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Contests & Programs AMC and other contests, summer programs, etc.
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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
Day Before Tips
elasticwealth   59
N a few seconds ago by elasticwealth
Hi Everyone,

USA(J)MO is tomorrow. I am a Junior, so this is my last chance. I made USAMO by ZERO points but I've actually been studying oly seriously since JMO last year. I am more stressed than I was before AMC/AIME because I feel Olympiad is more unpredictable and harder to prepare for. I am fairly confident in my ability to solve 1/4 but whether I can solve the rest really leans on the topic distribution.

Anyway, I'm just super stressed and not sure what to do. All tips are welcome!

Thanks everyone! Good luck tomorrow!
59 replies
+1 w
elasticwealth
Mar 19, 2025
elasticwealth
a few seconds ago
Prove a polynomial has a nonreal root
KevinYang2.71   29
N an hour ago by ihatemath123
Source: USAMO 2025/2
Let $n$ and $k$ be positive integers with $k<n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$ is zero. Prove that $P(x)$ has a nonreal root.
29 replies
KevinYang2.71
Yesterday at 12:00 PM
ihatemath123
an hour ago
F=ma USAPhO qualification
RabtejKalra   2
N an hour ago by Alex-131
Which would be more benificial to USAPhO qualification, doing the AoPS physics courses (Intro to Physics all the way to F=ma Problem Series) or doing the first half of HRK?
2 replies
RabtejKalra
an hour ago
Alex-131
an hour ago
USA Canada math camp
Bread10   17
N an hour ago by eugenewang1
How difficult is it to get into USA Canada math camp? What should be expected from an accepted applicant in terms of the qualifying quiz, essays and other awards or math context?
17 replies
3 viewing
Bread10
Mar 2, 2025
eugenewang1
an hour ago
No more topics!
Erecting Rectangles
franchester   101
N Mar 15, 2025 by Ilikeminecraft
Source: 2021 USAMO Problem 1/2021 USAJMO Problem 2
Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that \[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent.
101 replies
franchester
Apr 15, 2021
Ilikeminecraft
Mar 15, 2025
Erecting Rectangles
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 USAMO Problem 1/2021 USAJMO Problem 2
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lian_the_noob12
173 posts
#101
Y by
$\color{magenta} \boxed{\textbf{SOLUTION P1}}$
$\color{red} \textbf{Geo Marabot Solve 8}$

The angle condition implies the circumcircles of the three rectangles intersect at some point $P$
We have, $\angle B_2PC+\angle A_1PC=90+90=180$ and similar for other three we are done that the lines concur at $P\blacksquare$
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peace09
5411 posts
#102
Y by
Claim. The rectangles' circumcircles concur.
Proof. Let $(BCC_1B_2)$ and $(CAA_1C_2)$ meet again at $O$. Note $\sum_{\text{cyc}}\angle BOC=360^\circ$ and \[(\angle BOC,\angle COA)=(180^\circ-\angle BC_1C,180^\circ-\angle CA_1A),\]whence the angle condition gives $\angle AOB=180^\circ-\angle AB_1B$. Then $O\in(ABB_1A_2)$ as claimed.

The rest is easy: $\angle APB_1=\angle APC_2=90^\circ$ and so $O\in B_1C_2$ and so on. In all, $O$ is the desired point of concurrency. $\square$
This post has been edited 1 time. Last edited by peace09, Feb 25, 2025, 8:43 PM
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HamstPan38825
8855 posts
#103
Y by
Let $P = (ABB_1A_2) \cap (AC_1CA)$; the given angle condition then yields that $P$ lies on $(BCB_2C_1)$ too. It follows that $\angle BPC_1 +\angle BPA_2 = 90^\circ + 90^\circ = 180^\circ$, so $P$ lies on $\overline{A_2C_1}$ and cyclic permutations.
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setts
3 posts
#105
Y by
This problem is trivial when you notice that

<ABD = 180 - <AB_1 B
<BDC = 360 - <ADB - <ADC.
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alsk
28 posts
#106
Y by
Slightly different finish.

If we let $A \neq P = (ABB_1A_2) \cap (AC_1C_2A_1)$, the given angle condition gives us that $P \in (BCC_1B_2)$, so all three circumcircles of the rectangles concur.

