$$ACEG$$ 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, 1 month ago

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

- 5 years, 1 month ago

Oh hi pranay. Thanks for your encouragement.

- 5 years, 1 month 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, 1 month ago

LOL thx

- 5 years, 1 month 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, 1 month ago

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

- 5 years, 1 month 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, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

- 5 years, 1 month ago

AHHHHHHHHHHHHHHHHHH

- 5 years, 1 month ago

- 5 years, 1 month ago

Where ?

- 5 years, 1 month ago

Wait what !!!

- 5 years, 1 month ago

See abhays response above :(

- 5 years, 1 month ago

So we have found the answer and how to get it

Should i delete this discussion ?

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Okay XD

- 5 years, 1 month ago

Phew thanks!

- 5 years, 1 month ago

@Abhiram Rao @Abhay Tiwari help plz

- 5 years, 1 month ago

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

- 5 years, 1 month 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, 1 month 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, 1 month 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, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

No, answer is 15.988.I am sure.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

- 5 years, 1 month ago

I got my mistake 1 min 50 seconds ago XD

- 5 years, 1 month ago

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

- 5 years, 1 month 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, 1 month ago

Okay, give me a good title

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Sure...,, nice

- 5 years, 1 month ago

Okay, thx guys XD

- 5 years, 1 month ago

Post it fast :) :) :)

- 5 years, 1 month ago

- 5 years, 1 month ago

Nice representation, post the solution!

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Kill me XD

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month 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, 1 month ago

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

- 5 years, 1 month ago

I got this equation but isnt it too tedious?

- 5 years, 1 month ago

$x=0$...lol

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

okay then $x = 15.9876...$

Mystery Solved ;)

- 5 years, 1 month ago

Looks like I am late

- 5 years, 1 month ago

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

- 5 years, 1 month 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, 1 month ago

But how is $x = 15.9876...$

- 5 years, 1 month 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, 1 month 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, 1 month ago

- 5 years, 1 month ago

Yes delete it and post a similar question.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

- 5 years, 1 month ago

What the.... noooo.

- 5 years, 1 month ago

Yesss

- 5 years, 1 month ago

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

- 5 years, 1 month 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, 1 month ago

Nooooooo, never

- 5 years, 1 month ago

Using wolfram alpha. ;)

- 5 years, 1 month ago

Cheats, cheater everywhere XD

- 5 years, 1 month ago

I spent hours simplifying and he used W/A

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Lets see

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

LOL, yeah, so many people give up

- 5 years, 1 month ago

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

- 5 years, 1 month ago

AYS

- 5 years, 1 month ago

What is AYS?

- 5 years, 1 month ago

Are you sure XD

- 5 years, 1 month ago

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

- 5 years, 1 month ago

good luck LOL

- 5 years, 1 month ago

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

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

- 5 years, 1 month 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, 1 month ago

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

- 5 years, 1 month ago

I find the $SOLUTION$ !!!!

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity

- 5 years, 1 month ago

See told ya pythagoras would come handy

- 5 years, 1 month ago

YEAH XD

- 5 years, 1 month ago

@Nihar Mahajan @Svatejas Shivakumar Help usssss

- 5 years, 1 month ago

- 5 years, 1 month ago

Use coordinate geometry

- 5 years, 1 month 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, 1 month ago

Is there any other solution

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Okay, LOL

- 5 years, 1 month 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, 1 month ago

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

- 5 years, 1 month ago

That leads to $x\approx 40$.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Should i tag Calvin Lin here XD

- 5 years, 1 month ago

- 5 years, 1 month 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, 1 month 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, 1 month ago

I'll draw some line inside the rectangle

- 5 years, 1 month ago

Maybe or outside... ;)

- 5 years, 1 month ago

We need some senior here

- 5 years, 1 month ago

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

- 5 years, 1 month ago

- 5 years, 1 month ago

13

- 5 years, 1 month 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, 1 month ago

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

- 5 years, 1 month ago

Yes, I uaed all three XD

- 5 years, 1 month ago

yeah, pythagoras useless in this case

- 5 years, 1 month ago

Same

- 5 years, 1 month ago

Hahaha

- 5 years, 1 month ago

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

- 5 years, 1 month ago

That is entirely your wish, maybe rishabh cool first

- 5 years, 1 month ago

K @Rishabh Cool @Calvin Lin Help us XD

- 5 years, 1 month ago

Also @Paola Ramírez

- 5 years, 1 month ago

LOL

- 5 years, 1 month 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, 1 month ago

Ohh Okay my friend

- 5 years, 1 month ago

Hmmm...try it when you are free.

- 5 years, 1 month 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, 1 month ago

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

- 5 years, 1 month ago

Ahh come on, i'm stuck on simplyfying

My algebra level = 2

- 5 years, 1 month 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, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Okay

- 5 years, 1 month ago

I'm still simplyfying

- 5 years, 1 month ago

You mean $x$ is $AC$ ?

- 5 years, 1 month ago

x is CE

- 5 years, 1 month ago

Okay

- 5 years, 1 month ago

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, 1 month ago

So the answer is approach to $16$ ?

So what formula we use ? Both Pythagorean and Trigonometry ?

- 5 years, 1 month ago