# 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
2 years 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).

- 2 years ago

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

So what formula we use ? Both Pythagorean and Trigonometry ?

- 2 years ago

Comment deleted May 13, 2016

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!

- 2 years ago

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

- 2 years ago

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

- 2 years ago

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

- 2 years ago

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

- 2 years ago

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

- 2 years 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.

- 2 years ago

LOL thx

- 2 years ago

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

- 2 years ago

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

- 2 years ago

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

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity

- 2 years ago

See told ya pythagoras would come handy

- 2 years ago

YEAH XD

- 2 years ago

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

- 2 years ago

Oh hi pranay. Thanks for your encouragement.

- 2 years 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?

- 2 years ago

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

- 2 years 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.

- 2 years ago

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

- 2 years ago

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

- 2 years ago

- 2 years ago

AHHHHHHHHHHHHHHHHHH

- 2 years ago

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

- 2 years ago

Where ?

- 2 years ago

Wait what !!!

- 2 years ago

See abhays response above :(

- 2 years ago

So we have found the answer and how to get it

Should i delete this discussion ?

- 2 years ago

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

- 2 years ago

Okay XD

- 2 years ago

Phew thanks!

- 2 years ago

@Abhiram Rao @Abhay Tiwari help plz

- 2 years ago

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

- 2 years ago

I have found the solution, use the rule of Pythagoras and Similarity

I need someone to FIND $$x$$this expression below ( $$x = CE$$)

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

I'll explain how to get the expression above after someone find x

- 2 years ago

Okay, that says that the area of the whole rectangle = area of bottom 15 of rectangle (15x) plus that same 15x area divided by the height where the line segment of length 30 meets the left side of the rectangle. That can't be right. In any case, we have good equations in x that "just need to be solved/simplified". The problem is that it is not at all clear that they can be.

Edit: you've changed the equation since I wrote my comment. Anyway, to spend more energy on this I would want some stronger reason to believe that there is a simple solution to be found.

- 2 years ago

Assume that : $$CE = x$$, $$DE = y$$

That's mean $$CD = x-y$$

Using Pythagoras :

$$BC = \sqrt{900 - x^2}$$

$$EG = \sqrt{1600 - x^2}$$

Using Similiar Triangles :

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

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

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

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

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

- 2 years ago

Sorry guys answer approximately remains the asme now correct answer is $$\boxed{15.988}$$

- 2 years ago

What i still got $$\sqrt{700}$$ using W/A

- 2 years ago

No, answer is 15.988.I am sure.

- 2 years ago

But E/A still shows $$\pm \sqrt{700}$$

- 2 years ago

Nahhh the answer is 15.98...

- 2 years ago

I got my mistake 1 min 50 seconds ago XD

- 2 years ago

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

- 2 years 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 ;)

- 2 years ago

Okay, give me a good title

- 2 years ago

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

- 2 years ago

"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)

- 2 years ago

Sure...,, nice

- 2 years ago

Okay, thx guys XD

- 2 years ago

Post it fast :) :) :)

- 2 years ago

- 2 years ago

Nice representation, post the solution!

- 2 years ago

Comment deleted May 14, 2016

The answer still the same LOL

- 2 years ago

What?

- 2 years ago

How is $$\sqrt{700}$$ ? W/A ?

- 2 years ago

Calculated by hand man.

- 2 years ago

500000% sure that this joker is joking.

- 2 years ago

You simplify it by ur hand ?

- 2 years ago

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

- 2 years ago

LOL

- 2 years ago

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

- 2 years ago

How can you say that?

- 2 years ago

What? Why?

- 2 years ago

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

- 2 years ago

Don't believe on answers they can be wrong.

- 2 years ago

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

- 2 years ago

Kill me XD

- 2 years ago

Nah we are too close, we can still find the answer,

- 2 years ago

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

- 2 years ago

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

- 2 years ago

Yep, this queetion may be wrong but lets see if we can get some easier method. Oh! I forgot @Jon Haussmann maybe you can help :)

- 2 years ago

$$x$$ also $$15.9876...$$, try to input it

- 2 years ago

I got this equation but isnt it too tedious?

- 2 years ago

$$x=0$$...lol

- 2 years ago

But $$x$$ also $$15.9876$$, try to input it !!

