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100 Selected Problems Handout
Asjmaj   35
N 14 minutes ago by CBMaster
Happy New Year to all AoPSers!
 :clap2:

Here’s my modest gift to you all. Although I haven’t been very active in the forums, the AoPS community contributed to an immense part of my preparation and left a huge impact on me as a person. Consider this my way of giving back. I also want to take this opportunity to thank Evan Chen—his work has consistently inspired me throughout my olympiad journey, and this handout is no exception.



With 2025 drawing near, my High School Olympiad career will soon be over, so I want to share a compilation of the problems that I liked the most over the years and their respective detailed write-ups. Originally, I intended it just as a personal record, but I decided to give it some “textbook value” by not repeating the topics so that the selection would span many different approaches, adding hints, and including my motivations and thought process.

While IMHO it turned out to be quite instructive, I cannot call it a textbook by any means. I recommend solving it if you are confident enough and want to test your skills on miscellaneous, unordered, challenging, high-quality problems. Hints will allow you to not be stuck for too long, and the fully motivated solutions (often with multiple approaches) should help broaden your perspective. 



This is my first experience of writing anything in this format, and I’m not a writer by any means, so please forgive any mistakes or nonsense that may be written here. If you spot any typos, inconsistencies, or flawed arguments whatsoever (no one is immune :blush: ), feel free to DM me. In fact, I welcome any feedback or suggestions.

I left some authors/sources blank simply because I don’t know them, so if you happen to recognize where and by whom a problem originated, please let me know. And quoting the legend: “The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.” 



I’ll likely keep a separate file to track all the typos, and when there’s enough, I will update the main file. Some problems need polishing (at least aesthetically), and I also have more remarks to add.

This content is only for educational purposes and is not meant for commercial usage.



This is it! Good luck in 45^2, and I hope you enjoy working through these problems as much as I did!

Here's a link to Google Drive because of AoPS file size constraints: Selected Problems
35 replies
Asjmaj
Dec 31, 2024
CBMaster
14 minutes ago
trigonometric inequality
MATH1945   11
N 35 minutes ago by sqing
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
11 replies
1 viewing
MATH1945
May 26, 2016
sqing
35 minutes ago
Centrally symmetric polyhedron
genius_007   0
37 minutes ago
Source: unknown
Does there exist a convex polyhedron with an odd number of sides, where each side is centrally symmetric?
0 replies
genius_007
37 minutes ago
0 replies
Combinatorial Game
Cats_on_a_computer   2
N an hour ago by Cats_on_a_computer

Let n>1 be odd. A row of n spaces is initially empty. Alice and Bob alternate moves (Alice first); on each turn a player may either
1. Place a stone in any empty space, or
2. Remove a stone from a non-empty space S, then (if they exist) place stones in the nearest empty spaces immediately to the left and to the right of S.

