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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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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0 replies
jlacosta
Yesterday at 11:16 PM
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
existence of a circle tangent to AB and AC
NicoN9   0
2 minutes ago
Source: Japan Junior MO Preliminary 2020 P10
Let $ABC$ be a triangle with integer side lengths. Let $D, E$ be points on segment $BC$ such that $B, D, E,C$ are in this order, $BD=4$, and $EC=7$.
Suppose that there exists a circle which is tangent to sides $AB$ and $AC$, passes through $D, E$. Find the minimum of the perimeter of triangle $ABC$.
0 replies
+1 w
NicoN9
2 minutes ago
0 replies
filling tiles again?
NicoN9   0
4 minutes ago
Source: Japan Junior MO Preliminary 2020 P9
There is a board with regular hexagon shape with side length $1$. As shown below, we dessert the board into $24$ of equilateral triangle, with side length $1/2$. We call the $19$ points of $\circ$ is good in the figure.

IMAGE
There are $12$ of tiles with side length $\frac{1}{2}$, $\frac{\sqrt{3}}{2}$, $1$ (thus the tile is right-angled). How many ways are there to fill the board with these tiles such that
$\bullet$ Each vertex of the tiles are on good points, and
$\bullet$ There doesn't exist $2$ tiles, such that it forms a equilateral triangle of side length $1$.
0 replies
NicoN9
4 minutes ago
0 replies
3 variables NT
NicoN9   0
8 minutes ago
Source: Japan Junior MO Preliminary 2020 P8
Find all triples $(l, m, n)$ such that \[
l^2+mn=m^2+ln,\quad  n^2+lm=2020,\quad  l\le m\le n.
\]
0 replies
NicoN9
8 minutes ago
0 replies
Filling with tiles
NicoN9   0
10 minutes ago
Source: Japan Junior MO Preliminary 2020 P7
Consider the following tiles, created by using three and five unitsquare, respectively.
IMAGE
There are twelve of L, and four of X. We fill the following gray region created by $56$ unitsquare, using L and X.

IMAGE
Find the number of ways to do so.
0 replies
NicoN9
10 minutes ago
0 replies
3D combo puzzle
NicoN9   0
12 minutes ago
Source: Japan Junior MO Preliminary 2020 P6
As shown below, there is a figure $Q$ created by removing the unitcube at the cornor of the cube with side length $5$. Also, there are infinitely many figure $L$ created with four unitcube, and infinitely many unitcubes.

IMAGE
We paste together $L$ and unitcubes to create $Q$.
What is the maximum possible number of $L$ that we can use?
0 replies
NicoN9
12 minutes ago
0 replies
quadrilateral ABCD and mid of AC, BD
NicoN9   0
14 minutes ago
Source: Japan Junior MO Preliminary 2020 P5
Suppose $ABCD$ is a convex quadrilateral such that $AB=CD=7$, $DA=6$, $\angle B=72^\circ$, and $\angle C=48^\circ$. Let $P$, and $Q$ be the midpoint of segment $AC$, and $BD$, respectively. Find the value of $PQ$.
0 replies
NicoN9
14 minutes ago
0 replies
isosceles right triangle with four squares inscribed
NicoN9   0
16 minutes ago
Source: Japan Junior MO Preliminary 2020 P4
Four squares with side length $1$ are inscribed in isosceles right triangle, as shown below. Find the area of the isosceles right triangle.
IMAGE
0 replies
NicoN9
16 minutes ago
0 replies
number of palindromes divisible by 11
NicoN9   0
18 minutes ago
Source: Japan Junior MO Preliminary 2020 P3
We call an integer palindrome if it's the same value when read it backwards, and the unit digit is nonzero. Find the number of positive integers less than or equal to $10000$ such that it is a palindrome, and divisible by $11$.
0 replies
NicoN9
18 minutes ago
0 replies
Consecutive sum of integers sum up to 2020
NicoN9   0
20 minutes ago
Source: Japan Junior MO Preliminary 2020 P2
Let $a$ and $b$ be positive integers. Suppose that the sum of integers between $a$ and $b$, including $a$ and $b$, are equal to $2020$.
All among those pairs $(a, b)$, find the pair such that $a$ achieves the minimum.
0 replies
NicoN9
20 minutes ago
0 replies
3 right-angled triangle area
NicoN9   0
21 minutes ago
Source: Japan Junior MO Preliminary 2020 P1
Right angled triangle $ABC$, and a square are drawn as shown below. Three numbers written below implies each of the area of shaded small right angled triangle. Find the value of $AB/AC$.

