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IMO Shortlist 2014 G6
hajimbrak   30
N 18 minutes ago by awesomeming327.
Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$ . Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and $T$ , respectively. We call the pair $(E, F )$ $\textit{interesting}$, if the quadrilateral $KSAT$ is cyclic.
Suppose that the pairs $(E_1 , F_1 )$ and $(E_2 , F_2 )$ are interesting. Prove that $\displaystyle\frac{E_1 E_2}{AB}=\frac{F_1 F_2}{AC}$
Proposed by Ali Zamani, Iran
30 replies
hajimbrak
Jul 11, 2015
awesomeming327.
18 minutes ago
4th grader qual JMO
HCM2001   23
N 2 hours ago by nitride
i mean.. whattttt??? just found out about this.. is he on aops? (i'm sure he is) where are you orz lol..
https://www.mathschool.com/blog/results/celebrating-success-douglas-zhang-is-rsm-s-youngest-usajmo-qualifier
23 replies
+3 w
HCM2001
Yesterday at 12:53 AM
nitride
2 hours ago
[TEST RELEASED] OMMC Year 5
DottedCaculator   107
N 2 hours ago by ethan2011
Test portal: https://ommc-test-portal-2025.vercel.app/

Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.

Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
107 replies
DottedCaculator
Apr 26, 2025
ethan2011
2 hours ago
for the contest high achievers, can you share your math path?
HCM2001   25
N 3 hours ago by KnowingAnt
Hi all
Just wondering if any orz or high scorers on contests at young age (which are a lot of u guys lol) can share what your math path has been like?
- school math: you probably finish calculus in 5th grade or something lol then what do you do for the rest of the school? concurrent enrollment? college class? none (focus on math competitions)?
- what grade did you get honor roll or higher on AMC 8, AMC 10, AIME qual, USAJMO qual, etc?
- besides aops do you use another program to study? (like Mr Math, Alphastar, etc)?

You're all great inspirations and i appreciate the answers.. you all give me a lot of motivation for this math journey. Thanks
25 replies
HCM2001
Wednesday at 7:50 PM
KnowingAnt
3 hours ago
another diophantine about primes
AwesomeYRY   133
N 3 hours ago by EpicBird08
Source: USAMO 2022/4, JMO 2022/5
Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares.
133 replies
AwesomeYRY
Mar 24, 2022
EpicBird08
3 hours ago
[CASH PRIZES] IndyINTEGIRLS Spring Math Competition
Indy_Integirls   40
N 4 hours ago by OGMATH
[center]IMAGE

Greetings, AoPS! IndyINTEGIRLS will be hosting a virtual math competition on May 25,
2024 from 12 PM to 3 PM EST.
Join other woman-identifying and/or non-binary "STEMinists" in solving problems, socializing, playing games, winning prizes, and more! If you are interested in competing, please register here![/center]

----------

[center]Important Information[/center]

Eligibility: This competition is open to all woman-identifying and non-binary students in middle and high school. Non-Indiana residents and international students are welcome as well!

Format: There will be a middle school and high school division. In each separate division, there will be an individual round and a team round, where students are grouped into teams of 3-4 and collaboratively solve a set of difficult problems. There will also be a buzzer/countdown/Kahoot-style round, where students from both divisions are grouped together to compete in a MATHCOUNTS-style countdown round! There will be prizes for the top competitors in each division.

Problem Difficulty: Our amazing team of problem writers is working hard to ensure that there will be problems for problem-solvers of all levels! The middle school problems will range from MATHCOUNTS school round to AMC 10 level, while the high school problems will be for more advanced problem-solvers. The team round problems will cover various difficulty levels and are meant to be more difficult, while the countdown/buzzer/Kahoot round questions will be similar to MATHCOUNTS state to MATHCOUNTS Nationals countdown round in difficulty.

Platform: This contest will be held virtually through Zoom. All competitors are required to have their cameras turned on at all times unless they have a reason for otherwise. Proctors and volunteers will be monitoring students at all times to prevent cheating and to create a fair environment for all students.

Prizes: At this moment, prizes are TBD, and more information will be provided and attached to this post as the competition date approaches. Rest assured, IndyINTEGIRLS has historically given out very generous cash prizes, and we intend on maintaining this generosity into our Spring Competition.

Contact & Connect With Us: Email us at indy@integirls.org.

