JOMO 5, Short 3
Evaluate, to the nearest integer, the sum of the perimeter and the area of the figure when x 2 5 0 0 0 0 0 + y 2 5 0 0 0 0 0 ≤ 2 2 5 0 0 0 0 0 is drawn on a Cartesian plane.
Details and Assumptions :
The powers are all two million and five hundred thousand.
The answer is 32.
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1 solution
I really liked this problem! One can see that the geometry of this equation approaches a square of side length 4, which coincidentally has area = perimeter = 16. I didn't read the question carefully the first time, thinking it only wanted the area.
Awesome problem..... Keep posting!!!!
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Thank you very much @Abhinav Raichur , you may get all our Past contest problems on the website ( JOMO ).
how to know that is a square? because if x^2 + y^2 = _ _ is a circle. because i get the radius 2 then cant guess the real shape of the graph already..
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It can be written this way: y = ( 2 a − x a ) a 1 Assume a is always even.
As x approaches 2 or − 2 , 2 a − x a approaches 0 .
Also note that a z approaches 1 regardless of what value z is as a approaches infinity.
Therefore, y = ( 2 a − x a ) a 1 approaches 1 when x approaches 2 or − 2 and a approaches infinity.
This can be done again for y = − ( 2 a − x a ) a 1 to finally form a square.
There is an intuitive way to solve this. If you imagine the figure x a + b a = 2 a you will see that as a → ∞ or in this case 2 5 0 0 0 0 0 , the figure approaches a square of radius 2 . Meaning that the perimeter is 4 + 4 + 4 + 4 and the area 4 × 4 therefore the answer is 2 × 4 × 4 = 3 2 .
To rigorously show that the approximation is good enough would require some work but for an a > > 1 0 0 this should be good enough.