# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

• ## Meta

Sumsine. Topic: Trigonometry/Geometry. Level: AMC. April 1st, 2007

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 .

### 2 Responses to “Sumsine. Topic: Trigonometry/Geometry. Level: AMC.”

1. ch1n353ch3s54a1l Says:

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?

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