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9 Did I make the right choice?
Martin2001   33
N Today at 2:11 AM by happypi31415
If you were in 8th grade, would you rather go to MOP or mc nats? I chose to study the former more and got in so was wondering if that was valid given that I'll never make mc nats.
33 replies
Martin2001
Apr 29, 2025
happypi31415
Today at 2:11 AM
late ross acceptance
basketball314159   4
N Today at 2:07 AM by NoSignOfTheta
Hi everyone,

I got rejected from Ross a couple weeks ago and then today got an acceptance letter, but I didn't to my knowledge get put on a waitlist. Did anyone experience something similar?

Also, for those who have done Ross, honest thoughts? I did HCSSiM and absolutely loved it, but I realize every math program is different.

Thanks!
4 replies
basketball314159
Yesterday at 12:30 AM
NoSignOfTheta
Today at 2:07 AM
Mathcounts state
happymoose666   36
N Today at 1:47 AM by happypi31415
Hi everyone,
I just have a question. I live in PA and I sadly didn't make it to nationals this year. Is PA a competitive state? I'm new into mathcounts and not sure
36 replies
happymoose666
Mar 24, 2025
happypi31415
Today at 1:47 AM
[$10K+ IN PRIZES] Poolesville Math Tournament (PVMT) 2025
qwerty123456asdfgzxcvb   10
N Today at 12:47 AM by qwerty123456asdfgzxcvb
Hi everyone!

After the resounding success of the first three years of PVMT, the Poolesville High School Math Team is excited to announce the fourth annual Poolesville High School Math Tournament (PVMT)! The PVMT team includes a MOPper and multiple USA(J)MO and AIME qualifiers!

PVMT is open to all 6th-9th graders in the country (including rising 10th graders). Students will compete in teams of up to 4 people, and each participant will take three subject tests as well as the team round. The contest is completely free, and will be held virtually on June 7, 2025, from 10:00 AM to 4:00 PM (EST).

Additionally, thanks to our sponsors, we will be awarding approximately $10K+ worth of prizes (including gift cards, Citadel merch, AoPS coupons, Wolfram licenses) to top teams and individuals. More details regarding the actual prizes will be released as we get closer to the competition date.

Further, newly for this year we might run some interesting mini-events, which we will announce closer to the competition date, such as potentially a puzzle hunt and integration bee!

If you would like to register for the competition, the registration form can be found at https://pvmt.org/register.html or https://tinyurl.com/PVMT25.

Additionally, more information about PVMT can be found at https://pvmt.org

If you have any questions not answered in the below FAQ, feel free to ask in this thread or email us at falconsdomath@gmail.com!

We look forward to your participation!

FAQ
10 replies
qwerty123456asdfgzxcvb
Apr 5, 2025
qwerty123456asdfgzxcvb
Today at 12:47 AM
USAMO Medals
YauYauFilter   9
N Yesterday at 11:35 PM by Alex-131
YauYauFilter
Apr 24, 2025
Alex-131
Yesterday at 11:35 PM
everyone will get zero marx on this
Th3Numb3rThr33   48
N Yesterday at 9:50 PM by Blast_S1
Source: JMO 2018 Problem 6
Karl starts with $n$ cards labeled $1,2,3,\dots,n$ lined up in a random order on his desk. He calls a pair $(a,b)$ of these cards swapped if $a>b$ and the card labeled $a$ is to the left of the card labeled $b$. For instance, in the sequence of cards $3,1,4,2$, there are three swapped pairs of cards, $(3,1)$, $(3,2)$, and $(4,2)$.

He picks up the card labeled 1 and inserts it back into the sequence in the opposite position: if the card labeled 1 had $i$ card to its left, then it now has $i$ cards to its right. He then picks up the card labeled $2$ and reinserts it in the same manner, and so on until he has picked up and put back each of the cards $1,2,\dots,n$ exactly once in that order. (For example, the process starting at $3,1,4,2$ would be $3,1,4,2\to 3,4,1,2\to 2,3,4,1\to 2,4,3,1\to 2,3,4,1$.)

Show that, no matter what lineup of cards Karl started with, his final lineup has the same number of swapped pairs as the starting lineup.
48 replies
Th3Numb3rThr33
Apr 19, 2018
Blast_S1
Yesterday at 9:50 PM
Find the radius of circle O
TheMaskedMagician   3
N Yesterday at 8:38 PM by fruitmonster97
Source: 1976 AHSME Problem 18
IMAGE

In the adjoining figure, $AB$ is tangent at $A$ to the circle with center $O$; point $D$ is interior to the circle; and $DB$ intersects the circle at $C$. If $BC=DC=3$, $OD=2$, and $AB=6$, then the radius of the circle is

$\textbf{(A) }3+\sqrt{3}\qquad\textbf{(B) }15/\pi\qquad\textbf{(C) }9/2\qquad\textbf{(D) }2\sqrt{6}\qquad \textbf{(E) }\sqrt{22}$
3 replies
TheMaskedMagician
May 18, 2014
fruitmonster97
Yesterday at 8:38 PM
MathPath
PatTheKing806   13
N Yesterday at 8:05 PM by Nora2021
Is anybody else going to MathPath?

I haven't gotten in. its been 3+ weeks since they said my application was done.
13 replies
PatTheKing806
Mar 24, 2025
Nora2021
Yesterday at 8:05 PM
ARMl Local 2025 Final Results
PaulDreyer   27
N Yesterday at 5:50 PM by llddmmtt1
Results, problems, and solutions are here. Congratulations to SFBA ARML / AlphaStar: Foxes and Friends and Leading Aces Academy for placing 1st and 2nd overall and to Liam Reddy (Utah Rookies) for their perfect score on the individual round and for being the only student with a perfect score to answer the tiebreaker correctly.
27 replies
PaulDreyer
Saturday at 5:25 PM
llddmmtt1
Yesterday at 5:50 PM
MathILy 2025 Decisions Thread
mysterynotfound   28
N Yesterday at 5:49 PM by mysterynotfound
Discuss your decisions here!
also share any relevant details about your decisions if you want
28 replies
mysterynotfound
Apr 21, 2025
mysterynotfound
Yesterday at 5:49 PM
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
4186 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
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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
1607 posts
#38
Y by
inellipse!
solution
remark
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Leo.Euler
577 posts
#39
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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
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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_
281 posts
#41
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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
1324 posts
#42
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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
794 posts
#43
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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
105 posts
#44
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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
616 posts
#45
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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
1906 posts
#46
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????????????
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
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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
1265 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
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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
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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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