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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
6 months, 4 weeks 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). · 6 months, 4 weeks ago

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

So what formula we use ? Both Pythagorean and Trigonometry ? · 6 months, 4 weeks ago

Comment deleted 6 months ago

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! · 6 months, 4 weeks ago

I know this is a hard question, you said CE is approach to 16 right ? · 6 months, 4 weeks ago

Yes, I think it was 15.98 or something like that. · 6 months, 4 weeks ago

This discussion need more people to join. Why only 2 people :( ? · 6 months, 4 weeks ago

I don't know. It's an interesting problem! · 6 months, 4 weeks ago

Bring other people to join, maybe they can solve it :) · 6 months, 4 weeks 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. · 6 months, 3 weeks ago

LOL thx · 6 months, 3 weeks ago

Jason, where did you find this problem? Are you sure there's the sort of clean solution we've been looking for? · 6 months, 3 weeks ago

It's from a piece of paper that have been torn off by someone. I found it on school. · 6 months, 3 weeks ago

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

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity · 6 months, 3 weeks ago

See told ya pythagoras would come handy · 6 months, 3 weeks ago

YEAH XD · 6 months, 3 weeks ago

I loved the way you solved the problem and thx.....guys for such a nice sol. Btw nice prob . Rishabh · 6 months, 3 weeks ago

Oh hi pranay. Thanks for your encouragement. · 6 months, 3 weeks 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? · 6 months, 3 weeks ago

We solved it! With a little help of W/A · 6 months, 3 weeks 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. · 6 months, 3 weeks ago

Yeah you have told Pythagorean methi. We got several equations but not the answer, so,... · 6 months, 3 weeks ago

Guys, i make a problem based on this discussion :) · 6 months, 3 weeks ago

Your answer is wrong.Mistake founder-Abhay kumar. · 6 months, 3 weeks ago

AHHHHHHHHHHHHHHHHHH · 6 months, 3 weeks ago

Unfortunately u had made a mistake so the answer is wrong :( · 6 months, 3 weeks ago

Where ? · 6 months, 3 weeks ago

Wait what !!! · 6 months, 3 weeks ago

See abhays response above :( · 6 months, 3 weeks ago

So we have found the answer and how to get it

Should i delete this discussion ? · 6 months, 3 weeks ago

Nooooo, let it be there, we have worked very hard on this. · 6 months, 3 weeks ago

Okay XD · 6 months, 3 weeks ago

Phew thanks! · 6 months, 3 weeks ago

@Abhiram Rao @Abhay Tiwari help plz · 6 months, 3 weeks ago

Use Pythagoras Theorem. But the solution would be a very complex one. · 6 months, 3 weeks 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 · 6 months, 3 weeks 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. · 6 months, 3 weeks 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 :

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

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

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

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

$$\LARGE \text{SOMEONE FIND}\space \Huge x \space \LARGE \text{PLZZZZ !!!!!! }$$ · 6 months, 3 weeks ago

Sorry guys answer approximately remains the asme now correct answer is $$\boxed{15.988}$$ · 6 months, 3 weeks ago

What i still got $$\sqrt{700}$$ using W/A · 6 months, 3 weeks ago

No, answer is 15.988.I am sure. · 6 months, 3 weeks ago

But E/A still shows $$\pm \sqrt{700}$$ · 6 months, 3 weeks ago

Nahhh the answer is 15.98... · 6 months, 3 weeks ago

I got my mistake 1 min 50 seconds ago XD · 6 months, 3 weeks ago

LOL, okay now i'm gonna post this as a problem · 6 months, 3 weeks 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 ;) · 6 months, 3 weeks ago

Okay, give me a good title · 6 months, 3 weeks ago

Hmmm lets see how is "Geometry with so many tricks" or "Geometry mania" or "When lines come togegher in a rectangle".. · 6 months, 3 weeks ago

Because it might look simple but actually it's hard ( took us 3 day to solve LOL) · 6 months, 3 weeks ago

