[/quote]
,
such that 
.
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
and 
meet at 
and 
meet at 
and 
denote the circumcircle of triangle 
the midpoint of 
.
and 
,
.
and 
similarly.
such that for all positive integers 
and 
,![\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]](http://latex.artofproblemsolving.com/8/b/9/8b90a9326c0fa448a5a574809c184f43513f768f.png)
Jaedon Whyte
,
not dividing any of them, and an integer 
such that if one considers the infinite sequence 
and looks at the highest power of 
(which may depend on 
)
.
,
be the least prime that does not divide 
,
to be the product of all primes less than 
For 
,
.
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
.
.
,
,
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
,
is an integer. Prove that there is a positive integer 
for all 
.
.
such that 
be the midpoint of 
and the circumcircle of 
.
passes through a fixed point.
and 
intersect each other at 
and 
.
and 
respectively. Let 
and 
be the reflection of 
intersect circles 
and 
(both distinct from 
is the second tangent to 
to 
face-down on a table. On each move, the Detective can point to two cards and inquire if the numbers on them are consecutive. The Magician replies truthfully. After a finite number of moves, the Detective points to two cards. She wins if the numbers on these two cards are consecutive, and loses otherwise.
questions.
= 2∠AMP.
is the midpoint of the arc 
,
,
,
be the 
intersects 
.
is perpendicular to 
,
is the incenter of 
and the circle 
Let 
The line 
intersects 
Let 
Let 
and 
Y by
Let 
be an acute scalene triangle with circumcenter 
. Let 
be the feet of the altitudes from 
to 
respectively. Let 
be the midpoint of 
. Let 
be the circle with diameter 
. Let 
be the intersection of 
and 
. Let 
be the orthocenter of . Let 
be the intersection of 
and . Let 
be the lines through 
tangent to 
respectively. Let 
be the intersection of 
and 
. Let 
be the intersection of 
and 
. Let 
be the line through parallel to 
and let 
be the reflection of across . Prove that 
is tangent to 
.
be an acute scalene triangle with circumcenter 
. Let 
be the feet of the altitudes from 
to 
respectively. Let 
be the midpoint of 
. Let 
be the circle with diameter 
. Let 
be the intersection of 
and 
. Let 
be the orthocenter of . Let 
be the intersection of 
and . Let 
be the lines through 
tangent to 
respectively. Let 
be the intersection of 
and 
. Let 
be the intersection of 
and 
. Let 
be the line through parallel to 
and let 
be the reflection of across . Prove that 
is tangent to 
.This post has been edited 1 time. Last edited by vincentwant, Apr 30, 2025, 3:55 PM
in the unit circle, so that
Now we find the equation of 
.
![$$\frac{4\left[w(ab+ac+2bc) - bc(2a+b+c)\right]}{2a^2b+2a^2c+ab^2-2abc+ac^2-2b^2c-2bc^2} = \frac{4a\left[bc\overline{w}(2a+b+c) - (ab+ac+2bc)\right]}{2a^2b+2a^2c-ab^2+2abc-ac^2-2b^2c-2bc^2}$$](http://latex.artofproblemsolving.com/1/2/7/127f163658e70ed84d262f1710aaabe23db35187.png)
 = a\left[bc\overline{w}(2a+b+c) - (ab+ac+2bc)\right](2a^2b+2a^2c+ab^2-2abc+ac^2-2b^2c-2bc^2)$$](http://latex.artofproblemsolving.com/2/4/5/245a0c06c057cff192d28d48e1e572e13517b517.png)
(ab+ac+2bc) = a\left[bc\overline{w}(2a+b+c) - (ab+ac+2bc)\right](2a+b+c)(ab+ac-2bc)$$](http://latex.artofproblemsolving.com/2/e/8/2e82d8a317ba262c68c14f43fc88efc844cece61.png)
The equation of line 
We intersect these to find the coordinate of 
)
Solving for 
The denominator factors as![$$\left[(2a+b+c)(ab+ac-2bc) + (2a-b-c)(ab+ac+2bc)\right]\left[(2a+b+c)(ab+ac-2bc) - (2a-b-c)(ab+ac+2bc)\right]$$](http://latex.artofproblemsolving.com/6/e/5/6e50e6f41f3becd29122a7a0cd2bfcc9df8d26a9.png)
which we simplify to
Meanwhile we factor out 
in the numerator, so that![$$j = \frac{(2a+b+c)\left[(2a-b-c)(ab+ac+2bc)(a^2b+a^2c-4abc+b^2c+bc^2) - (2a+b+c)(ab+ac-2bc)(a^2b+a^2c-2ab^2-2ac^2+b^2c+bc^2)\right]}{8a(b-c)^2(b+c)(a^2-bc)}$$](http://latex.artofproblemsolving.com/3/e/9/3e9047873ea75597736d067d68551b4600b86814.png)
We write the expression in brackets as![$$(a^2b+a^2c-4abc+b^2c+bc^2)\left[(2a-b-c)(ab+ac+2bc) - (2a+b+c)(ab+ac-2bc)\right] + (2ab^2-4abc+2ac^2)(2a+b+c)(ab+ac-2bc)$$](http://latex.artofproblemsolving.com/f/2/4/f24efeb52b65f87a76efd38db826fc9a85c6e3dc.png)
which we simplify to![$$2a(b-c)^2\left[(2a+b+c)(ab+ac-2bc) - (a^2b+a^2c-4abc+b^2c+bc^2)\right] = 2a(b-c)^2(a^2b+a^2c+ab^2+2abc+ac^2-3b^2c-3bc^2)$$](http://latex.artofproblemsolving.com/f/b/8/fb8872a082271443710a5765ad0f68a8c09c7d93.png)
So factoring out 
(note that 
since the triangle is acute) we have
Now we find the equation of 
.
