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foldina a rectangle paper 3 times
parmenides51   1
N an hour ago by TheBaiano
Source: 2023 May Olympiad L2 p4
Matías has a rectangular sheet of paper $ABCD$, with $AB<AD$.Initially, he folds the sheet along a straight line $AE$, where $E$ is a point on the side $DC$ , so that vertex $D$ is located on side $BC$, as shown in the figure. Then folds the sheet again along a straight line $AF$, where $F$ is a point on side $BC$, so that vertex $B$ lies on the line $AE$; and finally folds the sheet along the line $EF$. Matías observed that the vertices $B$ and $C$ were located on the same point of segment $AE$ after making the folds. Calculate the measure of the angle $\angle DAE$.
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1 reply
parmenides51
Mar 24, 2024
TheBaiano
an hour ago
Trigo + Series
P162008   0
Today at 11:59 AM
$F(r,x) = \tan\left(\frac{\pi}{3} + 3^r x\right)$ and $G(r,x) = \cot\left(\frac{\pi}{6} + 3^r x\right)$

Let $A = \sum_{r=0}^{24} \frac{d(F(r,x))}{dx} \left[\frac{1}{F(r,x) + \frac{1}{F(r,x)}}\right]$

$B = \sum_{r=0}^{24} \frac{-d(G(r,x))}{dx} \left[\frac{1}{G(r,x) + \frac{1}{G(r,x)}}\right]$

$P = \prod_{r=0}^{24} F(r,x), Q = \prod_{r=0}^{24} G(r,x)$

and, $R = \frac{A - B}{\tan x\left[PQ - \frac{1}{3^{25}}\right]}$ where $x \in R$

Find the remainder when $R^{2025}$ is divided by $11.$
0 replies
P162008
Today at 11:59 AM
0 replies
D1026 : An equivalent
Dattier   2
N Today at 9:39 AM by Dattier
Source: les dattes à Dattier
Let $u_0=1$ and $\forall n \in \mathbb N, u_{2n+1}=\ln(1+u_{2n}), u_{2n+2}=\sin(u_{2n+1})$.

Find an equivalent of $u_n$.
2 replies
Dattier
Yesterday at 1:39 PM
Dattier
Today at 9:39 AM
Integration Bee in Czechia
Assassino9931   2
N Today at 9:35 AM by pi_quadrat_sechstel
Source: Vojtech Jarnik IMC 2025, Category II, P3
Evaluate the integral $\int_0^{\infty} \frac{\log(x+2)}{x^2+3x+2}\mathrm{d}x$.
2 replies
Assassino9931
May 2, 2025
pi_quadrat_sechstel
Today at 9:35 AM
Alternating series and integral
jestrada   4
N Today at 4:06 AM by bakkune
Source: own
Prove that for all $\alpha\in\mathbb{R}, \alpha>-1$, we have
$$ \frac{1}{\alpha+1}-\frac{1}{\alpha+2}+\frac{1}{\alpha+3}-\frac{1}{\alpha+4}+\cdots=\int_0^1 \frac{x^{\alpha}}{x+1}  \,dx. $$
4 replies
jestrada
Yesterday at 10:56 PM
bakkune
Today at 4:06 AM
Find all continuous functions
bakkune   3
N Today at 3:58 AM by bakkune
Source: Own
Find all continuous function $f, g\colon\mathbb{R}\to\mathbb{R}$ satisfied
$$
(x - k)f(x) = \int_k^x g(y)\mathrm{d}y 
$$for all $x\in\mathbb{R}$ and all $k\in\mathbb{Z}$.
3 replies
bakkune
Yesterday at 6:02 AM
bakkune
Today at 3:58 AM
Equivalent condition of the uniformly continuous fo a function
Alphaamss   2
N Today at 2:05 AM by Alphaamss
Source: Personal
Let $f_{a,b}(x)=x^a\cos(x^b),x\in(0,\infty)$. Get all the $(a,b)\in\mathbb R^2$ such that $f_{a,b}$ is uniformly continuous on $(0,\infty)$.
2 replies
Alphaamss
Yesterday at 7:35 AM
Alphaamss
Today at 2:05 AM
A problem in point set topology
tobylong   2
N Today at 12:00 AM by cosmicgenius
Source: Basic Topology, Armstrong
Let $f:X\to Y$ be a closed map with the property that the inverse image of each point in $Y$ is a compact subset of $X$. Prove that $f^{-1}(K)$ is compact whenever $K$ is compact in $Y$.
2 replies
tobylong
Yesterday at 3:14 AM
cosmicgenius
Today at 12:00 AM
Does the sequence log(1+sink)/k converge?
tom-nowy   7
N Yesterday at 9:25 PM by GreenKeeper
Source: Question arising while viewing https://artofproblemsolving.com/community/c7h3556569
Does the sequence $$ \frac{\ln(1+\sin k)}{k} \;\;\;(k=1,2,3,\ldots) $$converge?
7 replies
tom-nowy
Apr 30, 2025
GreenKeeper
Yesterday at 9:25 PM
s(I)=2019
math90   8
N Yesterday at 8:39 PM by MathSaiyan
Source: IMC 2019 Day 2 P8
Let $x_1,\ldots,x_n$ be real numbers. For any set $I\subset\{1,2,…,n\}$ let $s(I)=\sum_{i\in I}x_i$. Assume that the function $I\to s(I)$ takes on at least $1.8^n$ values where $I$ runs over all $2^n$ subsets of $\{1,2,…,n\}$. Prove that the number of sets $I\subset \{1,2,…,n\}$ for which $s(I)=2019$ does not exceed $1.7^n$.