Furthermore, no move may produce a position that has appeared earlier. The player loses when they cannot make a legal move.
Assuming optimal play, which move(s) can Alice make on her first turn?
2 replies
Cats_on_a_computer
an hour ago
Cats_on_a_computer
an hour ago
2-var inequality
sqing   4
N an hour ago by sqing
Source: Own
Let $ a,b> 0 $ and $2a+2b+ab=5. $ Prove that
$$ \frac{a^2}{b^2}+\frac{1}{a^2}-a^2\geq  1$$$$ \frac{a^3}{b^3}+\frac{1}{a^3}-a^3\geq  1$$
4 replies
sqing
2 hours ago
sqing
an hour ago
A game of digits and seventh powers
v_Enhance   28
N an hour ago by blueprimes
Source: Taiwan 2014 TST3 Quiz 1, P2
Alice and Bob play the following game. They alternate selecting distinct nonzero digits (from $1$ to $9$) until they have chosen seven such digits, and then consider the resulting seven-digit number by concatenating the digits in the order selected, with the seventh digit appearing last (i.e. $\overline{A_1B_2A_3B_4A_6B_6A_7}$). Alice wins if and only if the resulting number is the last seven decimal digits of some perfect seventh power. Please determine which player has the winning strategy.
28 replies
v_Enhance
Jul 18, 2014
blueprimes
an hour ago
n containers distribute gas to a car for a single loop along the track
parmenides51   16
N an hour ago by v_Enhance
Source: Spanish Mathematical Olympiad 1997 P6
The exact quantity of gas needed for a car to complete a single loop around a track is distributed among $n$ containers placed along the track. Prove that there exists a position starting at which the car, beginning with an empty tank of gas, can complete a loop around the track without running out of gas. The tank of gas is assumed to be large enough.
16 replies
1 viewing
parmenides51
Jul 31, 2018
v_Enhance
an hour ago
Factorial: n!|a^n+1
Nima Ahmadi Pour   68
N an hour ago by blueprimes
Source: IMO Shortlist 2005 N4, Iran preparation exam
Find all positive integers $ n$ such that there exists a unique integer $ a$ such that $ 0\leq a < n!$ with the following property:
\[ n!\mid a^n + 1
\]

