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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, 1 month 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, 1 month ago

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

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

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

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

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

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

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

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LOL thx · 1 year, 1 month 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? · 1 year, 1 month ago

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

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

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity · 1 year, 1 month ago

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See told ya pythagoras would come handy · 1 year, 1 month ago

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YEAH XD · 1 year, 1 month 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 · 1 year, 1 month ago

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

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

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

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

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Your answer is wrong.Mistake founder-Abhay kumar. · 1 year, 1 month ago

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AHHHHHHHHHHHHHHHHHH · 1 year, 1 month ago

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Unfortunately u had made a mistake so the answer is wrong :( · 1 year, 1 month ago

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Where ? · 1 year, 1 month ago

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Wait what !!! · 1 year, 1 month ago

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See abhays response above :( · 1 year, 1 month ago

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So we have found the answer and how to get it

Should i delete this discussion ? · 1 year, 1 month ago

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

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Okay XD · 1 year, 1 month ago

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Phew thanks! · 1 year, 1 month ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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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, 1 month ago

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Okay, give me a good title · 1 year, 1 month ago

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

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"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) · 1 year, 1 month ago

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Sure...,, nice · 1 year, 1 month ago

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Okay, thx guys XD · 1 year, 1 month ago

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Post it fast :) :) :) · 1 year, 1 month ago

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Done, click here · 1 year, 1 month ago

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Nice representation, post the solution! · 1 year, 1 month ago

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Comment deleted May 14, 2016

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The answer still the same LOL · 1 year, 1 month ago

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What? · 1 year, 1 month ago

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

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Calculated by hand man. · 1 year, 1 month ago

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

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You simplify it by ur hand ? · 1 year, 1 month ago

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

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LOL · 1 year, 1 month ago

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

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How can you say that? · 1 year, 1 month ago

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What? Why? · 1 year, 1 month ago

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

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

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

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Kill me XD · 1 year, 1 month ago

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

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

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

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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, 1 month ago

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

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

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$$x=0$$...lol · 1 year, 1 month ago

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

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

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

Mystery Solved ;) · 1 year, 1 month ago

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Looks like I am late · 1 year, 1 month ago

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

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

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

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

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Using wolfram alpha. ;) · 1 year, 1 month ago

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Cheats, cheater everywhere XD · 1 year, 1 month ago

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

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

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Lets see · 1 year, 1 month ago

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

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

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

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

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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, 1 month ago

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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, 1 month ago

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

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

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

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

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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 · 1 year, 1 month ago

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He made a mistake. · 1 year, 1 month ago

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What the.... noooo. · 1 year, 1 month ago

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Yesss · 1 year, 1 month ago

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Nooooooo, never · 1 year, 1 month ago

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AYS · 1 year, 1 month ago

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What is AYS? · 1 year, 1 month ago

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Are you sure XD · 1 year, 1 month ago

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

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good luck LOL · 1 year, 1 month ago

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

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

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@Nihar Mahajan @Svatejas Shivakumar Help usssss · 1 year, 1 month ago

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@ahmad saad · 1 year, 1 month ago

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Use coordinate geometry · 1 year, 1 month 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. · 1 year, 1 month ago

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Is there any other solution · 1 year, 1 month ago

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

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Okay, LOL · 1 year, 1 month 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. · 1 year, 1 month ago

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

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

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

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

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

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Should i tag Calvin Lin here XD · 1 year, 1 month ago

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

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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, 1 month ago

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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, 1 month ago

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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. · 1 year, 1 month ago

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

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Yes, I uaed all three XD · 1 year, 1 month ago

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yeah, pythagoras useless in this case · 1 year, 1 month ago

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Same · 1 year, 1 month ago

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Hahaha · 1 year, 1 month ago

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

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

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

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Also @Paola Ramírez · 1 year, 1 month ago

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

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Maybe or outside... ;) · 1 year, 1 month ago

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We need some senior here · 1 year, 1 month ago

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

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

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13 · 1 year, 1 month ago

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LOL · 1 year, 1 month ago

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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, 1 month ago

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Ohh Okay my friend · 1 year, 1 month ago

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

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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, 1 month ago

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

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Ahh come on, i'm stuck on simplyfying

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

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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, 1 month ago

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

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

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Okay · 1 year, 1 month ago

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I'm still simplyfying · 1 year, 1 month ago

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

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x is CE · 1 year, 1 month ago

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Okay · 1 year, 1 month 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 :).) · 1 year, 1 month ago

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Darn, i'm still stuck on sin law

I'm waiting till tomorrow okay :)

BTW, thx · 1 year, 1 month ago

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I think that's enough XD · 1 year, 1 month ago

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

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Because i don't know the answer · 1 year, 1 month ago

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