Sure...,, nice · 6 months, 3 weeks ago

Okay, thx guys XD · 6 months, 3 weeks ago

Post it fast :) :) :) · 6 months, 3 weeks ago

Nice representation, post the solution! · 6 months, 3 weeks ago

Comment deleted 6 months ago

The answer still the same LOL · 6 months, 3 weeks ago

What? · 6 months, 3 weeks ago

How is $$\sqrt{700}$$ ? W/A ? · 6 months, 3 weeks ago

Calculated by hand man. · 6 months, 3 weeks ago

500000% sure that this joker is joking. · 6 months, 3 weeks ago

You simplify it by ur hand ? · 6 months, 3 weeks ago

Haha, do you think abhay would have done that much donkeys work of simplifying it? · 6 months, 3 weeks ago

LOL · 6 months, 3 weeks ago

but this doesn't make any sense, try to draw it · 6 months, 3 weeks ago

How can you say that? · 6 months, 3 weeks ago

What? Why? · 6 months, 3 weeks ago

You wolfram alpha user I was just about to comment but answer key says it is 15.9.. · 6 months, 3 weeks ago

Don't believe on answers they can be wrong. · 6 months, 3 weeks ago

Jason here in second last step it sould be 1600 - x^2 under the root and not 900-x^2 · 6 months, 3 weeks ago

Kill me XD · 6 months, 3 weeks ago

Nah we are too close, we can still find the answer, · 6 months, 3 weeks ago

MISTAKE FOUND sqrt(1600) not sqrt(900)...You are wrong. · 6 months, 3 weeks ago

Now, i am pretty sure this question is wrong...You didn't make any mistake. · 6 months, 3 weeks 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 :) · 6 months, 3 weeks ago

$$x$$ also $$15.9876...$$, try to input it · 6 months, 3 weeks ago

I got this equation but isnt it too tedious? · 6 months, 3 weeks ago

$$x=0$$...lol · 6 months, 3 weeks ago

But $$x$$ also $$15.9876$$, try to input it !! · 6 months, 3 weeks ago

No, putting $$x=15.9876$$ results in $$0.0005609440415437967\approx0.00$$ · 6 months, 3 weeks ago

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

Mystery Solved ;) · 6 months, 3 weeks ago

Looks like I am late · 6 months, 3 weeks ago

Hmmm...Great..keep it up. :) · 6 months, 3 weeks ago

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

From the expression above, prove that $$x= 15.9876...$$ ? · 6 months, 3 weeks ago

But how is $$x = 15.9876...$$ · 6 months, 3 weeks ago

Using wolfram alpha. ;) · 6 months, 3 weeks ago

Cheats, cheater everywhere XD · 6 months, 3 weeks ago

I spent hours simplifying and he used W/A · 6 months, 3 weeks ago

Well, uhh is there any else solution for simplifying without W/A ? · 6 months, 3 weeks ago

Lets see · 6 months, 3 weeks ago

Man we did it, we solved it XD, i'm so proud of you guys , i'm gonna cry · 6 months, 3 weeks ago

Whoohoo we invited many people who just turned their backs :/ Only me, you, Abhay, Mark and Abhiram took part. :/ · 6 months, 3 weeks ago

LOL, yeah, so many people give up · 6 months, 3 weeks ago

You cheeky fellow XD. Why didnt I think of that? I was simplifying. · 6 months, 3 weeks 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. ;) · 6 months, 3 weeks 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 · 6 months, 3 weeks ago

Should i delete this discussion, we have already found the answer · 6 months, 3 weeks ago

Yes delete it and post a similar question. · 6 months, 3 weeks ago

Dont listen to this cheeky fellow, donr delete this plz dont. Abhay I shall deal with you later. XD :P · 6 months, 3 weeks ago

LOL, i won't delete it. This discussion is meant to honor our sacrifices of time XD · 6 months, 3 weeks 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 · 6 months, 3 weeks ago

He made a mistake. · 6 months, 3 weeks ago

What the.... noooo. · 6 months, 3 weeks ago

Yesss · 6 months, 3 weeks ago

Nooooooo, never · 6 months, 3 weeks ago

AYS · 6 months, 3 weeks ago

What is AYS? · 6 months, 3 weeks ago

Are you sure XD · 6 months, 3 weeks ago

I have done almost the same method, I have just woken up, sill catch up with u in a jiffy. · 6 months, 3 weeks ago

good luck LOL · 6 months, 3 weeks ago

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

I find the expression but can't find $$x$$ · 6 months, 3 weeks ago

@Nihar Mahajan @Svatejas Shivakumar Help usssss · 6 months, 3 weeks ago

Use coordinate geometry · 6 months, 4 weeks 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. · 6 months, 4 weeks ago

