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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
Infinite Sum
P162008   1
N 2 minutes ago by cazanova19921
Find $\Omega = \lim_{n \to \infty} \frac{1}{n^2} \left(\sum_{i + j + k + l = n} ijkl\right) \left(\sum_{i + j + k = n} ijk\right)^{-1}.$
1 reply
P162008
3 hours ago
cazanova19921
2 minutes ago
4 variables with quadrilateral sides
mihaig   0
19 minutes ago
Source: VL
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$4\left(abc+abd+acd+bcd\right)\geq3\left(a+b+c+d\right)+4.$$
0 replies
mihaig
19 minutes ago
0 replies
Tangent and orthogonal circumcircles
Vorsch   2
N 23 minutes ago by s27_SaparbekovUmar
Let $A$, $B$, $C$ and $D$ be four points on a line, in that order. Let $P$ be one of the intersections of the circle with diameter $AC$, and of the circle with diameter $BD$.

If $X$, $Y$ and $Z$ are three points, I will call their circumcircle $C_{XYZ}$.

Show that $C_{PAB}$ and $C_{PCD}$ are tangent, that $C_{PAD}$ and $C_{PBC}$ are tangent, and that the first two circumcircles are orthogonal to the last two circumcircles.
2 replies
Vorsch
Sep 23, 2013
s27_SaparbekovUmar
23 minutes ago
sequence positive
malinger   36
N an hour ago by Maximilian113
Source: ISL 2006, A2, VAIMO 2007, P4, Poland 2007
The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$.

Proposed by Mariusz Skalba, Poland
36 replies
malinger
Apr 22, 2007
Maximilian113
an hour ago
No more topics!
Gamma is the second-largest angle
Martin N.   5
N May 24, 2012 by yugrey
Source: Austrian Federal Competition for Advanced Students 2012, #4
Let $ABC$ be a scalene (i.e. non-isosceles) triangle. Let $U$ be the center of the circumcircle of this triangle and $I$ the center of the incircle. Assume that the second point of intersection different from $C$ of the angle bisector of $\gamma = \angle ACB$ with the circumcircle of $ABC$ lies on the perpendicular bisector of $UI$.
Show that $\gamma$ is the second-largest angle in the triangle $ABC$.
5 replies
Martin N.
May 23, 2012
yugrey
May 24, 2012
Gamma is the second-largest angle
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Source: Austrian Federal Competition for Advanced Students 2012, #4
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Martin N.
434 posts
#1 • 1 Y
Y by Adventure10
Let $ABC$ be a scalene (i.e. non-isosceles) triangle. Let $U$ be the center of the circumcircle of this triangle and $I$ the center of the incircle. Assume that the second point of intersection different from $C$ of the angle bisector of $\gamma = \angle ACB$ with the circumcircle of $ABC$ lies on the perpendicular bisector of $UI$.
Show that $\gamma$ is the second-largest angle in the triangle $ABC$.
This post has been edited 1 time. Last edited by Martin N., May 24, 2012, 7:11 AM
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yugrey
2326 posts
#2 • 1 Y
Y by Adventure10
By "the intersection," do you mean "one of the intersections?"

Does the wording imply that there is exactly one intersection, that is, the angle bisector is tangent to the circumcircle?

Please clarify or get someone who speaks German to see the original text of the problem (this was given in German, I presume, as it is the national olympiad of Austria).

EDIT: Oops, yeah that is pretty obvious. It should say "other than C" though.
This post has been edited 1 time. Last edited by yugrey, May 23, 2012, 3:57 PM
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mavropnevma
15142 posts
#3 • 1 Y
Y by Adventure10
Oh, my ... The angle bisector of $\gamma = \angle ACB$ intersects the circumcircle of $ABC$ in two points, $C$ and "the other point" of intersection, which is clearly meant by the statement, and in fact is the midpoint of the arc $AB$ not containing $C$.
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yugrey
2326 posts
#4 • 2 Y
Y by Adventure10, Mango247
OK. I should read geometry problems more carefully, and perhaps draw a diagram, before I comment on the ambiguities in such problems.

Sorry.

Anyway,

It is well known that $2Rr$ is the power of the incenter with respect to the circumcircle.

Also, C,I, and D are collinear.

If D lies on the perpendicular bisector UI, DI=R.

Then CI=2r.

Also, we draw IS perpendicular to AC, so IS=r.

As CIS (pun- The Commonwealth of Independent States, the former Soviet Union :P) is a 30-60-90 triangle in that order, <ACI=30 then. Similarly, <ICB=30, so <ACB=60, and gamma is obviously the middle angle.

:)

Nice and elegant, but I don't think it's the best way as it relies a lot on the power of the incenter being $2Rr$.
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Martin N.
434 posts
#5 • 1 Y
Y by Adventure10
I edited the problem statement, now it should be 100% clear.

An outline of my soltion: Let $D$ be the mentioned second point of intersection of the angle bisector of $\gamma$ and the circumcirle. It is well-known that $|DI|=|DA|=|DB|$ and the condition in the problem statement gives $|DI|=|DU|$. Therefore, $ABIU$ is cyclic and hence $\angle AIB=\angle AUB$, which is eqivalent to $\alpha+\beta=2\gamma$, from where the conclusion follows.
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yugrey
2326 posts
#6 • 2 Y
Y by Adventure10, Mango247
Ah, another solution!

Note of course that alpha+beta+gamma=2gamma+gamma=180, so gamma=60 as in my solution.

This is one of those problems with a stronger result (namely that gamma is 60).
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