iTeachChem•16mo ago

Olympiad related problem

Question 2. (I don't have much experience in inequalities. :sweaty: )

WhatsApp_Image_2024-09-02_at_21.31.07_acf423ce.jpg

iTeachChem Helper•9/3/24, 1:17 PM

@Apu

iTeachChem Helper•9/3/24, 1:17 PM

Note for OP

+solved @user1 @user2...

+solved @user1 @user2...

to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.

SSirLancelotDuLac Question 2. (I don't have much experience in inequalities. :sweaty: )

iTeachChem•9/17/24, 5:53 AM

Did you try a reverse image search on this one?

SSirLancelotDuLac Question 2. (I don't have much experience in inequalities. :sweaty: )

Deleted User•9/18/24, 7:34 AM

50?

SirLancelotDuLacOP•9/18/24, 10:10 AM

50 is correct.

SirLancelotDuLacOP•9/18/24, 10:10 AM

Can you share your solution?

SSirLancelotDuLac Can you share your solution?

Deleted User•9/18/24, 11:39 AM

Multiply Denominator on both sides
After multiplying
Shift 100a1 + 100a2 + 100a3 part to LHS

Now your eq looks like
a1^2 - 100a1 + a2^2 - 100a2.........+ a100^2 - 100a100 = 0
Now if you observe we can write it as perfect squares

(a1 - 50)^2 + ( a2 - 50)^2.....+ (a100 - 50)^2 -100•(50²)= 0
Now
(a1 -50)^2 + (a2 - 50)^2 +..... + (a100 - 50)^2 = 100•(50²)
now
a1 will be maximum when all other squared terms are zero
And since it isn't said that all integers are same or distinct so
a2 to a100 is 50 so (50 - 50)² becomes 0
So leaving us with
(a1 - 50)^2 = 100•(50²)
Sq root both sides
a1 = 10•50 + 50
a1 = 550
550/11 = 50

DDeleted User Multiply Denominator on both sides After multiplying Shift 100a1 + 100a2 + 100a3...

SirLancelotDuLacOP•9/18/24, 12:06 PM

Oh nice.

SirLancelotDuLacOP•9/18/24, 12:06 PM

Thanks a lot @Deleted User .

SirLancelotDuLacOP•9/18/24, 12:07 PM

+solved @Deleted User

iTeachChem Helper•9/18/24, 12:07 PM

Post locked and archived successfully!

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@1075951732460376214 (1075951732460376214)

Time

September 18, 2024

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@783277857463599115 (783277857463599115)

Olympiad related problem

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