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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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Area of triangle inside regular hexagon
smartvong   1
N 13 minutes ago by tom-nowy
The diagram shows a regular hexagon $ABCDEF$. $P$ is a point on the line segment $CD$, such that the areas of $\triangle PAF$ and $\triangle PEF$ are $64$ and $48$ respectively. Find the area of $\triangle PAB$.

IMAGE
1 reply
smartvong
an hour ago
tom-nowy
13 minutes ago
J
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Geometry Proof
strongstephen   14
N 37 minutes ago by strongstephen
Proof that choosing four distinct points at random has an equal probability of getting a convex quadrilateral vs a concave one.
not cohesive proof alert!

NOTE: By choosing four distinct points, that means no three points lie on the same line on the Gaussian Plane.
NOTE: The probability of each point getting chosen don’t need to be uniform (as long as it is symmetric about the origin), you just need a way to choose points in the infinite plane (such as a normal distribution)

Start by picking three of the four points. Next, graph the regions where the fourth point would make the quadrilateral convex or concave. In diagram 1 below, you can see the regions where the fourth point would be convex or concave. Of course, there is the centre region (the shaded triangle), but in an infinite plane, the probability the fourth point ends up in the finite region approaches 0.

Next, I want to prove to you the area of convex/concave, or rather, the probability a point ends up in each area, is the same. Referring to the second diagram, you can flip each concave region over the line perpendicular to the angle bisector of which the region is defined. (Just look at it and you'll get what it means.) Now, each concave region has an almost perfect 1:1 probability correspondence to another convex region. The only difference is the finite region (the triangle, shaded). Again, however, the actual significance (probability) of this approaches 0.

If I call each of the convex region's probability P(a), P(c), and P(e) and the concave ones P(b), P(d), P(f), assuming areas a and b are on opposite sides (same with c and d, e and f) you can get:
P(a) = P(b)
P(c) = P(d)
P(e) = P(f)

and P(a) + P(c) + P(e) = P(convex)
and P(b) + P(d) + P(f) = P(concave)

therefore:
P(convex) = P(concave)
14 replies
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strongstephen
Tuesday at 4:54 AM
strongstephen
37 minutes ago
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Inequalities
sqing   2
N 44 minutes ago by sqing
Let $ a,b>0 $ and $\frac{a}{a^2+3}+ \frac{b}{b^2+ 3} \geq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{a}{a^3+3}+ \frac{b}{b^3+ 3}\geq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
2 replies
sqing
Yesterday at 12:59 PM
sqing
44 minutes ago
J
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Unknown triangle area
smartvong   0
an hour ago
The diagram shows a convex quadrilateral $ABCD$. The points $E$ and $F$ divide $AB$ into three equal parts while the points $G$ and $H$ divide $CD$ into three equal parts. The line segments $AH$ and $ED$ intersect at $I$. The line segments $CF$ and $BG$ intersect at $J$. Given that the areas of the triangles $AID$, $EHI$ and $FJG$ are $154$, $112$, and $99$ respectively, find the area of the triangle $BJC$.

IMAGE
0 replies
smartvong
an hour ago
0 replies
J
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[CMC ARML 2020 I3] Unique Sequence
franchester   2
N Apr 28, 2025 by CubeAlgo15
There is a unique nondecreasing sequence of positive integers $a_1$, $a_2$, $\ldots$, $a_n$ such that \[\left(a_1+\frac1{a_1}\right)\left(a_2+\frac1{a_2}\right)\cdots\left(a_n+\frac1{a_n}\right)=2020.\]Compute $a_1+a_2+\cdots+a_n$.

Proposed by lminsl
2 replies
franchester
May 29, 2020
CubeAlgo15
Apr 28, 2025
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[CMC ARML 2020 I3] Unique Sequence
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franchester
1487 posts
franchester
#1 h  May 29, 2020, 11:52 PM
Y by
There is a unique nondecreasing sequence of positive integers $a_1$, $a_2$, $\ldots$, $a_n$ such that \[\left(a_1+\frac1{a_1}\right)\left(a_2+\frac1{a_2}\right)\cdots\left(a_n+\frac1{a_n}\right)=2020.\]Compute $a_1+a_2+\cdots+a_n$.

Proposed by lminsl
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thinkmath_
67 posts
thinkmath_
#2 h  May 30, 2020, 2:17 AM
Y by
amazing problem; just realize they all have to be like 1, 2, 10 or smth along those lines
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CubeAlgo15
42 posts
CubeAlgo15
#3 h  Apr 28, 2025, 6:09 PM
Y by
the sequence is 1 1 1 1 1 2 2 10
which gives the answer of 19 (not sure if this is correct but should be)
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