Difference between revisions of "1996 USAMO Problems/Problem 1"

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<math>\Box</math>
 
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===Solution 2 (but elaborated)-hashtagmath===
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===Solution 3 -hashtagmath===
 
We know that the average of a list of numbers is the sum of all the terms, divided by the number of terms. So we will set up an average. This average will average to <math>\cot 1^\circ</math>. So we can set it equal to <math>\frac{\cos 1^\circ}{\sin 1^\circ}</math>. Doing so, gives us <cmath>\frac{2 \sin 2^\circ + 4 \sin 4^\circ + 6 \sin 6^\circ + 8 \sin 8^\circ + 10 \sin 10^\circ + 12 \sin 12^\circ ... + 180 \sin 180^\circ}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}.(1)</cmath>
 
We know that the average of a list of numbers is the sum of all the terms, divided by the number of terms. So we will set up an average. This average will average to <math>\cot 1^\circ</math>. So we can set it equal to <math>\frac{\cos 1^\circ}{\sin 1^\circ}</math>. Doing so, gives us <cmath>\frac{2 \sin 2^\circ + 4 \sin 4^\circ + 6 \sin 6^\circ + 8 \sin 8^\circ + 10 \sin 10^\circ + 12 \sin 12^\circ ... + 180 \sin 180^\circ}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}.(1)</cmath>
  

Revision as of 15:34, 23 February 2019

Problem

Prove that the average of the numbers $n\sin n^{\circ}\; (n = 2,4,6,\ldots,180)$ is $\cot 1^\circ$.

Solution

Solution 1

First, as $180\sin{180^\circ}=0,$ we omit that term. Now, we multiply by $\sin 1^\circ$ to get, after using product to sum, $(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos 5^\circ)+\cdots +89(\cos 177^\circ-\cos 179^\circ)$. This simplifies to $\cos 1^\circ+\cos 3^\circ +\cos 5^\circ+\cos 7^\circ+...+\cos 177^\circ-89\cos 179^\circ$. Since $\cos x=-\cos(180-x),$ this simplifies to $90\cos 1^\circ$. We multiplied by $\sin 1^\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\cot 1^\circ$, so the average is $\cot 1^\circ$, as desired.

$\Box$

Solution 2

Notice that for every $n\sin n^\circ$ there exists a corresponding pair term $(180^\circ - n)\sin{180^\circ - n} = (180^\circ - n)\sin n^\circ$, for $n$ not $90^\circ$. Pairing gives the sum of all $n\sin n^\circ$ terms to be $90(\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ)$, and thus the average is \[S = (\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ). (*)\] We need to show that $S = \cot 1^\circ$. Multiplying (*) by $2\sin 1^\circ$ and using sum-to-product and telescoping gives $2\sin 1^\circ S = \cos 1^\circ - \cos 179^\circ = 2\cos 1^\circ$. Thus, $S = \frac{\cos 1^\circ}{\sin 1^\circ} = \cot 1^\circ$, as desired.

$\Box$

Solution 3 -hashtagmath

We know that the average of a list of numbers is the sum of all the terms, divided by the number of terms. So we will set up an average. This average will average to $\cot 1^\circ$. So we can set it equal to $\frac{\cos 1^\circ}{\sin 1^\circ}$. Doing so, gives us \[\frac{2 \sin 2^\circ + 4 \sin 4^\circ + 6 \sin 6^\circ + 8 \sin 8^\circ + 10 \sin 10^\circ + 12 \sin 12^\circ ... + 180 \sin 180^\circ}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}.(1)\]

We should try to simplify the numerator as it looks pretty messy.

Now we know that $\sin x = \sin (180-x)$, so maybe we can use this to clear things up a bit. Applying this to $(1)$ gives us \[\frac{2 \sin 178^\circ + 4 \sin 176^\circ + 6 \sin 174^\circ + 8 \sin 172^\circ + 10 \sin 170^\circ + 12 \sin 168^\circ ... + 180 \sin 0^\circ}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}. (2)\]

Now, we know that $\sin 2^\circ = \sin 178^\circ$, and $\sin 4^\circ = \sin 176^\circ$ and so on. We also know that $\sin 0^\circ = 0$, so we can omit the last term. Also, we know that $\cos 90 = 1$. Thus $90 \cos 90 = 90$. So we will use all these when we further simplify. Another thing that we can simplify is to notice that when we reach the term $92\sin 88^\circ$, we can rewrite it as $92\sin 92^\circ$. Thus we can rewrite everything in the form $x\sin y^\circ$ as $x \sin x^\circ$, starting with the term $92\sin 88^\circ$, then $94\sin 86^\circ$, and so on.

