$$ACEG$$ is a rectangle. If segment $$\color {blue}{BE}$$ is $$30$$, segment $$\color{red}{CG}$$ is $$40$$, segment $$\color{green}{DF}$$ is $$15$$ and $$\angle FDE = 90^\circ$$. Find $$CE$$

This problem has been solved

Note by Jason Chrysoprase
5 years, 2 months ago

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I'm sorry for my previous incorrect comment. The approach I suggested leads to a trigonometric equation that I don't see an easy way to solve. Same for just using Pythagorean identities:

$CD = 15\frac{w}{\sqrt{40^2-w^2}} = w-15\frac{w}{\sqrt{30^2-w^2}}$

Wolfram Alpha has the same numerical solution for both, though: just under 16 (15.9876490085686).

- 5 years, 2 months ago

So the answer is approach to $16$ ?

So what formula we use ? Both Pythagorean and Trigonometry ?

- 5 years, 2 months ago

You guys rocked it.! Nice team work..& hats off to ur efforts..!! Me & my friends couldn't solve it.. I really need to learn a lot from people like you guys...congrats.

- 5 years, 2 months ago

LOL thx

- 5 years, 2 months ago

Jason, where did you find this problem? Are you sure there's the sort of clean solution we've been looking for?

- 5 years, 2 months ago

I find the $SOLUTION$ !!!!

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity

- 5 years, 2 months ago

See told ya pythagoras would come handy

- 5 years, 2 months ago

YEAH XD

- 5 years, 2 months ago

It's from a piece of paper that have been torn off by someone. I found it on school.

- 5 years, 2 months ago

Hello Jason, thanks for inviting to this problem XD. Somehow I got the answer for GF/FE . Trying to solve further.

- 5 years, 2 months ago

Okay, LOL

- 5 years, 2 months ago

Yeah, I think I got the way to solve. I have got an equation in x, but a lengthy one, I have to solve that. An you help me with that.

- 5 years, 2 months ago

${\left(\dfrac{-x^4 + 1600x^2 + 50625}{9000}\right)}^{2} = \dfrac{x^2}{1600 - x^2}$

- 5 years, 2 months ago

Okay

- 5 years, 2 months ago

You mean $x$ is $AC$ ?

- 5 years, 2 months ago

x is CE

- 5 years, 2 months ago

Ashish, are you sure the equation above is true ?

I actually know the answer but dunno how to get it. $CE \approx 16$

- 5 years, 2 months ago

Yes damn sure that it is true did yoi simplify it?

- 5 years, 2 months ago

I'm still simplyfying

- 5 years, 2 months ago

Ahh come on, i'm stuck on simplyfying

My algebra level = 2

- 5 years, 2 months ago

Yeah sure I will try simplifying that equation. I just have a couple of minutes left before I head out, so doing it fast ;) but will surely ping u as soon as I am back,

- 5 years, 2 months ago

Okay

- 5 years, 2 months ago

BTW, you can write how you simplify it so i can learn XD

- 5 years, 2 months ago

Sure I am heading out, anyways I shall write my solution if I get it. :)

- 5 years, 2 months ago

That leads to $x\approx 40$.

- 5 years, 2 months ago

Yeah I know that something like that will come you know the exact value? This expression is nice

- 5 years, 2 months ago

But when i draw the figure, $CE$ isn't $40$, $CE \approx 16$

- 5 years, 2 months ago

Yrah, I am just back yes so my approach is wrong but why wrong? I am figurong out.

- 5 years, 2 months ago

- 5 years, 2 months ago

I assumed CE as x. Then we know that CG is 40. Applying pythagoras' theorem, we get GE = $\sqrt{1600 - x^2}$. Now $\triangle{FDC} ~ \triangle{GEC}$ So, $\dfrac{GE}{FD} = \dfrac{CE}{CD}$.. and so.. oh I get my mistake.

- 5 years, 2 months ago

I got too many things but not the answer.
1)450=BC×FE
2)600=CF×EG
3)4/3=FG/FE
4)600=BC×FG
5)BC/CF=EG/FG
6)1200=GE×BC
7)8/3=GE/FE
8)2CF=BC
9)2FG=GE
10)225=CF×FE
11)300=CF×FG

- 5 years, 2 months ago

Yes yes , I have got the same observations. I have got 16 of them XD. But not the answer XD. Thats why this question is nice. CE is not coming.

- 5 years, 2 months ago

Same

- 5 years, 2 months ago

Hahaha

- 5 years, 2 months ago

I should tag Calvin Lin and Rishabh Cool here or should i ?

- 5 years, 2 months ago

That is entirely your wish, maybe rishabh cool first

- 5 years, 2 months ago

K @Rishabh Cool @Calvin Lin Help us XD

- 5 years, 2 months ago

Also @Paola Ramírez

- 5 years, 2 months ago

Lol....What i used is sine rule and similarity and i think pythagoras is useless here.Need some construction.

