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Geometry Olympiad 2016

\(ACEG\) 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, 6 months 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).

Mark C - 1 year, 6 months ago

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

So what formula we use ? Both Pythagorean and Trigonometry ?

Jason Chrysoprase - 1 year, 6 months ago

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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!

Mark C - 1 year, 6 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, 6 months ago

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@Jason Chrysoprase Yes, I think it was 15.98 or something like that.

Mark C - 1 year, 6 months ago

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@Mark C This discussion need more people to join. Why only 2 people :( ?

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase I don't know. It's an interesting problem!

Mark C - 1 year, 6 months ago

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@Mark C Bring other people to join, maybe they can solve it :)

Jason Chrysoprase - 1 year, 6 months ago

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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.

Rishabh Tiwari - 1 year, 6 months ago

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LOL thx

Jason Chrysoprase - 1 year, 6 months ago

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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?

Mark C - 1 year, 6 months ago

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

Jason Chrysoprase - 1 year, 6 months ago

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

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity

Jason Chrysoprase - 1 year, 6 months ago

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See told ya pythagoras would come handy

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva YEAH XD

Jason Chrysoprase - 1 year, 6 months 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

Pranay Agrawal - 1 year, 6 months ago

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Oh hi pranay. Thanks for your encouragement.

Ashish Siva - 1 year, 6 months ago

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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?

Mark C - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 months ago

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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.

Mark C - 1 year, 6 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, 6 months ago

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

Jason Chrysoprase - 1 year, 6 months ago

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Your answer is wrong.Mistake founder-Abhay kumar.

Abhay Kumar - 1 year, 6 months ago

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AHHHHHHHHHHHHHHHHHH

Jason Chrysoprase - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 months ago

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Where ?

Jason Chrysoprase - 1 year, 6 months ago

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Wait what !!!

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase See abhays response above :(

Ashish Siva - 1 year, 6 months ago

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So we have found the answer and how to get it

Should i delete this discussion ?

Jason Chrysoprase - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 months ago

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Okay XD

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Phew thanks!

Ashish Siva - 1 year, 6 months ago

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@Abhiram Rao @Abhay Tiwari help plz

Ashish Siva - 1 year, 6 months ago

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

Abhiram Rao - 1 year, 6 months ago

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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

Jason Chrysoprase - 1 year, 6 months ago

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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.

Mark C - 1 year, 6 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, 6 months ago

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@Jason Chrysoprase Sorry guys answer approximately remains the asme now correct answer is \(\boxed{15.988}\)

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar What i still got \(\sqrt{700}\) using W/A

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva No, answer is 15.988.I am sure.

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar But E/A still shows \(\pm \sqrt{700}\)

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Nahhh the answer is 15.98...

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase I got my mistake 1 min 50 seconds ago XD

Ashish Siva - 1 year, 6 months ago

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

Jason Chrysoprase - 1 year, 6 months ago

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@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, 6 months ago

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@Ashish Siva Okay, give me a good title

Jason Chrysoprase - 1 year, 6 months ago

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@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, 6 months ago

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@Jason Chrysoprase "Don't Trust Your Eye " How about that ?

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

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Sure...,, nice

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Okay, thx guys XD

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Post it fast :) :) :)

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Done, click here

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Nice representation, post the solution!

Ashish Siva - 1 year, 6 months ago

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Comment deleted May 14, 2016

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@Abhay Kumar The answer still the same LOL

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase What?

Ashish Siva - 1 year, 6 months ago

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@Abhay Kumar How is \(\sqrt{700}\) ? W/A ?

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Calculated by hand man.

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar 500000% sure that this joker is joking.

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva You simplify it by ur hand ?

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Haha, do you think abhay would have done that much donkeys work of simplifying it?

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva LOL

Jason Chrysoprase - 1 year, 6 months ago

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@Abhay Kumar but this doesn't make any sense, try to draw it

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase How can you say that?

Abhay Kumar - 1 year, 6 months ago

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@Jason Chrysoprase What? Why?

Ashish Siva - 1 year, 6 months ago

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@Abhay Kumar You wolfram alpha user I was just about to comment but answer key says it is 15.9..

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Don't believe on answers they can be wrong.

Abhay Kumar - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Kill me XD

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Nah we are too close, we can still find the answer,

Ashish Siva - 1 year, 6 months ago

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

Abhay Kumar - 1 year, 6 months ago

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

Abhay Kumar - 1 year, 6 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, 6 months ago

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@Ashish Siva \(x\) also \( 15.9876... \), try to input it

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase I got this equation but isnt it too tedious?

Ashish Siva - 1 year, 6 months ago

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@Jason Chrysoprase \(x=0\)...lol

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar But \(x\) also \(15.9876\), try to input it !!