Thus we can further simplify our equation. Doing so, gives us \[\frac{2 \sin 178^\circ + 4 \sin 176^\circ + 6 \sin 174^\circ + 8 \sin 172^\circ + 10 \sin 170^\circ + 12 \sin 168^\circ ...+ 176 \sin 176^\circ + 178 \sin 178^\circ + 90}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}. (3)\]

Then, to make things easier, we can rewrite all the numbers with degrees less than $90^\circ$ and combine like terms. Doing so gives us \[\frac{180 \sin 2^\circ + 180 \sin 4^\circ + 180 \sin 6^\circ +180 \sin 8^\circ +180 \sin 10^\circ ... +180 \sin 88^\circ + (90)}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}. (4)\]

Now we can factor out $180$ to get \[\frac{180(\sin 2^\circ+ \sin 4^\circ + \sin 6^\circ + \sin 8^\circ + \sin 10^\circ ... + \sin 88^\circ) + 90}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}. (5)\]

Then we can simplify it by getting rid of the denominator to get to \[2(\sin 2^\circ+ \sin 4^\circ + \sin 6^\circ + \sin 8^\circ + \sin 10^\circ ... + \sin 88^\circ) + 1 = \frac{\cos 1^\circ}{\sin 1^\circ}. (6)\]

Now we notice that we can multiply both sides by $\sin 1^\circ$. Doing so gives us \[\sin 1^\circ(2 (\sin 2^\circ+ \sin 4^\circ + \sin 6^\circ + \sin 8^\circ + \sin 10^\circ ... + \sin 88^\circ) + 1) = \cos 1^\circ. (7)\]

Now, simplifying $\sin 2^\circ+ \sin 4^\circ + \sin 6^\circ + \sin 8^\circ + \sin 10^\circ ... + \sin 88^\circ$ doesn't look too promising. So maybe if we expand again, we can maybe somehow use our product to sum formulas.

Doing so, gives us \[(\sin 1^\circ)(2 \sin 2^\circ)+(\sin 1^\circ)(2\sin 4^\circ) +(\sin 1^\circ)(2\sin 6^\circ) + (\sin 1^\circ)(2\sin 8^\circ) +(\sin 1^\circ)(2 \sin 10^\circ) ... + (\sin 1^\circ)(2\sin 88^\circ) + \sin 1^\circ = \cos 1^\circ. (8)\]

Now we can use our product of sines to sum formula and see if we can find a pattern.

We will start with expanding $(\sin 1^\circ)(2 \sin 2^\circ)$. Because the format of our formula for the product of sines is $\sin \alpha + \sin \beta$, we can factor out the $2$ and find the product, then multiply by $2$. So, we are at $2(\sin 1^\circ)(\sin 2^\circ)$. Our formula for the product of sines is $\sin \alpha + \sin \beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)]$. Plugging in our values into the formula and simplifying gives us $2\left(\frac{1}{2}(\cos 1^\circ - \cos 3^\circ)\right)$. We know that $2\left(\frac{1}{2}\right)$ will just cancel out to $1$. So we are left with $\cos 1^\circ - \cos 3^\circ$. $(9)$

Next we expand $(\sin 1^\circ)(2\sin 4^\circ)$. We use the same steps or strategy as we did above and get $\cos 3^\circ - \cos 5^\circ$. $(10)$

We may be noticing a pattern. Just to make sure it is true, we will expand $(\sin 1^\circ)(2\sin 6^\circ)$. When we expand this using the same strategy above, we get $\cos 5^\circ - \cos 7^\circ$. $(11)$ So our pattern is that $(\sin 1^\circ)(2\sin \mu^\circ) = \cos (\mu-1)^\circ - \cos (\mu+1)^\circ. (12)$

Now we notice that when we add $(9)$ and $(10)$, $-\cos 3^\circ$ and $+\cos 3^\circ$ cancel themselves out to just $\cos 1^\circ - \cos 5^\circ$. Then, when we add $(11)$ to it, $+ \cos 5^\circ$ and $- \cos 5^\circ$ also cancel themselves out.

Now, we need to figure out when we should stop canceling out. We can use formula $(12)$ and replace $88$ with $\mu$ since $88$ is the last one in $(8)$. Applying this formula and simplifying gives us $\cos 87^\circ - \cos 89$, so the last term would be $- \cos 89$.

Thus we know that everything we cancel themselves out except the first and the last term. So, we are left with \[(\cos 1^\circ - \cos 89^\circ) + \sin 1^\circ = \cos 1^\circ. (13)\]

Now we notice that we have $\cos 1^\circ$ on both sides. Thus we can subtract $\cos 1^\circ$ from both sides to get $- \cos 89^\circ + \sin 1^\circ = 0$. Now we can add $\cos 89^\circ$ to both sides and get $\sin 1^\circ = \cos 89^\circ$. Now we also know that $\sin (\omega)^\circ = \cos(90-\omega)$. Thus we know that $\sin 1^\circ = \cos 89^\circ$. Hence, our proof is complete. $\square$

See Also

1996 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6
All USAMO Problems and Solutions

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