\(ACEG\) 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 \(CE\)
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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=15402−w2w=w−15302−w2w
Wolfram Alpha has the same numerical solution for both, though: just under 16 (15.9876490085686).
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.
@Jason Chrysoprase
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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,
@Jason Chrysoprase
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I assumed CE as x. Then we know that CG is 40. Applying pythagoras' theorem, we get GE = 1600−x2. Now △FDC△GEC So, FDGE=CDCE.. and so.. oh I get my mistake.
@Ashish Menon
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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
@A Former Brilliant Member
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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.
Okay, that says that the area of the whole rectangle = area of bottom 15 of rectangle (15x) plus that same 15x area divided by the height where the line segment of length 30 meets the left side of the rectangle. That can't be right. In any case, we have good equations in x that "just need to be solved/simplified". The problem is that it is not at all clear that they can be.
Edit: you've changed the equation since I wrote my comment. Anyway, to spend more energy on this I would want some stronger reason to believe that there is a simple solution to be found.
@Jason Chrysoprase
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Nice, just update the note to "this problem has been solved". A very nice question . Claps for you for posting this, Atlast its pythagoras which solved this question and not sine rule or similarity XD
@Jason Chrysoprase
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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 ;)
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.
Easy Math Editor
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
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=15402−w2w=w−15302−w2w
Wolfram Alpha has the same numerical solution for both, though: just under 16 (15.9876490085686).
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So the answer is approach to 16 ?
So what formula we use ? Both Pythagorean and Trigonometry ?
You guys rocked it.! Nice team work..& hats off to ur efforts..!! Me & my friends couldn't solve it.. I really need to learn a lot from people like you guys...congrats.
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LOL thx
Jason, where did you find this problem? Are you sure there's the sort of clean solution we've been looking for?
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I find the SOLUTION !!!!
I just need someone to simplify an expression
Use Pythagoras and the rule of similarity
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See told ya pythagoras would come handy
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It's from a piece of paper that have been torn off by someone. I found it on school.
Hello Jason, thanks for inviting to this problem XD. Somehow I got the answer for GF/FE . Trying to solve further.
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Okay, LOL
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Yeah, I think I got the way to solve. I have got an equation in x, but a lengthy one, I have to solve that. An you help me with that.
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(9000−x4+1600x2+50625)2=1600−x2x2
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x is AC ?
You meanLog in to reply
I actually know the answer but dunno how to get it. CE≈16
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My algebra level = 2
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x≈40.
That leads toLog in to reply
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CE isn't 40, CE≈16
But when i draw the figure,Log in to reply
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1600−x2. Now △FDC △GEC So, FDGE=CDCE.. and so.. oh I get my mistake.
I assumed CE as x. Then we know that CG is 40. Applying pythagoras' theorem, we get GE =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
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@Rishabh Cool @Calvin Lin Help us XD
KLog in to reply
@Paola Ramírez
AlsoLog in to reply
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PS:Maybe @Nihar Mahajan could be the right guy to consult here as his geometry skills are very good...
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Use coordinate geometry
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Is there any other solution
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I think we should invite some seniors at brilliant to look for a solution to this problem..
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.
@ahmad saad
@Nihar Mahajan @Svatejas Shivakumar Help usssss
I have found the solution, use the rule of Pythagoras and Similarity
I need someone to FIND xthis expression below ( x=CE)
0=1600−x2×(x−900−x215x)−15x
I'll explain how to get the expression above after someone find x
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Okay, that says that the area of the whole rectangle = area of bottom 15 of rectangle (15x) plus that same 15x area divided by the height where the line segment of length 30 meets the left side of the rectangle. That can't be right. In any case, we have good equations in x that "just need to be solved/simplified". The problem is that it is not at all clear that they can be.
Edit: you've changed the equation since I wrote my comment. Anyway, to spend more energy on this I would want some stronger reason to believe that there is a simple solution to be found.
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I'm serious, input x=15.9876, the expression above
I find the expression but can't find x
Assume that : CE=x, DE=y
That's mean CD=x−y
Using Pythagoras :
BC=900−x2
EG=1600−x2
Using Similiar Triangles :
BCFD=CEDE900−x215=xyy=900−x215x
EGFD=CECD1600−x215=xx−y1600−x215=xx−900−x215x
15x=1600−x2×(x−900−x215x)
0=1600−x2×(x−900−x215x)−15x
SOMEONE FIND x PLZZZZ !!!!!!
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@Mark C @Abhay Kumar @Ashish Siva @Rishabh Tiwari @Rohit Udaiwal
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x=0...lol
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x also 15.9876, try to input it !!
ButLog in to reply
x=15.9876 results in 0.0005609440415437967≈0.00
No, puttingLog in to reply
x=15.9876...
okay thenMystery Solved ;)
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x=15.9876...
But how isLog in to reply
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0=1600−x2×(x−900−x215x)−15x
From the expression above, prove that x=15.9876... ?
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@Jon Haussmann maybe you can help :)
Yep, this queetion may be wrong but lets see if we can get some easier method. Oh! I forgotLog in to reply
x also 15.9876..., try to input it
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15.988
Sorry guys answer approximately remains the asme now correct answer isLog in to reply
700 using W/A
What i still gotLog in to reply
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±700
But E/A still showsLog in to reply
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Because it might look simple but actually it's hard ( took us 3 day to solve LOL)
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click here
Done,Log in to reply
@Abhiram Rao @Abhay Tiwari help plz
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Use Pythagoras Theorem. But the solution would be a very complex one.
So we have found the answer and how to get it
Should i delete this discussion ?
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Nooooo, let it be there, we have worked very hard on this.
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Okay XD
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Guys, i make a problem based on this discussion :)
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Unfortunately u had made a mistake so the answer is wrong :(
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Wait what !!!
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Where ?
Your answer is wrong.Mistake founder-Abhay kumar.
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AHHHHHHHHHHHHHHHHHH
I'm confused by all the many comments. Has someone found a solution that simplifies, or are we still stuck with equations we need a computer to solve?
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We solved it! With a little help of W/A
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But we'd (I'd) already done that two days ago. Sort by top comments & the first is mine from two days ago. I still wonder if there is a W/A-free way to do it.
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I loved the way you solved the problem and thx.....guys for such a nice sol. Btw nice prob . Rishabh
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Oh hi pranay. Thanks for your encouragement.