Everyone should love algebra more than geometry

Geometry Level 5

A square having sides parallel to the coordinate axes is inscribed in the region $\{(x,y): x,y > 0,$ $y< -x^{3} + 3x\}$ . If the area of the square is written as $A^{1/3} + B^{1/3}$ where $A, B$ are integers and $A>B$ , then what is the circum-radius of triangle $OPQ$ where $O$ is the origin, $P(A^{1/3},0)$ and $Q(0, B^{1/3})$ ?

The answer is 3.83.

1 solution

Archit Tripathi
Sep 18, 2016

Let two x-coordinates of the square's vertices be $a$ and $b$ with $a > b$ , then $a^{3} - 3a = b^{3} - 3b = a - b$

so, $a = b^{3} - 2b$ and $b = 4a - a^{3}$

$\Rightarrow$ $a^{3} = 4a - b$ and $b^{3} = a + 2b$

$\Rightarrow$ $(a - b)(a^{2} + b^{2} + ab) = 3(a - b)$

$\Rightarrow$ $a^{2} + b^{2} + ab = 3$

Solving these equations, we get

$b^{2} = 1 + 2^{1/3}$

hence, area of square will be $(216)^{1/3} + (-108)^{1/3}$ So, $A = 216, B = -108$ .

The circum-radius of triangle OPQ is about 3.83, unless I am missing something.

Maria Kozlowska - 4 years, 7 months ago

I got the same values, the circumradius is 3.83.

Siva Bathula - 4 years, 7 months ago

but how do we solve to get b^2=1+2^(1/3)

Tanushree Aditya - 4 years, 6 months ago

https://youtu.be/5J7v8W5e_CA

For more such questions one can use the math app called doubtnut

Tanushree Aditya - 4 years, 6 months ago

How do we get the values of A and B?

Nisha Saxena - 2 years, 12 months ago

Everyone should love algebra more than geometry

The answer is 3.83.

1 solution

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