- 5 years, 2 months ago

Yes, I uaed all three XD

- 5 years, 2 months ago

yeah, pythagoras useless in this case

- 5 years, 2 months ago

Yeah, this problem is so nice i would throw my self out of my window

- 5 years, 2 months ago

- 5 years, 2 months ago

13

- 5 years, 2 months ago

We need some senior here

- 5 years, 2 months ago

I'll draw some line inside the rectangle

- 5 years, 2 months ago

Maybe or outside... ;)

- 5 years, 2 months ago

LOL

- 5 years, 2 months ago

I would try my best only after I get some time.... But I am very very busy at the moment... So please forgive me.... ("_")

PS:Maybe @Nihar Mahajan could be the right guy to consult here as his geometry skills are very good...

- 5 years, 2 months ago

Hmmm...try it when you are free.

- 5 years, 2 months ago

Ohh Okay my friend

- 5 years, 2 months ago

Should i tag Calvin Lin here XD

- 5 years, 2 months ago

Use coordinate geometry

- 5 years, 2 months ago

Is there any other solution

- 5 years, 2 months ago

I think we should invite some seniors at brilliant to look for a solution to this problem..

- 5 years, 2 months ago

Can u please post your solution ? It would be very nice because this problem is making very lengthy and tedious eq. So plz help us.

- 5 years, 2 months ago

- 5 years, 2 months ago

@Nihar Mahajan @Svatejas Shivakumar Help usssss

- 5 years, 2 months ago

I have found the solution, use the rule of Pythagoras and Similarity

I need someone to FIND $x$this expression below ( $x = CE$)

$\Huge 0 = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}}) - 15x$

I'll explain how to get the expression above after someone find x

- 5 years, 2 months ago

Okay, that says that the area of the whole rectangle = area of bottom 15 of rectangle (15x) plus that same 15x area divided by the height where the line segment of length 30 meets the left side of the rectangle. That can't be right. In any case, we have good equations in x that "just need to be solved/simplified". The problem is that it is not at all clear that they can be.

Edit: you've changed the equation since I wrote my comment. Anyway, to spend more energy on this I would want some stronger reason to believe that there is a simple solution to be found.

- 5 years, 2 months ago

I'm serious, input $x= 15.9876$, the expression above

I find the expression but can't find $x$

- 5 years, 2 months ago

Assume that : $CE = x$, $DE = y$

That's mean $CD = x-y$

Using Pythagoras :

$BC = \sqrt{900 - x^2}$

$EG = \sqrt{1600 - x^2}$

Using Similiar Triangles :

$\LARGE \frac{FD}{BC} = \frac{DE}{CE} \\ \LARGE \frac{15}{\sqrt{900-x^2}} = \frac{y}{x} \\ \LARGE y =\frac{15x}{\sqrt{900-x^2}}$

$\LARGE \frac{FD}{EG} = \frac{CD}{CE} \\ \LARGE \frac{15}{\sqrt{1600-x^2}} = \frac{x-y}{x} \\ \LARGE \frac{15}{\sqrt{1600-x^2}} = \frac{x-\frac{15x}{\sqrt{900-x^2}}}{x}$

$\LARGE 15x = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}})$

$\LARGE 0 = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}}) - 15x$

$\LARGE \text{SOMEONE FIND}\space \Huge x \space \LARGE \text{PLZZZZ !!!!!! }$

- 5 years, 2 months ago

I have done almost the same method, I have just woken up, sill catch up with u in a jiffy.

- 5 years, 2 months ago

good luck LOL

- 5 years, 2 months ago

$x=0$...lol

- 5 years, 2 months ago

AYS

- 5 years, 2 months ago

What is AYS?

- 5 years, 2 months ago

Are you sure XD

- 5 years, 2 months ago

But $x$ also $15.9876$, try to input it !!

- 5 years, 2 months ago

No, putting $x=15.9876$ results in $0.0005609440415437967\approx0.00$

- 5 years, 2 months ago

okay then $x = 15.9876...$

Mystery Solved ;)

- 5 years, 2 months ago

Hmmm...Great..keep it up. :)

- 5 years, 2 months ago

But how is $x = 15.9876...$

- 5 years, 2 months ago

Using wolfram alpha. ;)

- 5 years, 2 months ago

You cheeky fellow XD. Why didnt I think of that? I was simplifying.

- 5 years, 2 months ago

Cheats, cheater everywhere XD

- 5 years, 2 months ago

I spent hours simplifying and he used W/A

- 5 years, 2 months ago

Well, uhh is there any else solution for simplifying without W/A ?

