25 of EGMO, we must show the following expression is 
:![\[0 = a^{2}\left(\frac{bt+a^{2}-bc-c^{2}}{b+c+t} + \frac{b^{2} + bc -ct - a^{2}}{b+t+c}\right)+ b^{2}\left(\frac{c-b}{b(b+c+t)}(a^{2} + bt + ct) + \frac{t(b^{2} + bc - ct - a^{2})}{b(b+c+t)}\right)+ c^{2}\left(\frac{c-b}{c(b+c+t)}(a^{2} + bt + ct) + \frac{t(bt + a^{2}-bc-c^{2})}{c(b+c+t)}\right)\]](http://latex.artofproblemsolving.com/6/0/f/60f511fff89eed41009f1f4a962dfb299cff675b.png)
![\[\Rightarrow 0 = a^{2}(b^{2}-c^{2} + bt - ct) + b((c-b)(a^{2} + bt + ct) + b^{2}t + bct - ct^{2} - a^{2}t)+ c((c-b)(a^{2}+bt+ct) + bt^{2} + bct - ct^{2} - at^{2})\]](http://latex.artofproblemsolving.com/3/1/b/31b60be603f083d37eb8ad650ee0cab02e8cdcad.png)
![\[\Rightarrow 0 = a^{2}(b^{2} - c^{2} + bt - ct) + b[a^{2}c + bct + c^{2}t - a^{2}b - ct^{2} - a^{2}t] + c[a^{2}c - a^{2}b - b^{2}t + bt^{2} + a^{2}t - bct]\]](http://latex.artofproblemsolving.com/a/1/f/a1ffa5681a7b7f6c82c8393fcf2e3eca0b12e51f.png)
![\[= a^{2}b^{2} - a^{2}c^{2} + a^{2}bt - a^{2}ct + a^{2}bc + b^{2}ct + bc^{2}t - a^{2}b^{2} - bct^{2}-a^{2}tb + a^{2}c^{2} - a^{2}bc - b^{2}ct + bct^{2} + a^{2}ct - bc^{2}t = 0\]](http://latex.artofproblemsolving.com/8/e/c/8ec2b0e01f8d82d026aa22a202776566b5211544.png)
Therefore, this expression is , which means .
be a point outside and . By Jacobi, are concurrent at and lies on . Hence, . Let be the foot of perpendicular from to respectively. By Jacobi, are concurrent.Then Jacobi , but (). Therefore, ., " Why'd have this perpendicularity?
, " Why'd have this perpendicularity?
.By Jacobi, are concurrent 
are collinear. Then Jacobi with point 
. We get 
. By spiral sim 
and 
are cyclic. Under an inversion at with radius 
and a reflection about the angle bisector of 
, goes to 
. Since 
, if we let 
be the point such that 
is a parallelogram, 
are isogonal in , so 
. Let be the center of 
and let be the center of 
. Then the perpendicular from to 
, the perpendicular from to 
, and the perpendicular from to 
meet at . Thus 
are orthologic, as desired.
. Since 
, we have 
, so 
(the spiral similarity at sending to has a rotation of angle 
). Thus, 
, so .
. Note that 
. Similarly, 
.
which is clearly symmetric. Thus, 
, and we are done.
and 
are rotations about . In particular, and are cyclic. It follows that 
. Call this common angle . Now let 
be congruent to and , and let and 
be on 
so that 
. Let be the foot from to . For points in the plane, define 
. We have
and are the midpoints of 
and 
, this is
is parallel to the radical axis of and the point circle at , which is perpendicular to , as desired.
, so by spiral similarity facts and are cyclic. Obviously 
bisects as well. Thus we may view the problem with reference triangle 
, restating it as follows.
be a triangle with circumcenter and be a point on its internal 
-bisector. Let 
and 
intersect 
and 
at and respectively. Then 
.
. If the equation of is 
, then plugging in and implies 
, and then plugging in implies 
. To compute , we solve for 
given that 
lies on the circle: this will yield (after dividing out 
, which would correspond to itself) 
. Likewise, if 
then 
.
so we're done. 
satisfy 
, define 
similarly. Consider ellipse with foci at through 
such that 
and 
are tangents (not sure if it'll come in handy later). centered at which is tangent to 
. This entire ellipse configuration comes from djmathman's handout, yay!
