is clearly not true. Even a 1st grader can see that. And what does "a number not a product" mean? A product is a number? Literally makes 0 sense.
.
as 
?
if you do go around, acting like some kind of god again, "anticodon." Everyone will worship you finally for being able to copy from Google correctly!...
on the real line such that for all 
,![\[ \left( f(x) \right)^2 = \displaystyle\int_0^x \left[ \left( f(t) \right)^2 + \left( f'(t) \right)^2 \right] \, \mathrm{d}t + 1990. \]](http://latex.artofproblemsolving.com/e/7/9/e79b1d8b17010c15a0f6397c8775e4e268a66a2e.png)
2. 
3. 
Bracket 2
2. 
3. 
Bracket 3![$$\int_0^{1}\frac{dx}{1-\sqrt[3]{1-\sqrt{x}}}$$](http://latex.artofproblemsolving.com/d/5/e/d5e8db00985944b2eca341e37a94e34661cb5c86.png)
2. 
3. 
Bracket 4
2. 
3. 
Bracket 5
2. 
3. 
Bracket 6
2. 
3. 
Bracket 7
2. 
3. 
Bracket 8
2. 
3. 
![$$\int_0^{2\pi}\sqrt[4]{\cos^4(x)-\cos(2x)}dx$$](http://latex.artofproblemsolving.com/7/e/e/7eebb4a488fec702e0130a253ef87f0f6dfedb2d.png)
2. ![$$\int_0^1 \frac{x^5}{\sqrt[3]{1-x^3}}dx$$](http://latex.artofproblemsolving.com/6/0/5/605bb9d81d0ccff54a276c9f7bf1c1726c136398.png)
3. 
Bracket 2
2. 
3. 
Bracket 3
2. ![$$\int_0^1 \frac{dx}{(\sqrt{x}+\sqrt[3]{x})^2}$$](http://latex.artofproblemsolving.com/6/9/8/69869c0ed754c0dfacf412ac6411742ffa164a6d.png)
3. 
Bracket 4
2. 
3. 
2. 
3. 
Bracket 2
2. 
3. 
2. 
3. 
2. 
3. 
4. 
5. 
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
,
,
and 
could be chosen to satisfy 
and 
.
and the distribution function is 
then
and find the expression of 
be a function whose derivative is continuous on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
.![$$\lim_{n \to \infty} \sum^n_{k = 1}\left[f\left(\frac{k}{n}\right) - f\left(\frac{2k - 1}{2n}\right)\right] = \frac{f(1) - f(0)}{2}.$$](http://latex.artofproblemsolving.com/3/1/4/314a1597acae030381f980e6e26c12432f31e069.png)
be a positive integer greater than 
.
where each of 
is either 
.
Y by gyluo, megarnie, Jc426, Lionking212, sleepypuppy, mathleticguyyy, ImSh95, Adventure10, Mango247
Circles 
and 
intersect at points 
and 
. Line 
is tangent to and at 
and 
, respectively, with line 
closer to point than to . Circle 
passes through and intersecting again at 
and intersecting again at 
. The three points 
, , 
are collinear, 
, 
, and 
. Find 
.
and 
intersect at points 
and 
. Line 
is tangent to and at 
and 
, respectively, with line 
closer to point than to . Circle 
passes through and intersecting again at 
and intersecting again at 
. The three points 
, , 
are collinear, 
, 
, and 
. Find 
.
![[asy]
import olympiad;
size(300);
defaultpen(linewidth(0.8));
pair P = dir(100), D = dir(220), C = dir(320);
pair M = (D+C)/2, pt = bisectorpoint(D,P,C), Z = reflect(P,pt) * M, Y = extension(P,Z,D,C);
draw(P--D--C--P--Y);
pair T1 = P + Y - D, T2 = P + Y - C, A = intersectionpoint(P--D, T2--Y), B = intersectionpoint(T1--Y, P--C);
draw(A--B);
pair[] in = intersectionpoints(circumcircle(A,D,Y), circumcircle(B,C,Y));
pair X = in[0];
pair Q = 2 * A - P, R = 2 * B - P;
draw(A--X--B^^C--X--D,linetype("2 2"));
draw(D--Q--R,linetype("4 4"));
draw(circumcircle(A,D,Y)^^circumcircle(B,C,Y));
label("$A$", A, dir(120));
label("$B$", B, dir(60));
label("$C$", C, dir(X--C));
label("$D$", D, dir(200));
label("$X$", X, 2.7 * dir(195));
label("$Y$", Y, 1.5*S);
label("$M$", (A+B)/2, dir(140));
label("$P$", P, N);
label("$Q$", Q, dir(Y--Q));
label("$R$", R, dir(Y--R));
[/asy]](http://latex.artofproblemsolving.com/c/4/d/c4d7b0e1e0ffa3e2990a8568f6ff9aab277bfc65.png)
.
also passes through 
.
to be cyclic, we need 
.
