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
4 years, 6 months ago

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1 vote

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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=15w402w2=w15w302w2CD = 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 - 4 years, 6 months ago

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So the answer is approach to 1616 ?

So what formula we use ? Both Pythagorean and Trigonometry ?

Jason Chrysoprase - 4 years, 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 - 4 years, 6 months ago

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LOL thx

Jason Chrysoprase - 4 years, 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 - 4 years, 6 months ago

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I find the SOLUTIONSOLUTION !!!!

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity

Jason Chrysoprase - 4 years, 6 months ago

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See told ya pythagoras would come handy

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon YEAH XD

Jason Chrysoprase - 4 years, 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 - 4 years, 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 Menon - 4 years, 6 months ago

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Okay, LOL

Jason Chrysoprase - 4 years, 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 Menon - 4 years, 6 months ago

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@Ashish Menon (x4+1600x2+506259000)2=x21600x2{\left(\dfrac{-x^4 + 1600x^2 + 50625}{9000}\right)}^{2} = \dfrac{x^2}{1600 - x^2}

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon Okay

Jason Chrysoprase - 4 years, 6 months ago

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@Ashish Menon You mean xx is ACAC ?

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase x is CE

Ashish Menon - 4 years, 6 months ago

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

I actually know the answer but dunno how to get it. CE16CE \approx 16

Jason Chrysoprase - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon I'm still simplyfying

Jason Chrysoprase - 4 years, 6 months ago

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

My algebra level = 2

Jason Chrysoprase - 4 years, 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 Menon - 4 years, 6 months ago

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@Ashish Menon Okay

Jason Chrysoprase - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 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 Menon - 4 years, 6 months ago

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@Ashish Menon That leads to x40x\approx 40.

Rohit Udaiwal - 4 years, 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 Menon - 4 years, 6 months ago

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@Ashish Menon But when i draw the figure, CECE isn't 4040, CE16CE \approx 16

Jason Chrysoprase - 4 years, 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 Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 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 = 1600x2\sqrt{1600 - x^2}. Now FDC GEC\triangle{FDC} ~ \triangle{GEC} So, GEFD=CECD\dfrac{GE}{FD} = \dfrac{CE}{CD}.. and so.. oh I get my mistake.

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon 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 - 4 years, 6 months ago

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@A Former Brilliant Member 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 Menon - 4 years, 6 months ago

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@Ashish Menon Same

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase Hahaha

Ashish Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase Also @Paola Ramírez

Jason Chrysoprase - 4 years, 6 months ago

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

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member Yes, I uaed all three XD

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon yeah, pythagoras useless in this case

Jason Chrysoprase - 4 years, 6 months ago

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@A Former Brilliant Member Yeah, this problem is so nice i would throw my self out of my window

Jason Chrysoprase - 4 years, 6 months ago

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

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member 13

Jason Chrysoprase - 4 years, 6 months ago

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@A Former Brilliant Member We need some senior here

Jason Chrysoprase - 4 years, 6 months ago

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@A Former Brilliant Member I'll draw some line inside the rectangle

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase Maybe or outside... ;)

A Former Brilliant Member - 4 years, 6 months ago

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@Ashish Menon LOL

Jason Chrysoprase - 4 years, 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 Jain - 4 years, 6 months ago

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

A Former Brilliant Member - 4 years, 6 months ago

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@Rishabh Jain Ohh Okay my friend

Jason Chrysoprase - 4 years, 6 months ago

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@Ashish Menon Should i tag Calvin Lin here XD

Jason Chrysoprase - 4 years, 6 months ago

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Use coordinate geometry

Aakash Khandelwal - 4 years, 6 months ago

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Is there any other solution

Jason Chrysoprase - 4 years, 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 - 4 years, 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 - 4 years, 6 months ago

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@ahmad saad

A Former Brilliant Member - 4 years, 6 months ago

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@Nihar Mahajan @Svatejas Shivakumar Help usssss

Jason Chrysoprase - 4 years, 6 months ago

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

I need someone to FIND xxthis expression below ( x=CEx = CE)

0=1600x2×(x15x900x2)15x\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 - 4 years, 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 - 4 years, 6 months ago

