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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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Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. 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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[*]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
Warning!
VivaanKam   40
N 6 minutes ago by ayeshaaq
This problem will try to trick you! :!:

40 replies
VivaanKam
May 5, 2025
ayeshaaq
6 minutes ago
Worst Sillies of All Time
pingpongmerrily   50
N 2 hours ago by bbojy
Share the worst sillies you have ever made!

Mine was probably on the 2024 MathCounts State Target Round Problem 8, where I wrote my answer as a fraction instead of a percent, which cost me a trip to Nationals that year.
50 replies
pingpongmerrily
Yesterday at 12:34 PM
bbojy
2 hours ago
MathDash help
Spacepandamath13   8
N 2 hours ago by Yiyj
AkshajK ORZ by the way invited me to do MathDash a few months ago and I did try it one day but haven't done it much after (Sorry). Now, I'm getting back into it and finding the format kind of weird. When selecting certain problem type sometimes it lets me pick immediately, other times not. Any fixes?
8 replies
Spacepandamath13
May 29, 2025
Yiyj
2 hours ago
MIT PRIMES STEP
pingpongmerrily   5
N 5 hours ago by pingpongmerrily
Anyone else applying? How cooked am I for the placement test... (106.5 AMC 10, 5 AIME, 36/27 States/Nationals)
5 replies
1 viewing
pingpongmerrily
Yesterday at 9:01 PM
pingpongmerrily
5 hours ago
No more topics!
k Wrong Answers Only Pt.2
MathRook7817   72
N Apr 10, 2025 by MathRook7817
Problem: What is the area of a triangle with side lengths 13,14, and 15?
WRONG ANSWERS ONLY!

other one got locked for some reason
72 replies
MathRook7817
Apr 9, 2025
MathRook7817
Apr 10, 2025
Wrong Answers Only Pt.2
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G H BBookmark kLocked kLocked NReply
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MathRook7817
751 posts
#1 • 2 Y
Y by Exponent11, jkim0656
Problem: What is the area of a triangle with side lengths 13,14, and 15?
WRONG ANSWERS ONLY!

other one got locked for some reason
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Soupboy0
499 posts
#3
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$\sqrt{13^2+14^2-15^2} = 80$
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unicornlover9763
45 posts
#4
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(13+14)/15=1.8
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Mathdreams
1472 posts
#5
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$\sqrt(21 * 8 * 7 * 6) = 1$
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BossLu99
1402 posts
#7
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like 6-7
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Soupboy0
499 posts
#8
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Using the 1434 Geo Lemma, we find the area of the triangle to be $\frac{[\binom{13}{2}-\binom{13}{4}+\binom{14}{2}-\binom{14}{4}+\binom{15}{2}-\binom{15}{4}][14-15+13]}{13^2+15^2-\frac{14}{2}}(-1)^{\frac{1}{2}(13+14+15)}$


