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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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[*]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
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the same prime factors
andria   5
N 16 minutes ago by bin_sherlo
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
5 replies
andria
Sep 6, 2015
bin_sherlo
16 minutes ago
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A sharp estimation of the product
mihaig   0
19 minutes ago
Source: VL
Let $n\geq4$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$a_1+\cdots+a_n+2^{n-1}\geq n+2^{n-1}\cdot\prod_{i=1}^{n}{a_i}.$$
0 replies
mihaig
19 minutes ago
0 replies
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xf(x) + f ^2(y) +2 xf(y) perfect square for all positive integers x,y
parmenides51   9
N 22 minutes ago by Gaunter_O_Dim_of_math
Source: Balkan BMO Shortlist 2017 N2
Find all functions $f :Z_{>0} \to Z_{>0}$ such that the number $xf(x) + f ^2(y) + 2xf(y)$ is a perfect square for all positive integers $x,y$.
9 replies
parmenides51
Aug 1, 2019
Gaunter_O_Dim_of_math
22 minutes ago
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2024 Japan Mathematical Olympiad Preliminary
parkjungmin   0
25 minutes ago
2024 Japan Mathematical Olympiad Preliminary
0 replies
parkjungmin
25 minutes ago
0 replies
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No more topics!
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Easy perpendicularity
a_507_bc   1
N Apr 15, 2025 by zaidova
Source: Caucasus MO 2024, Juniors P2
The rhombuses $ABDK$ and $CBEL$ are arranged so that $B$ lies on the segment $AC$ and $E$ lies on the segment $BD$. Point $M$ is the midpoint of $KL$. Prove that $\angle DME=90^{\circ}$.
1 reply
a_507_bc
Mar 15, 2024
zaidova
Apr 15, 2025
J
Easy perpendicularity
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Reveal topic tags
geometry
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G H BBookmark kLocked kLocked NReply
Source: Caucasus MO 2024, Juniors P2
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a_507_bc
678 posts
a_507_bc
#1 h  Mar 15, 2024, 12:21 PM
Y by
The rhombuses $ABDK$ and $CBEL$ are arranged so that $B$ lies on the segment $AC$ and $E$ lies on the segment $BD$. Point $M$ is the midpoint of $KL$. Prove that $\angle DME=90^{\circ}$.
This post has been edited 1 time. Last edited by a_507_bc, Mar 15, 2024, 12:21 PM
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zaidova
87 posts
zaidova
#2 h  Apr 15, 2025, 1:31 PM
Y by
Nice problem, can be solved with an easy construction. Just construct a parallelogram with diagonal $KL$. Let the vertex on $AK$ be $P$, and Let the vertex $KD \cap LC=Q$. $PLKQ$ is a parallelogram. Let side of the $BCLE$ rhombus be $x$ and let $PK=y=ED=LQ$. Then we get $AB=AK=BD=KD=PE=x+y$ Now draw a line passing through point $M$, parallel to $PL$. Say that $X$, (parallel line passing through $M$) intersects $ED$. Then it is easy to see $EX=XD$. ===> $XE=XD=PK/2=y/2$ We have to show it is also equal to $MX$. Let the line passing through $M$ also intersect $AK$ at $Y$. Then $MY=PL/2=(PE+EL)/2=(AB+BC)/2=((x+y)+x)/2=x+y/2$. $PE=AB=x+y=XY$==>$MX=XY-MY=x+y-x-y/2=y/2$.
So, we get $MX=XE=XD$, $ \angle DME=90$. Done!
This post has been edited 3 times. Last edited by zaidova, Apr 15, 2025, 1:33 PM
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