I have a Discord now
by djmathman, Aug 9, 2020, 3:14 AM
oh no
time for my entire social life to disappear?
Russia 1996. A binary operation
on real numbers has the property that 
for all 
, 
, 
. Prove that 
.
AoPS. Let
be real numbers. Prove that
![\[
\left(\sum_{k=1}^{n} \frac{a_{k}}{k}\right)^{2} \leq \sum_{k=1}^{n} \sum_{j=1}^{n} \frac{a_{k} a_{j}}{k+j-1}
\]](//latex.artofproblemsolving.com/c/4/2/c427021147b6a4b4df929b6469be046646987031.png)
Spain 2020 P5. In an acute-angled triangle
, let 
be the midpoint of 
and 
the foot of the altitude to 
. Prove that if 
, then the circumcircle of triangle 
is tangent to 
.
USOMO 2020 P1, Zuming Feng. Let be a fixed acute triangle inscribed in a circle 
with center 
. A variable point 
is chosen on minor arc of , and segments 
and meet at 
. Denote by 
and 
the circumcenters of triangles 
and 
, respectively. Determine all points for which the area of triangle 
is minimized.
CNCM Online R2P7, Albert Wang. A circle is centered at point in the plane. Distinct pairs of points 
and 
are diametrically opposite on this circle. Point is chosen on line segment 
such that line 
hits the circle again at and line at such that is the midpoint of 
. Now, the point 
is taken for 
. IF 
, find the measure of 
.
Putnam 1985. Evaluate
You may assume that 
AoPS. Let
be continuous and let 
. Consider the sequence 
defined by 
suppose that the sequence 
converges. Show that 
has a fixed point.
Van de Corput Lemma. Let
be 
. Suppose that, for some 
, we have 
for any ![$x\in[a,b]$](//latex.artofproblemsolving.com/4/9/4/49432e6cff349a39878727bec9bfcb0efc9fbb7c.png)
, with 
monotonic when 
. Then
![\[
\left|\int_a^be^{i\lambda\phi(x)}\,dx\right|\leq c_k\lambda^{-\frac1k},
\]](//latex.artofproblemsolving.com/4/8/e/48eb9f73f7de506419135ffb74fac4fe77a1626d.png)
where the constant 
is independent of and .
time for my entire social life to disappear?
Russia 1996. A binary operation
on real numbers has the property that 
for all 
, 
, 
. Prove that 
.
are real numbers such that 
,
.![\[
(a*b)*0 = a+b+0 = c+d+0 = (c*d)*0,
\]](http://latex.artofproblemsolving.com/4/8/c/48c3f7e7f1128097429bbdd45f0292eb0a9446e7.png)
and so![\[
a*b = ((a*b)*0)*0 = ((c*d)*0)*0 = c*d. \quad\square
\]](http://latex.artofproblemsolving.com/b/0/a/b0a315f3cfa5970b94e98b1f7a525a32800bf23a.png)
such that 
.![\[
f(f(a+b)+c) = a+b+c\quad\text{for all }a,b,c\in\mathbb R.
\]](http://latex.artofproblemsolving.com/c/a/d/cadfeea22e8bf2dd7cdfbe508f0a455a462ed8ad.png)
We may furthermore reduce this to the simple two-variable functional equation 
.
to conclude 
for all ![\[
f(a+b) = f(f(f(a)+b)) = f(a) + b.
\]](http://latex.artofproblemsolving.com/d/3/5/d35fe39183842df2599446f870b05789addab446.png)
Now setting 
yields 
.
,
for all 
,
for all 
.
for all AoPS. Let
be real numbers. Prove that![\[
\left(\sum_{k=1}^{n} \frac{a_{k}}{k}\right)^{2} \leq \sum_{k=1}^{n} \sum_{j=1}^{n} \frac{a_{k} a_{j}}{k+j-1}
\]](http://latex.artofproblemsolving.com/c/4/2/c427021147b6a4b4df929b6469be046646987031.png)
.![\[
\left(\sum_{k=1}^n\frac{a_k}{k}\right)^2 = \left(\int_0^1 f(x)\,dx\right)^2\leq \int_0^1 f(x)^2\,dx = \sum_{k=1}^n\sum_{j=1}^n\frac{a_ka_j}{k+j-1}.
