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Cool Integral, Cooler Solution
Existing_Human1   2
N Yesterday at 11:33 AM by ysharifi
Source: https://youtu.be/YO38MCdj-GM?si=DCn6DaQTeX8RXhl0
$$\int_{0}^{\infty} \! e^{-x^2}\cos(5x) \,dx$$
Bonus points if you can do it without Feynman
2 replies
Existing_Human1
Yesterday at 2:15 AM
ysharifi
Yesterday at 11:33 AM
Putnam 2016 A1
Kent Merryfield   15
N Yesterday at 10:51 AM by anudeep
Find the smallest positive integer $j$ such that for every polynomial $p(x)$ with integer coefficients and for every integer $k,$ the integer
\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\](the $j$-th derivative of $p(x)$ at $k$) is divisible by $2016.$
15 replies
Kent Merryfield
Dec 4, 2016
anudeep
Yesterday at 10:51 AM
Determinant is 1
Entrepreneur   2
N Yesterday at 8:27 AM by Entrepreneur
If a determinant is of $n^{\text{th}}$ order, and if the constituents of its first, second, ..., $n^{\text{th}}$ rows are the first $n$ figurate numbers of the first, second, ..., $n^{\text{th}}$ orders respectively, show that it's value is $1.$
2 replies
Entrepreneur
Monday at 7:14 PM
Entrepreneur
Yesterday at 8:27 AM
Can cos(√2 t) be expressed as a polynomial in cost?
tom-nowy   1
N Yesterday at 7:17 AM by Aiden-1089
Source: Question arising while viewing https://artofproblemsolving.com/community/c51293h3562250
Can $\cos ( \sqrt{2}\,  t )$ be expressed as a polynomial in $\cos t$ with real coefficients?
1 reply
tom-nowy
Yesterday at 7:10 AM
Aiden-1089
Yesterday at 7:17 AM
36x⁴ + 12x² - 36x + 13 > 0
fxandi   2
N Yesterday at 7:14 AM by MeKnowsNothing
Prove that for any real $x \geq 0$ holds inequality $36x^4 + 12x^2 - 36x + 13 > 0.$
2 replies
fxandi
Monday at 10:02 PM
MeKnowsNothing
Yesterday at 7:14 AM
2024 Putnam A1
KevinYang2.71   20
N Yesterday at 5:50 AM by thelateone
Determine all positive integers $n$ for which there exists positive integers $a$, $b$, and $c$ satisfying
\[
2a^n+3b^n=4c^n.
\]
20 replies
KevinYang2.71
Dec 10, 2024
thelateone
Yesterday at 5:50 AM
2025 OMOUS Problem 4
enter16180   2
N Monday at 8:57 PM by Acridian9
Source: Open Mathematical Olympiad for University Students (OMOUS-2025)
Find all matrices $M \in M_{n}(\mathbb{C})$ such that following equality holds

$$
\operatorname{rank}(M)+\operatorname{rank}\left(M^{2023}-M^{2025}\right)=\operatorname{rank}\left(M-M^{2}\right)+\operatorname{rank}\left(M^{2023}+M^{2024}\right)
$$
2 replies
enter16180
Apr 18, 2025
Acridian9
Monday at 8:57 PM
Poker hand
Aksudon   1
N Monday at 6:32 PM by lucaminiati
Problem: In a standard 52-card deck, how many different five-card poker hands are there of 'two pairs'?

Can someone please explain what is logically wrong with the following solution? (It gives double of the right solution which supposed to be 123552).

