Circles and Polynomials

Geometry Level pending

The sides of a triangle $ABC$ are the roots of the equation $x^3 - px^2 + qx - r = 0$ . The inradius and circumradius of triangle $ABC$ are the roots of the equation $2x^2 - dx + 27 = 0$ .

If the minimum value of $q^3$ can be represented as $(m^v)$ , where $m>1$ and $v>1$ are integers and $m$ is minimized, determine the value of $(m+v)$ .

The answer is 18.

1 solution

Yashas Ravi
Apr 27, 2019

The product of the roots of the quadratic equation would be the constant term. Since the inradius and circumradius are the roots and their product is $13.5$ , $(abc) = 27(a+b+c)$ if $a$ , $b$ , and $c$ are the sides. Using the $AM-GM$ inequality, $(a+b+c)>27$ so $(abc)>27^2$ . Again, using the $AM-GM$ inequality, we can set the minimum value of $(ab+bc+ac)$ which is the value of $q$ :

By substituting $abc>27^2$ and cubing both sides to get the minimum value of $(ab+bc+ac)^3$ , the minimum value is $3$ to the power of $15$ . This is in the desired form so $(m+v)=3+15=18$ which is the final answer. In fact, Triangle $ABC$ has to be equilateral with side length $8$ when the minimum value is satisfied.

How do you get a+b+c>27? And also abc>27^2?

A Former Brilliant Member - 2 years, 1 month ago

(abc) = 27(a+b+c) is derived from multiplying the inradius and circumradius formulas together. I then substituted (a+b+c)=D and abc=27(a+b+c)=27D into the AM-GM inequality and I got D>27, meaning (a+b+c)>27 and (abc)=27(a+b+c) so (abc)>27^2.

Here is a link to my work: https://docs.google.com/document/d/1RW-SVv6ba0qf_uWMWvbiWyn8oNkHIsA34SjNBuLryNg/edit?usp=sharing

Yashas Ravi - 2 years, 1 month ago

Circles and Polynomials

The answer is 18.

1 solution

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