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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
An important lemma of isogonal conjugate points
buratinogigle   6
N 13 minutes ago by buratinogigle
Source: Own
Let $P$ and $Q$ be two isogonal conjugate with respect to triangle $ABC$. Let $S$ and $T$ be two points lying on the circle $(PBC)$ such that $PS$ and $PT$ are perpendicular and parallel to bisector of $\angle BAC$, respectively. Prove that $QS$ and $QT$ bisect two arcs $BC$ containing $A$ and not containing $A$, respectively, of $(ABC)$.
6 replies
buratinogigle
Mar 23, 2025
buratinogigle
13 minutes ago
A difficult problem [tangent circles in right triangles]
ThAzN1   48
N 21 minutes ago by Autistic_Turk
Source: IMO ShortList 1998, geometry problem 8; Yugoslav TST 1999
Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$.

Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$.

comment
48 replies
ThAzN1
Oct 17, 2004
Autistic_Turk
21 minutes ago
IMO 2008, Question 2
delegat   63
N 29 minutes ago by ezpotd
Source: IMO Shortlist 2008, A2
(a) Prove that
\[\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1\] for all real numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$.

(b) Prove that equality holds above for infinitely many triples of rational numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$.

Author: Walther Janous, Austria
63 replies
delegat
Jul 16, 2008
ezpotd
29 minutes ago
USAMO 2003 Problem 1
MithsApprentice   69
N 35 minutes ago by de-Kirschbaum
Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.
69 replies
MithsApprentice
Sep 27, 2005
de-Kirschbaum
35 minutes ago
No more topics!
Quadric function
soryn   4
N Apr 21, 2025 by soryn
If f(x)=ax^2+bx+c, a,b,c integers, |a|>=3, and M îs the set of integers x for which f(x) is a prime number and f has exactly one integer solution,prove that M has at most three elements.
4 replies
soryn
Apr 18, 2025
soryn
Apr 21, 2025
Quadric function
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soryn
5346 posts
#1
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If f(x)=ax^2+bx+c, a,b,c integers, |a|>=3, and M îs the set of integers x for which f(x) is a prime number and f has exactly one integer solution,prove that M has at most three elements.
This post has been edited 1 time. Last edited by soryn, Apr 18, 2025, 3:14 AM
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soryn
5346 posts
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My idea: If r îs the integer solution,let s the non integer solution of f(x)=0. But,f(x)=a(x-r)(x-s). f(x)=p,
p prime,implies that |a|•|x-r|•|x-s|=|p|=>|x-s| is in the interval [-p/3, p/3], where |x-r|=1,since p isca prime number,and |a|>=3. But how implies that a,b,c integers=>p is 2,3 or 5? Who can help me,pls?
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soryn
5346 posts
#5
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Anyone ? ....
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RagvaloD
4918 posts
#6 • 1 Y
Y by soryn
Case 1: $f(x)=a(x-r)^2$ with $r$ is integer
Then it is easy to see, that $f(x)$ can be prime only if $|x-r|=1$ so $M$ has at most two elements

Case 2: $f(x)$ has integer root $r$ and non integer root. Then $f(x)=(x-r)(ax-s)$ where $\frac{s}{a}$ is second non-integer root
If $f(x)$ is prime then $|x-r|=1$ or $|ax-s|=1$
So if $f(x)$ is prime, then $x$ can be $r-1,r+1, \frac{s+1}{a}$ or $\frac{s-1}{a}$
But as $|a| \geq 3$ then both $\frac{s+1}{a}$ and $\frac{s-1}{a}$ can not be integer at same time, because their difference is less than $1$ so $M$ has at most three elements

Let $f(x)=-4x^2+9x$
Then $f(x)$ has integer root $x=0$ and $f(1)=5,f(-1)=13, f(2)=2$ are prime values, so $M$ can have $3$ values
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soryn
5346 posts
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Excellent,thx!
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