Circles may help?

Algebra Level 4

If the maximum value of S = ( 1 − x 1 ) ( 1 − y 1 ) + ( 1 − x 2 ) ( 1 − y 2 ) S = (1 - x_1)(1 - y_1) + (1 - x_2)(1 - y_2) is ( c + k ) 2 (c+\sqrt k)^2 and x 1 2 + x 2 2 = y 1 2 + y 2 2 = c 2 x_1^{2}+x_2^{2}=y_1^{2}+y_2^{2}=c^2 , where c c is some positive number.

Find k ! k! .


The answer is 2.

Harsh Shrivastava by Harsh Shrivastava , 17, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

4 solutions

Shinya Kogami
Jun 22, 2018

Set z 1 = 1 + i , z 2 = x 1 + i x 2 , z 3 = y 1 + i y 2 , z 0 = ( z 1 − z 2 ) ( z 1 − z 3 ) ‾ z_1 = 1+i , z_2 = x_1 + ix_2 , z_3 = y_1+i y_2 , z_0 = (z_1- z_2)\overline{(z_1-z_3)} Then S = Re ( z 0 ) ≤ ∣ z 0 ∣ ≤ ( c + 2 ) 2 S = \text{Re}(z_0) \le |z_0| \le \left(c+\sqrt{2}\right)^2 and equality occurs when z 2 = z 3 = − ( c 2 + i c 2 ) z_2=z_3 = -\left({\dfrac{c}{\sqrt{2}}} +i{\dfrac{c}{\sqrt{2}}}\right) . Inequality above follows from triangle inequality.

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ritabrata Roy
Jun 30, 2018

My solution needs no help of circle.

Just use the following facts for all** a and b belong to real to find the maximum

Max(ab)=(a^2+b^2)/2

Max(a+b)=√2*√(a^2+b^2)

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Shaurya Agarwal
Jun 23, 2018

So, k ! = k = 2 k!=k=2

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Suhas Sheikh
Jun 19, 2018

Setting X1 as Ccostheta and X2 as Csintheta And Y1 as Ccosalpha and Y2 as Csinalpha Easily yields the maximum value

1 Helpful 0 Interesting 0 Brilliant 0 Confused

It's easy said than done. Could you please show your "easy" solution for everyone to see? Fan of Fujinuma huh?

Shinya Kogami - 2 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...