- 2 years ago

No, putting $$x=15.9876$$ results in $$0.0005609440415437967\approx0.00$$

- 2 years ago

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

Mystery Solved ;)

- 2 years ago

Looks like I am late

- 2 years ago

Hmmm...Great..keep it up. :)

- 2 years ago

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

From the expression above, prove that $$x= 15.9876...$$ ?

- 2 years ago

But how is $$x = 15.9876...$$

- 2 years ago

Using wolfram alpha. ;)

- 2 years ago

Cheats, cheater everywhere XD

- 2 years ago

I spent hours simplifying and he used W/A

- 2 years ago

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

- 2 years ago

Lets see

- 2 years ago

Man we did it, we solved it XD, i'm so proud of you guys , i'm gonna cry

- 2 years ago

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

- 2 years ago

LOL, yeah, so many people give up

- 2 years ago

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

- 2 years 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. ;)

- 2 years ago

Actually, I know how to use W/A and I have seen that there has been an attempt to find such an answer. :P

- 2 years ago

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

- 2 years ago

Yes delete it and post a similar question.

- 2 years ago

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

- 2 years ago

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

- 2 years ago

Nice, just update the note to "this problem has been solved". A very nice question . Claps for you for posting this, Atlast its pythagoras which solved this question and not sine rule or similarity XD

- 2 years ago

He made a mistake.

- 2 years ago

What the.... noooo.

- 2 years ago

Yesss

- 2 years ago

Nooooooo, never

- 2 years ago

AYS

- 2 years ago

What is AYS?

- 2 years ago

Are you sure XD

- 2 years ago

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

- 2 years ago

good luck LOL

- 2 years ago

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

I find the expression but can't find $$x$$

- 2 years ago

- 2 years ago

Use coordinate geometry

- 2 years 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.

- 2 years ago

Is there any other solution

- 2 years ago

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

- 2 years ago

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

- 2 years ago

Okay, LOL

- 2 years 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.

- 2 years ago

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

- 2 years ago

That leads to $$x\approx 40$$.

- 2 years ago

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

- 2 years ago

But when i draw the figure, $$CE$$ isn't $$40$$, $$CE \approx 16$$

- 2 years ago

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

- 2 years ago

Should i tag Calvin Lin here XD

- 2 years ago

About your expression, how did you get it ?

- 2 years 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.

- 2 years ago

I got too many things but not the answer.
1)450=BC×FE
2)600=CF×EG
3)4/3=FG/FE
4)600=BC×FG
5)BC/CF=EG/FG
6)1200=GE×BC
7)8/3=GE/FE
8)2CF=BC
9)2FG=GE
10)225=CF×FE
11)300=CF×FG

- 2 years 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.

- 2 years ago

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

- 2 years ago

Yes, I uaed all three XD

- 2 years ago

yeah, pythagoras useless in this case

- 2 years ago

Same

- 2 years ago

Hahaha

- 2 years ago

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

- 2 years ago

That is entirely your wish, maybe rishabh cool first

- 2 years ago

K @Rishabh Cool @Calvin Lin Help us XD

- 2 years ago

I'll draw some line inside the rectangle

- 2 years ago

Maybe or outside... ;)

- 2 years ago

We need some senior here

- 2 years ago

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

- 2 years ago

- 2 years ago

13

- 2 years ago

LOL

- 2 years ago

I would try my best only after I get some time.... But I am very very busy at the moment... So please forgive me.... ("_")

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

- 2 years ago

Ohh Okay my friend

- 2 years ago

Hmmm...try it when you are free.

- 2 years ago

Ashish, are you sure the equation above is true ?

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

- 2 years ago

Yes damn sure that it is true did yoi simplify it?

- 2 years ago

Ahh come on, i'm stuck on simplyfying

My algebra level = 2

- 2 years 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,

- 2 years ago

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

- 2 years ago

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

- 2 years ago

Okay

- 2 years ago

I'm still simplyfying

- 2 years ago

You mean $$x$$ is $$AC$$ ?

- 2 years ago

x is CE

- 2 years ago

Okay

- 2 years ago

Comment deleted May 12, 2016

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 :).)

- 2 years ago

Darn, i'm still stuck on sin law

I'm waiting till tomorrow okay :)

BTW, thx

- 2 years ago

I think that's enough XD

- 2 years ago

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

- 2 years ago

Because i don't know the answer

- 2 years ago