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

I find the expression but can't find xx

Jason Chrysoprase - 4 years, 6 months ago

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

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

Using Pythagoras :

BC=900x2 BC = \sqrt{900 - x^2}

EG=1600x2 EG = \sqrt{1600 - x^2}

Using Similiar Triangles :

FDBC=DECE15900x2=yxy=15x900x2 \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}}

FDEG=CDCE151600x2=xyx151600x2=x15x900x2x\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}

15x=1600x2×(x15x900x2)\LARGE 15x = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}})

0=1600x2×(x15x900x2)15x\LARGE 0 = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}}) - 15x

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

Jason Chrysoprase - 4 years, 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 Menon - 4 years, 6 months ago

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@Ashish Menon good luck LOL

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase x=0x=0...lol

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member AYS

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon What is AYS?

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member Are you sure XD

Ashish Menon - 4 years, 6 months ago

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@A Former Brilliant Member But xx also 15.987615.9876, try to input it !!

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase No, putting x=15.9876x=15.9876 results in 0.00056094404154379670.000.0005609440415437967\approx0.00

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member okay then x=15.9876...x = 15.9876...

Mystery Solved ;)

Jason Chrysoprase - 4 years, 6 months ago

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

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member But how is x=15.9876...x = 15.9876...

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase Using wolfram alpha. ;)

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member You cheeky fellow XD. Why didnt I think of that? I was simplifying.

Ashish Menon - 4 years, 6 months ago

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@A Former Brilliant Member Cheats, cheater everywhere XD

Jason Chrysoprase - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase Lets see

Ashish Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 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 Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 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. ;)

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member 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 Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase Nooooooo, never

Ashish Menon - 4 years, 6 months ago

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

A Former Brilliant Member - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 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 Menon - 4 years, 6 months ago

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@Ashish Menon He made a mistake.

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member What the.... noooo.

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon Yesss

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member 0=1600x2×(x15x900x2)15x\LARGE 0 = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}}) - 15x

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

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase Looks like I am late

Ashish Menon - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member 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 Menon - 4 years, 6 months ago

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@Ashish Menon xx also 15.9876... 15.9876... , try to input it

Jason Chrysoprase - 4 years, 6 months ago

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

A Former Brilliant Member - 4 years, 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 Menon - 4 years, 6 months ago

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@Ashish Menon Kill me XD

Jason Chrysoprase - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member What i still got 700\sqrt{700} using W/A

Ashish Menon - 4 years, 6 months ago

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

A Former Brilliant Member - 4 years, 6 months ago

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@A Former Brilliant Member But E/A still shows ±700\pm \sqrt{700}

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon Nahhh the answer is 15.98...

Jason Chrysoprase - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 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 Menon - 4 years, 6 months ago

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@Ashish Menon Okay, give me a good title

Jason Chrysoprase - 4 years, 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 - 4 years, 6 months ago

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@Jason Chrysoprase Sure...,, nice

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon Okay, thx guys XD

Jason Chrysoprase - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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@Ashish Menon Done, click here

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase Nice representation, post the solution!

Ashish Menon - 4 years, 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 Menon - 4 years, 6 months ago

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@Abhiram Rao @Abhay Tiwari help plz

Ashish Menon - 4 years, 6 months ago

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

Abhiram Rao - 4 years, 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 - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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Okay XD

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase Phew thanks!

Ashish Menon - 4 years, 6 months ago

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

Jason Chrysoprase - 4 years, 6 months ago

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

Ashish Menon - 4 years, 6 months ago

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Wait what !!!

Jason Chrysoprase - 4 years, 6 months ago

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@Jason Chrysoprase See abhays response above :(

Ashish Menon - 4 years, 6 months ago

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Where ?

Jason Chrysoprase - 4 years, 6 months ago

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Your answer is wrong.Mistake founder-Abhay kumar.

A Former Brilliant Member - 4 years, 6 months ago

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AHHHHHHHHHHHHHHHHHH

Jason Chrysoprase - 4 years, 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 - 4 years, 6 months ago

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

Ashish Menon - 4 years, 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 - 4 years, 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 Menon - 4 years, 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 - 4 years, 6 months ago

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Oh hi pranay. Thanks for your encouragement.

Ashish Menon - 4 years, 6 months ago

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