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$$ACEG$$ 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
1 year 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). · 1 year ago

So the answer is approach to $$16$$ ?

So what formula we use ? Both Pythagorean and Trigonometry ? · 1 year ago

Comment deleted May 13, 2016

If there is a way to find a simple expression for CE, I haven't found it. That certainly doesn't mean that there isn't one! I did find two not-so-simple equations in only CE ($$w$$ in the equation above) or only one angle in the diagram, and used the website wolframalpha.com to get solutions, which were produced numerically (computationally) instead of analytically (reducing the equation to simpler ones). I hope someone else here shows us a clever solution! · 1 year ago

I know this is a hard question, you said CE is approach to 16 right ? · 1 year ago

Yes, I think it was 15.98 or something like that. · 1 year ago

This discussion need more people to join. Why only 2 people :( ? · 1 year ago

I don't know. It's an interesting problem! · 1 year ago

Bring other people to join, maybe they can solve it :) · 1 year 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. · 1 year ago

LOL thx · 1 year ago

Jason, where did you find this problem? Are you sure there's the sort of clean solution we've been looking for? · 1 year ago

It's from a piece of paper that have been torn off by someone. I found it on school. · 1 year ago

I find the $$SOLUTION$$ !!!!

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity · 1 year ago

See told ya pythagoras would come handy · 1 year ago

YEAH XD · 1 year ago

I loved the way you solved the problem and thx.....guys for such a nice sol. Btw nice prob . Rishabh · 1 year ago

Oh hi pranay. Thanks for your encouragement. · 1 year 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? · 1 year ago

We solved it! With a little help of W/A · 1 year 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. · 1 year ago

Yeah you have told Pythagorean methi. We got several equations but not the answer, so,... · 1 year ago

Guys, i make a problem based on this discussion :) · 1 year ago

AHHHHHHHHHHHHHHHHHH · 1 year ago

Where ? · 1 year ago

Wait what !!! · 1 year ago

See abhays response above :( · 1 year ago

So we have found the answer and how to get it

Should i delete this discussion ? · 1 year ago

Nooooo, let it be there, we have worked very hard on this. · 1 year ago

Okay XD · 1 year ago

Phew thanks! · 1 year ago

@Abhiram Rao @Abhay Tiwari help plz · 1 year ago

Use Pythagoras Theorem. But the solution would be a very complex one. · 1 year 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 · 1 year 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. · 1 year 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 :

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

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

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

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

$$\LARGE \text{SOMEONE FIND}\space \Huge x \space \LARGE \text{PLZZZZ !!!!!! }$$ · 1 year ago

Sorry guys answer approximately remains the asme now correct answer is $$\boxed{15.988}$$ · 1 year ago

What i still got $$\sqrt{700}$$ using W/A · 1 year ago

No, answer is 15.988.I am sure. · 1 year ago

But E/A still shows $$\pm \sqrt{700}$$ · 1 year ago

Nahhh the answer is 15.98... · 1 year ago

I got my mistake 1 min 50 seconds ago XD · 1 year ago

LOL, okay now i'm gonna post this as a problem · 1 year 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 ;) · 1 year ago

Okay, give me a good title · 1 year ago

Hmmm lets see how is "Geometry with so many tricks" or "Geometry mania" or "When lines come togegher in a rectangle".. · 1 year ago

Because it might look simple but actually it's hard ( took us 3 day to solve LOL) · 1 year ago

Sure...,, nice · 1 year ago

Okay, thx guys XD · 1 year ago

Post it fast :) :) :) · 1 year ago

Nice representation, post the solution! · 1 year ago

Comment deleted May 14, 2016

The answer still the same LOL · 1 year ago

What? · 1 year ago

How is $$\sqrt{700}$$ ? W/A ? · 1 year ago

Calculated by hand man. · 1 year ago

500000% sure that this joker is joking. · 1 year ago

You simplify it by ur hand ? · 1 year ago

Haha, do you think abhay would have done that much donkeys work of simplifying it? · 1 year ago

LOL · 1 year ago

but this doesn't make any sense, try to draw it · 1 year ago

How can you say that? · 1 year ago

What? Why? · 1 year ago

You wolfram alpha user I was just about to comment but answer key says it is 15.9.. · 1 year ago

Don't believe on answers they can be wrong. · 1 year ago

Jason here in second last step it sould be 1600 - x^2 under the root and not 900-x^2 · 1 year ago

Kill me XD · 1 year ago

Nah we are too close, we can still find the answer, · 1 year ago

MISTAKE FOUND sqrt(1600) not sqrt(900)...You are wrong. · 1 year ago

Now, i am pretty sure this question is wrong...You didn't make any mistake. · 1 year 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 :) · 1 year ago

$$x$$ also $$15.9876...$$, try to input it · 1 year ago

I got this equation but isnt it too tedious? · 1 year ago

$$x=0$$...lol · 1 year ago

But $$x$$ also $$15.9876$$, try to input it !! · 1 year ago

No, putting $$x=15.9876$$ results in $$0.0005609440415437967\approx0.00$$ · 1 year ago

okay then $$x = 15.9876...$$

Mystery Solved ;) · 1 year ago

Looks like I am late · 1 year ago

Hmmm...Great..keep it up. :) · 1 year ago

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

From the expression above, prove that $$x= 15.9876...$$ ? · 1 year ago

But how is $$x = 15.9876...$$ · 1 year ago

Using wolfram alpha. ;) · 1 year ago

Cheats, cheater everywhere XD · 1 year ago

I spent hours simplifying and he used W/A · 1 year ago

Well, uhh is there any else solution for simplifying without W/A ? · 1 year ago

Lets see · 1 year ago

Man we did it, we solved it XD, i'm so proud of you guys , i'm gonna cry · 1 year ago