Is there any other solution · 6 months, 4 weeks ago

I think we should invite some seniors at brilliant to look for a solution to this problem.. · 6 months, 4 weeks ago

Hello Jason, thanks for inviting to this problem XD. Somehow I got the answer for GF/FE . Trying to solve further. · 6 months, 4 weeks ago

Okay, LOL · 6 months, 4 weeks 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. · 6 months, 4 weeks ago

$${\left(\dfrac{-x^4 + 1600x^2 + 50625}{9000}\right)}^{2} = \dfrac{x^2}{1600 - x^2}$$ · 6 months, 4 weeks ago

That leads to $$x\approx 40$$. · 6 months, 3 weeks ago

Yeah I know that something like that will come you know the exact value? This expression is nice · 6 months, 3 weeks ago

But when i draw the figure, $$CE$$ isn't $$40$$, $$CE \approx 16$$ · 6 months, 3 weeks ago

Yrah, I am just back yes so my approach is wrong but why wrong? I am figurong out. · 6 months, 3 weeks ago

Should i tag Calvin Lin here XD · 6 months, 3 weeks ago

About your expression, how did you get it ? · 6 months, 3 weeks 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. · 6 months, 3 weeks 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 · 6 months, 3 weeks 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. · 6 months, 3 weeks ago

Lol....What i used is sine rule and similarity and i think pythagoras is useless here.Need some construction. · 6 months, 3 weeks ago

Yes, I uaed all three XD · 6 months, 3 weeks ago

yeah, pythagoras useless in this case · 6 months, 3 weeks ago

Same · 6 months, 3 weeks ago

Hahaha · 6 months, 3 weeks ago

I should tag Calvin Lin and Rishabh Cool here or should i ? · 6 months, 3 weeks ago

That is entirely your wish, maybe rishabh cool first · 6 months, 3 weeks ago

K @Rishabh Cool @Calvin Lin Help us XD · 6 months, 3 weeks ago

Also @Paola Ramírez · 6 months, 3 weeks ago

I'll draw some line inside the rectangle · 6 months, 3 weeks ago

Maybe or outside... ;) · 6 months, 3 weeks ago

We need some senior here · 6 months, 3 weeks ago

Yeah, this problem is so nice i would throw my self out of my window · 6 months, 3 weeks ago

Anyways....What's your age? · 6 months, 3 weeks ago

13 · 6 months, 3 weeks ago

LOL · 6 months, 3 weeks 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... · 6 months, 3 weeks ago

Ohh Okay my friend · 6 months, 3 weeks ago

Hmmm...try it when you are free. · 6 months, 3 weeks ago

Ashish, are you sure the equation above is true ?

I actually know the answer but dunno how to get it. $$CE \approx 16$$ · 6 months, 4 weeks ago

Yes damn sure that it is true did yoi simplify it? · 6 months, 4 weeks ago

Ahh come on, i'm stuck on simplyfying

My algebra level = 2 · 6 months, 4 weeks 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, · 6 months, 4 weeks ago

BTW, you can write how you simplify it so i can learn XD · 6 months, 4 weeks ago

Sure I am heading out, anyways I shall write my solution if I get it. :) · 6 months, 4 weeks ago

Okay · 6 months, 4 weeks ago

I'm still simplyfying · 6 months, 4 weeks ago

You mean $$x$$ is $$AC$$ ? · 6 months, 4 weeks ago

x is CE · 6 months, 4 weeks ago

Okay · 6 months, 4 weeks ago

Comment deleted 6 months ago

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 :).) · 6 months, 4 weeks ago

Darn, i'm still stuck on sin law

I'm waiting till tomorrow okay :)

BTW, thx · 6 months, 4 weeks ago

I think that's enough XD · 6 months, 4 weeks ago

I actually will post it on problem after i know what is the answer and the solution · 6 months, 4 weeks ago