.
.
is a parallelogram, so letting 
we have 
.
and 
at 
and 
respectively. Since 
,
.
,
is cyclic. This means ![\[DY\cdot YC = QY\cdot YR = AB^2.\]](http://latex.artofproblemsolving.com/3/8/b/38bc4c0bfda28a02df658113e9289a498ca67bd1.png)
To compute 
,![\[\angle YXC=\angle YBC=\angle DPC=\angle DAY=\angle DXY,\]](http://latex.artofproblemsolving.com/e/0/9/e09b2b45222d1889fa4d5fb7f841f1ff09f17ec5.png)
so 
.
and 
for some 
.![\[37^2\cdot 67x+37\cdot 67^2x = 47^2\cdot 104x + 37x\cdot 67x\cdot 104x.\]](http://latex.artofproblemsolving.com/f/3/a/f3a14d18ec12f0015f9ed50de758db12caf7297e.png)
Note that the LHS can be rewritten as 
,
from both sides gives ![\[37\cdot 67 = 47^2 + (37x)(67x) = 47^2+DY\cdot YC.\]](http://latex.artofproblemsolving.com/8/c/2/8c26aed984416b8218e6b0eccf9ce109fb57df26.png)
Hence ![\[AB^2=DY\cdot YC = 37\cdot 67 - 47^2 = \boxed{270}.\]](http://latex.artofproblemsolving.com/7/3/9/73990f77e907f8f23f3c8281500f53a3d07daeba.png)
passes through the external center of homothety of the two circles.
concur. We also know that 
.
collinear in mind, we have ![\[\angle BAP=\angle CDB=\angle YDB=\angle YBA\]](http://latex.artofproblemsolving.com/8/3/5/835ce14df02b2948f17396bfea8f3cc222d5f69b.png)
This means 
and similarly 
,
is a parallelogram.
from more angle chasing. Thus we can obtain 
and 
.![\[AB^2=4AM^2=4MX\cdot MY=4(\frac {\sqrt {37\cdot 67}+47}{2}\cdot \frac {\sqrt {37\cdot 67}-47}{2})=\boxed {270}\]](http://latex.artofproblemsolving.com/1/1/d/11d92da347da5e66a35a4270984c4ab8cf631a02.png)
.
,
and 
(again resulting from the homothety).
,
Hence, 
Similarly, 
Then since 
,
Similarly, 
Because 
by Power of a Point, multiplying these relations gives 
![[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 = 25.561273455869426, xmax = 251.43455114359386, ymin = 45.70873344820468, ymax = 159.632334179233; /* image dimensions */
Label laxis; laxis.p = fontsize(10);
xaxis(xmin, xmax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true);
yaxis(ymin, ymax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */
/* draw figures */
draw(circle((123.90174878101861,80.37090307120728), 23.786991826712317));
draw((xmin, 0.030438670316427052*xmin + 52.80148983853563)--(xmax, 0.030438670316427052*xmax + 52.80148983853563)); /* line */
draw(circle((165.45131811180127,94.34666730069631), 36.49215802761939));
draw(circle((137.59059581752737,108.42508624472585), 42.923890684144), linetype("2 2") + green);
draw((46.11140346430864,54.20505964641348)--(174.93682540909933,129.5844701383543), red);
draw((142.70941966835346,65.80750731479884)--(115.91289192359766,145.4728600692278), red);
draw(circle((144.3889574163432,96.80650504615912), 44.80588392709177), dotted + blue);
/* dots and labels */
dot((124.62545799779946,56.59492306756442),linewidth(3.pt) + dotstyle);
label("$A$", (124.1164676529477,54.02741221445552), S * labelscalefactor);
dot((142.70941966835346,65.80750731479884),linewidth(3.pt) + dotstyle);
label("$X$", (142.44575984982922,63.051063757507265), S * labelscalefactor);
dot((130.08476718136293,103.3402579516797),linewidth(3.pt) + dotstyle);
label("$Y$", (127.21834787088149,105.06744125484197), NW * labelscalefactor);
dot((166.56157666392102,57.87140275799297),linewidth(3.pt) + dotstyle);
label("$B$", (165.9918505950539,55.296363212697166), S * labelscalefactor);
dot((46.11140346430864,54.20505964641348),linewidth(3.pt) + dotstyle);
