\(ACEG\) 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:

So what formula we use ? Both Pythagorean and Trigonometry ?
–
Jason Chrysoprase
·
1 year, 4 months ago

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Comment deleted
May 13, 2016

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@Jason Chrysoprase
–
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!
–
Mark C
·
1 year, 4 months ago

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@Mark C
–
I know this is a hard question, you said CE is approach to 16 right ?
–
Jason Chrysoprase
·
1 year, 4 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.
–
Rishabh Tiwari
·
1 year, 4 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?
–
Mark C
·
1 year, 4 months ago

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@Mark C
–
We solved it! With a little help of W/A
–
Ashish Siva
·
1 year, 4 months ago

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@Ashish Siva
–
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.
–
Mark C
·
1 year, 4 months ago

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@Mark C
–
Yeah you have told Pythagorean methi. We got several equations but not the answer, so,...
–
Ashish Siva
·
1 year, 4 months ago

I'll explain how to get the expression above after someone find x
–
Jason Chrysoprase
·
1 year, 4 months ago

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@Jason Chrysoprase
–
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.
–
Mark C
·
1 year, 4 months ago

@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 ;)
–
Ashish Siva
·
1 year, 4 months ago

@Jason Chrysoprase
–
Hmmm lets see how is "Geometry with so many tricks" or "Geometry mania" or "When lines come togegher in a rectangle"..
–
Ashish Siva
·
1 year, 4 months ago

@Jason Chrysoprase
–
Now, i am pretty sure this question is wrong...You didn't make any mistake.
–
Abhay Kumar
·
1 year, 4 months ago

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@Abhay Kumar
–
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 :)
–
Ashish Siva
·
1 year, 4 months ago

@Ashish Siva
–
Man we did it, we solved it XD, i'm so proud of you guys , i'm gonna cry
–
Jason Chrysoprase
·
1 year, 4 months ago

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@Jason Chrysoprase
–
Whoohoo we invited many people who just turned their backs :/ Only me, you, Abhay, Mark and Abhiram took part. :/
–
Ashish Siva
·
1 year, 4 months ago

@Abhay Kumar
–
You cheeky fellow XD. Why didnt I think of that? I was simplifying.
–
Ashish Siva
·
1 year, 4 months ago

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@Jason Chrysoprase
–
Actually, x=-15.9876,15.9876 and 0....but 0 and -15.9876 can not be the answer .Therefore correct answer is 15.9876. ;)
–
Abhay Kumar
·
1 year, 4 months ago

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@Abhay Kumar
–
Actually, I know how to use W/A and I have seen that there has been an attempt to find such an answer. :P
–
Ashish Siva
·
1 year, 4 months ago

@Abhay Kumar
–
Dont listen to this cheeky fellow, donr delete this plz dont. Abhay I shall deal with you later. XD :P
–
Ashish Siva
·
1 year, 4 months ago

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@Ashish Siva
–
LOL, i won't delete it. This discussion is meant to honor our sacrifices of time XD
–
Jason Chrysoprase
·
1 year, 4 months ago

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@Jason Chrysoprase
–
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
–
Ashish Siva
·
1 year, 4 months ago

@Aakash Khandelwal
–
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.
–
Rishabh Tiwari
·
1 year, 4 months ago

@Jason Chrysoprase
–
I think we should invite some seniors at brilliant to look for a solution to this problem..
–
Rishabh Tiwari
·
1 year, 4 months ago

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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.
–
Ashish Siva
·
1 year, 4 months ago

@Jason Chrysoprase
–
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.
–
Ashish Siva
·
1 year, 4 months ago

@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
·
1 year, 4 months ago

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@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
·
1 year, 4 months ago

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@Abhay Kumar
–
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.
–
Ashish Siva
·
1 year, 4 months ago

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@Ashish Siva
–
Lol....What i used is sine rule and similarity and i think pythagoras is useless here.Need some construction.
–
Abhay Kumar
·
1 year, 4 months ago

@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,
–
Ashish Siva
·
1 year, 4 months ago

@Mark C
–
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 :).)
–
Mark C
·
1 year, 4 months ago

## Comments

Sort by:

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). – Mark C · 1 year, 4 months ago

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So what formula we use ? Both Pythagorean and Trigonometry ? – Jason Chrysoprase · 1 year, 4 months ago

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Log in to reply

– Mark C · 1 year, 4 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!Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

I know this is a hard question, you said CE is approach to 16 right ?Log in to reply

– Mark C · 1 year, 4 months ago

Yes, I think it was 15.98 or something like that.Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