Furthermore, notice that $BA_2, BC_1$ are diameters of their respective circles, and both circles go through $B, P$. Thus, $A_2C_1$ goes through $P$ (can be shown by similar triangles with circumcenters), and we can use the same logic to find that $P$ lies on all three lines.
This post has been edited 1 time. Last edited by alsk, Mar 12, 2024, 2:09 AM
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Mr.Sharkman
487 posts
#107
Y by
Solution
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xHypotenuse
739 posts
#108
Y by
redacted
This post has been edited 1 time. Last edited by xHypotenuse, Apr 23, 2024, 6:01 PM
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Markas
105 posts
#109
Y by
Let $\angle BC_1C = \alpha$, $\angle CA_1A = \beta$ and $\angle AB_1B = 180 - \alpha - \beta$. Every rectangle is a cyclic quadrilateral. Lets draw the circumcircles around the rectangles $BCC_1B_2$, $CAA_1C_2$, $ABB_1A_2$ and name them $\omega$, $\omega_1$ and $\omega_2$, respectively.

Let $\omega$ $\cap$ $\omega_1$ = P, where P is a point other than C.

Claim: P $\in$ $\omega_2$

Proof: $\angle APB = 360 - \angle APC - \angle BPC$. Since $A_1APC$ is cyclic, $\angle APC = 180 - \angle AA_1C = 180 - \beta$. Since $BC_1CP$ is cyclic, $\angle BPC = 180 - \angle BC_1C = 180 - \alpha$ $\Rightarrow$ $\angle APB = 360 - (180 - \beta) - (180 - \alpha) = \alpha + \beta$. Now $\angle APB + \angle AB_1B = \alpha + \beta + 180 - \alpha - \beta = 180$ $\Rightarrow$ $AB_1BP$ is cyclic $\Rightarrow$ P $\in$ $\omega_2$.

Now we want to show that P lies on $B_1C_2$, $A_2C_1$ and $A_1B_2$. We have $\angle A_1PC + \angle CPB_2 = 90^\circ + 90^\circ = 180^\circ$, since $A_1C$ is a diameter in $\omega$ and $B_2C$ is a diameter in $\omega_1$ $\Rightarrow$ P lies on $A_1B_2$. Also $\angle B_1PA + \angle APC_2 =  90^\circ + 90^\circ = 180^\circ$, since $AB_1$ is a diameter in $\omega_2$ and $AC_2$ is a diameter in $\omega$ $\Rightarrow$ P lies on $B_1C_2$. Finally $\angle A_2PB + \angle BPC_1 = 90^\circ + 90^\circ = 180^\circ$, since $A_2B$ is a diameter in $\omega_2$ and $BC_1$ is a diameter in $\omega_1$ $\Rightarrow$ P lies on $A_2C_1$. This means that $B_1C_2$, $A_2C_1$ and $A_1B_2$ are concurrent at point P $\Rightarrow$ we are ready.
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shendrew7
792 posts
#110
Y by
First notice the circumcircles of the rectangles are forced to meet at a single point, say $K$, from the angle condition. Then we have
\[\angle B_1KX_2 = \angle B_1KA + \angle AKC_2 = \angle AA_2B_1 + \angle AA_1C_2 = 90 + 90 = 180,\]
so $X$ lies on $B_1C_2$, and analogously the other two lines. $\blacksquare$
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eg4334
605 posts
#111
Y by
Immediately constructed circumcircles for potential radical axis, turned out to be a lot different.

Let $(AA_2B_1B) \cap (BB_2C_1C) = D$. $$\angle ADC = 360 - \angle ADB - \angle CDB  = \angle AB_1B + \angle CC_1B = 180 - \angle AA_1C$$so $A_1ADC$ is cyclic and the three circumcircles of the rectangles concur. Also, $B_1DC_2$ are collinear and cyclic variations hold by angle chasing: $$\angle B_1DC_2 = \angle B_1DA + \angle ADC_2 = 180$$so done.
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trolled767
3 posts
#112
Y by
Let the circumcircles of rectangles intersect at a point O.
Proof:
let circles ABB1A2 and AA1CC2 intersect at point O,
angle BOC is 360 - 180 - 180 + angleCA1A + angle AB1B
which = angle BC1C (mod 180)
Now trivially angle BOA2 = 90; and so is angle BOC1, cyclically all those lines pass through O.
hence proved
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Saucepan_man02
1299 posts
#113
Y by
(Easy P1 tho):

Let $P$ denote the intersection of $(ABB_1A_2), (BCC_1B_2)$. We will prove that $P$ lies on $(AA_1C_2C)$.
Notice that: $\angle BPC = 180^\circ - \angle B C_1 C$ and $\angle APB = 180^\circ - \angle A B_1 B$ which forces $\angle APC = 180^\circ - \angle A A_1 C$ and therefore $P$ lies on $(AA_1C_2C)$.