label("$Z$", (45.58250031707846,51.34851566261202), S * labelscalefactor);
dot((174.93682540909933,129.5844701383543),linewidth(3.pt) + dotstyle);
label("$C$", (175.43848580421593,130.446461219675), NE * labelscalefactor);
dot((100.84847094295199,86.2332712859296),linewidth(3.pt) + dotstyle);
label("$D$", (98.31446402195296,86.17417083657737), NW * labelscalefactor);
dot((115.91289192359766,145.4728600692278),linewidth(3.pt) + dotstyle);
label("$M$", (113.6828705562613,146.37884597537575), NW * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/f/5/b/f5b767157857f084590bee09b7951399e5dacbd3.png)
and 
,
By cyclic quadrilaterals, we infer 
Hence, if 
for a second time at 
and 
,
Then by Power of a Point,
![[asy]
import olympiad;
size(300);
defaultpen(linewidth(0.8));
pair P = dir(100), D = dir(220), C = dir(320);
pair M = (D+C)/2, pt = bisectorpoint(D,P,C), Z = reflect(P,pt) * M, Y = extension(P,Z,D,C);
pair T1 = P + Y - D, T2 = P + Y - C, A = intersectionpoint(P--D, T2--Y), B = intersectionpoint(T1--Y, P--C);
draw(A--B--C--D--cycle);
pair[] in = intersectionpoints(circumcircle(A,D,Y), circumcircle(B,C,Y));
pair X = in[0];
pair Q = 2 * A - P, R = 2 * B - P;
draw(circumcircle(A,D,Y)^^circumcircle(B,C,Y));
label("$A$", A, dir(120));
label("$B$", B, dir(60));
label("$C$", C, dir(X--C));
label("$D$", D, dir(200));
label("$X$", X, 2.7 * dir(195));
label("$Y$", Y, 1.5*S);
draw(C--X--Y);
draw(D--X);
draw(A--Y--B);
draw(circumcircle(X,C,D));
pair K = 0.5*C + 0.5*D;
pair L = K+dir(90)*(K-D);
pair N = extension(X,Y,K,L);
draw(Y--N);
draw(C--N--D);
label("$M$",N,S);
[/asy]](http://latex.artofproblemsolving.com/d/c/0/dc0f43fe730375fb4ee25cc402464eb1006ddcd6.png)
and similarly 
.
and 
.
,
,
are all similar to each other. Furthermore the ratio of similarity of 
and the ratio of similarity of 
.
.
.
and 
,
.
be the incenter of 
and 
and let its radius be 
again at 
then 
by symmetry. Also, the circumcircle of 
is tangent to 
because 
.
,
and 
to intersect at a point 
.
is a cyclic quad because of Miquel's Theorem. Now, let 
and 
.
.
,
.
.
is a parallelogram. Now from this, we have that 
(last equality from 
cyclic). Thus, 
.
.
is similar to 
,
,
.
,
.
,
.
.
and 
?
?
,
concur, say at 
,
are swapped. Then 
are fixed. Since 
is also tangent to these circles. Then radical axis on 
implies 
is the symmedian of 
.
,
,
,
is similar to 
so 
.
,
,
.
,
.
,
and 
,
and so we get that 
and 
.
.
.
and 
are cyclic. Now, let the circumcircle of 
.
. By power of a point (about 
.
Thus, we get that 
and 
hence, 
and 
.
and similarly, 
Multiplying both expressions, we get that 
which gives us 
.
.
and 
from the tangency condition and that 
from the fact that 
,
.![\[\frac{AB}{DY} \stackrel{(*)}= \frac{BY}{YA} \stackrel{(\dagger)}= \frac{CY}{AB},\]](http://latex.artofproblemsolving.com/a/4/5/a458d170434065ed252e4d4d2fcd511c8ea851d9.png)
so 
.
because 
.
.
past 
so 
,
,
.
are collinear
Let 
.
.
.
so 
.
.
.
so multiplying the two ratios gives 
,
.
is cyclic, since 
.
,
and using the three similar triangles we found earlier gives 
So, 
,
by radical axis. Furthermore, let 
.
hence the result. 
by the parallelogram in the previous claim. 
,
at 
.
is cyclic because 
so 
by power of a point. Set 
via the angle bisector theorem. Then Stewart's theorem implies 
directly. Observe the similarity 
,
so we can compute 
without the construction: note by angle chasing that 
and 
.
by 
.
,
,
so 
and then radical axis on 
by definition.
.