![$$\frac{2\left[w(ab+ac+2bc) - bc(2a+b+c)\right]}{a(b+c)(2a+b+c)} = -\frac{2a\left[bc\overline{w}(2a+b+c) - (ab+ac+2bc)\right]}{(b+c)(ab+ac+2bc)}$$](http://latex.artofproblemsolving.com/4/0/3/40324167520307d87f5548012224f07d775b3510.png)
![$$\overline{w} = \frac{(ab+ac+2bc)\left[(a^2+bc)(2a+b+c) - w(ab+ac+2bc)\right]}{a^2bc(2a+b+c)^2}$$](http://latex.artofproblemsolving.com/9/a/e/9ae499e7996d40b6c55a3b316a8c7afed53fbf0a.png)
The equation of line 
We intersect these to get![$$\frac{(ab+ac+2bc)\left[(a^2+bc)(2a+b+c) - p(ab+ac+2bc)\right]}{a^2bc(2a+b+c)^2} = \frac{(a^2b+a^2c-ab^2+2abc-ac^2+b^2c+bc^2) - 2bcp}{2a^2bc}$$](http://latex.artofproblemsolving.com/4/d/0/4d077690f82c56b4bc45c580cf85b1d74d01cc5c.png)
![$$p = \frac{(2a+b+c)\left[2(a^2+bc)(ab+ac+2bc) - (2a+b+c)(a^2b+a^2c-ab^2+2abc-ac^2+b^2c+bc^2)\right]}{2(ab+ac+2bc)^2 - 2bc(2a+b+c)^2}$$](http://latex.artofproblemsolving.com/0/f/9/0f943d5e3450eb68e6685e36caaafd4b1c2aeae1.png)
Now we find the equation of 
.
Then plugging in 
,
and so
Then we find the vectors![$$p - o' = \frac{(2a+b+c)\left[(a^2+ab+ac-bc) + (a-b)(a-c)\right]}{2(a^2-bc)} = \frac{a^2(2a+b+c)}{a^2-bc}$$](http://latex.artofproblemsolving.com/4/4/1/4411962da8d2711fe8e9b54bd8e3de1502de4a7d.png)
![$$p - q = \frac{(2a+b+c)\left[(ab+ac+2bc)(a^2+ab+ac-bc) - 2bc(a^2-bc)\right]}{2(a^2-bc)(ab+ac+2bc)} = \frac{a(a+b)(a+c)(b+c)(2a+b+c)}{2(a^2-bc)(ab+ac+2bc)}$$](http://latex.artofproblemsolving.com/1/5/2/1520941094d5aa46ccdd90748bb4879c608b4467.png)
![$$k-q = (p-q) - (p-k) = \frac{a(a+b)(a+c)\left[(b+c)(2a+b+c) - 2(ab+ac+2bc)\right]}{2(a^2-bc)(ab+ac+2bc)} = \frac{a(a+b)(a+c)(b-c)^2}{2(a^2-bc)(ab+ac+2bc)}$$](http://latex.artofproblemsolving.com/f/f/5/ff587aea99716e9295485a23dd6c887b0fc36333.png)
And then
All the factors we canceled out are nonzero. 