13\binom{4}{2}*12\binom{4}{2}*44=247104

Thanks
1 reply
Aksudon
Monday at 5:14 PM
lucaminiati
Monday at 6:32 PM
Sequence with GCD involved
mathematics2004   3
N Monday at 5:54 PM by anudeep
Source: 2021 Simon Marais, A2
Define the sequence of integers $a_1, a_2, a_3, \ldots$ by $a_1 = 1$, and
\[ a_{n+1} = \left(n+1-\gcd(a_n,n) \right) \times a_n \]for all integers $n \ge 1$.
Prove that $\frac{a_{n+1}}{a_n}=n$ if and only if $n$ is prime or $n=1$.
Here $\gcd(s,t)$ denotes the greatest common divisor of $s$ and $t$.
3 replies
mathematics2004
Nov 2, 2021
anudeep
Monday at 5:54 PM
Putnam 2000 B2
ahaanomegas   20
N Monday at 5:05 PM by reni_wee
Prove that the expression \[ \dfrac {\text {gcd}(m, n)}{n} \dbinom {n}{m} \] is an integer for all pairs of integers $ n \ge m \ge 1 $.
20 replies
ahaanomegas
Sep 6, 2011
reni_wee
Monday at 5:05 PM
Collinearity with orthocenter
liberator   180
N Apr 26, 2025 by joshualiu315
Source: IMO 2013 Problem 4
Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand
180 replies
liberator
Jan 4, 2016
joshualiu315
Apr 26, 2025
Collinearity with orthocenter
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G H BBookmark kLocked kLocked NReply
Source: IMO 2013 Problem 4
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peace09
5419 posts
#184
Y by
peace09 wrote:
Click to reveal hidden text
Silly. Here's an actual solution: let $P$ be the Miquel point of $(AMN)$, $(BNW)$, and $(CMW)$. Clearly $A$ is the radical center of $(BCMN)$, $(BNPW)$, and $(CMPW)$; in particular $A,P,W$ are collinear. Then
  • $HP\perp AW$ since $P\in(AMN)$ of diameter $AH$,
  • $XP\perp AW$ since $P\in(BNW)$ of diameter $WX$, and
  • $YP\perp AW$ since $P\in(CMW)$ of diameter $WY$;
and so $H,X,Y$ are collinear. $\square$
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kinnikuma
9 posts
#185
Y by
Consider $D = (BNW) \cap (WMC)$, and let us demonstrate how powerful this point is.

- $X, D, Y$ are collinear : it's because $\angle XDY = \angle XDW + \angle WDY = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.
- $A, D, W$ are collinear : we just need to show that $A$ lies on the radical axis of $(BNM)$ and $(WMC)$. That's true because $AN \cdot AB = AM \cdot AC$ by power of point (here, $B, N, M, C$ are concyclic).

Hence $\angle ADH = \angle HDW = \frac{\pi}{2}$. Consider finally $(AHD)$ and $(HDW)$. If we show that $(XY)$ is their radical axis than we are done. Let us remember that a radical axis is the line perpendicular to the line formed by the centers, and, if a common point exist between the circles, passing through this common point. Here, we already showed that $(XY)$ contains $D$, a common point between the circles. In addition, it is perpendicular to $(AW)$, which is parallel to the line formed by the centers (why : the centers, since there are only right triangles, are midpoints, so the parallelism is not mysterious anymore, by Thales) $\huge \blacksquare$.
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fasttrust_12-mn
118 posts
#186
Y by
let $\omega_3$ be the circumcircle of $\triangle ANM$ donate $E$ be the miquel point now we have that $A-E-W$ as a fast result we get $X-E-H-Y$



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MagicalToaster53
159 posts
#187
Y by
We use directed angles $\measuredangle$ modulo $\pi$. Let $Z$ be the second point of intersection between $\omega_1$ and $\omega_2$. Then observe that $X, Z, Y$ are collinear as $\measuredangle XZW = \measuredangle YZW = 90^{\circ}$. Let $D$ be the point where the altitude from $A$ meets $\overline{BC}$. Then make the observation that $BX \parallel HD$, as $\measuredangle XBW = \measuredangle 90^{\circ} = \measuredangle HDW$. We now make a claim:

Claim: $X, H, Z$ are collinear.
Proof: We have the following angle chase: \[\measuredangle BXH = \measuredangle DHZ = \measuredangle DWZ = \measuredangle BWZ = \measuredangle BXZ. \square\]
Hence as $XZ$ is the same line passing simultaneously through both $H$ and $Y$, we find that $X, H, Z, Y$ are collinear, as desired. $\blacksquare$
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ezpotd
1263 posts
#188
Y by
Draw in $(BWN) \cap (CWM) = K$, then Miquel tells us $(AMNHK)$ is cyclic. Direct angles. We then see $\angle NKX = \angle NBX = 90 - \angle ABC = \angle NAH = \angle NKH$, so $K,H,X$ are collinear, and so are $K,H,Y$, finishing.
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legogubbe
19 posts
#189
Y by
Let $Z$ be the second intersection point of $(BWN)$ and $(CWN)$.

