USA Math Olympiad

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.


\sqrt {1 +\cos\theta} = \cos\theta + \sin \theta \Rightarrow \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}) \Rightarrow \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}


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 {{12}\choose {5}}(2x)^5 {{7}\choose{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.


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.

By Ashani Dasgupta

Ph.D. in Mathematics from University of Wisconsin Milwaukee (USA)
Founder - Faculty at Cheenta

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