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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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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Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Shortest number theory you might've seen in your life
AlperenINAN   13
N 3 minutes ago by lksb
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
13 replies
+2 w
AlperenINAN
May 11, 2025
lksb
3 minutes ago
n-term Sequence
MithsApprentice   15
N 34 minutes ago by Ilikeminecraft
Source: USAMO 1996, Problem 4
An $n$-term sequence $(x_1, x_2, \ldots, x_n)$ in which each term is either 0 or 1 is called a binary sequence of length $n$. Let $a_n$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to 0, 1, 0 in that order. Let $b_n$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that $b_{n+1} = 2a_n$ for all positive integers $n$.
15 replies
MithsApprentice
Oct 22, 2005
Ilikeminecraft
34 minutes ago
Drawing Triangles Against Your Clone
pieater314159   19
N an hour ago by Ilikeminecraft
Source: 2019 ELMO Shortlist C1
Elmo and Elmo's clone are playing a game. Initially, $n\geq 3$ points are given on a circle. On a player's turn, that player must draw a triangle using three unused points as vertices, without creating any crossing edges. The first player who cannot move loses. If Elmo's clone goes first and players alternate turns, who wins? (Your answer may be in terms of $n$.)

Proposed by Milan Haiman
19 replies
pieater314159
Jun 27, 2019
Ilikeminecraft
an hour ago
Odd digit multiplication
JuanDelPan   12
N an hour ago by Ilikeminecraft
Source: Pan-American Girls' Mathematical Olympiad 2021, P4
Lucía multiplies some positive one-digit numbers (not necessarily distinct) and obtains a number $n$ greater than 10. Then, she multiplies all the digits of $n$ and obtains an odd number. Find all possible values of the units digit of $n$.

$\textit{Proposed by Pablo Serrano, Ecuador}$
12 replies
JuanDelPan
Oct 6, 2021
Ilikeminecraft
an hour ago
No more topics!
Constructing triangles holding many similarities
WakeUp   4
N Apr 15, 2025 by Nari_Tom
Source: Baltic Way 2011
Let $E$ be an interior point of the convex quadrilateral $ABCD$. Construct triangles $\triangle ABF,\triangle BCG,\triangle CDH$ and $\triangle DAI$ on the outside of the quadrilateral such that the similarities $\triangle ABF\sim\triangle DCE,\triangle BCG\sim \triangle ADE,\triangle CDH\sim\triangle BAE$ and $ \triangle DAI\sim\triangle CBE$ hold. Let $P,Q,R$ and $S$ be the projections of $E$ on the lines $AB,BC,CD$ and $DA$, respectively. Prove that if the quadrilateral $PQRS$ is cyclic, then
\[EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.\]
4 replies
WakeUp
Nov 6, 2011
Nari_Tom
Apr 15, 2025
Constructing triangles holding many similarities
G H J
Source: Baltic Way 2011
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WakeUp
1347 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $E$ be an interior point of the convex quadrilateral $ABCD$. Construct triangles $\triangle ABF,\triangle BCG,\triangle CDH$ and $\triangle DAI$ on the outside of the quadrilateral such that the similarities $\triangle ABF\sim\triangle DCE,\triangle BCG\sim \triangle ADE,\triangle CDH\sim\triangle BAE$ and $ \triangle DAI\sim\triangle CBE$ hold. Let $P,Q,R$ and $S$ be the projections of $E$ on the lines $AB,BC,CD$ and $DA$, respectively. Prove that if the quadrilateral $PQRS$ is cyclic, then
\[EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.\]
Z K Y
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skytin
418 posts
#2 • 2 Y
Y by Adventure10, Mango247
Hints :
Let E' is isogonal to E wrt ABC
E'BCD ~ AFEI , AFBE , IAED , EBGC , DECH are cycic
Z K Y
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KRIS17
134 posts
#3 • 1 Y
Y by Adventure10
Based on the given similarities of the triangles, it is easy to see that quadrilaterals $AFBE$ and $DECH$ are similar.
Thus, we have:
$AB/DC$ = $FE/EH$
or $AB.EH$ = $FE.DC$

Same way, since quadrilaterals $DIAE$ and $CEBG$ are similar, we get:
$EG.DA$ = $EI.BC$

It remains to be proven that:
$EG.DA$ = $AB.EH$ as the fourth eqiality follows directly form these 3 eualities.