---------
[center]Help Us Out

Please help us in sharing the news of this competition! Our amazing team of officers has worked very hard to provide this educational opportunity to as many students as possible, and we would appreciate it if you could help us spread the word!
40 replies
Indy_Integirls
May 11, 2025
OGMATH
4 hours ago
MAA finally wrote sum good number theory
IAmTheHazard   96
N Yesterday at 4:54 PM by megahertz13
Source: 2021 AIME I P14
For any positive integer $a,$ $\sigma(a)$ denotes the sum of the positive integer divisors of $a.$ Let $n$ be the least positive integer such that $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a.$ Find the sum of the prime factors in the prime factorization of $n.$
96 replies
IAmTheHazard
Mar 11, 2021
megahertz13
Yesterday at 4:54 PM
Integer Functional Equation
msinghal   99
N Yesterday at 3:55 PM by DeathIsAwe
Let $\mathbb{Z}$ be the set of integers. Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that \[xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))\] for all $x, y \in \mathbb{Z}$ with $x \neq 0$.
99 replies
msinghal
Apr 29, 2014
DeathIsAwe
Yesterday at 3:55 PM
mathpath: how much do recommendations matter
mm999aops   25
N Yesterday at 2:09 PM by ethan2011
See question^

I'm hoping only the QT matters : P
25 replies
mm999aops
Feb 3, 2023
ethan2011
Yesterday at 2:09 PM
Segment has Length Equal to Circumradius
djmathman   74
N Yesterday at 12:19 PM by amirhsz
Source: 2014 USAMO Problem 5
Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.
74 replies
djmathman
Apr 30, 2014
amirhsz
Yesterday at 12:19 PM
Essentially, how to get good at olympiad math?
gulab_jamun   11
N Yesterday at 2:13 AM by oinava
Ok, so I'm posting this as an anynonymous user cuz I don't want to get flamed by anyone I know for my goals but I really do want to improve on my math skill.

Basically, I'm alright at computational math (10 AIME, dhr stanford math meet twice) and I hope I can get good enough at olympiad math over the summer to make MOP next year (I will be entering 10th as after next year, it becomes much harder :( )) Essentially, I just want to get good at olympiad math. If someone could, please tell me how to study, like what books (currently thinking of doing EGMO) but I don't know how to get better at the other topics. Also, how would I prepare? Like would I study both proof geometry and proof number theory concurrently or just study each topic one by one?? Would I do mock jmo/amo or js prioritize olympiad problems in each topic. I have the whole summer ahead of me, and intend to dedicate it to olympiad math, so any advice would be really appreciated. Thank you!
11 replies
gulab_jamun
May 18, 2025
oinava
Yesterday at 2:13 AM
Lines AD, BE, and CF are concurrent
orl   48
N Apr 7, 2025 by Avron
Source: IMO Shortlist 2000, G3
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.
48 replies
orl
Aug 10, 2008
Avron
Apr 7, 2025
Lines AD, BE, and CF are concurrent
G H J
Source: IMO Shortlist 2000, G3
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john0512
4190 posts
#36
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Let $H_1,H_2,H_3$ be the reflections of $H$ over $BC,CA,AB$ respectively. Let $OH_1,OH_2,OH_3$ intersect $BC,CA,AB$ at $D,E,F$ respectively.

We claim that these points $D,E,F$ are the desired points in the problem.

Let $\angle A=\alpha,\angle B=\beta,\angle C=\gamma$.

Note that $\angle BAH_1=90-\beta$, so $\angle BOH_1=180-2\beta$. Since $OB=OH_1$, $\angle DH_1B=\beta$. Similarly, $\angle DH_1C=\gamma.$

Additionally, note that $\triangle ABH_1$ and $\triangle ACH_1$ have the same circumradius, so $$\frac{BH_1}{CH_1}=\frac{\sin(90-\beta)}{\sin(90-\gamma)}.$$Then, by Ratio Lemma, $$\frac{BD}{CD}=\frac{BH_1}{CH_1}\cdot \frac{\sin \beta}{\sin\gamma}=\frac{\sin(90-\beta)\sin\beta}{\sin(90-\gamma)\sin\gamma}.$$Taking the cyclic product of this over $\alpha,\beta,\gamma$ yields 1 as everything cancels, so by Ceva's, $AD,BE,CF$ concur so we are done.
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huashiliao2020
1292 posts
#37
Y by
We look for an obvious placement for D,E,F, because this condition is not so convenient. First reflect H to obtain H_a, H_b, H_c. Then the condition is that H_aD+DO=..., so we make this a straight line by placing D as the intersection of H_aO and BC, and cyclically, so that the sum is equal to the circumradius. Now employ Ceva's to get the desired result.
We can express these lengths as law of sines. We have <H_aBD=<H_aBC=<H_aAC=90-<C, and <DH_aB=<OH_aB=90-<H_aOB/2=90-H_aAB=<B. Then BD/DH_a (by law of sines) = sinB/cosC, and an analogous argument gives CD/DH_a=sinC/cosB. Dividing, we have BD/CD=sinBcosB/sinCcosC and cyclically. Then by Ceva's we're done $\blacksquare$