Whoohoo we invited many people who just turned their backs :/ Only me, you, Abhay, Mark and Abhiram took part. :/ · 1 year ago

LOL, yeah, so many people give up · 1 year ago

You cheeky fellow XD. Why didnt I think of that? I was simplifying. · 1 year 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. ;) · 1 year 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 · 1 year ago

Should i delete this discussion, we have already found the answer · 1 year ago

Yes delete it and post a similar question. · 1 year ago

Dont listen to this cheeky fellow, donr delete this plz dont. Abhay I shall deal with you later. XD :P · 1 year ago

LOL, i won't delete it. This discussion is meant to honor our sacrifices of time XD · 1 year 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 · 1 year ago

He made a mistake. · 1 year ago

What the.... noooo. · 1 year ago

Yesss · 1 year ago

Nooooooo, never · 1 year ago

AYS · 1 year ago

What is AYS? · 1 year ago

Are you sure XD · 1 year ago

I have done almost the same method, I have just woken up, sill catch up with u in a jiffy. · 1 year ago

good luck LOL · 1 year ago

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

I find the expression but can't find $$x$$ · 1 year ago

Use coordinate geometry · 1 year 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. · 1 year ago

Is there any other solution · 1 year ago

I think we should invite some seniors at brilliant to look for a solution to this problem.. · 1 year ago

Hello Jason, thanks for inviting to this problem XD. Somehow I got the answer for GF/FE . Trying to solve further. · 1 year ago

Okay, LOL · 1 year 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. · 1 year ago

$${\left(\dfrac{-x^4 + 1600x^2 + 50625}{9000}\right)}^{2} = \dfrac{x^2}{1600 - x^2}$$ · 1 year ago

That leads to $$x\approx 40$$. · 1 year ago

Yeah I know that something like that will come you know the exact value? This expression is nice · 1 year ago

But when i draw the figure, $$CE$$ isn't $$40$$, $$CE \approx 16$$ · 1 year ago

Yrah, I am just back yes so my approach is wrong but why wrong? I am figurong out. · 1 year ago

Should i tag Calvin Lin here XD · 1 year ago

About your expression, how did you get it ? · 1 year 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. · 1 year 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 · 1 year 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. · 1 year ago

Lol....What i used is sine rule and similarity and i think pythagoras is useless here.Need some construction. · 1 year ago

Yes, I uaed all three XD · 1 year ago

yeah, pythagoras useless in this case · 1 year ago

Same · 1 year ago

Hahaha · 1 year ago

I should tag Calvin Lin and Rishabh Cool here or should i ? · 1 year ago

That is entirely your wish, maybe rishabh cool first · 1 year ago

K @Rishabh Cool @Calvin Lin Help us XD · 1 year ago

Also @Paola Ramírez · 1 year ago

I'll draw some line inside the rectangle · 1 year ago

Maybe or outside... ;) · 1 year ago

We need some senior here · 1 year ago

Yeah, this problem is so nice i would throw my self out of my window · 1 year ago

Anyways....What's your age? · 1 year ago

13 · 1 year ago

LOL · 1 year 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... · 1 year ago

Ohh Okay my friend · 1 year ago

Hmmm...try it when you are free. · 1 year ago

Ashish, are you sure the equation above is true ?

I actually know the answer but dunno how to get it. $$CE \approx 16$$ · 1 year ago

Yes damn sure that it is true did yoi simplify it? · 1 year ago

Ahh come on, i'm stuck on simplyfying

My algebra level = 2 · 1 year 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, · 1 year ago

BTW, you can write how you simplify it so i can learn XD · 1 year ago

Sure I am heading out, anyways I shall write my solution if I get it. :) · 1 year ago

Okay · 1 year ago

I'm still simplyfying · 1 year ago

You mean $$x$$ is $$AC$$ ? · 1 year ago

x is CE · 1 year ago

Okay · 1 year ago

Comment deleted May 12, 2016

Ah! Use the fact I mentioned and focus on the angles $$\angle FCD$$ and $$\angle FED$$. Get an expression for the sine of one in terms of the sine of the other (using the law of sines), and then get an expression for $$CD+DE$$ in terms of DF and the sines of the angles. Let me know if that's not enough information and I'll be happy to put together a full solution tomorrow. (Game time in USA for Warriors basketball :).) · 1 year ago

Darn, i'm still stuck on sin law

I'm waiting till tomorrow okay :)

BTW, thx · 1 year ago

I think that's enough XD · 1 year ago

I actually will post it on problem after i know what is the answer and the solution · 1 year ago