**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 $a-d,a,a+d$

ReplyDelete$3a=\frac{4}{7}$, 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?

Close. The product of the roots is actually $ -\frac{K}{7} $, so you need to multiply by a factor of $ -7 $.

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