extra sol for storage
This post has been edited 1 time. Last edited by huashiliao2020, Jun 12, 2023, 5:09 PM
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starchan
1610 posts
#38
Y by
inellipse!
solution
remark
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Leo.Euler
577 posts
#39
Y by
Note that by the existence of the MacBeath inconic, there is an ellipse with foci at $O$ and $H$ inscribed in $\triangle  ABC$. Take $D, E, F$ to be the tangency points of the ellipse with sides $BC, AC, AB$, respectively. By Brianchon theorem, $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ concur. By ellipse properties $OD + DH = OE + EH = OF + FH$, so we are done.

To be fair, this is more or less cheating. However, proving that the MacBeath inconic exists can be done easily using phantom point + angle chase, but I guess I am too lazy to write it up.
This post has been edited 1 time. Last edited by Leo.Euler, Aug 20, 2023, 10:55 PM
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peace09
5419 posts
#40
Y by
If $H_A,H_B,H_C$ are the reflections of $H$ about $BC,CA,AB$, $(D,E,F):=(OH_A\cap BC,OH_B\cap CA,OH_C\cap AB)$ satisfy the length condition with a constant sum of $R_{(ABC)}$. We prove concurrency by Ceva's: by the Law of Sines,
\begin{align*}
BD=\sin\angle BOH_A\cdot\tfrac{OD}{\sin\angle OBC}&=OD\cdot\tfrac{\cos2B}{\cos A}\\
CD=\sin\angle COH_A\cdot\tfrac{OD}{\sin\angle OCB}&=OD\cdot\tfrac{\cos2C}{\cos A}.\\
\end{align*}So $\tfrac{BD}{CD}=\tfrac{\cos2B}{\cos2C}$ and analogously $(\tfrac{CE}{AE},\tfrac{AF}{BF})=(\tfrac{\cos2C}{\cos2A},\tfrac{\cos2A}{\cos2B})$, whence multiplying gives $\tfrac{BD\cdot CE\cdot AF}{CD\cdot AE\cdot BF}=1$ as desired. $\square$
This post has been edited 2 times. Last edited by peace09, Aug 29, 2023, 1:58 PM
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peppapig_
280 posts
#41
Y by
Let $A'$ be the reflection of $H$ across $BC$ and let $D$ be $OA'\cap BC$. Construct $E$ and $F$ similarly. Note that by orthocenter properties, we have that $A'\in (ABC)$, so we have that
\[OD+DH=OD+DA'=OA'=R,\]and similarly, we also find that $OE+EH=OF+FH=R$ also. Now we must prove that the $AD$, $BE$, and $CF$ are concurrent.