\]](http://latex.artofproblemsolving.com/5/2/4/5240691df03776d115dc41b270d072c73b1f5626.png)
Spain 2020 P5. In an acute-angled triangle
, let 
be the midpoint of 
and 
the foot of the altitude to 
. Prove that if 
, then the circumcircle of triangle 
is tangent to 
.![[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A = origin, B = (2,0);
path El = ellipse((1,0), sqrt(2),1);
pair C = relpoint(El, 0.36);
pair M = (A+B)/2, P = foot(A,C,B);
real r = abs(B-C), s = abs(A-C);
pair T = C * (r+s)/s, N = (A+T)/2, D = s/(r+s)*B;
draw(A--B--C--cycle,rgb(0.1,0.1,0.8));
draw(C--N,rgb(0.1,0.5,0.1));
draw(C--D^^N--M,rgb(0.9,0.1,0.1));
draw(circumcircle(B,M,P),orange+linetype("3 3"));
draw(A--P^^rightanglemark(A,P,C,2),rgb(0.6,0.1,0.6));
clip((-2,-1)--(-2,1.6)--(2.2,1.6)--(2.2,-1)--cycle);
dot("$A$",A,SW);
dot("$B$",B,SE);
dot("$C$",C,NW);
dot("$D$",D,S);
dot("$P$",P,dir(A--P));
dot("$M$",M,S);
dot("$N$",N,NW);
[/asy]](http://latex.artofproblemsolving.com/8/2/8/82837c16b10c60f37621b4f091dcc4d3898f00bb.png)
be the point on ray 
(not pictured above for space reasons), and let 
be the midpoint of 
.![\[
AN = \frac{AC+CB}2 = \frac{AB}{\sqrt 2},
\]](http://latex.artofproblemsolving.com/4/3/c/43cdd9f20d046d60f127129dc4e2e911b9bcc8b8.png)
or 
.
is tangent to 
.
is cyclic. To prove this, let 
.
across the external angle bisector of 
,
.
Thus quadrilateral 
is cyclic, as desired.USOMO 2020 P1, Zuming Feng. Let
be a fixed acute triangle inscribed in a circle 
with center 
. A variable point 
is chosen on minor arc of , and segments 
and meet at 
. Denote by 
and 
the circumcenters of triangles 
and 
, respectively. Determine all points for which the area of triangle 
is minimized.![\[
\angle BO_2D = 2\angle CXB = 2\angle CAB = \angle COB,
\]](http://latex.artofproblemsolving.com/a/1/3/a136dd9fd19c93666c6247e750005427b7bf15ed.png)
so 
,
.
.![[asy]
//base diagram is from Evan, I just modified things to make my solution work
size(200); defaultpen(linewidth(0.7)); pair C = dir(130); pair A = dir(200); pair B = dir(340); pair X = dir(280); pair D = extension(C, X, A, B); pair O_1 = circumcenter(A, D, X); pair O_2 = circumcenter(B, D, X); pair O = origin; draw(A--O_1--D--O_2--B, orange); filldraw(A--X--B--cycle, invisible, red); filldraw(A--B--C--cycle, invisible, blue); draw(X--C, red); draw(A--O--C^^O--B, darkcyan); draw(unitcircle, grey); draw(O_2--O--O_1,brown);
dot("$C$", C, dir(C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$X$", X, dir(X)); dot("$D$", D, dir(245)); dot("$O_1$", O_1, S); dot("$O_2$", O_2, S); dot("$O$", O, dir(45));
[/asy]](http://latex.artofproblemsolving.com/6/c/f/6cf6937f61a7f465a99217f3a4a6512719c444e8.png)
This has two consequences. First, some angle chasing reveals![\[
\angle O_1OO_2 = \angle AOB - (\angle AOO_1 + \angle BOO_2) = 2\angle ACB - (\angle ACX + \angle XCB) = \angle ACB.
\]](http://latex.artofproblemsolving.com/5/6/2/5620ccd8a463b55f7287f1affd2898d1784528cd.png)
Second, explicitly writing out the similarity ratios yields![\[
\frac{O_2O}{DC} = \frac{BO}{BC}=:\alpha\quad\text{and}\quad \frac{O_1O}{DC} = \frac{AO}{AC}=:\beta,
\]](http://latex.artofproblemsolving.com/b/4/d/b4de7fec4419e2b0b6ae76c21abf4d936eae2303.png)
so 
and 
are both proportional to 
.![$[O_1OO_2]\propto CD^2$](http://latex.artofproblemsolving.com/f/6/5/f65dd95b44ff605650f2a8fb2e5d649754c44aaa.png)
,
is minimized, i.e. 
