https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Matrix+math&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T21:30:22ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_3&diff=956462005 AIME II Problems/Problem 32018-06-27T14:59:13Z<p>Matrix math: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
An [[infinite]] [[geometric series]] has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common [[ratio]] of the original series is <math> \frac mn </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[integer]]s. Find <math> m+n. </math><br />
<br />
== Solution 1 ==<br />
Let's call the first term of the original [[geometric series]] <math>a</math> and the common ratio <math>r</math>, so <math>2005 = a + ar + ar^2 + \ldots</math>. Using the sum formula for [[infinite]] geometric series, we have <math>(*)\;\;\frac a{1 -r} = 2005</math>. Then we form a new series, <math>a^2 + a^2 r^2 + a^2 r^4 + \ldots</math>. We know this series has sum <math>20050 = \frac{a^2}{1 - r^2}</math>. Dividing this equation by <math>(*)</math>, we get <math>10 = \frac a{1 + r}</math>. Then <math>a = 2005 - 2005r</math> and <math>a = 10 + 10r</math> so <math>2005 - 2005r = 10 + 10r</math>, <math>1995 = 2015r</math> and finally <math>r = \frac{1995}{2015} = \frac{399}{403}</math>, so the answer is <math>399 + 403 = \boxed{802}</math>. <br />
<br />
(We know this last fraction is fully reduced by the [[Euclidean algorithm]] -- because <math>4 = 403 - 399</math>, <math>\gcd(403, 399) | 4</math>. But 403 is [[odd integer | odd]], so <math>\gcd(403, 399) = 1</math>.)<br />
<br />
==Solution 2==<br />
We can write the sum of the original series as <math>a + a\left(\dfrac{m}{n}\right) + a\left(\dfrac{m}{n}\right)^2 + \ldots = 2005</math>, where the common ratio is equal to <math>\dfrac{m}{n}</math>. We can also write the sum of the second series as <math>a^2 + a^2\left(\dfrac{m}{n}\right)^2 + a^2\left(\left(\dfrac{m}{n}\right)^2\right)^2 + \ldots = 20050</math>. Using the formula for the sum of an infinite geometric series <math>S=\dfrac{a}{1-r}</math>, where <math>S</math> is the sum of the sequence, <math>a</math> is the first term of the sequence, and <math>r</math> is the ratio of the sequence, the sum of the original series can be written as <math>\dfrac{a}{1-\frac{m}{n}}=\dfrac{a}{\frac{n-m}{n}}=\dfrac{a \cdot n}{n-m}=2005\;\text{(1)}</math>, and the second sequence can be written as <math>\dfrac{a^2}{1-\frac{m^2}{n^2}}=\dfrac{a^2}{\frac{n^2-m^2}{n^2}}=\dfrac{a^2\cdot n^2}{(n+m)(n-m)}=20050\;\text{(2)}</math>. Dividing <math>\text{(2)}</math> by <math>\text{(1)}</math>, we obtain <math>\dfrac{a\cdot n}{m+n}=10</math>, which can also be written as <math>a\cdot n=10(m+n)</math>. Substitute this value for <math>a\cdot n</math> back into <math>\text{(1)}</math>, we obtain <math>10\cdot \dfrac{n+m}{n-m}=2005</math>. Dividing both sides by 10 yields <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> we can now write a system of equations with <math>n+m=401</math> and <math>n-m=2</math>, but this does not output integer solutions. However, we can also write <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> as <math>\dfrac{n+m}{n-m}=\dfrac{802}{4}</math>. This gives the system of equations <math>m+n=802</math> and <math>n-m=4</math>, which does have integer solutions. Our answer is therefore <math>m+n=\boxed{802}</math> (Solving for <math>m</math> and <math>n</math> gives us <math>399</math> and <math>403</math>, respectively, which are co-prime).<br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=2|num-a=4}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_18&diff=902552011 AMC 10A Problems/Problem 182018-02-05T04:01:21Z<p>Matrix math: /* Problem 18 */</p>