Notice that: $CP \perp A_1P$ and $CP \perp B_2P$. Thus, $A_1 B_2$ passes through point $P$. Similarly, the other two lines pass through $P$ and we are done.
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Mathandski
715 posts
#114
Y by
Note: This was transcribed by ChatGPT so there may be typoes

Let \( BCC_1B_2 \) and \( CAA_1C_2 \) meet at \( P \). Let the intersections of \( \overline{A_2 C} \cap \overline{CA_1} = O_B \) and \( \overline{B C_1} \cap \overline{C B_2} = O_A \).

Since \( C A A_1 C_2 \), \( A B B_1 A_2 \) are rectangles, we have \( O_B, O_C \) are the centers of the respective circumcircles.

We use directed angles to show \( P \in \odot(A B B_1 A_2) \).

Since \( \angle B C_1 C \), \( \angle C_1 A A \), and \( \angle A B_1 B \) are in the same direction as each other as the rectangles are all erected outside \( \triangle A B C \),

\[
\angle B C_1 C + \angle C A_1 A + \angle A B_1 B = 180^\circ
\]
\[
\Rightarrow \measuredangle B C_1 C + \measuredangle C A_1 A + \measuredangle A B_1 B = 0
\]
\[
\Rightarrow \measuredangle B P C + \angle C P A + \measuredangle A B_1 B = 0
\]
\[
\Rightarrow \measuredangle B P A + \measuredangle A B_1 B = 0
\]
\[
\Rightarrow \measuredangle B P A = \measuredangle B B_1 A \Rightarrow (A P B B_1) \text{ cyclic.}
\]
It now suffices to show \( P \in \overline(A_1 B_2) \) as the other two follow analogously.

Indeed, the midpoint \( M \) of \( \overline{C P} \) satisfies \( M \in \overline{O_A O_B} \) since \( \overline{C P} \) is the shared chord of \( \odot(B C C_1 B_2) \) and \( \odot(C A A_1 C_2) \).

A homothety at \( C \) of scale \( 2 \) suffices.
This post has been edited 2 times. Last edited by Mathandski, Feb 28, 2025, 12:19 AM
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bjump
973 posts
#116 • 2 Y
Y by KenWuMath, imagien_bad
Consider drawing the circles $(ABB_1A_2)$, $(BCC_1B_2)$, $(CAA_1C_2)$. Let $(ABB_1A_2) \cap  (BCC_1B_2) = X_{\text{onk}}$. Now we have
$$\angle AX_{\text{onk}}B = 360^{\circ}- \angle BX_{\text{onk}}C - \angle CX_{\text{onk}}A = \angle AB_1B + \angle BC_1C = 180^\circ - \angle AA_1C  $$Therefore the $3$ drawn circles concur at $X_{\text{onk}}$.

Now the line between the centers of $(AA_2B_1B)$, and $(AA_1C_2C)$, intersects $AX_{\text{onk}}$ at the midpoint of $AX_{\text{onk}}$ since $B_1$, and $C_2$, are the reflections of $A$ over the centers of the circles respectively. By homothety $X_{\text{onk}} \in B_1C_2$. By a symmetry all the desired lines concur at $X_{\text{onk}}$.
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Ilikeminecraft
300 posts
#117
Y by
pretty easy???

Let $P = (ABB_1A_2)\cap (AA_1C_2C).$ We have that $\angle BPC = 360 - \angle BPA - \angle APC = \angle AB_1B + \angle CA_1A = 180 - \angle BC_1C,$ so all 3 circles concur at $P$.
Thus, $\angle APC = \angle APB + \angle BPC_1 = 180$
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