This discussion need more people to join. Why only 2 people :( ?Log in to reply

– Mark C · 1 year, 4 months ago

I don't know. It's an interesting problem!Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Bring other people to join, maybe they can solve it :)Log in to reply

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. – Rishabh Tiwari · 1 year, 4 months ago

Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

LOL thxLog in to reply

Jason, where did you find this problem? Are you sure there's the sort of clean solution we've been looking for? – Mark C · 1 year, 4 months ago

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– Jason Chrysoprase · 1 year, 4 months ago

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

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity – Jason Chrysoprase · 1 year, 4 months ago

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– Ashish Siva · 1 year, 4 months ago

See told ya pythagoras would come handyLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

YEAH XDLog in to reply

I loved the way you solved the problem and thx.....guys for such a nice sol. Btw nice prob . Rishabh – Pranay Agrawal · 1 year, 4 months ago

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– Ashish Siva · 1 year, 4 months ago

Oh hi pranay. Thanks for your encouragement.Log in to reply

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? – Mark C · 1 year, 4 months ago

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– Ashish Siva · 1 year, 4 months ago

We solved it! With a little help of W/ALog in to reply

– Mark C · 1 year, 4 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.Log in to reply

– Ashish Siva · 1 year, 4 months ago

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

Guys, i make a problem based on this discussion :) – Jason Chrysoprase · 1 year, 4 months ago

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– Abhay Kumar · 1 year, 4 months ago

Your answer is wrong.Mistake founder-Abhay kumar.Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

AHHHHHHHHHHHHHHHHHHLog in to reply

– Ashish Siva · 1 year, 4 months ago

Unfortunately u had made a mistake so the answer is wrong :(Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Where ?Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Wait what !!!Log in to reply

– Ashish Siva · 1 year, 4 months ago

See abhays response above :(Log in to reply

So we have found the answer and how to get it

Should i delete this discussion ? – Jason Chrysoprase · 1 year, 4 months ago

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– Ashish Siva · 1 year, 4 months ago

Nooooo, let it be there, we have worked very hard on this.Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Okay XDLog in to reply

– Ashish Siva · 1 year, 4 months ago

Phew thanks!Log in to reply

@Abhiram Rao @Abhay Tiwari help plz – Ashish Siva · 1 year, 4 months ago

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– Abhiram Rao · 1 year, 4 months ago

Use Pythagoras Theorem. But the solution would be a very complex one.Log in to reply

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 – Jason Chrysoprase · 1 year, 4 months ago

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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. – Mark C · 1 year, 4 months ago

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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 :

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

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

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

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

\(\LARGE \text{SOMEONE FIND}\space \Huge x \space \LARGE \text{PLZZZZ !!!!!! }\) – Jason Chrysoprase · 1 year, 4 months ago

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– Abhay Kumar · 1 year, 4 months ago

Sorry guys answer approximately remains the asme now correct answer is \(\boxed{15.988}\)Log in to reply

– Ashish Siva · 1 year, 4 months ago

What i still got \(\sqrt{700}\) using W/ALog in to reply

– Abhay Kumar · 1 year, 4 months ago

No, answer is 15.988.I am sure.Log in to reply

– Ashish Siva · 1 year, 4 months ago

But E/A still shows \(\pm \sqrt{700}\)Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Nahhh the answer is 15.98...Log in to reply

– Ashish Siva · 1 year, 4 months ago

I got my mistake 1 min 50 seconds ago XDLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

LOL, okay now i'm gonna post this as a problemLog in to reply

– Ashish Siva · 1 year, 4 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 ;)Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Okay, give me a good titleLog in to reply

– Ashish Siva · 1 year, 4 months ago

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

Because it might look simple but actually it's hard ( took us 3 day to solve LOL) – Jason Chrysoprase · 1 year, 4 months ago

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– Ashish Siva · 1 year, 4 months ago

Sure...,, niceLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Okay, thx guys XDLog in to reply

– Ashish Siva · 1 year, 4 months ago

Post it fast :) :) :)Log in to reply

click here – Jason Chrysoprase · 1 year, 4 months ago

Done,Log in to reply

– Ashish Siva · 1 year, 4 months ago

Nice representation, post the solution!Log in to reply

Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

The answer still the same LOLLog in to reply

– Ashish Siva · 1 year, 4 months ago

What?Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

How is \(\sqrt{700}\) ? W/A ?Log in to reply

– Abhay Kumar · 1 year, 4 months ago

Calculated by hand man.Log in to reply

– Ashish Siva · 1 year, 4 months ago

500000% sure that this joker is joking.Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