<hr />
<div>== Problem 18 ==<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<asy> pathpen = linewidth(.7); pointpen = black; pair A=(-1,0), B=-A, C=(0,1); fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5)); D(CR(D("A",A,SW),1)); D(CR(D("B",B,SE),1)); D(CR(D("C",C,N),1)); </asy><br />
<br />
<math> \textbf{(A)}\ 3 - \frac{\pi}{2} \qquad\textbf{(B)}\ \frac{\pi}{2} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{3\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\pi}{2} </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
== Solution ==<br />
<br />
Not specific: Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shaded area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{ \mathbf{(C)} 2}</math>.<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2011|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_4&diff=900972011 AMC 10A Problems/Problem 42018-01-31T03:06:01Z<p>Matrix math: /* Solution 2 */</p>
<hr />
<div>Let X and Y be the following sums of arithmetic sequences: <br />
<br />
<cmath> \begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*} </cmath><br />
<br />
What is the value of Y - X? <br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
==Solution 1==<br />
We see that both sequences have equal numbers of terms, so reformat the sequence to look like: <br />
<br />
<cmath>\begin{align*}<br />
Y = \ &12 + 14 + \cdots + 100 + 102\\<br />
X = 10 \ + \ &12 + 14 + \cdots + 100\\<br />
\end{align*}</cmath><br />
From here it is obvious that Y - X = 102 - 10 = <math>\boxed{92 \ \mathbf{(A)}}</math>.<br />
<br />
===Note===<br />
Another way to see this is to let the sum <math>12+14+16+...+100=x.</math> So, the sequences become<br />
<cmath>\begin{align*}<br />
X = 10+x \\<br />
Y= x+102 \\<br />
\end{align*}</cmath><br />
<br />
Like before, the difference between the two sequences is <math>Y-X=102-12=90.</math><br />
<br />
==Solution 2==<br />
<br />
We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be:<br />
<math>46\cdot 2=\boxed{92}</math><br />
<br />
== See Also ==<br />
{{AMC10 box|year=2011|ab=A|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_3&diff=898652005 AIME II Problems/Problem 32018-01-17T21:19:11Z<p>Matrix math: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
An [[infinite]] [[geometric series]] has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common [[ratio]] of the original series is <math> \frac mn </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[integer]]s. Find <math> m+n. </math><br />
<br />
== Solution 1 ==<br />
Let's call the first term of the original [[geometric series]] <math>a</math> and the common ratio <math>r</math>, so <math>2005 = a + ar + ar^2 + \ldots</math>. Using the sum formula for [[infinite]] geometric series, we have <math>(*)\;\;\frac a{1 -r} = 2005</math>. Then we form a new series, <math>a^2 + a^2 r^2 + a^2 r^4 + \ldots</math>. We know this series has sum <math>20050 = \frac{a^2}{1 - r^2}</math>. Dividing this equation by <math>(*)</math>, we get <math>10 = \frac a{1 + r}</math>. Then <math>a = 2005 - 2005r</math> and <math>a = 10 + 10r</math> so <math>2005 - 2005r = 10 + 10r</math>, <math>1995 = 2015r</math> and finally <math>r = \frac{1995}{2015} = \frac{399}{403}</math>, so the answer is <math>399 + 403 = \boxed{802}</math>. <br />
<br />
(We know this last fraction is fully reduced by the [[Euclidean algorithm]] -- because <math>4 = 403 - 399</math>, <math>\gcd(403, 399) | 4</math>. But 403 is [[odd integer | odd]], so <math>\gcd(403, 399) = 1</math>.)<br />
<br />
==Solution 2==<br />