.CNCM Online R2P7, Albert Wang. A circle is centered at point
in the plane. Distinct pairs of points 
and 
are diametrically opposite on this circle. Point is chosen on line segment 
such that line 
hits the circle again at and line at such that is the midpoint of 
. Now, the point 
is taken for 
. IF 
, find the measure of 
.
,
is a 45-45-90 triangle. The fact that 
implies that 
.
and 
,
is a 45-45-90 triangle as well. Now the 
condition implies![\[
\angle YXB = \angle(BX,CD) = \frac{\widehat{MC}+\widehat{BD}}2 = \frac{\widehat{MD} + \widehat{BD}}2 = \angle MAB.\qquad(*)
\]](http://latex.artofproblemsolving.com/0/5/2/052245c165594d5ce25925b48bd8b07dd25d7e75.png)
In turn, letting 
.
so that 
is a square.![[asy]
size(300);
defaultpen(linewidth(0.7)+fontsize(10));
real r = 25;
pair A = (-1,0), B = (1,0), O = origin, D = dir(r), C = -1*D, M = dir(r+90), P = intersectionpoint(M--B,A--D), X = 2*M-P, Q = 2*M - A, Y = reflect(B,Q)*X;
draw(M--O,gray(0.7));
draw(A--B^^C--D,rgb(0.1,0.1,0.7));
draw(unitcircle,rgb(0.1,0.6,0.1));
draw(C--X--B^^A--M,rgb(0.5,0.1,0.5));
draw(A--D,rgb(0.1,0.3,0.1));
draw(X--Y--B,brown);
draw(X--Q--P^^M--Q,red);
draw(B--Q--Y,orange);
dot("$A$",A,W,linewidth(3.3));
dot("$B$",B,E,linewidth(3.3));
dot("$C$",C,dir(O--C),linewidth(3.3));
dot("$D$",D,dir(O--D),linewidth(3.3));
dot("$X$",X,NW,linewidth(3.3));
dot("$Q$",Q,dir(100),linewidth(3.3));
dot("$Y$",Y,dir(70),linewidth(3.3));
dot("$P$",P,S,linewidth(3.3));
dot("$M$",M,dir(O--M));
dot("$O$",O,S);
[/asy]](http://latex.artofproblemsolving.com/b/2/2/b224d3076b7d903d9aa9677f5f9c600ab27829fb.png)
and 
,
.
and 
are similar isosceles triangles. But, actually, they are spirally similar; the dual spiral similarity implies 
.
yields that, in fact, these triangles are congruent, so 
;
.
-- the former from 
,![\[
\angle QXY = \angle XYQ = \angle PYB = 10^\circ.
\]](http://latex.artofproblemsolving.com/4/5/c/45cef796eb94fe0b09fcfb22a40b9b493f7d77bf.png)
But this when combined with 
yields 
as well, so![\[
\angle XCM = \angle MBA = 45^\circ - \angle PAB = \boxed{35^\circ}.
\]](http://latex.artofproblemsolving.com/c/0/9/c09bb724b67762609a1f0bf74a028673d3d7260f.png)
Putnam 1985. Evaluate
You may assume that 
.
implies![\[
I = \int_\infty^0 t^{1/2}e^{-1985\left(t+t^{-1}\right)}\cdot -t^{-2}\,dt = \int_0^\infty t^{-3/2}e^{-1985\left(t+t^{-1}\right)}\,dt.
\]](http://latex.artofproblemsolving.com/8/0/e/80ec2366e6bf623f504df17551c419f06234ceac.png)
This means![\begin{align*}
2I &= \int_0^\infty (t^{-1/2} + t^{-3/2})e^{-1985\left(t+t^{-1}\right)}\,dt \\
&= e^{-3970}\int_0^\infty (t^{-1/2} + t^{-3/2})e^{-1985\left(t^{1/2}-t^{-1/2}\right)^2}\,dt\\\
&\stackrel{\mathmakebox[\widthof{=}]{(*)}}= 2e^{-3970}\int_0^\infty e^{-1985u^2}\,du = 2e^{-3970}\sqrt{\frac{\pi}{1985}},
\end{align*}](http://latex.artofproblemsolving.com/2/6/8/268fc4b9affbb4c5bc0696cd060266a4c311761e.png)
where in 
.