Through angle chasing, we get that
\[\angle DHA'=\angle OAH=\angle BAC-\angle BAH-\angle OAC=a-(90-b)-(90-b)=180-(2b+a)=b-c,\]by orthocenter properties. Additionally, we also get that $\angle BCH=90-a$, which gives that
\[\angle DHC=90-\angle DHA'-\angle DCH=c.\]Now using Law of Sines, we get that
\[BD=DH*\frac{\sin b}{\cos c},\]and
\[CD=DH*\frac{\sin c}{\cos b},\]which gives that
\[\frac{BD}{CD}=\frac{\sin b\cos b}{\sin c\cos c}=\frac{2\sin b\cos b}{2\sin c\cos c}=\frac{\sin 2b}{\sin 2c}.\]Similarly, we get that $\frac{CE}{EA}=\frac{\sin 2c}{\sin 2a}$ and $\frac{AF}{FB}=\frac{\sin 2a}{\sin 2b}$. This gives us that
\[\frac{BD}{CD}*\frac{CE}{EA}*\frac{AF}{FB}=1,\]which means that $AD$, $BE$, and $CF$ are concurrent by Ceva's Theorem, finishing the problem.
This post has been edited 1 time. Last edited by peppapig_, Jan 14, 2024, 6:52 PM
Reason: Is it concurrent or collinear v2
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dolphinday
1328 posts
#42
Y by
Note that $D$, $E$, and $F$ must be the tangency points of $\triangle ABC$ and the Macbeath ellipse, since the ellipse has foci $O$ and $H$. Then Brianchon's finishes.
This post has been edited 2 times. Last edited by dolphinday, Jul 2, 2024, 2:53 PM
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shendrew7
799 posts
#43
Y by
We claim finding the smallest possible value of the quantity $OD+DH$ works. It's a classic result that this point $D$ is just $OH_A \cap BC$, where $H_A$ is the reflection of $H$ over $BC$, so
\[OD+DH = OD+DH_A = R,\]
which is obviously cyclically symmetric. Ceva's gives confirms the final requirement, as
\[\prod \frac{BD}{DC} = \prod \frac{BH_A \cdot \sin \angle BH_AD}{CH_A \cdot \sin \angle CH_AD} = \prod \frac{\sin 
B \cos B}{\sin C \cos C} = 1. \quad \blacksquare\]
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Markas
150 posts
#44
Y by
Let $(ABC) = \Gamma$. Let $AH \cap \Gamma = A_1$, $BH \cap \Gamma = B_1$ and $CH \cap \Gamma = C_1$. Obviously $A_1$ is a reflection of H over BC, because of famous orthocenter property, similarly for $B_1$ and $C_1$. Now let $OA_1 \cap BC = D$, $OB_1 \cap AC = E$ and $OC_1 \cap AB = F$. We will show that the D, E and F we defined satisfy the condition of the problem. Firstly $DO + DH = DO + DA_1 = OA_1 = R$, similarly $EO + EH = EO + EB_1 = OB_1 = R$ and $FO + FH = FO + FC_1 = OC_1 = R$ $\Rightarrow$ DO + DH = EO + EH = FO + FH = R. So it is only left to show that AD, BE and CF are concurrent. Now using the ratio lemma $\frac{AF}{FB} = \frac{AC_1}{BC_1} \cdot \frac{\sin \angle AC_1F}{\sin \angle BC_1F} = \frac{\sin \alpha \cdot \cos \alpha}{\sin \beta \cdot \cos \beta}$, similarly $\frac{BD}{DC} = \frac{BA_1}{CA_1} \cdot \frac{\sin \angle BA_1D}{\sin \angle CA_1D} = \frac{\sin \beta \cdot \cos \beta}{\sin \gamma \cdot \cos \gamma}$ and $\frac{CE}{EA} = \frac{CB_1}{EB_1} \cdot \frac{\sin \angle CD_1E}{\sin \angle AD_1E} = \frac{\sin \gamma \cdot \cos \gamma}{\sin \alpha\cdot \cos \alpha}$ $\Rightarrow$ by Ceva's $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} =  \frac{\sin \alpha \cdot \cos \alpha}{\sin \beta \cdot \cos \beta} \cdot \frac{\sin \beta \cdot \cos \beta}{\sin \gamma \cdot \cos \gamma} \cdot \frac{\sin \gamma \cdot \cos \gamma}{\sin \alpha\cdot \cos \alpha} = 1$ it follows that AD, BE and CF are concurrent, which was what we needed to show $\Rightarrow$ the D, E and F we choose work and we are ready.
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cursed_tangent1434
639 posts
#45
Y by
Pretty interesting problem. Actually the proof of the problem generalizes for an arbitrary pair of isogonal conjugates $P$ and $Q$ in $\triangle ABC$. We define by $H_a$ , $H_b$ and $H_c$ the reflections of the orthocenter $H$ across sides $BC$ , $AC$ and $AB$ of $\triangle ABC$. Then, let $D$ , $E$ and $F$ be the intersections of $\overline{OH_a}$ , $\overline{OH_b}$ and $\overline{OH_c}$ with sides $BC$ , $AC$ and $AB$ respectively. The following is the pith of the problem.

Claim : Points $D$ , $E$ and $F$ lie on an inellipse $\mathcal E$ of $\triangle ABC$ with foci $O$ and $H$.

Proof : First note that since $H_a$ is the reflection of the orthocenter $H$ across side $BC$ , $DH_a=DH$. Thus,
\[OD+DH = OD + DH_a = R \]where $R$ is the circumradius of $\triangle ABC$. Similarly we can compute that,
\[OD+DH = OE + EH = OF + FH = R\]which implies that $D$ , $E$ and $F$ lie on an ellipse $\mathcal E$ with foci $O$ and $H$. Next, note that due to the reflection,
\[\measuredangle ODC = \measuredangle H_a DB = \measuredangle BDH\]which implies that side $BC$ is in fact tangent to $\mathcal E$ at $D$. A similar argument shows that $\mathcal E$ is also tangent to $\triangle ABC$ on sides $AB$ and $AC$ at points $F$ and $E$ respectively, which proves the claim.