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase No, putting \(x=15.9876\) results in \(0.0005609440415437967\approx0.00\)

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar okay then \(x = 15.9876...\)

Mystery Solved ;)

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Looks like I am late

Ashish Siva - 1 year, 6 months ago

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@Jason Chrysoprase Hmmm...Great..keep it up. :)

Abhay Kumar - 1 year, 6 months ago

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

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

Jason Chrysoprase - 1 year, 6 months ago

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@Abhay Kumar But how is \(x = 15.9876...\)

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Using wolfram alpha. ;)

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar Cheats, cheater everywhere XD

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase I spent hours simplifying and he used W/A

Ashish Siva - 1 year, 6 months ago

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

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Lets see

Ashish Siva - 1 year, 6 months ago

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@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, 6 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, 6 months ago

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@Ashish Siva LOL, yeah, so many people give up

Jason Chrysoprase - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 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, 6 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, 6 months ago

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@Ashish Siva Should i delete this discussion, we have already found the answer

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Yes delete it and post a similar question.

Abhay Kumar - 1 year, 6 months ago

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@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, 6 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, 6 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, 6 months ago

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@Ashish Siva He made a mistake.

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar What the.... noooo.

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Yesss

Abhay Kumar - 1 year, 6 months ago

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@Jason Chrysoprase Nooooooo, never

Ashish Siva - 1 year, 6 months ago

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@Abhay Kumar AYS

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva What is AYS?

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar Are you sure XD

Ashish Siva - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva good luck LOL

Jason Chrysoprase - 1 year, 6 months ago

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

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

Jason Chrysoprase - 1 year, 6 months ago

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@Nihar Mahajan @Svatejas Shivakumar Help usssss

Jason Chrysoprase - 1 year, 6 months ago

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@ahmad saad

Abhay Kumar - 1 year, 6 months ago

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Use coordinate geometry

Aakash Khandelwal - 1 year, 6 months ago

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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.

Rishabh Tiwari - 1 year, 6 months ago

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Is there any other solution

Jason Chrysoprase - 1 year, 6 months ago

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

Rishabh Tiwari - 1 year, 6 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, 6 months ago

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Okay, LOL

Jason Chrysoprase - 1 year, 6 months ago

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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.

Ashish Siva - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva That leads to \(x\approx 40\).

Rohit Udaiwal - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva But when i draw the figure, \(CE\) isn't \(40\), \(CE \approx 16\)

Jason Chrysoprase - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Should i tag Calvin Lin here XD

Jason Chrysoprase - 1 year, 6 months ago

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@Ashish Siva About your expression, how did you get it ?

Jason Chrysoprase - 1 year, 6 months ago

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@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, 6 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, 6 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, 6 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, 6 months ago

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@Abhay Kumar Yes, I uaed all three XD

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva yeah, pythagoras useless in this case

Jason Chrysoprase - 1 year, 6 months ago

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@Ashish Siva Same

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Hahaha

Ashish Siva - 1 year, 6 months ago

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

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase That is entirely your wish, maybe rishabh cool first

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva K @Rishabh Cool @Calvin Lin Help us XD

Jason Chrysoprase - 1 year, 6 months ago

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@Abhay Kumar I'll draw some line inside the rectangle

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Maybe or outside... ;)

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar We need some senior here

Jason Chrysoprase - 1 year, 6 months ago

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

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Anyways....What's your age?

Abhay Kumar - 1 year, 6 months ago

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@Abhay Kumar 13

Jason Chrysoprase - 1 year, 6 months ago

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@Ashish Siva LOL

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase 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...

Rishabh Cool - 1 year, 6 months ago

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@Rishabh Cool Ohh Okay my friend

Jason Chrysoprase - 1 year, 6 months ago

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@Rishabh Cool Hmmm...try it when you are free.

Abhay Kumar - 1 year, 6 months ago

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@Ashish Siva Ashish, are you sure the equation above is true ?

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

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase Yes damn sure that it is true did yoi simplify it?

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Ahh come on, i'm stuck on simplyfying

My algebra level = 2

Jason Chrysoprase - 1 year, 6 months ago

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@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, 6 months ago

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

Jason Chrysoprase - 1 year, 6 months ago

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

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Okay

Jason Chrysoprase - 1 year, 6 months ago

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@Ashish Siva I'm still simplyfying

Jason Chrysoprase - 1 year, 6 months ago

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@Ashish Siva You mean \(x\) is \(AC\) ?

Jason Chrysoprase - 1 year, 6 months ago

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@Jason Chrysoprase x is CE

Ashish Siva - 1 year, 6 months ago

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@Ashish Siva Okay

Jason Chrysoprase - 1 year, 6 months ago

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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 :).)

Mark C - 1 year, 6 months ago

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Darn, i'm still stuck on sin law

I'm waiting till tomorrow okay :)

BTW, thx

Jason Chrysoprase - 1 year, 6 months ago

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I think that's enough XD

Jason Chrysoprase - 1 year, 6 months ago

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

Jason Chrysoprase - 1 year, 6 months ago

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Because i don't know the answer

Jason Chrysoprase - 1 year, 6 months ago

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