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
so we can rewrite  as
 .
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  form an arithmetic sequence. Let  , where  and  are relatively prime positive integers. Find the value of  .
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April 1st, 2007 at 8:03 pm
Zeroes are [tex]a-d,a,a+d[/tex]
[tex]3a=\frac{4}{7}[/tex], and we are trying to find [tex]a^3-ad^2[/tex]
We’re also given [tex]a(a-d)+a(a+d)+(a-d)(a+d)=3a^2-d^2=0[/tex] so [tex]3a^2=d^2[/tex]
Then we want [tex]a^3-3a^3=-2a^3[/tex] but we already know [tex]a[/tex]…the answer is [tex]\frac{(-2)(4^3)}{21^3}[/tex] whatever that is…?…is this correct?
April 2nd, 2007 at 3:55 pm
Close. The product of the roots is actually [tex] -\frac{K}{7} [/tex], so you need to multiply by a factor of [tex] -7 [/tex].