Claim 1. Points $X$, $Y$ and $Z$ are collinear.
Proof. We see by constructions that $BXZW$ and $CYZW$ are cyclic. This implies
\[ \measuredangle WZX = \measuredangle WBX = 90^{\circ} = \measuredangle WCY = \measuredangle WZY, \]which means $X$, $Y$ and $Z$ are indeed collinear.

Claim 2. Quadrilateral $AHZM$ is cyclic.
Proof. By Miquel's theorem, $ANZM$ is cyclic. However, by the properties of the orthocenter, $ANHM$ is cyclic. From this we deduce that in fact, $ANHZM$ is cyclic, thereby $AHZM$ is cyclic.

Let $K$ be the intersection point of lines $AH$ and $BC$.

Claim 3. Quadrilateral $KHZW$ is cyclic.
Proof. From cyclic quadrilaterals,
\[ \measuredangle KHZ = \measuredangle AHZ = \measuredangle AMZ = \measuredangle CMZ = \measuredangle CWZ = \measuredangle KWZ, \]proving our claim.

Finally, $\measuredangle WZH = \measuredangle WKH = \measuredangle CKA = 90^{\circ}$. Hence $X$, $Y$, $Z$ and $H$ are collinear.
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cursed_tangent1434
622 posts
#190
Y by
We let $T$ denote the second intersection of circles $(BNW)$ and $(CMW)$. We start ff by noting that since,
\[\measuredangle WTX = \measuredangle WBX = \frac{\pi}{2}\]and
\[\measuredangle WTY = \measuredangle WCY = \frac{\pi}{2}\]which implies that points $X$ , $T$ and $Y$ are collinear. Next, by Radical Center Theorem on $(BNTW)$ , $(CMTW)$ and $(BNMC)$, lines $\overline{BN}$ , $\overline{TW}$ and $\overline{MC}$ concur. It is clear that $A= BN \cap CM$, so
\[AT \cdot AW = AN \cdot AB = AH \cdot AD\]which implies that $DHTW$ is also cyclic. Now this finishes since,
\[\measuredangle WTH = \measuredangle WDH = \frac{\pi}{2} = \measuredangle WTX\]which implies that $X$ , $H$ and $T$ are collinear. Combining this with our previous observation concludes that points $X$ , $Y$ and $H$ are collinear, as desired.
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Saucepan_man02
1331 posts
#191
Y by
Storage
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cj13609517288
1908 posts
#192 • 1 Y
Y by peace09
Let $Z$ be the other intersection of $\omega_1$ and $\omega_2$. By radax on $\omega_1,\omega_2,(BNMC)$ we get that $A,Z,W$ are collinear. Since $\angle XZW=\angle YZW=90^{\circ}$, we get that $X,Y,Z$ are collinear. Thus it suffices to show that $\angle HZW=90^{\circ}$, which is equivalent to $Z$ being on $(AH)$.

Let $f$ denote an inversion centered at $A$ with radius $\sqrt{AN\cdot AB}$. Then $f$ swaps $Z$ and $W$. Unsurprisingly, $f$ also swaps $(AH)$ and $BC$, so we win. $\blacksquare$
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ItsBesi
146 posts
#194
Y by
Nothing new but posting it for storage.

Let $D$ be the feet of the altitude from $A$ to $BC$, let $\omega_1 \cap \omega_2=\{Z\}$.

Claim: Points $\overline{X-Z-Y}$ are collinear.

Proof:

$\angle XZW \stackrel{\omega_1}{=}=90$ and $\angle YZW \stackrel{\omega_1}{=}=90$

Hence $\angle XZY=\angle XZW+\angle YZW=180  \implies \angle XZY=180  \implies$ Points $\overline{X-Z-Y}$ are collinear. $\square$

Claim: Points $\overline{A-Z-W}$ are collinear.

