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

Latest revision as of 11:01, 11 August 2020

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 $\sin 90 = 1$ . Thus $90 \sin 90 = 90$ . 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. $\square$

@@ Line 4: / Line 4: @@
 ==Solution==
 ===Solution 1===
-First, as <math>180\sin{180^\circ}=0,</math> we omit that term. Now, we multiply by <math>\sin 1^\circ</math> to get, after using product to sum, <math>(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos5)+\cdots +89(\cos 177^\circ-\cos 179^\circ)</math>.
+First, as <math>180\sin{180^\circ}=0,</math> we omit that term. Now, we multiply by <math>\sin 1^\circ</math> to get, after using product to sum, <math>(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos 5^\circ)+\cdots +89(\cos 177^\circ-\cos 179^\circ)</math>.
 This simplifies to <math>\cos 1^\circ+\cos 3^\circ +\cos 5^\circ+\cos 7^\circ+...+\cos 177^\circ-89\cos 179^\circ</math>. Since <math>\cos x=-\cos(180-x),</math> this simplifies to <math>90\cos 1^\circ</math>. We multiplied by <math>\sin 1^\circ</math> in the beginning, so we must divide by it now, and thus the sum is just <math>90\cot 1^\circ</math>, so the average is <math>\cot 1^\circ</math>, as desired.
@@ Line 14: / Line 14: @@
 <math>\Box</math>
-===Solution 3 (Very long and detailed)===
+===Solution 3 -hashtagmath===
-Make the sum of the numbers equal <math>x</math>. Now, <math>2\sin{2^\circ}+4\sin{4^\circ}+...+180\sin{180^\circ}=x</math> the average of these numbers is <math>\frac{x}{90}</math>.
+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 <math>180\sin{180^\circ}=0</math>, so we can eliminate that term and use the identity, <math>\sin(\theta)=\sin(180-\theta)</math> to get
+We should try to simplify the numerator as it looks pretty messy.
+Now we know that <math>\sin x = \sin (180-x)</math>, so maybe we can use this to clear things up a bit. Applying this to <math>(1)</math> gives us <cmath>\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)</cmath>
-<math>x=(2\sin{2^\circ}+178\sin{2^\circ})+...+(88\sin{88^\circ}+92\sin{22^\circ})+90\sin{90^\circ}</math>
+Now, we know that <math>\sin 2^\circ = \sin 178^\circ</math>, and <math>\sin 4^\circ = \sin 176^\circ</math> and so on. We also know that <math>\sin 0^\circ = 0</math>, so we can omit the last term. Also, we know that <math>\sin 90 = 1</math>. Thus <math>90 \sin 90 = 90</math>. 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 <math>92\sin 88^\circ</math>, we can rewrite it as <math>92\sin 92^\circ</math>. Thus we can rewrite everything in the form <math>x\sin y^\circ</math> as <math>x \sin x^\circ</math>, starting with the term <math>92\sin 88^\circ</math>, then <math>94\sin 86^\circ</math>, and so on.
-Or, <math>x=180(\sin{2^\circ}+\sin{4^\circ}+\sin{6^\circ}+...+\sin{88^\circ})+90</math>
+Thus we can further simplify our equation. Doing so, gives us <cmath>\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)</cmath>
-Pairing the terms up, using sum-product identity, and simplifying, yields: <math>x=180\sqrt 2 (\sin47^\circ + \sin49^\circ + \ldots + \sin89^\circ)+ 90</math>.
+Then, to make things easier, we can rewrite all the numbers with degrees less than <math>90^\circ</math> and combine like terms. Doing so gives us  <cmath>\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)</cmath>
-After dividing both sides by <math>90</math>, you get: <math>\frac{x}{90}=2\sqrt 2 (\sin47^\circ + \sin49^\circ + \ldots + \sin89^\circ) + 1=\frac{\cos{1^\circ}}{\sin{1^\circ}}</math>.
+Now we can factor out <math>180</math> to get <cmath>\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)</cmath>
-Now we have to prove that <math>2\sqrt 2 (\sin47^\circ + \sin49^\circ + \ldots + \sin89^\circ) + 1=\frac{\cos{1^\circ}}{\sin{1^\circ}}</math>.
+Then we can simplify it by getting rid of the denominator to get to <cmath>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)</cmath>
-Multiply both sides by <math>\sin{1^\circ}</math> to get <math>\cos{1^\circ}=2\sqrt{2}(\sin{1^\circ}\sin{47^\circ}+\sin{1^\circ}\sin{49^\circ}+...+\sin{1^\circ}\sin{89^\circ})+\sin{1^\circ}</math>.