Proposed by Fedor Part and Fedor Petrov, St. Petersburg State University
8 replies
math90
Jul 31, 2019
MathSaiyan
Yesterday at 8:39 PM
Cauchy's functional equation with f({max{x,y})=max{f(x),f(y)}
tom-nowy   1
N Yesterday at 7:04 PM by Filipjack
Source: https://x.com/D_atWork/status/1788496152855560470, Problem 4
Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the following two conditions for all $x,y \in \mathbb{R}$:
\[ f(x+y)=f(x)+f(y), \;\;\; f \left( \max \{x, y \} \right) = \max \left\{ f(x),f(y) \right\}. \]
1 reply
tom-nowy
Yesterday at 2:23 PM
Filipjack
Yesterday at 7:04 PM
Two circles, a tangent line and a parallel
Valentin Vornicu   104
N Friday at 9:58 PM by cubres
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
104 replies
Valentin Vornicu
Oct 24, 2005
cubres
Friday at 9:58 PM
Two circles, a tangent line and a parallel
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
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Math_.only.
22 posts
#96 • 1 Y
Y by cubres
arqady wrote:
let MN and AB intesect at K. Then $KA^2=KM*KN=KB^2.$ Hence KA=KB.
CD||AB. Hence PM=QM.
$\angle AEB=\angle ECD=\angle MAB,$ $\angle EBA=\angle EDC=\angle MBA.$
hence$\triangle AMB$ and$\triangle AEB$ congruity.
Hence $EM\perp AB.$ Hence $EM\perp PQ.$
Hence $EP=EQ.$

Why angles AEB and ECD are equal ?
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MagicalToaster53
159 posts
#97 • 1 Y
Y by cubres
Notice that $MN$ bisects $\overline{AB},$ as the point where they intersect has equivalent powers to the respective circles. From this we discover that $M$ is the midpoint of $\overline{PQ}$ due to the homothety at $N$ mapping $\overline{PQ} \mapsto \overline{AB}.$ It then suffices to show $EM \perp PQ$.

Claim: $\triangle EAB \cong \triangle MAB.$
Proof: Observe that $AB$ bisects $\angle EAM:$ \[\angle EAB = \angle ECM = \angle ACM = \angle BAM. \]Similarly by symmetry, we find that $AB$ bisects $\angle EBM.$ Then by SAS, we conclude that $\triangle EAB \cong \triangle MAB,$ as was to be shown. $\square$

