Not really Cauchy

Algebra Level 5

Let $a, b, c, d$ be real numbers such that ${ a }^{ 2 }+{ b }^{ 2 }+2a-4b+4=0$ and ${ c }^{ 2 }+{ d }^{ 2 }+4d-4c+4=0$ . Let $m$ and $M$ be the minimum and maximum values of $\sqrt { (a-c)^{ { 2 } }+(b-d)^{ 2 } }$ , respectively. What is $M\times m$ (to 2 decimal places)?

The answer is 16.00.

1 solution

Kushal Bose
Jun 5, 2017

Equation 1: $a^2+b^2+2a-4b+4=0 \\ (a+1)^2+(b-2)^2=1$

Equation 2: $c^2+d^2+4d-4c+4=0 \\ (c-2)^2+(d+2)^2=2^2$

These are two circles non-intersecting.The expression $(a-c)^2+(b-d)^2$ is the square of the distance between two points $(a,b)$ and $(c,d)$ .

The minimum distance will be if two points will be on $M$ and $N$

Now $MN=A_1A_2-r_1-r_2=2=m$

The maximum willbe if the points are $P$ and $R$

Now $PR=A_1A_2+r_1+r_2=5+2+1=8=M$

So, $Mm=8 \times 2=16$

Note: I think there is a square root is missing

Good solution!

Steven Jim - 4 years ago

But I think there will be a square root on the (a-c)^2+(b-d)^2 or answer will be 256

Kushal Bose - 4 years ago

Thanks. I forgot that "The expression $(a-c)^2+(b-d)^2$ is the square of the distance..." I've updated the problem.

Steven Jim - 4 years ago

We see that the centers are A1(2,-2), A2(-1,2).
So A1-A2=5.
For M we shall add both radii to this, M=5+1+2=8.
For m we shall subtract both radii to this, M=5-1-2=2.
M*m=16.
Through over sight I took $\sqrt8$ .

Niranjan Khanderia - 3 years, 6 months ago

Not really Cauchy

The answer is 16.00.

1 solution

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