I dare you to solve this question

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Mathkiddie

322 posts

Mathkiddie

#1 h Jan 30, 2024, 1:15 AM • 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

eduD_looC

#2 h Jan 30, 2024, 1:31 AM • 1 Y

Y by Spiritpalm

Click to reveal hidden text

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northstar47

330 posts

northstar47

#3 h Jan 30, 2024, 1:37 AM

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eduD_looC wrote:

Click to reveal hidden text

...what explain pls im stupid

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ostriches88

1523 posts

ostriches88

#4 h Jan 30, 2024, 1:38 AM

Y by

sol (cheese sorry)

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ostriches88

1523 posts

ostriches88

#5 h Jan 30, 2024, 1:39 AM

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eduD_looC wrote:

Click to reveal hidden text

epic geo proof but answer is negative </3

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paganiniana

211 posts

paganiniana

#6 h Jan 30, 2024, 1:42 AM

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

ostriches88

#7 h Jan 30, 2024, 1:45 AM

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

Mathkiddie

#8 h Jan 30, 2024, 2:10 AM

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

Mathkiddie

#9 h Jan 30, 2024, 2:11 AM

Y by

tbh this problem looks like something that would appear on the AIME exam

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fruitmonster97

2391 posts

fruitmonster97

#10 h Jan 30, 2024, 2:17 AM

Y by

Cool Problem
Why is this wrong?

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Mathkiddie

322 posts

Mathkiddie

#11 h Jan 30, 2024, 3:32 AM

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

ethanzhang1001

#12 h Jan 30, 2024, 4:01 AM

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

aleyang

#13 h Jan 30, 2024, 5:34 AM

Y by

i smell law of cosines

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VeritasTian

3128 posts

VeritasTian

#14 h Jan 30, 2024, 6:36 AM

Y by

why does it cinda familiar

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SomeonecoolLovesMaths

3137 posts

SomeonecoolLovesMaths

#15 h Jan 30, 2024, 1:16 PM

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

fruitmonster97

#16 h Jan 30, 2024, 1:18 PM

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

Mathkiddie

#17 h Jan 31, 2024, 2:56 AM • 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

Maheshwarananda

#18 h Jan 31, 2024, 10:00 AM

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

3137 posts

SomeonecoolLovesMaths

#19 h Jan 31, 2024, 5:47 PM

Y by

Mathkiddie wrote:

Solution

Can someone help me with this part?

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Maheshwarananda

259 posts

Maheshwarananda

#20 h Feb 1, 2024, 5:24 AM

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

ostriches88

#21 h Feb 1, 2024, 4:02 PM

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

3137 posts

SomeonecoolLovesMaths

#22 h Feb 2, 2024, 10:49 AM

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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