Now $EAMB$ is a kite, so that $EM \perp AB \implies EM \perp PQ,$ as $AB \parallel PQ$. Hence $\triangle EPQ$ is isosceles, as desired. $\blacksquare$
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lnzhonglp
120 posts
#98 • 1 Y
Y by cubres
Observe that $$\measuredangle EBA = \measuredangle EDC = \measuredangle ABM$$and $$\measuredangle EAB = \measuredangle ECM = \measuredangle MAB,$$so $\triangle EAB \cong \triangle MAB$ and $EM \perp AB$. From radical axis we see that $MN$ bisects $AB$, so by homothety at $N$ we find that $M$ is the midpoint of $PQ$. Then since $EM \perp PQ$ it follows that $EP = EQ$.
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Likeminded2017
391 posts
#99 • 1 Y
Y by cubres
Let $MN$ and $AB$ intersect at $K.$ As $K$ lies on the radical axis we have $KA=KB.$ Taking the homothety centered at $N$ that maps $M$ to $K$ we find that $MP=MQ.$ Then $\angle BAM=\angle MCA=\angle BAE$ and similarly $\angle EBA=\angle ABM$ so by ASA congruence $\triangle EAB \cong \triangle MAB.$ Thus $EM$ is perpendicular to $AB$ and thus $PQ$ so $EP=EQ.$
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ezpotd
1262 posts
#100 • 1 Y
Y by cubres
Letting $AB$ meet $MN$ at $K$, we see $KA^2 = KM \cdot KN = KB^2$, so $K$ is the midpoint of $AB$, taking the homothety from $AB$ to $PQ$ this gives $MP = MQ$, so it suffices to prove $EM \perp PQ$, or $EM \perp AB$. However, angle chasing gives $\angle EAB = \angle ECM = \angle AMC = \angle BAM, \angle EBA = \angle EDM = \angle BMD = \angle ABM$, so $\triangle EAB \cong \angle MAB$, so $E,M$ are reflections about $AB$, done.
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optimusprime154
21 posts
#101 • 2 Y
Y by SorPEEK, cubres
let \(Z \)= \(MN  \cap AB\) from PoP we know \(ZB^2 = ZM * ZN\) and \(ZA^2 = ZM * ZN\) which means \(ZA = ZB\) and since \(PQBA\) is a trapezoid, \(PM = MQ\)
now we prove \(EM \perp CD\) and we're done. label \(\angle ACM = x, \angle BDM = y\) we know from \(AB \parallel CD\) that \(\angle EAB = x\) and \(\angle EDC = y\)
we also know from \(AB\) being tangent that \(\angle MBA\ = y\) and \(\angle MAB = x\) the above facts indicate the similarity of \(\triangle MAB , \triangle EAB \) so \(EM \perp AB\) => \(EM \perp CD\) and we're done
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Maximilian113
575 posts
#102 • 1 Y
Y by cubres
Extend $MN$ to intersect $AB$ at $K,$ then clearly $AK=BK.$ Therefore $PM=MQ$ by the parallel lines.

Now, notice that $\angle EAB = \angle ACM = \angle BAM = \angle AMC.$ Therefore $AB$ bisects $\angle EAM$ and $AC=AM.$ Similarly $AB$ bisects $EBM$ and $BD=BM.$ Therefore, $\triangle EAB \cong \triangle MAB$ so $A, B$ are the circumcenters of $\triangle EMC, \triangle EMD$ respectively so by Thales' Theorem it follows that $\angle EMP = \angle EMQ = 90^\circ.$ Hence $EM$ is the perpendicular bisector of $PQ,$ and we are done. QED
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ehuseyinyigit
816 posts
#104 • 1 Y
Y by cubres
WELL-KNOWN. Since $AB\parallel CD$, we have $$\angle ACM=\angle AMC=\angle MAB=\angle BAE$$Thus, $AC=AM$ and $AB$ bisects $\angle EAM$. Similarly $BD=BM$ and $AB$ bisects $\angle EBM$ hence $ABME$ is a kite $\Rightarrow AE=AM=AC$ implying $EM\perp CD$. On the other hand, for point $F$ being intersection of lines $AB=MN$, $AF=BF$ implies $PM=MQ$. We obtain $EP=EQ$.

WELL-KNOWN. We will show $EN$ bisects $\angle CND$. This is true because $$\angle ENB=\angle EAB=\angle ENC$$and $\angle ANE=\angle ABE=\angle BND$. Thus, $EN$ bisects $\angle CND$.
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mathwiz_1207
96 posts
#105 • 1 Y
Y by cubres
We have
\[\measuredangle EAB = \measuredangle ACM = \measuredangle ANM = \measuredangle BAM = \measuredangle CMA\]\[\measuredangle EBA = \measuredangle BDM = \measuredangle BNM = \measuredangle ABM = \measuredangle DMB\], implying that $\triangle AMB \cong \triangle AEB$, and that $\triangle AMC, \triangle BDM$ are isosceles. Therefore,
\[AE = AM = AC\]so $A$ is the center of $(EMC)$, and similarly $B$ is the cetner of $(EMD)$. Thus, $EM \perp CD$. Now, let $L = MN \cap AB$, by Radical Axis,
\[LA = LB\]and since $\triangle NPQ \sim \triangle NAB$, we have $MP = MQ \implies EQ = EP$, so we are done.
This post has been edited 1 time. Last edited by mathwiz_1207, Feb 23, 2025, 11:53 PM
Reason: forgot to add last line
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study1126
554 posts
#106 • 1 Y
Y by cubres
All angles will be directed in this solution.