IMAGE
0 replies
NicoN9
21 minutes ago
0 replies
Function equation
LeDuonggg   3
N 24 minutes ago by luutrongphuc
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
3 replies
LeDuonggg
Yesterday at 2:59 PM
luutrongphuc
24 minutes ago
A sequence containing every natural number exactly once
Pomegranat   4
N 33 minutes ago by Pomegranat
Source: Own
Does there exist a sequence \( \{a_n\}_{n=1}^{\infty} \), which is a permutation of the natural numbers (that is, each natural number appears exactly once), such that for every \( n \in \mathbb{N} \), the sum of the first \( n \) terms is divisible by \( n \)?
4 replies
Pomegranat
3 hours ago
Pomegranat
33 minutes ago
hard square root problem
kjhgyuio   0
43 minutes ago
........
0 replies
kjhgyuio
43 minutes ago
0 replies
Queue geo
vincentwant   5
N an hour ago by Ilikeminecraft
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $Y, Z$ be the feet of the altitudes from $B, C$ to $AC, AB$ respectively. Let $D$ be the midpoint of $BC$. Let $\omega_1$ be the circle with diameter $AD$. Let $Q\neq A$ be the intersection of $(ABC)$ and $\omega$. Let $H$ be the orthocenter of $ABC$. Let $K$ be the intersection of $AQ$ and $BC$. Let $l_1,l_2$ be the lines through $Q$ tangent to $\omega,(AYZ)$ respectively. Let $I$ be the intersection of $l_1$ and $KH$. Let $P$ be the intersection of $l_2$ and $YZ$. Let $l$ be the line through $I$ parallel to $HD$ and let $O'$ be the reflection of $O$ across $l$. Prove that $O'P$ is tangent to $(KPQ)$.
5 replies
vincentwant
Wednesday at 3:54 PM
Ilikeminecraft
an hour ago
Reflection about Euler line
keyree10   9
N Jan 25, 2022 by Mahdi_Mashayekhi
Source: INMO 2010 Problem 5
Let $ ABC$ be an acute-angled triangle with altitude $ AK$. Let $ H$ be its ortho-centre and $ O$ be its circum-centre. Suppose $ KOH$ is an acute-angled triangle and $ P$ its circum-centre. Let $ Q$ be the reflection of $ P$ in the line $ HO$. Show that $ Q$ lies on the line joining the mid-points of $ AB$ and $ AC$.
9 replies
keyree10
Jan 18, 2010
Mahdi_Mashayekhi
Jan 25, 2022
Reflection about Euler line
G H J
Source: INMO 2010 Problem 5
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keyree10
249 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ ABC$ be an acute-angled triangle with altitude $ AK$. Let $ H$ be its ortho-centre and $ O$ be its circum-centre. Suppose $ KOH$ is an acute-angled triangle and $ P$ its circum-centre. Let $ Q$ be the reflection of $ P$ in the line $ HO$. Show that $ Q$ lies on the line joining the mid-points of $ AB$ and $ AC$.
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gilcu3
196 posts
#2 • 3 Y
Y by Sx763_, Adventure10, and 1 other user
Let $ R$ the center of the nine points circle. Let $ Q_1$, $ P_1$ and $ R_1$ the projections of $ Q$, $ P$ and $ R$ on the altitude $ AK$. Then for do the problem we have to prove that $ Q_1$ is the midpoint of $ AK$. But is easy to prove that $ Q_1P_1=2P_1R_1$.

