Difference between revisions of "1996 USAMO Problems/Problem 1"
m (→Solution 3 -hashtagmath) |
|||
(9 intermediate revisions by 4 users not shown) | |||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
− | 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-\ | + | ===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-\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. | 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. | ||
<math>\Box</math> | <math>\Box</math> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | 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> | ||
+ | |||
+ | ===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 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> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Now we can use our product of sines to sum formula and see if we can find a pattern. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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 == | ||
+ | {{USAMO newbox|year=1996|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Olympiad Algebra Problems]] |
Latest revision as of 12:01, 11 August 2020
Problem
Prove that the average of the numbers is .
Solution
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.
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 . So we can set it equal to . Doing so, gives us
We should try to simplify the numerator as it looks pretty messy.
Now we know that , so maybe we can use this to clear things up a bit. Applying this to gives us
Now, we know that , and and so on. We also know that , so we can omit the last term. Also, we know that . Thus . 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 , we can rewrite it as . Thus we can rewrite everything in the form as , starting with the term , then , and so on.
Thus we can further simplify our equation. Doing so, gives us
Then, to make things easier, we can rewrite all the numbers with degrees less than and combine like terms. Doing so gives us
Now we can factor out to get
Then we can simplify it by getting rid of the denominator to get to
Now we notice that we can multiply both sides by . Doing so gives us
Now, simplifying 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
Now we can use our product of sines to sum formula and see if we can find a pattern.
We will start with expanding . Because the format of our formula for the product of sines is , we can factor out the and find the product, then multiply by . So, we are at . Our formula for the product of sines is . Plugging in our values into the formula and simplifying gives us . We know that will just cancel out to . So we are left with .
Next we expand . We use the same steps or strategy as we did above and get .
We may be noticing a pattern. Just to make sure it is true, we will expand . When we expand this using the same strategy above, we get . So our pattern is that
Now we notice that when we add and , and cancel themselves out to just . Then, when we add to it, and also cancel themselves out.
Now, we need to figure out when we should stop canceling out. We can use formula and replace with since is the last one in . Applying this formula and simplifying gives us , so the last term would be .
Thus we know that everything we cancel themselves out except the first and the last term. So, we are left with
Now we notice that we have on both sides. Thus we can subtract from both sides to get . Now we can add to both sides and get . Now we also know that . Thus we know that . Hence, our proof is complete.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.