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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
My hardest algebra ever created (only one solve in the contest)
mshtand1   6
N 18 minutes ago by mshtand1
Source: Ukraine IMO TST P9
Find all functions \( f: (0, +\infty) \to (0, +\infty) \) for which, for all \( x, y > 0 \), the following identity holds:
\[
f(x) f(yf(x)) + y f(xy) = \frac{f\left(\frac{x}{y}\right)}{y} + \frac{f\left(\frac{y}{x}\right)}{x}
\]
Proposed by Mykhailo Shtandenko
6 replies
mshtand1
Saturday at 9:37 PM
mshtand1
18 minutes ago
Killer NT that nobody solved (also my hardest NT ever created)
mshtand1   4
N 26 minutes ago by mshtand1
Source: Ukraine IMO 2025 TST P8
A positive integer number \( a \) is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence \( \{b_k\}_{k=1}^{\infty} \), where
\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]
Proposed by Arsenii Nikolaev and Mykhailo Shtandenko
4 replies
mshtand1
Saturday at 9:31 PM
mshtand1
26 minutes ago
Advanced topics in Inequalities
va2010   22
N 44 minutes ago by Primeniyazidayi
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
22 replies
va2010
Mar 7, 2015
Primeniyazidayi
44 minutes ago
Funny easy transcendental geo
qwerty123456asdfgzxcvb   0
2 hours ago
Let $\mathcal{S}$ be a logarithmic spiral centered at the origin (ie curve satisfying for any point $X$ on it, line $OX$ makes a fixed angle with the tangent to $\mathcal{S}$ at $X$). Let $\mathcal{H}$ be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.

Prove that for a point $P$ on the spiral, the polar of $P$ wrt. $\mathcal{H}$ is tangent to the spiral.
0 replies
+2 w
qwerty123456asdfgzxcvb
2 hours ago
0 replies
No more topics!
ABC
Omid Hatami   1
N May 9, 2005 by yetti
Source: SRMO 2005
Assume $A,B,C$ are three collinear points that $B \in [AC]$. Suppose $AA'$ and $BB'$
are to parrallel lines that $A'$, $B'$ and $C$ are not collinear. Suppose $O_1$ is circumcenter of circle passing through $A$, $A'$ and $C$. Also $O_2$ is circumcenter of circle passing through $B$, $B'$ and $C$. If area of $A'CB'$ is equal to area of $O_1CO_2$, then find all possible values for $\angle CAA'$
1 reply
Omid Hatami
Apr 10, 2005
yetti
May 9, 2005
Source: SRMO 2005
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Omid Hatami
1275 posts
#1 • 2 Y
Y by Adventure10, Mango247
Assume $A,B,C$ are three collinear points that $B \in [AC]$. Suppose $AA'$ and $BB'$
are to parrallel lines that $A'$, $B'$ and $C$ are not collinear. Suppose $O_1$ is circumcenter of circle passing through $A$, $A'$ and $C$. Also $O_2$ is circumcenter of circle passing through $B$, $B'$ and $C$. If area of $A'CB'$ is equal to area of $O_1CO_2$, then find all possible values for $\angle CAA'$
Z K Y
The post below has been deleted. Click to close.
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yetti
2643 posts
#2 • 2 Y
Y by Adventure10, Mango247
Let $C'$ be the intersection of the circles $(O_1), (O_2)$ other than the point $C$. The radical axis $CC'$ of these 2 circles is perpendicular to their center line $O_1O_2$ and the center line cuts the segment $CC'$ at its midpoint $D$. Since the lines $AA' \parallel BB'$ are parallel, the angles $\angle CAA' = \angle CBB'$ are equal. Thus the angles spanning the major arcs $CA', CB'$ of the circles $(O_1), (O_2)$ are equal. In particular, the angles $\angle CC'A' = \angle CC'B'$ are equal, which means that the points $A', B', C'$ are collinear. The triangles $\triangle CC'O_1, \triangle CC'O_2$ are both isosceles. Since $O_1O_2 \perp CC'$, the center line $O_1O_2$ bisects the angles $\angle CO_1C', \angle CO_2C'$, i.e.,

$\angle CO_1O_2 = \frac{\angle CO_1C'}{2},\ \ \angle CO_2D = \frac{\angle CO_2C'}{2}$

The angles $\angle CA'C', \angle CB'C'$ span the major arcs $CC'$ of the circles $(O_1), (O_2)$ with the central angles $\angle CO_1C', \angle CO_2C'$. Hence, we also have

$\angle CA'B' \equiv CA'C' = \frac{\angle CO_1C'}{2},\ \ \angle CB'C' = \frac{\angle CO_2C'}{2}$

Thus the internal angles $\angle CA'B' = \angle CO_1O_2$ and the corresponding external angles $\angle CB'C' = \angle CO_2D$ of the triangles $\triangle A'CB', \triangle O_1CO_2$ are equal, which means that these 2 triangles are always similar, regardless of the angle $\angle CAA' =  \angle CBB'$.

The similar triangles $\triangle A'CB' \sim \triangle O_1CO_2$ are congruent, iff their corresponding sides $CO_1 = CA'$ are equal. The segment $CO_1$ is the circumradius of the triangle $\triangle CAA'$ and the segment $CA'$ its side against the angle $\angle CAA'$. Using the extended sine theorem for this triangle,

$\sin{\widehat{CAA'}} = \frac{CA'}{2 CO_1}$

Hence, $CO_1 = CA'$ iff $\sin{\widehat{CAA'}} = \frac 1 2$. i.e., iff $\angle CAA' = 30^o$ or $150^o$.
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