So:

$ Q_1K=Q_1P_1+P_1K=2P_1R_1+P_1K=2(R_1K-P_1K)+P_1K=2R_1K-P_1K=\frac{1}{2}AH+HK-P_1K=\frac{1}{2}AK$ as claimed.
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groshanlal
11 posts
#3 • 1 Y
Y by Adventure10
By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?
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Rijul saini
904 posts
#4 • 6 Y
Y by Wizard_32, SHREYAS333, Adventure10, Mango247, Math_DM, and 1 other user
Soution 1 (own)

Solution 2 (Due to Akashnil)
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madhusudan kale
23 posts
#5 • 2 Y
Y by Adventure10, Mango247
i have a very easy solution using pure geometry

Let ABC be an acute-angled triangle with altitude AK. Let H be its ortho-centre and O be its circum-centre. Suppose KOH is an acute-angled triangle and P its circum-centre. Let Q be the reflection of P in the line HO.
it is very easy to prove that
quadrilateral HPOQ is a rhombus.

Thus now consider the \triangle HKO
in this triangle
$ \angle HPO = 2\angle HKO$or $ 1/2\angle HPO = \angle HKO = \angle PQO    ------------------(1)$
since P is the circumcenter and if we draw the circle we will get that \angle HPO is subtended at the center and \angle HKO is subtended at any other point by the same arc .

now we know that AK\perp BC since it is the altitude and $ UN\perp BC$ thus $ UN \parallel AK$.
hence$ \angle HKO = \angle KOT$

but $ 1/2\angle HKO = \angle HPO$
hence $ 1/2\angle HPO = \angle KON$
since HPOQ is a rhombus ,
$ \angle HQP = \angle PQO$

we get $ \angle PQO = \angle KON$ OR $ \angle MQO = \angle KOT$
consider$ \triangle KOT$and$ \triangle OQM$
$ \angle MQO = \angle KON$
and $ \angle QMO = \angle OTK$
thus $ \triangle KOT$ and $ \triangle OQM$ are similar so we get
$ \angle OKT= \angle QOM$


now since UT is a straight line
$ \angle QOX + \angle QOM + \angle HOK  +  \angle KOT=180$
$ \angle QOX + \angle OKT + \angle HOK  +  \angle KOT=180$ since $ \angle OKT= \angle QOM$
$ \angle QOX +  \angle HOK  +90=180$
$ \angle QOX +  \angle HOK=90$
but $ \angle QOX=90-\angle OQX$
thus $ \angle OQX= \angle HOK$ $ --------------(2)$
now consider $ \triangle HLN$and $ \triangle NMQ$
$ \angle HLN= \angle NMQ$
and $ \angle HNL= \angle NQM$
thus $ \triangle HLN$ and $ \triangle NMQ$ are similar.
hence $ \angle LHN=\angle NQM$
but $ \angle LHN and \angle KOH$ are same.
hence we get $ \angle KOH= \angle NQM=\angle RQP$
we know that in a triangle sum of all angles is 180.
Thus in $ \triangle KOH$,

$ \angle HKO + \angle KOH + \angle KHO=180.$
or $ \angle PQO + \angle RQP + \angle OQX =180$ from (1) and (2),
this implies that RQX is a straight line or point Q lies on RS.
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Mathias_DK
1312 posts
#6 • 5 Y
Y by div5252, thewitness, mathetillica, Understandingmathematics, Adventure10
keyree10 wrote:
Let $ ABC$ be an acute-angled triangle with altitude $ AK$. Let $ H$ be its ortho-centre and $ O$ be its circum-centre. Suppose $ KOH$ is an acute-angled triangle and $ P$ its circum-centre. Let $ Q$ be the reflection of $ P$ in the line $ HO$. Show that $ Q$ lies on the line joining the mid-points of $ AB$ and $ AC$.
Let $ a,b,c,h,p,q,z_O,k$ be the complex numbers corresponding to the points $ A,B,C,H,P,Q,O,K$ in that order.