is nonzero since 
are nonzero since otherwise 
which implies 
.
hit 
at 
respectively then 
so it becomes clear that 
are orthogonal from orthocenter config or existence of Humpty point.
diameter on 
isosceles trapezoid and also notice that since 
we have projecting at 
that 
and as a result 
is also tangent to 
so 
and so if you reflect 
then 
so in fact since we can see that 
is the perpendicular bisector of 
we have that 
and thus reflecting we want to prove 
is tangent to 
however as 
and so since we have from the angles that 
which shows that 
is tangent to 
is tangent to 
is cyclic with diameter 
and thus 
which gives our tangency as desired thus we are done 
.
is an isosceles trapezoid, and thus cyclic.
be the circumcenter of 
and thus, the midpoint of 
Hence, 
to 
Thus, 
which implies 
is tangent to 
so 
bisects 
.
implying our result.
.
is tangent to 
is cyclic since 
is the miquel point of quadrilateral 
Hence, 
which finishes.
Thus, 
is cyclic. Thus, 
which finishes.![[asy]
unitsize(4.5cm);
import geometry;
pair O = (0, 0);
pair A = (-10/17, 3*sqrt(21)/17);
pair B = (-12/13, -5/13);
pair C = (12/13, -5/13);
pair D = foot(A, B, C);
pair EE = foot(B, C, A);
pair F = foot(C, A, B);
pair G = intersectionpoints(circumcircle(A, EE, F), circumcircle(A,B,C))[0];
pair H = extension(A, D, EE, B);
pair M = (B + C)/2;
pair T = extension(EE, F, B, C);
pair I = (T + H)/2;
pair X = (T + M)/2;
pair P = extension(G, X,T, EE);
pair Pp = intersectionpoint(tangent(circle(A, B, C), G), line(B, C));
pair Hp = -A;
draw(A -- B -- C -- cycle);
draw(A -- M);
draw(A -- D);
draw(T -- H);
draw(T -- B);
draw(M -- Hp -- A, dashed);
draw(T -- B -- EE);
draw(F -- EE);
draw(T -- EE);
draw(G -- I, red);
draw(circumcircle(A, G, M), red);
draw(G -- X, deepgreen);
draw(circumcircle(A,EE,F), deepgreen);
draw(I -- X, purple);
draw(G -- M, purple);
draw(Pp -- G, deepmagenta);
draw(unitcircle, deepmagenta);
draw(Pp -- P, orange);
draw(T -- A, orange);
draw(Pp -- O, blue);
draw(circle(T, G, P), blue);
dot(A);
dot(B);
dot(C);
dot(D);
dot(EE);
dot(F);
dot(H);
dot(M);
dot(T);
dot(G);
dot(I);
dot(X);
dot(P);
dot(Pp);
dot((0, 0));
dot(Hp);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, NE);
label("$E$", EE, NE);
label("$F$", F, W);
label("$G$", G, W);
label("$H$", H, SE);
label("$M$", M, NE);
label("$T$", T, W);
label("$I$", I, N);
label("$X$", X, S);
label("$P$", P, S);
label("$P'$", Pp, S);
label("$O$", O, S);
label("$H'$", Hp, SE);
[/asy]](http://latex.artofproblemsolving.com/7/9/5/7959251c7b51ce1debaa292654496243aec72a40.png)
.
,
,
.
and 
;
and 
.
and 
is tangent to 
.
.
is circumcenter of 
.
.
bisector of 
and so let 
is tangent to 
.
and 
is tangent to ![\[\angle Q_AX_AD=180^\circ-\angle Q_AHD=\angle Q_AHA=180^\circ-\angle PQ_AA=\angle PQ_AX_A=\angle Q_AX_AP'\]](http://latex.artofproblemsolving.com/c/3/e/c3ee69268753328efa4cec4d776f698e55d02c9f.png)
And now the second thing is just ![\[\angle P'Q_AA=180^\circ-\angle P'Q_AX_A=180^\circ-\angle Q_AX_AF=180 ^\circ-\angle Q_ABA\]](http://latex.artofproblemsolving.com/b/4/3/b43a4eb7527f1f87fb07a2cd2b4d3087d3b3ae8a.png)
And so we are done. 
then 
and hence see that ![\[\angle OP'Q_A=90^\circ-\angle Q_AAH'=\angle AX_AD=\angle Q_AX_AP'\]](http://latex.artofproblemsolving.com/6/8/f/68fa42816f9c6c451b085ee4c9050fef863bbecb.png)
And done.