Let 
and 
.
lies on 
.
is cyclic, so 
(where we consider the absolute value of all quantities).
,
is a parallelogram.
which yields 
.
finishes.
which implies 
which is easy to find via currently known lengths.
is the 
.
so 
.
by Radical Axis. Notice 
is cyclic since 
Hence, 
and similarly 
Therefore, 
and 
and 
Also, 
so 
and similarly 
Thus, 
is the midpoint of 
and 
Then, 
so 
,
meets at a same point, call it 
,
are concyclic. We do little angle chasing.
,
,
,
is a parallelogram. Then, 
bisect each, call their intersection 
.
.
,
in, we attain 
are all concurrent at one point, call it 
the fact that quadrilaterals 
and 
are cyclic implies 
is cyclic as well. Denote 
and 
.
lies on the radical axis of 
,![\[AZ^{2}=\text{Pow}_{\omega_{1}}(Z)=ZX\cdot ZY=\text{Pow}_{\omega_{2}}(Z)=ZB^{2}\implies AZ=ZB.\]](http://latex.artofproblemsolving.com/8/3/5/835825f8818eacf30b964169ab446b9f657923e6.png)
Also, notice that ![\[AZ\cdot ZB=\text{Pow}_{\omega_{3}}(Z)=ZX\cdot ZF\implies ZY=ZF.\]](http://latex.artofproblemsolving.com/a/7/2/a72c48864a787c2bde011b7cdea7299714fe996a.png)
The diagonals of quadrilateral 
bisect each other at 
,
.
are cyclic, ![\[\angle DFX=\angle AFX=\angle BYX=\angle BCX=\angle FCX~~\text{and}~~\angle XDF=\angle XDA=\angle XYA=\angle XFB=\angle XFC\]](http://latex.artofproblemsolving.com/5/d/4/5d4832eb3d8228af3052b127d813bca879896964.png)
so we have the pair of similar triangles 
.![\[\dfrac{37}{2u+47}=\dfrac{2u+47}{67}\implies 2u+47=\sqrt{37\cdot 67}\implies u=\dfrac{1}{2}\left(\sqrt{37\cdot 67}-47\right).\]](http://latex.artofproblemsolving.com/5/f/c/5fc4b9c6fdb7a8233ec31ab27ef34f52e49fe4b7.png)
Now compute ![\[AB^{2}=4AZ^{2}=4\cdot ZX\cdot ZY=4u(u+47)=37\cdot 67-47^{2}=\textbf{270}.\]](http://latex.artofproblemsolving.com/3/3/2/33215ce12dd16279badb093615343767b84abae9.png)
by the 
tangency and 
by 
or 
.![\[\angle BYC=180^\circ-\angle ABC=\angle ADC\iff BY\parallel AD.\]](http://latex.artofproblemsolving.com/e/0/6/e060676d558344d1f8b4ebf921e41c07b8a9dd42.png)
These parallelities and tangencies imply that 
and 
.
by AA Similarity, and additionally because 
,
.
i.e. 
for some 
,
But 
,
.
and 
to meet at point 
,
and 
,
and ![\[ |\text{Pow}(M, (PAXB))| = MX \cdot MP = MA^2 = |\text{Pow}(M, (ADY))| = MX \cdot MY, \]](http://latex.artofproblemsolving.com/e/5/9/e59c2730e33c252a8a4c327b75fd9312eb2312bc.png)
so 
.![\[ \angle DAY = \angle YBC, \]](http://latex.artofproblemsolving.com/d/a/2/da2d764a579592449e81d32a90cdf8f4413b9777.png)
so by the inscribed angle theorem 
.
.
is parallel to the radical axis 
.
while 
,
,
,
.
and 
.
is in fact a parallelogram.![\[ \angle DXY = \angle DAY = \angle DZC = \angle YBC = \angle YXC,\]](http://latex.artofproblemsolving.com/f/9/4/f94381b4eff9dcf7cc5764f8acbce482772750b7.png)
so 
.![\[ XD \cdot XC - XY^2 = YD \cdot YC, \]](http://latex.artofproblemsolving.com/c/e/3/ce34fc7cfac1480b7d8c1ea3e42f13d8078d2ff3.png)
so 
.
,
,
form a geometric sequence in that order, hence![\[ AB^2 = 67 \cdot 37 - 47^2 = \boxed{270}.\]](http://latex.artofproblemsolving.com/1/c/c/1cc2091f601f0ffa776499124f8cdf517077a067.png)
and 
.
which implies 
.
to extract 
.
Let 
.
is cyclic since 
is a parallelogram because 
we have 
or 
but 