We finish by simply noting that Brianchon's Theorem on $AFBDCE$ implies that $AD, BE$ and $CF$ concur, as desired.
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cj13609517288
1922 posts
#46
Y by
????????????
Since $O$ and $H$ are isogonal conjugates, there is an ellipse with them as foci that is tangent to all three sides of the triangle. Then Brianchon finishes.
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Mapism
19 posts
#47
Y by
Let the $A,B,C$ altitudes respectively intersect the sides and the circumcircle at $K,L,M$ and $P,Q,R$. Take $D=OP\cap BC, E=OQ\cap CA, F=OR\cap AB$. These clearly satisfy the length conditions since $OD+DH=OD+DP=r$ which is constant, where $r$ is the circumradius.

To prove that $AD,BE,CF$ concur we use Ceva's theorem.

First notice $PK$ and $PD$ are isogonal lines wrt $\angle BPC$. If we let $M$ be the intersection of $OP$ with the circumcircle we have,
$$\angle PAM=90 \implies AM\parallel BC \implies \angle MPC=\angle MAC=\angle ACB=\angle BPA$$which implies the isogonality. Now we can use Steiner's Ratio theorem and the law of sines to compute $\frac{DB}{DC},$
$$\frac{HB^2}{HC^2}=\frac{PB^2}{PC^2}=\frac{KB}{KC}\frac{DB}{DC}\implies \frac{DB}{DC}=\frac{HB^2}{HC^2}\frac{KC}{KB}=\frac{HB^2}{HC^2}\frac{tan\angle ABC}{tan\angle BCA}$$Applying a similar argument to the other points we obtain
$$\frac{DB}{DC}\frac{EC}{EA}\frac{FA}{FB}=\frac{HB^2}{HC^2}\frac{HC^2}{HA^2}\frac{HA^2}{HB^2}\frac{tan\angle ABC}{tan\angle BCA}\frac{tan\angle BCA}{tan\angle CAB}\frac{tan\angle CAB}{tan\angle ABC}=1$$and we're done.
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Yiyj1
1266 posts
#48
Y by
sayantanchakraborty wrote:
orl wrote:
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\]and the lines $AD$, $BE$, and $CF$ are concurrent.

This problem also appeared in one of the Indian TSTs!!!

yea. IMO Shortlist problems are not plublicized until the year after the actual imo, so some countries use shortlisted problems from the year before on their tst :)
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Maximilian113
575 posts
#49
Y by
Let $A', B', C'$ be the reflections of $H$ over $BC, AC, AB$ respectively. Then let $D=OA' \cap BC,$ and define $E, F$ similarly. Then $OD+DH=OE+EH=OF+FH=R.$

It now suffices to show that $AD, BE, CF$ are concurrent. By the Sine Law/Ratio Lemma, $$\frac{BD}{DC} = \frac{\sin \angle BOD}{\sin \angle COD} = \frac{\sin \angle A'OB}{\sin \angle A'OC},$$and a similar expression holds for the other ratios $\frac{CE}{EA}, \frac{AF}{FB}.$ But, $$\angle A'OB=2\angle A'CB = 2\angle HCB = 2\angle HAB = 2\angle C'AB = \angle C'OB,$$so $$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$$and we are done by Ceva's Theorem. QED
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Avron
37 posts
#50
Y by
Let $H_A$ be the reflection of $H$ over $BC$ and let the lines $OH_A$, $BC$ intersect at $D$. Clearly $OD+DH=OD+DH_A=R$. Define $E,F$ similarly, now the first condition is satisfied so we only need to show that they are concurrent.

By a quick angle chase we get that $\angle BH_AD=\angle B, \angle CH_AD=\angle C$, so by the law of sines
\[
\frac{BD}{CD}=\frac{\sin \angle B \cdot DH_A \cdot \sin \angle BDH_A}{\sin \angle C \cdot DH_A \cdot \sin 180-\angle BDH_A}=\frac{\sin \angle B}{\sin \angle C}
\]deriving the other ratios similarly we get that
\[
\frac{BC \cdot CE \cdot AF}{DC \cdot EA \cdot FB}=\frac{\sin \angle A \cdot \sin \angle B \cdot \sin \angle C}{\sin \angle A \cdot \sin \angle B \cdot \sin \angle C}=1
\]so we're done by Ceva.
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