Proposed by Carlos Caicedo, Colombia
68 replies
Nima Ahmadi Pour
Apr 24, 2006
blueprimes
an hour ago
Geometry problem
Whatisthepurposeoflife   1
N an hour ago by Royal_mhyasd
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
1 reply
1 viewing
Whatisthepurposeoflife
2 hours ago
Royal_mhyasd
an hour ago
\frac{x^2}{x+y}+\frac{y^2}{1-x}+\frac{(1-x-y)^2}{1-y}\geq\frac{1}{2} if 0<x,y<1
parmenides51   6
N 2 hours ago by JH_K2IMO
Source: KJMO 2009 p3
For two arbitrary reals $x, y$ which are larger than $0$ and less than $1.$ Prove that$$\frac{x^2}{x+y}+\frac{y^2}{1-x}+\frac{(1-x-y)^2}{1-y}\geq\frac{1}{2}.$$
6 replies
parmenides51
May 2, 2019
JH_K2IMO
2 hours ago
2-var inequality
sqing   6
N 2 hours ago by sqing
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1}  -\frac{b^4}{a+1} \leq 1 $$
6 replies
sqing
Yesterday at 2:15 AM
sqing
2 hours ago
Symmedian line
April   93
N May 18, 2025 by aidenkim119
Source: All Russian Olympiad - Problem 9.2, 10.2
Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.
93 replies
April
May 10, 2009
aidenkim119
May 18, 2025
Symmedian line
G H J
Source: All Russian Olympiad - Problem 9.2, 10.2
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April
1270 posts
#1 • 8 Y
Y by Amir Hossein, dave_mathsguru, 799786, itslumi, Adventure10, Mango247, fuzzball2023, and 1 other user
Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.
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mathVNpro
469 posts
#2 • 6 Y
Y by Olemissmath, Psyduck909, Adventure10, Mango247, fuzzball2023, ehuseyinyigit
Let $ M_b$ be midpoint of $ AC$.
We have $ \angle FAM_b = \angle FBC$, $ \angle AM_bF\equiv \angle DM_bF = \angle DEF = \angle BEF = \angle BCF$
$ \Rightarrow \triangle FAM_b\sim \triangle FBC$, $ (i)$. Thus, through the spiral similarity with center $ F$, denote this $ f$ then:
$ f: \triangle FAM_b\mapsto \triangle FBC$
$ \Rightarrow f: A\mapsto B$, $ M_b\mapsto C$
$ \Rightarrow f: A\mapsto M_b$, $ B\mapsto C$
$ \Rightarrow f: \triangle FAB\mapsto \triangle FM_bC$, $ (1)$
But it is easy to notice that: $ \triangle BAF\sim \triangle BM_bC$ (a.a), $ (2)$
$ (1),(2)\Rightarrow \triangle CM_bB\sim \triangle FM_bC$
$ \Rightarrow \frac {CB}{CF} = \frac {CM_b}{FM_b} = \frac {M_bB}{M_bC}$, $ (*)$.
With the same argument, we get $ \triangle BAM_b\sim \triangle BFC$
$ \Rightarrow \triangle BAM_b\sim \triangle AFM_b$ (thanks to $ (i)$).
$ \Rightarrow \frac {BA}{AF} = \frac {M_bB}{M_bA}$, $ (**)$
But $ M_bA = M_bC \Rightarrow \frac {BA}{BC} = \frac {FA}{FC}$ (Combine $ (*)\&(**)$)
$ \Longrightarrow ABCF$ is a harmonic quadrilateral. Therefore, $ BF$ passes through the Pole of $ AC$ wrt $ (O)$, where $ O$ is the circumcenter of $ \triangle ABC$. Then we will get the result of the problem.
Our proof is completed then :lol:
This post has been edited 1 time. Last edited by mathVNpro, May 11, 2009, 4:56 AM
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silouan
3952 posts
#3 • 6 Y
Y by ultralako, Adventure10, Mango247, fuzzball2023, and 2 other users
Let $ M$ the midpoint of $ AC$ , $ O$ the circumcenter of $ \Omega$ and let the tangent to $ \Omega$ at $ B$ meets $ AC$ at $ R$ .
Then $ BOMR$ is cyclic (let $ Q$ its circumcenter) . Then $ R,Q,O$ are collinear . So if the circumcircle of $ BOMR$ meets $ \Omega$ at $ F'$ , then $ BF'\bot RO$ . So by polars $ RF'$ is tangent to $ \Omega$ . By well known theorem we have that $ BF'$ is the $ B$ symmedian. Finally if $ BE$ meets the circumcircle of $ BOMR$ at $ S$ , then $ SM=SF$ . But also because $ RF'=RB$ we have that $ D$ is the incenter of $ F'BM$ so $ SD=SM=SF$ and because $ DM\bot ME$ we also have $ SD=SM=SF=SE$ . This shows that $ MDF'E$ is cyclic and so $ F\equiv F'$ and we are done .
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yetti
2643 posts
#4 • 5 Y
Y by Adventure10, Mango247, MS_asdfgzxcvb, and 2 other users
External bisector of $ \angle B$ cuts $ CA$ at $ Y.$ $ I, I_b$ are incenter and B-excenter of $ \triangle ABC.$ $ (E)$ is a circle with center $ E$ and radius $ EA = EC = EI = EI_b.$ The cross ratio $ (B, D, I, I_b)$ is harmonic and $ E$ the midpoint of $ II_b$ $ \Longrightarrow$ $ \overline{BD} \cdot \overline{BE} = \overline{BI}\cdot \overline{BI_b}$ $ \Longrightarrow$ $ BY \perp BDE$ is radical axis of $ \omega, (E).$ Pairwise radical axes $ EF, CA, BY$ of $ \Omega, \omega, (E)$ meet at their radical center $ Y.$ Since $ DF \perp EFY$ and $ BY \perp BDE,$ $ BDFY$ is cyclic. Its circumcircle is the B-Apollonius circle of $ \triangle ABC$ $ \Longrightarrow$ $ BF$ is the B-symmedian.
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Virgil Nicula
7054 posts
#5 • 7 Y
Y by Amir Hossein, Adventure10, Mango247, fuzzball2023, MS_asdfgzxcvb, and 2 other users
PP. Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$ . The line $ BD$ intersects the circumcircle $ \Omega$ of
$\triangle ABC$ at $ B$ and $ E$ . Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$ .