Proof: First note that $\omega_1 \cap \omega_2 =\{W,Z\}$ so $ZW$ is the radical axis of $\omega_1$ and $\omega_2$

Also its well known that points $B,N,M$ and $C$ are concyclic so by Power of the Point Theorem we get:

$Pow(A,\omega_1)=AN \cdot AB=Pow(A, \odot(BNMC))=AM \cdot AC=Pow(A,\omega_2) \implies Pow(A,\omega_1)=Pow(A,\omega_2)$

So $A$ lies on the radical axis of $\omega_1$ and $\omega_2$ hence $A$ lies on the line $WZ \therefore$ Points $\overline{A-Z-W}$ are collinear. $\square$

Claim: Points $W,D,H$ and $Z$ are concyclic.

Proof: It's also well known that points $D,N,B$ and $H$ are concyclic

So by POP we get:
$AH \cdot AD=Pow(A,\odot(DNBH))=AN \cdot AB=Pow(A,\omega_1)=AZ \cdot AW \implies AH \cdot AD= AZ \cdot AW$

which by the converse of POP we have that: Points $W,D,H$ and $Z$ are concyclic. $\square$ Let $\odot(WDHZ)=\Gamma$.

Claim: Points $\overline{X-H-Y}$ are collinear.

Proof:

From $\Gamma \implies \angle HZW \stackrel{\Gamma}{=} 180-\angle HDZ=180-90=90=\angle XZW \implies \angle HZW=\angle XZW \implies$
Points $\overline{X-H-Z}$ are collinear, combining with the first claim that Points $\overline{X-Z-Y}$ are collinear we get that $\overline{X-H-Z-Y}$ are all collinear

Hence points $\overline{X-H-Y}$ are collinear $\blacksquare$
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Maximilian113
575 posts
#195
Y by
Let $P$ be the second intersection of $w_1, w_2.$ Note that by Miquel's Theorem, $AMPN$ is cyclic, but clearly $H$ also lies on this circle so $AMPHN$ is cyclic. Therefore, $$\angle NPH = \angle NAH = 90^\circ - \angle ABC = \angle XBN = \angle XPN,$$so $X, H, P$ are collinear. Similarly, $H, P, Y$ are collinear as well and we are done. QED
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Ilikeminecraft
616 posts
#196
Y by
let $P$ be second intersection of $(BNW), (MWC).$
By Radax on those two circles and $(BNMC),$ we get $A,P,W$ are collinear.
furthermore, miquel theorem on $ABC$ with $M, P, W$ as the points, we get $ANHPM$ is cyclic. Hence, $\angle APH = 90.$
However, $\angle BHP = 90 = \angle HPW$ so $X, H, P$ are collinear.
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endless_abyss
45 posts
#197
Y by
Let $P$ denote the second intersection of the circumcircles $B N W$ and $C M W$,

Claim we claim that $X - P - H$ and $Y - P - H$ are collinear

Let $Q$ denote the foot of $A$ onto $B C$, then we know that -

$\angle B H Q = C$
$\angle B H X  = \angle P W B - C$
$\angle Q H P = \angle 180 - \angle P W B$

So, $X - P - H$ are collinear, similarly, $Y - P - H$ are collinear.
$\square$
:starwars:
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bjump
1020 posts
#198
Y by
Let $D$ be the second interssection of $\omega_1$, and $\omega_2$. Since $\angle XDW = \angle YDW = 90^\circ$, $X$, $D$, and $Y$ are collinear. Let $E$ be the foot from $A$, since $AM \cdot AC = AN \cdot AB$. $A$ lies on $WD$ by radical axis. Converse of Radical axis with lines $NB$, $MC$ and $HE$, and $\omega_1$, and $\omega_2$ gives $HEWD$ cyclic, which means that $\angle WDH = 90^\circ$. So the points are collinear.
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joshualiu315
2534 posts
#199
Y by
not necessarily the fastest solution + i took a ridiculous amount of time to find the miquel point


Let $P$ be the second intersection of $\omega_1$ and $\omega_2$. Since $\angle WPX = \angle WPY$, we have that $P$ lies on $\overline{XY}$. Thus, it suffices to show that $X, H, P$ are collinear since showing that $Y,H,P$ are collinear is analogous.

Note that $P$ is the Miquel point of $(WBN)$, $(WCM)$ and $(AMN)$. Then, note that

\[\angle NPX = \angle NWX = 90^\circ - \angle NXW = 90^\circ - \angle NBW = \angle NAH = \angle NPH,\]
so we are done. $\blacksquare$
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