and 
Prove that
.![\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]](http://latex.artofproblemsolving.com/b/3/2/b32e0bcba967ca847e878ac186c3168cd6c5fc66.png)
and 
such that 
+ 
+ 2 is a perfect square.
and 
be two positive integer numbers such that 
.
divides 
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
,
are all greater than 
and 
and 
intersect at 
,
and 
intersect at 
.
is twice the area of 
.
,
.
is a perfect square.
,
,
away. What is the least number of moves that Carina can make in order for triangle 
in the coordinate plane where 
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
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, 
sends 
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)
,
,
be orthocenter, 
.
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.