is basically showing is parallel to some radical axis. Let be the foot of the altitude from to , and define similarly.![\[PK^2-PR\cdot PC=QL^2-QR\cdot QB.\]](http://latex.artofproblemsolving.com/4/a/9/4a9a2dcf07cf81ee3055ecf2e47edc5dba9ccfac.png)
We're basically almost done I think. Notice that this equates to![\[(PK+QL)(PK-QL)=PR\cdot PC-QR\cdot QB\implies PC(PK-QL)=PR\cdot PC-QR\cdot PC\implies PK-QL=PR-QR\]](http://latex.artofproblemsolving.com/1/2/d/12dce25f6fda9be8b3b2c90f699f00ae3ddb691c.png)
or just![\[PK-PR=QL-QR\implies RK=RL\]](http://latex.artofproblemsolving.com/0/2/5/025839d165c2059d7ad01a3ad48371355cc0cb6a.png)
which is obviously true. Ta-da! 22 minutes! WOOOOOOOOOOOOOO
taking 
to 
. This means that is the Miquel point of 
, so 
is cyclic. This means that 
Thus, if we let denote the arc midpoint of not containing , we have that 
. Since they are both isosceles, we have 
as well.. Let 
. Furthermore, we let 
as another free variable as well (the problem has four degrees of freedom, so this decision is ok). to compute 
. We have 
which we can solve for to get 
Similarly, 
Thus, we have 
Finally, dividing this by gives 
, which is clearly imaginary after conjugating and multiplying top and bottom by 
, so we are done.
as the unit circle as usual, then note that there is an "angle of rotation" of some to get and , and setting , 
, 
, and 
as the four "free" variables to represent the four degrees of freedom. While this quickly gives and 
, the computation for 
is now very messy as it requires the general intersection formula, much to the detriment of this setup, as is required to compute the circumcenter. as the unit circle is that, rather than having to go through the tedious process of computing the circumcenter, this gives us the circumcenter "for free". Then, we can make our remaining free variable.
into a similarity condition, which is actually a much stronger condition than just the angle equivalence, as it effectively encodes two degrees of freedom rather than just one. If we were to use directly, we would actually need two equations to compute (the other one probably being 
collinear), which will be much less elegant. and , we used the fact that 
is similar to 
to indirectly compute .
as the origin, at 
, at 
, at , and at 
.
and are cyclic, say by considering the pair of congruent triangles. Letting 
, it follows that . In other words, we have 
, so 
, and thus 
Hence 
as required.
and 
are cyclic so 
.
and move with degree 
along the internal angle bisector of . Then all are spirally similar at , so has degree . Similarly has degree , so line has degree . Line has degree , so the conclusion has degree .
cases.
then is the reflection of over the external bisector, so taking isogonals of the desired result gives perpendicular to the altitude, which is clear.
then 
so 
and 
. Similarly this works when 
.
then 
so 
. We finish.
is a triangle and is an arbitrary point on the interior angle bisector of 
. Let 
meet 
at . Reflection of over 
is 
. Let be the midpoint of arc 
of . Prove that 
.
. Since 
we see that are collinear. Let 
be the foot of the altitude from to . Since 
, we get that 
is the angle bisector of 
. DDIT on 
gives 
is an involution and this must be reflection over angle bisector of thus, 
. Note that 
lie on the circle with diameter 
. It sufficies to show that 
are collinear.![\[180-\measuredangle DLP=\measuredangle PLC=\measuredangle DLA=\measuredangle DKA=180-\measuredangle PKD\]](http://latex.artofproblemsolving.com/9/6/6/966e87cb5f6642a4eb8f66d1072bd615e5d4f89a.png)
Hence 
. Since 
, we observe ![\[\measuredangle A'LD+\measuredangle DLP=\measuredangle A'KD+\measuredangle PKD=\measuredangle DKA+\measuredangle PKD=180\]](http://latex.artofproblemsolving.com/0/0/9/009b74de2aac186d4097b8fac1b3a61bae87aa02.png)
Which completes the proof of the lemma.
. Since is the center of spiral homothety sending to , we get that is the miquel point of quadrilateral 
. Note that 
. Invert at and set 
the main triangle.
is a triangle with 
and is any point on the interior angle bisector of . intersect at . Reflection of over is . Let be the midpoint of arc . Prove that 
.
where is the midpoint of arc , by the lemma we have proven the new statement as desired.
and some trigonometry solve the problem 
is cyclic
,
,
,
so 
and so 
(since O is the circumcenter of BRC)
and 
(actually, we only need the former).
such that 
and 
.
.
such that 
.
,
.
,
at 
,
at 
and let 
interset 
at 
at 
.
;
.
and 
.
- is this
.
and 
for some complex number 
on the unit circle. The point ![\[\frac{b-o}{c-o} = z^2 \implies o = \frac{b-z^2c}{1-z^2}.\]](http://latex.artofproblemsolving.com/e/5/b/e5bc3b909f5860c26d2f4119588e03fe44a9697f.png)
Then![\[\frac{a-o}{p-q} = \frac{a - \frac{b-z^2c}{1-z^2}}{\frac{b-a}{z} - z(c-a)} = \frac{-z}{1-z^2}\]](http://latex.artofproblemsolving.com/f/d/c/fdc3ad9513058358493df89be07d51f7ffac71e9.png)
which is purely imaginary.
![\[ \frac{b-o}{c-o} = t^2\]](http://latex.artofproblemsolving.com/b/e/b/beb72590e3d7c5a9d83a9f6fb5b276ae20334685.png)
gives one. The reason is that the second value of 
,
,
'