.AoPS. Let
be continuous and let 
. Consider the sequence 
defined by 
suppose that the sequence 
converges. Show that 
has a fixed point.
is a fixed point of 
;
.
for some 
.![$c\in[u_0,x]$](http://latex.artofproblemsolving.com/4/c/5/4c5d00008522a20d07356fffa9bfa49d264dfcb8.png)
satisfying 
.
for all 
for all 
.
.
for some 
.
for every 
,
for all such 
.
as 
,
,
for all 
converges to some real number 
.Van de Corput Lemma. Let
be 
. Suppose that, for some 
, we have 
for any ![$x\in[a,b]$](http://latex.artofproblemsolving.com/4/9/4/49432e6cff349a39878727bec9bfcb0efc9fbb7c.png)
, with 
monotonic when 
. Then![\[
\left|\int_a^be^{i\lambda\phi(x)}\,dx\right|\leq c_k\lambda^{-\frac1k},
\]](http://latex.artofproblemsolving.com/4/8/e/48eb9f73f7de506419135ffb74fac4fe77a1626d.png)
where the constant 
is independent of and .
.
.
for any 
is continuous.) Furthermore, since 
does not change sign; we may again assume without loss that 
for all ![\begin{align*}
\int_a^b e^{i\lambda \phi(x)}\,dx &= \int_a^b (e^{i\lambda\phi})'(x) \cdot\frac{1}{i\lambda\phi'(x)}\,dx \\
&= \left[\frac{e^{i\lambda\phi(x)}}{i\lambda\phi'(x)}\right]_a^b - \frac{1}{i\lambda}\int_a^be^{i\lambda\phi(x)}\left(\frac{1}{\phi'}\right)'(x)\,dx.
\end{align*}](http://latex.artofproblemsolving.com/c/6/e/c6ef88fb05e9ee31342b524907c0eff3b43047dd.png)
To bound the first term, use 
to write![\[
\left|\frac{e^{i\lambda\phi(b)}}{i\lambda\phi'(b)} - \frac{e^{i\lambda\phi(a)}}{i\lambda\phi'(a)}\right| \leq \frac{1}{\lambda\phi'(b)} + \frac{1}{\lambda\phi'(a)} \leq \frac{2}{\lambda}.
\]](http://latex.artofproblemsolving.com/6/e/f/6ef2d98b685237aeba8958fcd47099ab05533c67.png)
For the second term, use the Quotient Rule to write 
.
with 
which once again can be bounded above by 
.
;
for all 
is monotonic, and hence can vanish at most one point in ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
.
and ![\begin{align*}
A &= \left\{x\in[a,b]: |\phi^{(k-1)}(x)| < \lambda^{-\frac 1k}\right\},\\
B &= \left\{x\in[a,b]: |\phi^{(k-1)}(x)| \geq \lambda^{-\frac 1k}\right\}.
\end{align*}](http://latex.artofproblemsolving.com/e/2/2/e226953a22a544515567a097f89a6e168cbec01b.png)
Observe that the Fundamental Theorem of Calculus (or the Mean Value Theorem) implies 
.
.![\[
\int_a^b e^{i\lambda \phi(x)}\,dx = \int_A e^{i\lambda \phi(x)}\,dx + \int_B e^{i\lambda \phi(x)}\,dx.
\]](http://latex.artofproblemsolving.com/4/6/b/46bac19601e20a2f5e299ad7eee034331001bae0.png)
We bound each integral individually.![\[
\left|\int_A e^{i\lambda\phi(x)}\,dx\right| \leq m(A) \leq 2\lambda^{-\frac 1k}.
\]](http://latex.artofproblemsolving.com/5/a/c/5ac4fff06c1a426eb85eb74f0fa9f3272c3225e9.png)
satisfies 
in 
is monotonic. Therefore (implicitly using the fact that ![\[
\left|\int_B e^{i\lambda\phi(x)}\,dx\right| = \left|\int_B e^{i\lambda^{1-\frac 1k}\psi(x)}\,dx\right|\stackrel{\textrm{(IH)}}\leq c_{k-1}(\lambda^{\frac{k-1}k})^{-\frac{1}{k-1}} = c_{k-1}\lambda^{-\frac{1}{k}}.
\]](http://latex.artofproblemsolving.com/1/9/3/1936e57ccca4632b389a2eba3f07b57e33b97e68.png)
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