Since quadrilateral $PQRS$ is cyclic, we get that:
$\angle RSE$ = $\angle RDE$ = $\angle BAF$ (from similarities given)
and $\angle ESP$ = $\angle EAP$
Thus, we have $\angle RSP$ = $\angle EAF$.

In the same manner we can derive that $\angle PQR$ = $\angle FBE$.

Since, $\angle RSP$ and $\angle PQR$ are supplementry, so are $\angle EAF$ and $\angle FBE$.
Thus, quadrilateral $AFBE$ is cyclic. Since quadrilaterals $AFBE$ and $DECH$ are similar, $DECH$ is cyclic too.
In the same manner, we can derive that, quadrilaterals $DIAE$ and $CEBG$ are cyclic too.

Applying Ptolemy's theorem to $AFBE$, we get:
$AB.EF$ = $AF.EB$ + $FB.AE$
or $AB$ = $(AF/EF).EB + (FB/EF).AE$ = $(DE/EH).EB + (EC/EH).AE$ (from similarity of quadrilaterals $AFBE$ and $DECH$)
or $AB.EH$ = $DE.EB$ + $EC.AE ---> (1)$

In a similar manner, by applying Ptolemy's theorem to $DIAE$, we get:
$EG.AD$ = $DE.EB$ + $EC.AE ---> (2)$

From $(1)$ and $(2)$, we have $EG.DA$ = $AB.EH$ which completes the proof.
This post has been edited 1 time. Last edited by KRIS17, Aug 18, 2019, 10:03 PM
Z K Y
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rafaello
1079 posts
#4
Y by
Claim. $ABEF$, $BECG$, $CEDH$ and $AEDI$ are cyclic.
Proof. We have
\begin{align*}\measuredangle IAE&=\measuredangle SAE+\measuredangle EBQ\\&=\measuredangle SPE+\measuredangle EPQ\\&=\measuredangle SPQ\\&=\measuredangle SRQ\\&=\measuredangle SRE+\measuredangle ERQ\\&=\measuredangle SDE+\measuredangle ECQ\\&=\measuredangle SDE+\measuredangle IDA\\&=\measuredangle IDE,
\end{align*}we do the same for others, the claim follows.

Claim.
There exists $T$, the isogonal conjugate of $E$ wrt $ABCD$.
Proof. By six-point circle theorem, we conclude that there exists $T$, the isogonal conjugate of $E$.

Now, note that angle chasing implies that $\triangle GBE\sim\triangle CTD$ and $\triangle FBE\sim\triangle ATD$, thus
$$\frac{EF}{EG}=\frac{AD}{DC}.$$Similarly, $\dfrac{EG}{EH}=\dfrac{AB}{AD}$, $\dfrac{EH}{EI}=\dfrac{BC}{AB}$ and $\dfrac{EI}{EF}=\dfrac{CD}{BC}$.
This yields that
$$EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.$$
Z K Y
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Nari_Tom
117 posts
#5
Y by
Here is a solution using spiral similarity and avoiding isogonal conjugates.

$EF \cdot CD= EH \cdot AB$ is clear since there is similar quadrilaterals. So let's prove that $EF \cdot CD=EG \cdot DA$.

We have $\angle FEG=\angle FEB+\angle BEG=\angle FAB+\angle BCG= \angle EDC+\angle EDA=\angle ADC$. Now let's consider $S$, the center of the spiral similarity that takes $AB$ to $CD$. So it takes $F$ to $E$. We have $\frac{EF}{ES}=\frac{AD}{DS}$.
Since $S$ is also the center of the similarity that takes $AD$ to $BC$, we have $\frac{EG}{ES}=\frac{CD}{DS}$.
Combining these two relations we get $\frac{AD}{CD}=\frac{EF}{EG}$, So $\triangle FEG \sim \triangle ADC$. Conclusion follows.
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