We can write the sum of the original series as <math>a + a(\dfrac{m}{n}) + a(\dfrac{m}{n})^2 + \ldots = 2005</math>, where the common ratio is equal to <math>\dfrac{m}{n}</math>. We can also write the sum of the second series as <math>a^2 + a(\dfrac{m}{n})^2 + a((\dfrac{m}{n})^2)^2 + \ldots = 20050</math>. Using the formula for the sum of an infinite geometric series <math>S=\dfrac{a}{1-r}</math>, where <math>S</math> is the sum of the sequence, <math>a</math> is the first term of the sequence, and <math>r</math> is the ratio of the sequence, the sum of the original series can be written as <math>\dfrac{a}{1-\frac{m}{n}}=\dfrac{a}{\frac{n-m}{n}}=\dfrac{a \cdot n}{n-m}=2005\;\text{(1)}</math>, and the second sequence can be written as <math>\dfrac{a^2}{1-\frac{m^2}{n^2}}=\dfrac{a^2}{\frac{n^2-m^2}{n^2}}=\dfrac{a^2\cdot n^2}{(n+m)(n-m)}=20050\;\text{(2)}</math>. Dividing <math>\text{(2)}</math> by <math>\text{(1)}</math>, we obtain <math>\dfrac{a\cdot n}{m+n}=10</math>, which can also be written as <math>a\cdot n=10(m+n)</math>. Substitute this value for <math>a\cdot n</math> back into <math>\text{(1)}</math>, we obtain <math>10\cdot \dfrac{n+m}{n-m}=2005</math>. Dividing both sides by 10 yields <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> we can now write a system of equations with <math>n+m=401</math> and <math>n-m=2</math>, but this does not output integer solutions. However, we can also write <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> as <math>\dfrac{n+m}{n-m}=\dfrac{802}{4}</math>. This gives the system of equations <math>m+n=802</math> and <math>n-m=4</math>, which does have integer solutions. Our answer is therefore <math>m+n=\boxed{802}</math> (Solving for <math>m</math> and <math>n</math> gives us <math>399</math> and <math>403</math>, respectively, which are co-prime).<br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=2|num-a=4}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_3&diff=898522005 AIME II Problems/Problem 32018-01-17T00:59:23Z<p>Matrix math: /* Solution */</p>
<hr />
<div>== Problem ==<br />
An [[infinite]] [[geometric series]] has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common [[ratio]] of the original series is <math> \frac mn </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[integer]]s. Find <math> m+n. </math><br />
<br />
== Solution 1 ==<br />
Let's call the first term of the original [[geometric series]] <math>a</math> and the common ratio <math>r</math>, so <math>2005 = a + ar + ar^2 + \ldots</math>. Using the sum formula for [[infinite]] geometric series, we have <math>(*)\;\;\frac a{1 -r} = 2005</math>. Then we form a new series, <math>a^2 + a^2 r^2 + a^2 r^4 + \ldots</math>. We know this series has sum <math>20050 = \frac{a^2}{1 - r^2}</math>. Dividing this equation by <math>(*)</math>, we get <math>10 = \frac a{1 + r}</math>. Then <math>a = 2005 - 2005r</math> and <math>a = 10 + 10r</math> so <math>2005 - 2005r = 10 + 10r</math>, <math>1995 = 2015r</math> and finally <math>r = \frac{1995}{2015} = \frac{399}{403}</math>, so the answer is <math>399 + 403 = \boxed{802}</math>. <br />
<br />
(We know this last fraction is fully reduced by the [[Euclidean algorithm]] -- because <math>4 = 403 - 399</math>, <math>\gcd(403, 399) | 4</math>. But 403 is [[odd integer | odd]], so <math>\gcd(403, 399) = 1</math>.)<br />
<br />
==Solution 2==<br />
