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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Inequalities
sqing   4
N 25 minutes ago by sqing
Let $a,b,c\ge \frac{1}{2}$ and $\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\le 1. $ Prove that
$$a+b+c\geq 2$$Let $a,b,c\ge \frac{1}{2}$ and $ \left(a+\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(a+\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\le \frac{9}{2}. $ Prove that
$$a^2+b^2+c^2\geq 1$$Let $a,b\ge \frac{1}{2}$ and $ \left( \frac{1}{a}-\frac{1}{b}+2\right)\left( \frac{1}{b}-\frac{1}{a}+2\right) \le   \frac{20}{9}. $ Prove that
$$ a+b\geq 2$$Let $a,b\ge \frac{1}{2}$ and $a^2+b^2=1. $ Prove that
$$\left(\frac{2}{a}+\frac{1}{b}-1\right)\left(\frac{2}{a}-\frac{1}{b}+1\right)\ge \frac{13}{3}$$
4 replies
sqing
Mar 15, 2025
sqing
25 minutes ago
Binomial Summation
Saucepan_man02   0
2 hours ago
Given that: $l_n = \sum_{r=1}^{n-1} \frac{2r^2-r(n-2)+1}{(n-r) \binom{n}{r}}$, find $l_{15}$.
0 replies
Saucepan_man02
2 hours ago
0 replies
Inequality with integers and indices
Michael Niland   5
N 2 hours ago by ohiorizzler1434
Without using a calculator, show that $2^{\frac{1}{2}}$ $< 3^{\frac{1}{3}}$, and for a positive integer $n\geq3$, prove that

$n^{\frac{1}{n}}$ $>(n+1)^{\frac{1}{n+1}}$.
5 replies
Michael Niland
Yesterday at 6:52 AM
ohiorizzler1434
2 hours ago
Quadratic Equation Problems
Saucepan_man02   4
N 3 hours ago by Saucepan_man02
P1) Let $\alpha < \beta <	\gamma$ be the roots of $ax^3+bx^2+cx+d=0$ with $7a+3b>0, 3a+2b<0$ and $\alpha(\beta+1)+\beta(\gamma+1)+\gamma(1+\alpha)=0$.
If [.] denotes the greatest integer function, then the number of possible values of $[|3 \alpha |] + [|6 \beta|] +
 [|9 \gamma |]$ is equal to
4 replies
Saucepan_man02
Mar 3, 2025
Saucepan_man02
3 hours ago
No more topics!
I dare you to solve this question
Mathkiddie   21
N Feb 2, 2024 by SomeonecoolLovesMaths
Hi people, I dare you to solve this insanely hard question I came up with:

Given \begin{align*}x^2+xy+y^2&=25,\\ x^2+xz+z^2&=49,\\ y^2+yz+z^2&=64\\\end{align*}find $xy+xz+yz$. Please put your solution in [code]Click to reveal hidden text[/code] tags so you don't spoil the solution for others (I think this is a really nice problem!)
21 replies
Mathkiddie
Jan 30, 2024
SomeonecoolLovesMaths
Feb 2, 2024
I dare you to solve this question
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G H BBookmark kLocked kLocked NReply
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Mathkiddie
322 posts
#1 • 1 Y
Y by Spiritpalm
Hi people, I dare you to solve this insanely hard question I came up with:

Given \begin{align*}x^2+xy+y^2&=25,\\ x^2+xz+z^2&=49,\\ y^2+yz+z^2&=64\\\end{align*}find $xy+xz+yz$. Please put your solution in
[hide][/hide]
tags so you don't spoil the solution for others (I think this is a really nice problem!)
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eduD_looC
6609 posts
#2 • 1 Y
Y by Spiritpalm
Click to reveal hidden text
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northstar47
330 posts
#3
Y by
eduD_looC wrote:
Click to reveal hidden text

...what explain pls im stupid
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ostriches88
1523 posts
#4
Y by
sol (cheese sorry)
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ostriches88
1523 posts
#5
Y by
eduD_looC wrote:
Click to reveal hidden text

epic geo proof but answer is negative </3
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paganiniana
211 posts
#6
Y by
ostriches88 wrote:
sol (cheese sorry)

i used similar solution but i got Click to reveal hidden text

same answer anyways
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ostriches88
1523 posts
#7
Y by
paganiniana wrote:
ostriches88 wrote:
sol (cheese sorry)

i used similar solution but i got Click to reveal hidden text

same answer anyways

yeah that works too since the parity of $x^2$ and $z^2$ don't depend on the parity of $x$ and $z$ so really just $xz$ needs to be negative meaning that either $x$ or $z$ can be the negative, which somehow i didn't realize lol
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Mathkiddie
322 posts
#8
Y by
@eduD_looC congratulations click me is correct (I used the same method as you!) @ostriches88 close, you used a really clever method though. I'll probably reveal my solution tomorow.
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Mathkiddie
322 posts
#9
Y by
tbh this problem looks like something that would appear on the AIME exam
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fruitmonster97
2391 posts
#10
Y by
Cool Problem
Why is this wrong?
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Mathkiddie
322 posts
#11
Y by
@fruitmonster97 Can you explain how you got $\frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}=138-2a$?
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ethanzhang1001
1060 posts
#12
Y by
Actually, both answers of 40 and -40 are correct.
Sure, 40 works, but -40 also works. (5, 0, -8) satisfies the equations, so it must be an answer.
This post has been edited 1 time. Last edited by ethanzhang1001, Jan 30, 2024, 4:02 AM
Reason: typo
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aleyang
192 posts
#13
Y by
i smell law of cosines
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VeritasTian
3128 posts
#14
Y by
why does it cinda familiar
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SomeonecoolLovesMaths
3136 posts
#15
Y by
eduD_looC wrote:
Click to reveal hidden text

