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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
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Circumcenter lie on a line.
RootofUnityfilter   0
5 minutes ago
Source: my teacher
Let $ABC$ be a triangle with circumcircle $(O)$, incentre $I$. The second intersections of $AI$, $BI$, and $CI$ with $(O)$ are $X_A$, $X_B$, and $X_C$, respectively. Lines $AI$ and $BC$ intersect at $D$ and lines $BX_C$ and $CX_B$ intersect at $X$. Suppose $(XX_BX_C)$ and $(XBC)$ intersect again at $S \ne X$. Lines $BX$ and $CX$ intersect the circumcircle of triangle $SXX_A$ again at $P \ne X$ and $Q \ne X$, respectively.
Prove that the circumcentre of triangle $SID$ lies on $PQ$.
0 replies
RootofUnityfilter
5 minutes ago
0 replies
J
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ab + c + d = 3, bc + d + a = 5, cd + a + b = 2, da + b + c = 6
parmenides51   9
N 9 minutes ago by DensSv
Source: 2021 JBMO TST Bosnia and Herzegovina P1
Determine all real numbers $a, b, c, d$ for which
$$ab + c + d = 3$$$$bc + d + a = 5$$$$cd + a + b = 2$$$$da + b + c = 6$$
9 replies
parmenides51
Oct 7, 2022
DensSv
9 minutes ago
J
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Prime number and composite number
mingzhehu   3
N 9 minutes ago by mingzhehu
I have one topic on how to identify Prime Number and Composite Number quickly? Maybe the number is more than 100 or 1000.......!
If there are some formula that can be used to verify the number easily, it will be highly appreciated.
Does anybody has any good idea for that?

3 replies
mingzhehu
Today at 8:06 AM
mingzhehu
9 minutes ago
J
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Problems for v_p(n)
xytunghoanh   1
N 25 minutes ago by whwlqkd
Hello everyone. I need some easy problems for $v_p(n)$ (use in a problem for junior) to practice. Can anyone share to me?
Thanks :>
1 reply
1 viewing
xytunghoanh
an hour ago
whwlqkd
25 minutes ago
J
No more topics!
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J
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center of parallelogram equidistant from midpoints of side of AKD, AB=BK=KC
parmenides51   1
N Apr 23, 2022 by ythomashu
Source: Russian Olympiad 2022 HSO 8.2 High Standards Olympiad - Высшая проба
Point $K$ is marked in parallelogram $ABCD$ such that $AB=BK=KC$. Prove that the center of the parallelogram is equidistant from the midpoints of all sides of triangle $AKD$.
1 reply
parmenides51
Apr 23, 2022
ythomashu
Apr 23, 2022
J
center of parallelogram equidistant from midpoints of side of AKD, AB=BK=KC
G H J
Reveal topic tags
geometry
parallelogram
equal segments
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G H BBookmark kLocked kLocked NReply
Source: Russian Olympiad 2022 HSO 8.2 High Standards Olympiad - Высшая проба
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parmenides51
30629 posts
parmenides51
#1 h  Apr 23, 2022, 8:22 PM
Y by
Point $K$ is marked in parallelogram $ABCD$ such that $AB=BK=KC$. Prove that the center of the parallelogram is equidistant from the midpoints of all sides of triangle $AKD$.
Z K Y
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ythomashu
6322 posts
ythomashu
#2 h  Apr 23, 2022, 10:37 PM
Y by
Let $D'$ be the reflection of $D$ about the midpoint of $AK.$ Then $AD||D'K||BC$ so $D'B=KC=AB=KB.$ Then dilate $\triangle D'AK$ and its circumcenter $B$ from $D$ by a factor of $\frac12.$
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