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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Wednesday at 3:18 PM
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.

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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 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Wednesday at 3:18 PM
0 replies
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Inequalities
sqing   23
N 8 minutes ago by sqing
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that
$$a^2+b^2+ ab +21abc\leq\frac{512}{441}$$Equality holds when $a=b=\frac{38}{21},c=\frac{5}{214}.$
$$a^2+b^2+ ab +19abc\leq\frac{10648}{9747}$$Equality holds when $a=b=\frac{22}{57},c=\frac{13}{57}.$
$$a^2+b^2+ ab +22abc\leq\frac{15625}{13068}$$Equality holds when $a=b=\frac{25}{66},c=\frac{8}{33}.$
23 replies
+1 w
sqing
Mar 26, 2025
sqing
8 minutes ago
J
H
Inequalities
sqing   3
N 24 minutes ago by sqing
Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=11.$ Prove that
$$a+ab+abc\leq\frac{49}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=10.$ Prove that
$$a+ab+abc\leq\frac{169}{24}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=14.$ Prove that
$$a+ab+abc\leq\frac{63+5\sqrt 5}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=32.$ Prove that
$$a+ab+abc\leq48+\frac{64\sqrt{2}}{3}$$
3 replies
sqing
Apr 1, 2025
sqing
24 minutes ago
J
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Puzzling p&c question
Hunter87   0
3 hours ago
There are 9 numbered tickets (distinct) to be distributed (1-9), and 7 contestants. Each contestant must get atleast one ticket. Two PARTICULAR contestants (say, A and B) can never get tickets with adjacent numbers and any contestant can get more than 1 ticket. All tickets are to be distributed.
In how many ways can this be done?
0 replies
Hunter87
3 hours ago
0 replies
J
H
.problem.
Cobedangiu   1
N 3 hours ago by Cobedangiu
Find the integer coefficients after expanding Newton's binomial:
$$(\frac{3}{2}-\frac{2}{3}x^2)^n (n \in Z)$$
1 reply
Cobedangiu
6 hours ago
Cobedangiu
3 hours ago
J
No more topics!
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J
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computational, tangent circles (2013 Euler Teachers' MO I p9)
parmenides51   5
N Jul 22, 2020 by vanstraelen
Two circles of radii $4$ and $8$ touch at point $A$. Through point $A$, a line that intersects the larger circle at point $B$, and the smaller circle at point $C$. Find $AB$ if it is known that $BC = 6\sqrt2$.
5 replies
parmenides51
Jul 21, 2020
vanstraelen
Jul 22, 2020
J
computational, tangent circles (2013 Euler Teachers' MO I p9)
G H J
Reveal topic tags
geometry
circles
tangent circles
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G H BBookmark kLocked kLocked NReply
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parmenides51
30629 posts
parmenides51
#1 h  Jul 21, 2020, 12:21 AM
Y by
Two circles of radii $4$ and $8$ touch at point $A$. Through point $A$, a line that intersects the larger circle at point $B$, and the smaller circle at point $C$. Find $AB$ if it is known that $BC = 6\sqrt2$.
Z K Y
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UmbQbify
3 posts
UmbQbify
#2 h  Jul 21, 2020, 12:27 AM • 1 Y
Y by Mango247
Is that $4\sqrt{2}$?
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parmenides51
30629 posts
parmenides51
#3 h  Jul 21, 2020, 12:34 AM
Y by
that is the result to the first case, there is also another one case
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UmbQbify
3 posts
UmbQbify
#4 h  Jul 21, 2020, 12:48 AM
Y by
And what's that? I'm pretty sure I maintenaned the generality (except the impossible case when the line passes through centre)
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parmenides51
30629 posts
parmenides51
#5 h  Jul 21, 2020, 12:53 AM • 1 Y
Y by UmbQbify
there are 2 cases for 2 circles to be tangent
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vanstraelen
8946 posts
vanstraelen
#6 h  Jul 22, 2020, 10:56 AM
Y by
The second case, the circles are internally tangent, is impossible. Because the maximum for $BC$ is $8 < 6\sqrt{2}$. The Russian solution is wrong.
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