# Duke Math Meet 2009 : First Relay Round

1A. Find the lowest positive angle $$\theta$$ that satisfies the equation $$\sqrt {1+\cos \theta} = \sin \theta + \cos\theta$$ expressed in degrees.

Discussion:

$$\sqrt {1 +\cos\theta} = \cos\theta + \sin \theta \implies \sqrt{2\cos^2 \frac{\theta}{2} } = \sqrt2{\frac{1}{\sqrt2} \cos\theta + \frac{1}{\sqrt2} \sin\theta }$$

Now this gives

$$\sqrt2 \cos\frac{\theta}{2} = \sqrt2\cos(\theta – \frac{\pi}{4}) \implies \frac{\theta}{2} = \theta – \frac{\pi}{4}$$ or $$\frac{\theta}{2} = -\theta + \frac{\pi}{4}$$

Thus the possible values of $$\theta$$ are $$90^o$$ or $$30^o$$.

Since we require the smallest positive angle hence the answer is $$30^o$$.

1B Let n be two times the tens digit of TNYWR. Find the coefficient of the $$x^{n-1}y^{n+1}$$ term in the expansion of $$(2x + \frac{y}{2} + 3)^{2n}$$

Discussion:

TNYWR is 3. Hence n = 6 Thus we are required to find coefficient of $$x^5 y^7$$ term in the expansion of $$(2x + \frac{y}{2} + 3 )^{12}$$

This can be easily found from trinomial expansion. The required term is $$\binom{12}{5}(2x)^5 \binom{7}{7} (\frac{y}{2})^7 = 792 \times 32 \times \frac{1}{128} = 198$$

1C Let k be TNYWR, and let n = k/2. Find the smallest integer m greater than n such that 15
divides m and 12 divides the number of positive integer factors of m.

Discussion:

k = 198, hence n = 99.

So we have to look at multiples of 15 greater than 99. We want 12 to divide the number of positive divisors of m.

Suppose $$m = p_1^{\alpha_1} p_2^{\alpha_2} … p_k^{\alpha_k}$$. The number positive divisors of k is $$(\alpha_1 +1 )… (\alpha_k + 1)$$

The first multiple of 15 greater than 99 is $$105 = 15 \times 7$$ . By inspection we see that m = 150.