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Let M=[111abca2b2c2]M = \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix} where a,ba, b and cc are real numbers such that a+b+c=0a + b + c = 0 and abc0abc \neq 0. If detM=0\det M = 0, then the maximum possible value of a2+b2c2\frac{a^2+b^2}{c^2} is ___

Entered answer:

Solution

Correct Answer: 5

The determinant of this matrix has a known factored form:

detM=(ba)(ca)(cb)\det M = (b - a)(c - a)(c - b)

For detM=0\det M = 0, at least one factor must vanish, meaning at least two of a,b,ca, b, c must be equal.


Using a+b+c=0a + b + c = 0 along with abc0abc \neq 0, examine each case.

If a=ba = b, then c=2ac = -2a:

a2+b2c2=2a24a2=12\dfrac{a^2 + b^2}{c^2} = \dfrac{2a^2}{4a^2} = \dfrac{1}{2}


If b=cb = c, then a=2ca = -2c:

a2+b2c2=4c2+c2c2=5\dfrac{a^2 + b^2}{c^2} = \dfrac{4c^2 + c^2}{c^2} = 5


If a=ca = c, then b=2cb = -2c:

a2+b2c2=c2+4c2c2=5\dfrac{a^2 + b^2}{c^2} = \dfrac{c^2 + 4c^2}{c^2} = 5


The maximum value among the three cases is 55.

Answer =5= 5

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