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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Thursday at 11:16 PM
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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0 replies
jlacosta
Thursday at 11:16 PM
0 replies
Random modulos
m4thbl3nd3r   6
N a few seconds ago by GreekIdiot
Find all pair of integers $(x,y)$ s.t $x^2+3=y^7$
6 replies
m4thbl3nd3r
Apr 7, 2025
GreekIdiot
a few seconds ago
Concurrency in Parallelogram
amuthup   89
N a minute ago by happypi31415
Source: 2021 ISL G1
Let $ABCD$ be a parallelogram with $AC=BC.$ A point $P$ is chosen on the extension of ray $AB$ past $B.$ The circumcircle of $ACD$ meets the segment $PD$ again at $Q.$ The circumcircle of triangle $APQ$ meets the segment $PC$ at $R.$ Prove that lines $CD,AQ,BR$ are concurrent.
89 replies
amuthup
Jul 12, 2022
happypi31415
a minute ago
deleting multiple or divisor in pairs from 2-50 on a blackboard
parmenides51   1
N an hour ago by TheBaiano
Source: 2023 May Olympiad L2 p3
The $49$ numbers $2,3,4,...,49,50$ are written on the blackboard . An allowed operation consists of choosing two different numbers $a$ and $b$ of the blackboard such that $a$ is a multiple of $b$ and delete exactly one of the two. María performs a sequence of permitted operations until she observes that it is no longer possible to perform any more. Determine the minimum number of numbers that can remain on the board at that moment.
1 reply
parmenides51
Mar 24, 2024
TheBaiano
an hour ago
at everystep a, b, c are replaced by a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)
NJAX   9
N an hour ago by atdaotlohbh
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem 8
Three positive integers are written on the board. In every minute, instead of the numbers $a, b, c$, Elbek writes $a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)$ . Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
Note. $\gcd(x,y)$ - Greatest common divisor of numbers $x$ and $y$

Proposed by Sergey Berlov, Russia
9 replies
NJAX
May 31, 2024
atdaotlohbh
an hour ago
trigonometric functions
VivaanKam   11
N 3 hours ago by aok
Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!
11 replies
VivaanKam
Apr 29, 2025
aok
3 hours ago
1201 divides sum of powers
V0305   1
N 4 hours ago by vincentwant
(Source: me) Prove that for all positive integers $n$, $1201 \mid 2^{2^n} + 59^{2^n} + 61^{2^n}$.
1 reply
V0305
4 hours ago
vincentwant
4 hours ago
Interesting geometry
polarLines   5
N 5 hours ago by Mathworld314
Let $ABC$ be an equilateral triangle of side length $2$. Point $A'$ is chosen on side $BC$ such that the length of $A'B$ is $k<1$. Likewise points $B'$ and $C'$ are chosen on sides $CA$ and $AB$. with $CB'=AC'=k$. Line segments are drawn from points $A',B',C'$ to their corresponding opposite vertices. The intersections of these line segments form a triangle, labeled $PQR$. Prove that $\Delta PQR$ is an equilateral triangle with side length ${4(1-k) \over \sqrt{k^2-2k+4}}$.
5 replies
polarLines
May 20, 2018
Mathworld314
5 hours ago
Showing that certain number is divisible by 13
BBNoDollar   3
N 5 hours ago by Shan3t
Show that 3^(n+2) + 9^(n+1) + 4^(2n+1) + 4^(4n+1) is divisible by 13 for every n natural number.
3 replies
BBNoDollar
Today at 2:54 PM
Shan3t
5 hours ago
Inequality
tom-nowy   0
6 hours ago
Let $0<a,b,c,<1$. Show that
$$ \frac{3(a+b+c)}{a+b+c+3abc} > \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} .$$
0 replies
tom-nowy
6 hours ago
0 replies
Logarithm of a product
axsolers_24   2
N Today at 2:59 PM by axsolers_24
Let $x_1=97 ,$ $x_2=\frac{2}{x_1} ,$ $x_3=\frac{3}{x_2} ,$$... , $ $x_8=\frac{8}{x_7}$
then
$ \log_{3\sqrt{2}} \left(\prod_{i=1}^8 x_i-60\right)$
2 replies
axsolers_24
Today at 10:42 AM
axsolers_24
Today at 2:59 PM
Inequalities
sqing   1
N Today at 2:48 PM by sqing
Let $ a,b>0 , a^2 + 2b^2 =  a + 2b $. Prove that $$\sqrt{\frac{a}{b( a+2)}} + \sqrt{\frac{b}{a(2b+1)}}  \geq \frac {2}{\sqrt{3}} $$Let $ a,b>0 , a^3 + 2b^3 =  a + 2b $. Prove that $$\sqrt[3]{\frac{a}{b( a+2)}} + \sqrt[3]{\frac{b}{a(2b+1)}}  \geq \frac {2}{\sqrt[3]{3}} $$
1 reply
sqing
Today at 2:27 PM
sqing
Today at 2:48 PM
Coprime sequence
Ecrin_eren   4
N Today at 2:37 PM by Pal702004
"Let N be a natural number. Show that any two numbers from the following sequence are coprime:

2^1 + 1, 2^2 + 1, 2^4+ 1,2^8+1 ..., 2^(2^N )+ 1."
4 replies
Ecrin_eren
May 1, 2025
Pal702004
Today at 2:37 PM
Hard Inequality
William_Mai   0
Today at 2:13 PM
Given $a, b, c \in \mathbb{R}$ such that $a^2 + b^2 + c^2 = 1$.
Find the minimum value of $P = ab + 2bc + 3ca$.

Source: Pham Le Van
0 replies
William_Mai
Today at 2:13 PM
0 replies
Find the minimum
Ecrin_eren   5
N Today at 2:05 PM by Jackson0423
The polynomial is given by P(x) = x^4 + ax^3 + bx^2 + cx + d, and its roots are x1, x2, x3, x4. Additionally, it is stated that d ≥ 5.Find the minimum value of the product:

(x1^2 + 1)(x2^2 + 1)(x3^2 + 1)(x4^2 + 1).

5 replies
Ecrin_eren
Thursday at 9:03 PM
Jackson0423
Today at 2:05 PM
Two circles, a tangent line and a parallel
Valentin Vornicu   104
N Yesterday 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
Yesterday 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
574 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
813 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
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MOHS Hardness Scale
Daily IMO Problem #1 (May 2) - IMO 2001 p1
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