## Sunday, April 1, 2007

### Sumsine. Topic: Trigonometry/Geometry. Level: AMC.

Problem: (2007 MAO State - Gemini) Find the sum of the sines of the angles of a triangle whose perimeter is five times as large as its circumradius.

Solution: At first, this problem seems very strange because we do not expect a nice relationship between the perimeter and the circumradius to offer much, but take another look. Sines, circumradius... Law of Sines! In fact, we get

$\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} = 2R$

so we can rewrite $\sin{A}+\sin{B}+\sin{C}$ as

$\sin{A}+\sin{B}+\sin{C} = \frac{a}{2R}+\frac{b}{2R}+\frac{c}{2R} = \frac{a+b+c}{2R} = \frac{5R}{2R} = \frac{5}{2}$.

QED.

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Comment: Admittedly, MAO does not exactly have a plethora of cool problems, but this one was not bad. It wasn't too difficult, but it required some clever use of well-known identities to pull it off. Given that half of the problems were computationally intensive (think AIME I but worse), this was a nice relief in the middle of the contest.

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Practice Problem: (2007 MAO State - Gemini) The zeros of $f(x) = 7x^3-4x^2+K$ form an arithmetic sequence. Let $K = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find the value of $m+n$.

Zeroes are $a-d,a,a+d$
$3a=\frac{4}{7}$, and we are trying to find $a^3-ad^2$
We're also given $a(a-d)+a(a+d)+(a-d)(a+d)=3a^2-d^2=0$ so $3a^2=d^2$
Then we want $a^3-3a^3=-2a^3$ but we already know $a$...the answer is $\frac{(-2)(4^3)}{21^3}$ whatever that is...?...is this correct?
2. Close. The product of the roots is actually $-\frac{K}{7}$, so you need to multiply by a factor of $-7$.