edit: this totally isn't the correct answer
This post has been edited 1 time. Last edited by Soupboy0, Apr 9, 2025, 8:49 PM
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Lankou
1406 posts
#9
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$13^{14^{15}}\approx 10^{10^{17.24}}$$
This post has been edited 1 time. Last edited by Lankou, Apr 9, 2025, 6:20 PM
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komodraqon
150 posts
#10
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13+14+15 = 131415
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maxamc
585 posts
#11
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the inradius is $\frac{13 \cdot 14 \cdot 15}{4 Area}$. it is also $\frac{Area}{\frac{13+14+15}{2}},$ solving gives $Area=\frac{21\sqrt{130}}{2}$.
This post has been edited 3 times. Last edited by maxamc, Apr 9, 2025, 6:50 PM
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iwastedmyusername
185 posts
#12 • 10 Y
Y by komodraqon, donut_bear, Exponent11, MathRook7817, huajun78, aidan0626, legospartan, Rice_Farmer, EthanNg6, efx
We want to find the area of a triangle with side lengths 13, 14, and 15. Now this might seem quite difficult, but there's a neat trick to find the area of this specific triangle. First, split the side with length 14 into two sides of lengths 5 and 9. Now, we want to connect the point connecting the sides with lengths 13 and 15 to the center of the splitting (idk). Hmmm, its not looking promising. Then I realized. What rhymes with five? Hive. Beehive. A beehive is shaped like a bunch of hexagons. Hexagons have 6 sides. So I found that the number 6 has something to do with this. Because its a triangle and triangles have 3 sides, and multiplication is the third operation most students learn in elementary school, I know it has to be 6 times something. What could 6 possibly be multiplied by? Then I realized that I haven't used the 9 yet. So I searched up on google what happened in the year 9. According to Gemini AI which is a very credible source, there was a devastating defeat for the Roman army led by Publiius Quinticlius Varus. Wait. P, Q, V. How many letters are in between Q and V? Let's see, Q, R, S, T, U. There are 4 letters. Because it's a triangle and it has 1 less side than a square, we want to find the square root. So 2. 2 * 6 = 12. Wait a minute. 5-12-13 and 9-12-15 are both right triangles. So I just need to find the area of two right triangles. This is fairly simple. The area of a right triangle is the product of its two legs, divided by 2. So the area of 30+54=84 right? No actually. You see, when you consider the transformations of a function, f(x+b) is actually a shift b to the LEFT of f(x). f(x-b) is a shift to the RIGHT of f(x). Wait a minute. There are two right triangles. And shifting to the right is adding NEGATIVE b. So we have to multiply 84 by -1. So the answer is -84. Q.E.D.
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ChickensEatGrass
51 posts
#14
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$13^2+14^2=15^2$ using the Pythagorean $- 140$ Theorem, so $13 \cdot 7=91$.
But, we mustn’t forget to add the $140$ back to make up for lost aura—which I lose every day sob sob sob. Therefore, our final answer is $231$.
This post has been edited 3 times. Last edited by ChickensEatGrass, Apr 9, 2025, 7:23 PM
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ilikemath247365
267 posts
#15
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Remember that $13 + 14 + 15 = 42$. Now, the area is 4 times the perimeter so $4 * 42 = 168$.
This post has been edited 1 time. Last edited by ilikemath247365, Apr 9, 2025, 7:26 PM
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Runner1600
12 posts
#16
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We know that $13=1+1+1+1+1+1+1+1+1+1+1+1+1$, right and $14=1+1+1+1+1+1+1+1+1+1+1+1+1+1$ and $15=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$. So according to the 5th Symphony by Beethoven, you have to plug in any form of adding 1s into the quadratic formula. So $a=1+1+1+1+1+1+1+1+1+1+1+1+1, b=1+1+1+1+1+1+1+1+1+1+1+1+1+1 c=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$. Plugging in, we get

$\frac{-(1+1+1+1+1+1+1+1+1+1+1+1+1+1) \pm \sqrt (1+1+1+1+1+1+1+1+1+1+1+1+1+1)^{2}-4(1+1+1+1+1+1+1+1+1+1+1+1+1)(1+1+1+1+1+1+1+1+1+1+1+1+1+1+1)} {2(1+1+1+1+1+1+1+1+1+1+1+1+1)}$

= $\frac{-14 \pm \sqrt (196-780)} {26}$
=$\frac{-7 \pm \sqrt -274} {13}$
So our two solutions are $\frac{-7-\sqrt-274} {13}$ and $\frac{-7+\sqrt-274} {13}$
This post has been edited 2 times. Last edited by Runner1600, Apr 9, 2025, 7:45 PM
Reason: Sillied, forgot about i
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Runner1600
12 posts
#17
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These are the two different areas
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Existing_Human1
214 posts
#18 • 1 Y
Y by EthanNg6
Runner1600 wrote:
We know that $13=1+1+1+1+1+1+1+1+1+1+1+1+1$, right and $14=1+1+1+1+1+1+1+1+1+1+1+1+1+1$ and $15=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$. So according to the 5th Symphony by Beethoven, you have to plug in any form of adding 1s into the quadratic formula. So $a=1+1+1+1+1+1+1+1+1+1+1+1+1, b=1+1+1+1+1+1+1+1+1+1+1+1+1+1 c=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$. Plugging in, we get

$\frac{-(1+1+1+1+1+1+1+1+1+1+1+1+1+1) \pm \sqrt (1+1+1+1+1+1+1+1+1+1+1+1+1+1)^{2}-4(1+1+1+1+1+1+1+1+1+1+1+1+1)(1+1+1+1+1+1+1+1+1+1+1+1+1+1+1)} {2(1+1+1+1+1+1+1+1+1+1+1+1+1)}$

= $\frac{-14 \pm \sqrt (196-780)} {26}$ =$\frac{-7 \pm -\sqrt 274} {13}$
So our two solutions are $\frac{-7-\sqrt274} {13}$ and $\frac{-7+\sqrt274} {13}$

Wait, I just realized re-listening to Beethoven's 5th Symphony:

Lyrics:
Da da da da
duhn duhn duhn duhn
My name is Beethoven
And what you should do
Is, whenever you have to find the area of a 13-14-15 triangle, plug the numbers into the quadratic formula
Da da da da
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