You simplify it by ur hand ?Log in to reply

– Ashish Siva · 1 year, 4 months ago

Haha, do you think abhay would have done that much donkeys work of simplifying it?Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

LOLLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

but this doesn't make any sense, try to draw itLog in to reply

– Abhay Kumar · 1 year, 4 months ago

How can you say that?Log in to reply

– Ashish Siva · 1 year, 4 months ago

What? Why?Log in to reply

– Ashish Siva · 1 year, 4 months ago

You wolfram alpha user I was just about to comment but answer key says it is 15.9..Log in to reply

– Abhay Kumar · 1 year, 4 months ago

Don't believe on answers they can be wrong.Log in to reply

– Ashish Siva · 1 year, 4 months ago

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

– Jason Chrysoprase · 1 year, 4 months ago

Kill me XDLog in to reply

– Ashish Siva · 1 year, 4 months ago

Nah we are too close, we can still find the answer,Log in to reply

– Abhay Kumar · 1 year, 4 months ago

MISTAKE FOUND sqrt(1600) not sqrt(900)...You are wrong.Log in to reply

– Abhay Kumar · 1 year, 4 months ago

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

@Jon Haussmann maybe you can help :) – Ashish Siva · 1 year, 4 months ago

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

– Jason Chrysoprase · 1 year, 4 months ago

\(x\) also \( 15.9876... \), try to input itLog in to reply

– Ashish Siva · 1 year, 4 months ago

I got this equation but isnt it too tedious?Log in to reply

– Abhay Kumar · 1 year, 4 months ago

\(x=0\)...lolLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

But \(x\) also \(15.9876\), try to input it !!Log in to reply

– Abhay Kumar · 1 year, 4 months ago

No, putting \(x=15.9876\) results in \(0.0005609440415437967\approx0.00\)Log in to reply

Mystery Solved ;) – Jason Chrysoprase · 1 year, 4 months ago

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– Ashish Siva · 1 year, 4 months ago

Looks like I am lateLog in to reply

– Abhay Kumar · 1 year, 4 months ago

Hmmm...Great..keep it up. :)Log in to reply

From the expression above, prove that \(x= 15.9876...\) ? – Jason Chrysoprase · 1 year, 4 months ago

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– Jason Chrysoprase · 1 year, 4 months ago

But how is \(x = 15.9876...\)Log in to reply

– Abhay Kumar · 1 year, 4 months ago

Using wolfram alpha. ;)Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Cheats, cheater everywhere XDLog in to reply

– Ashish Siva · 1 year, 4 months ago

I spent hours simplifying and he used W/ALog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Well, uhh is there any else solution for simplifying without W/A ?Log in to reply

– Ashish Siva · 1 year, 4 months ago

Lets seeLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Man we did it, we solved it XD, i'm so proud of you guys , i'm gonna cryLog in to reply

– Ashish Siva · 1 year, 4 months ago

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

– Jason Chrysoprase · 1 year, 4 months ago

LOL, yeah, so many people give upLog in to reply

– Ashish Siva · 1 year, 4 months ago

You cheeky fellow XD. Why didnt I think of that? I was simplifying.Log in to reply

– Abhay Kumar · 1 year, 4 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. ;)Log in to reply

– Ashish Siva · 1 year, 4 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. :PLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Should i delete this discussion, we have already found the answerLog in to reply

– Abhay Kumar · 1 year, 4 months ago

Yes delete it and post a similar question.Log in to reply

– Ashish Siva · 1 year, 4 months ago

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

– Jason Chrysoprase · 1 year, 4 months ago

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

– Ashish Siva · 1 year, 4 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 XDLog in to reply

– Abhay Kumar · 1 year, 4 months ago

He made a mistake.Log in to reply

– Ashish Siva · 1 year, 4 months ago

What the.... noooo.Log in to reply

– Abhay Kumar · 1 year, 4 months ago

YesssLog in to reply

– Ashish Siva · 1 year, 4 months ago

Nooooooo, neverLog in to reply

– Ashish Siva · 1 year, 4 months ago

AYSLog in to reply

– Abhay Kumar · 1 year, 4 months ago

What is AYS?Log in to reply

– Ashish Siva · 1 year, 4 months ago

Are you sure XDLog in to reply

– Ashish Siva · 1 year, 4 months ago

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

– Jason Chrysoprase · 1 year, 4 months ago

good luck LOLLog in to reply

@Mark C @Abhay Kumar @Ashish Siva @Rishabh Tiwari @Rohit Udaiwal – Jason Chrysoprase · 1 year, 4 months ago