+Now we notice that we can multiply both sides by <math>\sin 1^\circ</math>. Doing so gives us <cmath> \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)</cmath>
-After applying product-sum identities, you get <math>\cos{1^\circ}=\sqrt{2}[(\cos{46^\circ}-\cos{48^\circ})+(\cos{48^\circ}-\cos{50^\circ})+...+(\cos{88^\circ}-\cos{90^\circ})]+\sin{1^\circ}</math>.
+Now, simplifying <math>\sin 2^\circ+ \sin 4^\circ + \sin 6^\circ + \sin 8^\circ + \sin 10^\circ ... + \sin 88^\circ</math> doesn't look too promising. So maybe if we expand again, we can maybe somehow use our product to sum formulas.
-This is just <math>\cos{1^\circ}=\sqrt{2}(\cos{(1^\circ+45^\circ}))+\sin{1^\circ}</math>.
+Doing so, gives us <cmath> (\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)</cmath>
-After applying angle addition formulas, you get: <math>\cos{1^\circ}=\sqrt{2}(\cos{1^\circ}\cos{45^\circ}-\sin{1^\circ}\sin{45^\circ})+\sin{1^\circ}</math>.
+Now we can use our product of sines to sum formula and see if we can find a pattern.
-Since the cosine and sine of <math>45^\circ</math> are <math>\frac{\sqrt{2}}{2}</math> you can simplify that to: <math>\cos{1^\circ}=(\cos{1^\circ}-\sin{1^\circ})+\sin{1^\circ}</math>. Or, <math>\cos{1^\circ}=\cos{1^\circ}</math>.
+We will start with expanding <math>(\sin 1^\circ)(2 \sin 2^\circ)</math>. Because the format of our formula for the product of sines is <math>\sin \alpha + \sin \beta</math>, we can factor out the <math>2</math> and find the product, then multiply by <math>2</math>. So, we are at <math>2(\sin 1^\circ)(\sin 2^\circ)</math>. Our formula for the product of sines is <math>\sin \alpha + \sin \beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)]</math>. Plugging in our values into the formula and simplifying gives us <math>2\left(\frac{1}{2}(\cos 1^\circ - \cos 3^\circ)\right)</math>. We know that <math>2\left(\frac{1}{2}\right)</math> will just cancel out to <math>1</math>. So we are left with <math>\cos 1^\circ - \cos 3^\circ</math>. <math>(9)</math>
-Therefore, the average of the numbers <math>n \sin n^\circ</math> (<math>n = 2, 4, 6, \ldots, 180</math>) is <math>\cot 1^\circ</math>.
+Next we expand <math>(\sin 1^\circ)(2\sin 4^\circ)</math>. We use the same steps or strategy as we did above and get <math>\cos 3^\circ - \cos 5^\circ</math>. <math>(10)</math>
-<cmath>\boxed{\frac{n\sin{n^\circ}(n=2,4,6,\ldots,180)}{90}=\cot{1^\circ}}</cmath>
+We may be noticing a pattern. Just to make sure it is true, we will expand <math>(\sin 1^\circ)(2\sin 6^\circ)</math>. When we expand this using the same strategy above, we get <math>\cos 5^\circ - \cos 7^\circ</math>. <math>(11)</math> So our pattern is that <math>(\sin 1^\circ)(2\sin \mu^\circ) = \cos (\mu-1)^\circ - \cos (\mu+1)^\circ. (12)</math>
-<math>\Box</math>
+Now we notice that when we add <math>(9)</math> and <math>(10)</math>, <math>-\cos 3^\circ</math> and <math>+\cos 3^\circ</math> cancel themselves out to just <math>\cos 1^\circ - \cos 5^\circ</math>. Then, when we add <math>(11)</math> to it, <math>+ \cos 5^\circ</math> and <math>- \cos 5^\circ</math> also cancel themselves out.
+Now, we need to figure out when we should stop canceling out. We can use formula <math>(12)</math> and replace <math>88</math> with <math>\mu</math> since <math>88</math> is the last one in <math>(8)</math>. Applying this formula and simplifying gives us <math>\cos 87^\circ - \cos 89</math>, so the last term would be <math>- \cos 89</math>.
+Thus we know that everything we cancel themselves out except the first and the last term. So, we are left with <cmath>(\cos 1^\circ - \cos 89^\circ) + \sin 1^\circ = \cos 1^\circ. (13)</cmath>
+Now we notice that we have <math>\cos 1^\circ</math> on both sides. Thus we can subtract <math>\cos 1^\circ</math> from both sides to get <math>- \cos 89^\circ + \sin 1^\circ = 0</math>. Now we can add <math>\cos 89^\circ</math> to both sides and get <math>\sin 1^\circ = \cos 89^\circ</math>. Now we also know that <math>\sin (\omega)^\circ = \cos(90-\omega)</math>. Thus we know that <math>\sin 1^\circ = \cos 89^\circ</math>. Hence, our proof is complete. <math>\square</math>
 == See Also ==

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

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