Wlog assume that $ z_O = 0, |a| = |b| = |c| = 1$. Then $ h = 3(g - z_O) + z_O = a + b + c$. $ k = \frac {1}{2} \left ( a + b + c - \overline{a}bc \right )$ is easy to verify.

A point $ Z(z)$ lies on the line joining the midpoints of $ AB$ and $ AC$ iff $ \frac {z - \frac {a + b}{2}}{\overline{z - \frac {a + b}{2}}} = \frac {\frac {a + c}{2} - \frac {a + b}{2}}{\overline{\frac {a + c}{2} - \frac {a + b}{2}}} \iff$
$ 2z + 2bc\overline{z} = a + b + c + \overline{a}bc$. $ bc = \frac {h - k}{\overline{h - k}}$ and $ a + b + c + \overline{a}bc = 2h - 2k$, so:
$ \iff z + \frac {h - k}{\overline{h - k}}\overline{z} = h - k$

$ P$ is the orthocenter of $ \triangle OKH$ so:
$ p = \frac {hk(\overline{h - k})}{\overline{h}k - h\overline{k}}$

$ Q$ is obtained by the reflection of $ P$ in $ OH$, so:
$ q = \overline{p} \frac {h}{\overline{h}} = \frac { - h\overline{k}(h - k)}{\overline{h}k - h\overline{k}}$

Inserting it is obvious that:
$ q + \frac {h - k}{\overline{h - k}}\overline{q} = (h - k) \frac {\overline{h}k - h\overline{k}}{\overline{h}k - h\overline{k}} = h - k$, and therefore $ Q$ lies on the line joining the midpoints of $ AB$ and $ AC$, and we are done :)
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Moonmathpi496
413 posts
#7 • 2 Y
Y by Adventure10, Mango247
keyree10 wrote:
Let $ ABC$ be an acute-angled triangle with altitude $ AK$. Let $ H$ be its ortho-centre and $ O$ be its circum-centre. Suppose $ KOH$ is an acute-angled triangle and $ P$ its circum-centre. Let $ Q$ be the reflection of $ P$ in the line $ HO$. Show that $ Q$ lies on the line joining the mid-points of $ AB$ and $ AC$.
Nice Problem!
Let the midpoints of $ AB,CD$ be $ Q,R$ respectively, and let the reflection of $ P$ wrt $ OH$ be $ P'$.
We consider a half turn wrt $ T$, the midpoint of $ OH$ (and the center of the nine point circle of $ \triangle ABC$). This maps $ P' \to P$, $ QR \to Q'R'$.
It is enough to prove that $ P$ lies on $ Q'R'$.
We know that $ Q'R' \parallel CB$. So $ Q'R'$ intersect $ HK$ at the midpoint, and also $ HK \perp Q'R'$. So $ P$ lies on $ Q'R'$.
groshanlal wrote:
By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?
Though I have not ever took part in InMO (I am from Bangladesh), I believe that it depends on many facts, such as: the quality of the contestants, difficulty level of the problems etc. In general, in Olympiads the contestants who solve half of the problem gets some sort of awards. I don't know if it true for InMO. :P
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kamallohia
40 posts
#8 • 2 Y
Y by Adventure10, Mango247
My goodness - Dr Rijul has posted his solution already. :lol:
My Solution:
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Shravu
131 posts
#9 • 2 Y
Y by Adventure10, Mango247
we have not used that triangle is acute in rijul's solution
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Mahdi_Mashayekhi
695 posts
#10
Y by
Let E be midpoint of AC and D be midpoint of BH. Note that BH = 2ON so DH = ON and we also have DH || ON because they're both perpendicular to AC so DHEO is parallelogram. Let S be midpoint of OH we know S is midpoint of ED and OH and PQ so DQEP is parallelogram so EQ || PD || BC so Q lies on the line made of midpoints of AC and AB.
we're Done.
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