First, $\angle EAB=\angle ECB=\angle BAM$, since $AB||CD$ and $AB$ is tangent to the two circles. Similarly, $\angle EBA=\angle ABM$. Combined with $AB=AB$ gives $\triangle EAB\cong \triangle MAB$.

This implies $EAMB$ is a kite, so $EM\perp AB$, which means $EM\perp CD$. Since $M$ is on the radical axis, extending $NM$ to $X$ on $AB$ gives $X$ is the midpoint of $AB$. Homothety shows $M$ is the midpoint of $PQ$. Thus, $MP=MQ$, so by congruent triangle $\triangle EPM\cong \triangle EQM$, $EP=EQ$, as desired.
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joshualiu315
2534 posts
#107 • 1 Y
Y by cubres
It is well-known that $\overline{MN}$ bisects $\overline{AB}$, and since $\overline{AB} \parallel \overline{CD}$, we also know that $\overline{MN}$ bisects $\overline{PQ}$. This implies $M$ is the midpoint of $\overline{PQ}$.

We have $\angle EAB = \angle ACM = \angle MAB$ and similarly, $\angle EBA = \angle BDM = \angle MBA$. This means $\triangle EAB \cong \triangle MAB$, which implies that $\overline{EM} \perp \overline{AB}$. Because $\overline{AB} \parallel \overline{CD}$, we have $\overline{EM} \perp \overline{PQ}$, so $\overline{EM}$ is the perpendicular bisector of $\overline{PQ}$. Thus, $EP=EQ$, as desired. $\blacksquare$
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LeYohan
41 posts
#108 • 1 Y
Y by cubres
First IMO #1 solve and it's geo!! :D

Claim: $PM = MQ$.
Let $NM$ intersect $AB$ at $X$. It's well known that the radical axis intersects the common tangent, hence $AX = XB$. By the Thales Theorem, we inmediately get the result. $\square$

Now doing some angle chase using the fact that $CD \parallel AB$ and $AB$ is tangent to both circles, we get:
$\angle EBA = \angle BDM = \angle BNM = \angle ABM = \angle BMD$, and $\angle EAB = \angle ACM = \angle ANM = \angle BAM = \angle AMC$. As a result, $\triangle ACM$ and $\triangle BMC$ are isosceles. Furthermore, $\triangle MAB \cong \triangle EAB \implies EB = BD, EA = AC$. This means that $\angle EMC = \angle EMD = 90$, and notice that we originally proved that $PM = MQ$, so $\triangle EPQ$ is isosceles and we're done. $\square$
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Avron
37 posts
#109 • 1 Y
Y by cubres
$MN$ is the radical axis of $G_1,G_2$ so it bisects $AB$ so we also know that $M$ is the midpoint of $PQ$, thus it is enough to prove $ME\perp CD$.
Now notice that $\angle MAB=\angle ACM = \angle EAB$ and similarly $\angle MBA = \angle ABE$ so $EAB \cong MAB$ and we're done
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zuat.e
55 posts
#110 • 1 Y
Y by cubres
Easy angle chasing yields the cyclicity of $(BEAN)$. We claim the following:

Claim: $EA=AM=AC$ and similarly $EB=BM=BD$
Proof: Let $O_1$ be the center of $G_1$. It is clear that $O_1A$ is the side bisector of $MC$, hence $AM=AC$.
Furthermore, note that $\measuredangle BAE=\measuredangle DCE= \measuredangle BMC=\measuredangle MAB$ and similarly $\measuredangle EBA =\measuredangle ABM$, hence $\triangle EAB\cong MAB$ (because they also share $AB$).

It now follows that $AC=AM=AE$ and analogously $BD=BM=BE$
Finally, as $\measuredangle MCA= \measuredangle MNA=\measuredangle ANC$, we have $AP\cdot AN=AC^2=AE^2$, hence $\measuredangle CEP =\measuredangle ENA=\measuredangle EBA=\measuredangle EDC$ and similarly $\measuredangle QED=\measuredangle PCE$, yielding $\measuredangle CPE =\measuredangle EQD$, as desired.
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cubres
118 posts
#111
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ACGN
MOHS Hardness Scale
Daily IMO Problem #1 (May 2) - IMO 2001 p1
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