- 5 years, 2 months ago

Lets see

- 5 years, 2 months ago

Man we did it, we solved it XD, i'm so proud of you guys , i'm gonna cry

- 5 years, 2 months ago

Whoohoo we invited many people who just turned their backs :/ Only me, you, Abhay, Mark and Abhiram took part. :/

- 5 years, 2 months ago

LOL, yeah, so many people give up

- 5 years, 2 months ago

Actually, x=-15.9876,15.9876 and 0....but 0 and -15.9876 can not be the answer .Therefore correct answer is 15.9876. ;)

- 5 years, 2 months ago

Actually, I know how to use W/A and I have seen that there has been an attempt to find such an answer. :P

- 5 years, 2 months ago

- 5 years, 2 months ago

Nooooooo, never

- 5 years, 2 months ago

Yes delete it and post a similar question.

- 5 years, 2 months ago

Dont listen to this cheeky fellow, donr delete this plz dont. Abhay I shall deal with you later. XD :P

- 5 years, 2 months ago

LOL, i won't delete it. This discussion is meant to honor our sacrifices of time XD

- 5 years, 2 months ago

Nice, just update the note to "this problem has been solved". A very nice question . Claps for you for posting this, Atlast its pythagoras which solved this question and not sine rule or similarity XD

- 5 years, 2 months ago

- 5 years, 2 months ago

What the.... noooo.

- 5 years, 2 months ago

Yesss

- 5 years, 2 months ago

$\LARGE 0 = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}}) - 15x$

From the expression above, prove that $x= 15.9876...$ ?

- 5 years, 2 months ago

Looks like I am late

- 5 years, 2 months ago

I got this equation but isnt it too tedious?

- 5 years, 2 months ago

Now, i am pretty sure this question is wrong...You didn't make any mistake.

- 5 years, 2 months ago

Yep, this queetion may be wrong but lets see if we can get some easier method. Oh! I forgot @Jon Haussmann maybe you can help :)

- 5 years, 2 months ago

$x$ also $15.9876...$, try to input it

- 5 years, 2 months ago

MISTAKE FOUND sqrt(1600) not sqrt(900)...You are wrong.

- 5 years, 2 months ago

Jason here in second last step it sould be 1600 - x^2 under the root and not 900-x^2

- 5 years, 2 months ago

Kill me XD

- 5 years, 2 months ago

Nah we are too close, we can still find the answer,

- 5 years, 2 months ago

Sorry guys answer approximately remains the asme now correct answer is $\boxed{15.988}$

- 5 years, 2 months ago

What i still got $\sqrt{700}$ using W/A

- 5 years, 2 months ago

No, answer is 15.988.I am sure.

- 5 years, 2 months ago

But E/A still shows $\pm \sqrt{700}$

- 5 years, 2 months ago

- 5 years, 2 months ago

I got my mistake 1 min 50 seconds ago XD

- 5 years, 2 months ago

LOL, okay now i'm gonna post this as a problem

- 5 years, 2 months ago

Along with your solution. I want you to post the solution because you were the one who held this contest. Copy and paste this solution above replace 900 with 1600 thats all Good luck ;)

- 5 years, 2 months ago

Okay, give me a good title

- 5 years, 2 months ago

Because it might look simple but actually it's hard ( took us 3 day to solve LOL)

- 5 years, 2 months ago

Sure...,, nice

- 5 years, 2 months ago

Okay, thx guys XD

- 5 years, 2 months ago

Post it fast :) :) :)

- 5 years, 2 months ago

- 5 years, 2 months ago

Nice representation, post the solution!

- 5 years, 2 months ago

Hmmm lets see how is "Geometry with so many tricks" or "Geometry mania" or "When lines come togegher in a rectangle"..

- 5 years, 2 months ago

@Abhiram Rao @Abhay Tiwari help plz

- 5 years, 2 months ago

Use Pythagoras Theorem. But the solution would be a very complex one.

- 5 years, 2 months ago

So we have found the answer and how to get it

Should i delete this discussion ?

- 5 years, 2 months ago

Nooooo, let it be there, we have worked very hard on this.

- 5 years, 2 months ago

Okay XD

- 5 years, 2 months ago

Phew thanks!

- 5 years, 2 months ago

Guys, i make a problem based on this discussion :)

- 5 years, 2 months ago

- 5 years, 2 months ago

Wait what !!!

- 5 years, 2 months ago

See abhays response above :(

- 5 years, 2 months ago

Where ?

- 5 years, 2 months ago

- 5 years, 2 months ago

AHHHHHHHHHHHHHHHHHH

- 5 years, 2 months ago

I'm confused by all the many comments. Has someone found a solution that simplifies, or are we still stuck with equations we need a computer to solve?

- 5 years, 2 months ago

We solved it! With a little help of W/A

- 5 years, 2 months ago

But we'd (I'd) already done that two days ago. Sort by top comments & the first is mine from two days ago. I still wonder if there is a W/A-free way to do it.

- 5 years, 2 months ago

Yeah you have told Pythagorean methi. We got several equations but not the answer, so,...

- 5 years, 2 months ago

I loved the way you solved the problem and thx.....guys for such a nice sol. Btw nice prob . Rishabh

- 5 years, 2 months ago

Oh hi pranay. Thanks for your encouragement.

- 5 years, 2 months ago