Log in to reply

I find the expression but can't find \(x\) – Jason Chrysoprase · 1 year, 4 months ago

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@Nihar Mahajan @Svatejas Shivakumar Help usssss – Jason Chrysoprase · 1 year, 4 months ago

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@ahmad saad – Abhay Kumar · 1 year, 4 months ago

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Use coordinate geometry – Aakash Khandelwal · 1 year, 4 months ago

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– Rishabh Tiwari · 1 year, 4 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.Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Is there any other solutionLog in to reply

– Rishabh Tiwari · 1 year, 4 months ago

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

Hello Jason, thanks for inviting to this problem XD. Somehow I got the answer for GF/FE . Trying to solve further. – Ashish Siva · 1 year, 4 months ago

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– Jason Chrysoprase · 1 year, 4 months ago

Okay, LOLLog in to reply

– Ashish Siva · 1 year, 4 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.Log in to reply

– Ashish Siva · 1 year, 4 months ago

\({\left(\dfrac{-x^4 + 1600x^2 + 50625}{9000}\right)}^{2} = \dfrac{x^2}{1600 - x^2}\)Log in to reply

– Rohit Udaiwal · 1 year, 4 months ago

That leads to \(x\approx 40\).Log in to reply

– Ashish Siva · 1 year, 4 months ago

Yeah I know that something like that will come you know the exact value? This expression is niceLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

But when i draw the figure, \(CE\) isn't \(40\), \(CE \approx 16\)Log in to reply

– Ashish Siva · 1 year, 4 months ago

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

– Jason Chrysoprase · 1 year, 4 months ago

Should i tag Calvin Lin here XDLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

About your expression, how did you get it ?Log in to reply

– Ashish Siva · 1 year, 4 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.Log in to reply

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 · 1 year, 4 months ago

Log in to reply

– Ashish Siva · 1 year, 4 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.Log in to reply

– Abhay Kumar · 1 year, 4 months ago

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

– Ashish Siva · 1 year, 4 months ago

Yes, I uaed all three XDLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

yeah, pythagoras useless in this caseLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

SameLog in to reply

– Ashish Siva · 1 year, 4 months ago

HahahaLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

I should tag Calvin Lin and Rishabh Cool here or should i ?Log in to reply

– Ashish Siva · 1 year, 4 months ago

That is entirely your wish, maybe rishabh cool firstLog in to reply

@Rishabh Cool @Calvin Lin Help us XD – Jason Chrysoprase · 1 year, 4 months ago

KLog in to reply

@Paola Ramírez – Jason Chrysoprase · 1 year, 4 months ago

AlsoLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

I'll draw some line inside the rectangleLog in to reply

– Abhay Kumar · 1 year, 4 months ago

Maybe or outside... ;)Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

We need some senior hereLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Yeah, this problem is so nice i would throw my self out of my windowLog in to reply

– Abhay Kumar · 1 year, 4 months ago

Anyways....What's your age?Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

13Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

LOLLog in to reply

PS:Maybe @Nihar Mahajan could be the right guy to consult here as his geometry skills are very good... – Rishabh Cool · 1 year, 4 months ago

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– Jason Chrysoprase · 1 year, 4 months ago

Ohh Okay my friendLog in to reply

– Abhay Kumar · 1 year, 4 months ago

Hmmm...try it when you are free.Log in to reply

I actually know the answer but dunno how to get it. \(CE \approx 16\) – Jason Chrysoprase · 1 year, 4 months ago

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– Ashish Siva · 1 year, 4 months ago

Yes damn sure that it is true did yoi simplify it?Log in to reply

My algebra level = 2 – Jason Chrysoprase · 1 year, 4 months ago

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– Ashish Siva · 1 year, 4 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,Log in to reply

– Jason Chrysoprase · 1 year, 4 months ago

BTW, you can write how you simplify it so i can learn XDLog in to reply

– Ashish Siva · 1 year, 4 months ago

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

– Jason Chrysoprase · 1 year, 4 months ago

OkayLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

I'm still simplyfyingLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

You mean \(x\) is \(AC\) ?Log in to reply

– Ashish Siva · 1 year, 4 months ago

x is CELog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

OkayLog in to reply

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– Mark C · 1 year, 4 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 :).)Log in to reply

I'm waiting till tomorrow okay :)

BTW, thx – Jason Chrysoprase · 1 year, 4 months ago

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– Jason Chrysoprase · 1 year, 4 months ago

I think that's enough XDLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

I actually will post it on problem after i know what is the answer and the solutionLog in to reply

– Jason Chrysoprase · 1 year, 4 months ago

Because i don't know the answerLog in to reply