How did you get this? Can you please show your working?
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fruitmonster97
2391 posts
#16
Y by
Mathkiddie wrote:
@fruitmonster97 Can you explain how you got $\frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}=138-2a$?
Stuff

Edit: This problem is very similar to 2005 Mildorf Mock AIME #10
This post has been edited 1 time. Last edited by fruitmonster97, Jan 30, 2024, 6:28 PM
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Mathkiddie
322 posts
#17 • 1 Y
Y by ESAOPS
Guys, I'm going to reveal the solution (however, you can continue trying to solve this problem without looking at it!). Solution Also @fruitmonster97, this problem is really similar to that lol
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Maheshwarananda
259 posts
#18
Y by
Yes, but we do not know if $x,y,z$ can be or not distances, i.e. they are positive or not
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SomeonecoolLovesMaths
3136 posts
#19
Y by
Mathkiddie wrote:
Solution
Can someone help me with this part?
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Maheshwarananda
259 posts
#20
Y by
Sure:
We write the cosine law in triangles $PAB,PBC,PCA$ $\cos (\widehat{APB})=\frac{{{x}^{2}}+{{y}^{2}}-25}{2xy}=\frac{-xy}{2xy}=-\frac{1}{2}$.
$\cos (\widehat{BPC})=\frac{{{y}^{2}}+{{z}^{2}}-64}{2yz}=\frac{-yz}{2yz}=-\frac{1}{2}$, $\cos (\widehat{APC})=\frac{{{x}^{2}}+{{z}^{2}}-25}{2xz}=\frac{-xz}{2xz}=-\frac{1}{2}$. Hence$\widehat{APB}=\widehat{BPC}=\widehat{CPA}=120{}^\circ $.
$T=T(ABC)=T(APB)+T(BPC)+T(CPA)$. $T(APB)=\frac{xy\sin {{120}^{\circ }}}{2},T(BPC)=\frac{yz\sin {{120}^{\circ }}}{2},$$T(CPA)=\frac{xz\sin {{120}^{\circ }}}{2}$. So $T=\frac{\sqrt{3}}{4}(xy+yz+zx)$. Using Heron formula we have $T=\sqrt{p(p-a)(p-b)(p-c)}=10\sqrt{3}$, because $a=8,b=7,c=5$. Hence $10\sqrt{3}=\frac{\sqrt{3}}{4}(xy+yz+zx)$, i.e. $xy+yz+zx=40$.
(T= area)
This post has been edited 1 time. Last edited by Maheshwarananda, Feb 1, 2024, 5:25 AM
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ostriches88
1523 posts
#21
Y by
@Mathkiddie i tried the LoC solution and got 40 as the answer, but I remember from doing problems similar to this that you have to check for parity, and so I tried a different solution and got -40.

if 40 is your answer, what are your values of $x, y, z$? because $(-5, 0, 8)$ and $(5, 0, -8)$ both work and give an answer of $-40$ so maybe there are two possible answers?
This post has been edited 2 times. Last edited by ostriches88, Feb 1, 2024, 4:04 PM
Reason: edited for clarity. solutions -> answers
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SomeonecoolLovesMaths
3136 posts
#22
Y by
Maheshwarananda wrote:
Sure:
We write the cosine law in triangles $PAB,PBC,PCA$ $\cos (\widehat{APB})=\frac{{{x}^{2}}+{{y}^{2}}-25}{2xy}=\frac{-xy}{2xy}=-\frac{1}{2}$.
$\cos (\widehat{BPC})=\frac{{{y}^{2}}+{{z}^{2}}-64}{2yz}=\frac{-yz}{2yz}=-\frac{1}{2}$, $\cos (\widehat{APC})=\frac{{{x}^{2}}+{{z}^{2}}-25}{2xz}=\frac{-xz}{2xz}=-\frac{1}{2}$. Hence$\widehat{APB}=\widehat{BPC}=\widehat{CPA}=120{}^\circ $.
$T=T(ABC)=T(APB)+T(BPC)+T(CPA)$. $T(APB)=\frac{xy\sin {{120}^{\circ }}}{2},T(BPC)=\frac{yz\sin {{120}^{\circ }}}{2},$$T(CPA)=\frac{xz\sin {{120}^{\circ }}}{2}$. So $T=\frac{\sqrt{3}}{4}(xy+yz+zx)$. Using Heron formula we have $T=\sqrt{p(p-a)(p-b)(p-c)}=10\sqrt{3}$, because $a=8,b=7,c=5$. Hence $10\sqrt{3}=\frac{\sqrt{3}}{4}(xy+yz+zx)$, i.e. $xy+yz+zx=40$.
(T= area)

Wow! Thanks a lot!
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