Difference between revisions of "1996 USAMO Problems/Problem 1"
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− | Notice that for every <math>n\sin n^\circ</math> there exists a corresponding pair term <math>(180^\circ - n)\sin{180^\circ - n} = (180^\circ - n)\sin n^\circ</math>, for <math>n</math> not <math>90^\circ</math>. Pairing gives the sum of all <math>n\sin n^\circ</math> terms to be <math>90(\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ)</math>, and thus the average is <cmath>S = (\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ). (*)</cmath> We need to show that <math>S = \cot 1^\circ</math>. Multiplying (*) by <math>2\sin | + | Notice that for every <math>n\sin n^\circ</math> there exists a corresponding pair term <math>(180^\circ - n)\sin{180^\circ - n} = (180^\circ - n)\sin n^\circ</math>, for <math>n</math> not <math>90^\circ</math>. Pairing gives the sum of all <math>n\sin n^\circ</math> terms to be <math>90(\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ)</math>, and thus the average is <cmath>S = (\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ). (*)</cmath> We need to show that <math>S = \cot 1^\circ</math>. Multiplying (*) by <math>2\sin 1^\circ</math> and using sum-to-product and telescoping gives <math>2\sin 1^\circ S = \cos 1^\circ - \cos 179^\circ = 2\cos 1^\circ</math>. Thus, <math>S = \frac{\cos 1^\circ}{\sin 1^\circ} = \cot 1^\circ</math>, as desired. |
<math>\Box</math> | <math>\Box</math> | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:55, 5 June 2014
Problem
Prove that the average of the numbers is
.
Solution 1
First, as we omit that term. Now, we multiply by
to get, after using product to sum,
.
This simplifies to
. Since
this simplifies to
. We multiplied by
in the beginning, so we must divide by it now, and thus the sum is just
, so the average is
, as desired.
Solution 2
Notice that for every there exists a corresponding pair term
, for
not
. Pairing gives the sum of all
terms to be
, and thus the average is
We need to show that
. Multiplying (*) by
and using sum-to-product and telescoping gives
. Thus,
, as desired.
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