\(ACEG\) 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 \(CE\)

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:

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!

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.

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.

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.

@Jason Chrysoprase
–
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 ;)

@Jason Chrysoprase
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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

@Jason Chrysoprase
–
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.

@Ashish Siva
–
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

@Abhay Kumar
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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.

@Jason Chrysoprase
–
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,

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 :).)

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## Comments

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TopNewestI'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).

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So the answer is approach to \(16\) ?

So what formula we use ? Both Pythagorean and Trigonometry ?

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Comment deleted May 13, 2016

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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!

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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.

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LOL thx

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

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

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I find the \(SOLUTION\) !!!!

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity

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See told ya pythagoras would come handy

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

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Oh hi pranay. Thanks for your encouragement.

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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?

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

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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.

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

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Your answer is wrong.Mistake founder-Abhay kumar.

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AHHHHHHHHHHHHHHHHHH

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Unfortunately u had made a mistake so the answer is wrong :(

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Where ?

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Wait what !!!

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So we have found the answer and how to get it

Should i delete this discussion ?

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

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Okay XD

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@Abhiram Rao @Abhay Tiwari help plz

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

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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

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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.

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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 :

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

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

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

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

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

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

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click here

Done,Log in to reply

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Comment deleted May 14, 2016

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@Jon Haussmann maybe you can help :)

Yep, this queetion may be wrong but lets see if we can get some easier method. Oh! I forgotLog in to reply

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Mystery Solved ;)

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From the expression above, prove that \(x= 15.9876...\) ?

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@Mark C @Abhay Kumar @Ashish Siva @Rishabh Tiwari @Rohit Udaiwal

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I'm serious, input \(x= 15.9876\), the expression above

I find the expression but can't find \(x\)

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@Nihar Mahajan @Svatejas Shivakumar Help usssss

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@ahmad saad

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Use coordinate geometry

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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.

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Is there any other solution

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

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

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Okay, LOL

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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.

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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

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@Rishabh Cool @Calvin Lin Help us XD

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@Paola Ramírez

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

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

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My algebra level = 2

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Comment deleted May 12, 2016

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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 :).)

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Darn, i'm still stuck on sin law

I'm waiting till tomorrow okay :)

BTW, thx

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I think that's enough XD

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

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Because i don't know the answer

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