Lemma. Let $ ABC$ be a triangle with the circumcircle $ w$ . Denote the following points : the middlepoint $ M$ of the side $ [BC]$ ;

the point $ D\in (BC)$ for which $ \angle DAB\equiv\angle DAC$ ; the point $ E$ for which $ \{A,E\} = AD\cap w$ ; the $ A$-symmedian $ AL$ ,

where $ L\in (BC)$ ; the point $ S$ for which $ \{A,S\} = AL\cap w$ . Then $ SD\perp SE$ , i.e. the quadrilateral $ MDSE$ is cyclically.


Proof. Denote $ XX$ - the tangent in $ X\in w$ to $ w$ . Denote $ T\in BB\cap CC$ . Observe that $ \overline {MET}\perp BC$ and $ \angle ASB\equiv\angle ACB$ .

From the first well-known property, $ T\in \overline {ALS}$ and the line $ \overline{ALST}$ is the $ A$-symmedian in $ \triangle ABC$ $ \Longrightarrow$ $ \left\|\begin{array}{c} \angle BAS\equiv\angle MAC \\
\ \angle SAE\equiv\angle DAM\end{array}\right\|$ .

From the second well-known property, the division $ \{A,S;L,T\}$ is harmonically. Since $ ML\perp MT$ results $ \angle DMA\equiv\angle DMS$ .

Show easily that $ \triangle BAS\sim\triangle MAC$ (a.a.) . Thus, $ \frac {AB}{AS} = \frac {AM}{AC}$ , i.e. $ AS\cdot AM = AB\cdot AC$ .

From the third well-known property $ AD\cdot AE = AB\cdot AC$, obtain $ \boxed {AS\cdot AM = AD\cdot AE = AB\cdot AC}$ , i.e. $ \frac {AS}{AE} = \frac {AD}{AM}$ .

Since $ \angle SAE\equiv\angle DAM$ obtain $ \triangle SAE\sim\triangle DAM$ and in consequence, $ \angle DMA\equiv\angle SEA$ . Since $ \angle DMA\equiv\angle DMS$

and $ \angle SEA\equiv\angle SED$ obtain $ \angle DMS\equiv\angle SED$ , i.e. the quadrilateral $ MDSE$ is cyclically. In conclusion, $ SD\perp SE$ .

Remark. Here are another interesting metrical relations : $ \left\|\begin{array}{ccc} ABS\sim CMS & \implies & \boxed {SB\cdot SC = SA\cdot SM} \\
 \\
AMC\sim CMS & \implies & \boxed {MA\cdot MS = MB^2}\end{array}\right\|$ .

The proposed problem is an immediate consequence of the above lemma.
See PP13 from here
This post has been edited 4 times. Last edited by Virgil Nicula, Aug 11, 2014, 1:54 AM
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jayme
9801 posts
#6 • 3 Y
Y by AlastorMoody, Adventure10, fuzzball2023
Dear Mathlinkers,
1. Let A' be the midpoint of BC, X, Y the second meetpoint of AA', EA' with the circumcircle of ABC and Te the tangent to this circumcircle at E.
2. Te // DA'
3. The circle with diameter ED passes through A'
4. By a converse of Reim's theorem applied to this circle and the circumcircle, FD goes through Y;
5. By a converse of Pascal's theorem applied to EAXFYTe, FX //DA' (BC)
and we are done...
Sincerely
Jean-Louis
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Agr_94_Math
881 posts
#7 • 3 Y
Y by Adventure10, Mango247, ehuseyinyigit
Very nice application of Reim's Theorem jayme.Though a very obvious result, it has great application in many synthetic geometry results.
Thanks.
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mathVNpro
469 posts
#8 • 2 Y
Y by Adventure10, Mango247
jayme wrote:
Dear Mathlinkers,
1. Let A' be the midpoint of BC, X, Y the second meetpoint of AA', EA' with the circumcircle of ABC and Te the tangent to this circumcircle at E.
2. Te // DA'
3. The circle with diameter ED passes through A'
4. By a converse of Reim's theorem applied to this circle and the circumcircle, FD goes through Y;
5. By a converse of Pascal's theorem applied to EAXFYTe, FX //DA' (BC)
and we are done...
Sincerely
Jean-Louis
Do you have any document about Reim's theorem, Jayme? I am so excited about this. :blush:
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jayme
9801 posts
#9 • 3 Y
Y by Adventure10, Mango247, fuzzball2023
See for example my website
http://perso.orange.fr/jl.ayme then: A propos
You will see all the possibility.
Sincerely
Jean-Louis
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ma 29
40 posts
#10 • 16 Y
Y by MexicOMM, Wizard_32, Durjoy1729, HolyMath, Siddharth03, Greenleaf5002, myh2910, snakeaid, centslordm, Jalil_Huseynov, Adventure10, Mango247, GustavoPudim, fuzzball2023, and 2 other users
Dear friends.
I have a short solution for the nice problem,here :) .
//cdn.artofproblemsolving.com/images/4a3755e6df266a6313f50f6a8a6c7c8845715061.jpg

