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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
Perpendicularity
April   32
N 22 minutes ago by zuat.e
Source: CGMO 2007 P5
Point $D$ lies inside triangle $ABC$ such that $\angle DAC = \angle DCA = 30^{\circ}$ and $\angle DBA = 60^{\circ}$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \perp EF$.
32 replies
April
Dec 28, 2008
zuat.e
22 minutes ago
Number theory
MuradSafarli   0
44 minutes ago
Prove that for any natural number \( n \) :

\[
1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n + 1) \mid (4n + 3)(4n + 5) \cdot \ldots \cdot (8n + 3).
\]
0 replies
MuradSafarli
44 minutes ago
0 replies
The number of integers
Fang-jh   16
N an hour ago by ihategeo_1969
Source: ChInese TST 2009 P3
Prove that for any odd prime number $ p,$ the number of positive integer $ n$ satisfying $ p|n! + 1$ is less than or equal to $ cp^\frac{2}{3}.$ where $ c$ is a constant independent of $ p.$
16 replies
Fang-jh
Apr 4, 2009
ihategeo_1969
an hour ago
functional equation interesting
skellyrah   12
N an hour ago by jasperE3
find all functions IR->IR such that $$xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2  + (x+1)f(x)$$
12 replies
skellyrah
Apr 24, 2025
jasperE3
an hour ago
No more topics!
Locus of incenter of quadrilateral
mofumofu   9
N Feb 12, 2024 by LLL2019
Source: CMO 2022 P1
Let $a$ and $b$ be two positive real numbers, and $AB$ a segment of length $a$ on a plane. Let $C,D$ be two variable points on the plane such that $ABCD$ is a non-degenerate convex quadrilateral with $BC=CD=b$ and $DA=a$. It is easy to see that there is a circle tangent to all four sides of the quadrilateral $ABCD$.
Find the precise locus of the point $I$.
9 replies
mofumofu
Dec 21, 2021
LLL2019
Feb 12, 2024
Locus of incenter of quadrilateral
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G H BBookmark kLocked kLocked NReply
Source: CMO 2022 P1
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mofumofu
179 posts
#1 • 3 Y
Y by jhu08, centslordm, Honestcobra
Let $a$ and $b$ be two positive real numbers, and $AB$ a segment of length $a$ on a plane. Let $C,D$ be two variable points on the plane such that $ABCD$ is a non-degenerate convex quadrilateral with $BC=CD=b$ and $DA=a$. It is easy to see that there is a circle tangent to all four sides of the quadrilateral $ABCD$.
Find the precise locus of the point $I$.
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sanyalarnab
930 posts
#2
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Is a,b fixed?
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gghx
1072 posts
#3
Y by
sanyalarnab wrote:
Is a,b fixed?

Yes.
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cadaeibf
701 posts
#5 • 1 Y
Y by khina
Since $ABCD$ is a deltoid, $I$ lies on its axis of symmetry $AC$. Therefore, by bisector theorem on $BCA$, we have $AB:BC=AI:IC$, which implies $AI=\frac{a}{a+b}AC$. Since $C$ lies on a fixed circle of radius $b$ and center $B$, it follows that the locus of $I$ is this circle scaled with center $A$ and factor $\frac{a}{a+b}$ (which in fact passws trough $B$).
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FishHeadTail
75 posts
#6
Y by
Well not exactly. First of all the problem says “non degenerate convex quadrilateral” but not all points on the circle satisfy this property.
This post has been edited 1 time. Last edited by FishHeadTail, Dec 21, 2021, 7:33 AM
Reason: Typo
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Jalil_Huseynov
439 posts
#7
Y by
@2 above since $ABCD$ is not degenrate, not all points on this circle work. My solution is same with at 2 above, but I find the locus as arc of this circle.
Let circle with center $B$ and radius $b$ be $\Gamma$ and let Homothety with center $A$ and radius $\frac{a}{a+b}$ maps $\Gamma$ to $\omega$. Then it's obvious that $B$ lies on $\omega$. Let $AX_1$ and $AX_2$ tangent to $\omega$ at $X_1,X_2$, rescpectively. Then locus of $I$ is $\text{arc}$ $X_1BX_2$ of $\omega $, except points $X_1,X_2$ and $B$.
This post has been edited 2 times. Last edited by Jalil_Huseynov, Dec 21, 2021, 7:44 AM
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cadaeibf
701 posts
#8
Y by
FishHeadTail wrote:
Well not exactly. First of all the problem says “non degenerate convex quadrilateral” but not all points on the circle satisfy this property.

Yes I omitted the fact that the two diametrically opposite on segment $AB$ don't work
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szjzc2018
75 posts
#9
Y by
By the way,I am curious about why the title is 2022 National Olympiad.Isn't that 2021?
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AshAuktober
997 posts
#10
Y by
Clearly $I$ lies on $AC$. Also, from the angle bisector theorem, $\frac{AI}{AC} = \frac{a}{a+b}$, so its obvious that the locus is the result of taking the locus of $C$ and performing a homothety $H\left(A, \frac{a}{a+b}\right)$ on it. As the locus of $C$ is clearly half of a circular arc, we are done.
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LLL2019
834 posts
#11
Y by
Not a single complete solution...

After obtaining $I$ is on that circular arc, as not all points must work, we need to still give some discussion.

Then, we have to split into multiple cases, and I summarize the result:
If $a=b$, the entire circle except two diametrically opposite points is the locus.
If $a<b$, we can see there are two arcs with central angle $\cos^{-1} \frac ab$
If $a>b$, we can see there are two arcs with central angle $\cos^{-1} \frac ba$

I hope somebody can provide a good solution: without a complete discussion, one can only obtain 15 marks, which the committee correctly noted was too high, but this mark was given due to the fact that locus is not often tested.

Indeed, the average mark is 18.6, for silver is >19, but still not 20 even for the training camp.
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