Denote by $ A'$ the midpoint of segment $ AC$.

$ EO$ meets $ (O)$ again at $ P$

By ${ \angle {DFE} = 90^0}$ ,we have $ F,D,P$ are collinear.

From $ BPEF$ is cyclic and $ BDA'P$ is cylic ,we have :

$ \angle {EBF} = \angle {EPF} = \angle{DBA'}$.

Therefor,$ BF$is the symmedian line of the triangle $ BAC$.
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armpist
527 posts
#11 • 3 Y
Y by Adventure10, Mango247, and 1 other user
All are very nice solutions indeed, I must admit.



T.Y.
M.T.
This post has been edited 1 time. Last edited by armpist, May 11, 2009, 8:29 PM
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Virgil Nicula
7054 posts
#12 • 2 Y
Y by Adventure10, Mango247
After the Ma29's shortest proof, this problem seems very simple ...
In the another proofs appeared many nice properties of this configuration.
Thanks to all for their interest and proofs.
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ma 29
40 posts
#13 • 2 Y
Y by Adventure10, Mango247
Hello my friends.
I think all proof above are very nice. :)
Here is another solution to this problem:
//cdn.artofproblemsolving.com/images/c51085c9ea4036fa17fa17def72321254472bf4e.jpg

Denote $ V = EF \cap AC$.

We have :$ \bar {VD} .\bar{VA'} = \bar{VF}.\bar{VE} = \bar{VA}.\bar{VC}$ (1)

But $ A'$ is the midpoint of segment $ AC$.(2)

From (1) ,(2) and Maclaurin's theorem , we obtain the division $ {VDAC}$ is harmonically,so $ V,B,P$ are collinear.

Denote $ Q = BC\cap {PA}$ and denote by $ H$ the intersection of two tangent to $ (O)$ at $ A,C$ ,respectively.

We have $ Q,D,H$ lie on the polar of $ V$ wrt $ (O)$,i.e $ Q,D,H$ are collinear.

By Pascal's theorem applied to six point $ A,A,P,C,B,F$ ,we obtain $ B,F,H$ are collinear.

Therefore ,$ BF$ is the symedian line of the triangle $ ABC$.
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bravado
19 posts
#14 • 2 Y
Y by Adventure10, Mango247
It can also be viewed as an easy application of : http://www.mathlinks.ro/viewtopic.php?p=1082899#1082899

Indeed, from the above problem it follows that the line $ FM$ (where $ M$ is the midpoint of $ AC$) meets the circumcircle of $ \triangle{ABC}$ at a point $ T$ such that $ BT||AC$. By symmetry,
\[ \angle{MBC} = \angle{MTA} = \angle{ABF}.
\]
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Virgil Nicula
7054 posts
#15 • 2 Y
Y by Adventure10, Mango247
We can use my found property at the proposed problem from here.
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