We can write the sum of the original series as <math>a + a(\dfrac{m}{n}) + a(\dfrac{m}{n})^2 + \ldots = 2005</math>, where the common ratio is equal to <math>\dfrac{m}{n}</math>. We can also write the sum of the second series as <math>a^2 + a(\dfrac{m}{n})^2 + a((\dfrac{m}{n})^2)^2 + \ldots = 20050</math> Using the formula for the sum of an infinite geometric series <math>S=\dfrac{a}{1-r}</math>, where <math>S</math> is the sum of the sequence, <math>a</math> is the first term of the sequence, and <math>r</math> is the ratio of the sequence, the sum of the original series can be written as <math>\dfrac{a}{1-\frac{m}{n}}=\dfrac{a}{\frac{n-m}{n}}=\dfrac{a \cdot n}{n-m}=2005\;\text{(1)}</math>, and the second sequence can be written as <math>\dfrac{a^2}{1-\frac{m^2}{n^2}}=\dfrac{a^2}{\frac{n^2-m^2}{n^2}}=\dfrac{a^2\cdot n^2}{(n+m)(n-m)}=20050\;\text{(2)}</math>. Dividing <math>\text{(2)}</math> by <math>\text{(1)}</math>, we obtain <math>\dfrac{a\cdot n}{m+n}=10</math>, which can also be written as <math>a\cdot n=10(m+n)</math>. Substitute this value for <math>a\cdot n</math> back into <math>\text{(1)}</math>, we obtain <math>10\cdot \dfrac{n+m}{n-m}=2005</math>. Dividing both sides by 10 yields <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> we can now write a system of equations with <math>n+m=401</math> and <math>n-m=2</math>, but this does not output integer solutions. However, we can also write <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> as <math>\dfrac{n+m}{n-m}=\dfrac{802}{4}</math>. This gives the system of equations <math>m+n=802</math> and <math>n-m=4</math>, which does have integer solutions. Our answer is therefore <math>m+n=\boxed{802}</math> (Solving for <math>m</math> and <math>n</math> gives us <math>399</math> and <math>403</math>, respectively, which are co-prime).<br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=2|num-a=4}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=Physics_scholarships&diff=89849Physics scholarships2018-01-16T23:07:32Z<p>Matrix math: /* National physics scholarships */</p>
<hr />
<div>The following list of '''physics scholarships''' is primarily for American [[physics]] students. This list can be reorganized to incorporate scholarship programs in other countries. Just make that reorganization as clear and as clean as possible.<br />
<br />
Additions to this list are welcomed and encouraged. Please don't be stingy about letting students in on how they can finance their educations! If you are unfamiliar with how to edit a Wiki and don't have time to learn, please contact [[Mathew Crawford]] using the email crawford@artofproblemsolving.com with the scholarship listing you wish to contribute.<br />
<br />
<br />
<br />
== National physics scholarships ==<br />
* [http://www.aapt.org/programs/grants/lotze.cfm Barbara Lotze Scholarship] of $2,000 for high school seniors or undergraduate students enrolled, or planning to enroll, in physics teacher preparation curricula. <br />
=== Minority-specific scholarships ===<br />
* [http://www.aps.org/programs/minorities/honors/scholarship/ APS Minority Scholarship] for minority high school seniors or college freshmen/sophomores majoring in or planning to major in physics.<br />
* [http://www.awis.affiniscape.com/displaycommon.cfm?an=1&subarticlenbr=340 Kirsten R. Lorentzen Award] for female sophomores and juniors in college who are pursuing a bachelors degree in physics or geosciences.<br />
<br />
=== Graduate school scholarship programs ===<br />
* Department of Energy [http://www.musc.edu/specialprograms/ Nuclear Engineering and Health Physics Scholarship and Fellowship Program] (NOTE: Suspended as of 2008)<br />
<br />
== See also ==<br />
* [[Mathematics, science, and technology scholarships]]<br />
* [[Academic scholarships]]<br />
* [[Physics competitions]]<br />
* [[Physics summer programs]]<br />
* [[Sigma Pi Sigma]]<br />
* [[Academic competitions]]</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_17&diff=898302016 AMC 10B Problems/Problem 172018-01-16T00:46:14Z<p>Matrix math: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
All the numbers <math>2, 3, 4, 5, 6, 7</math> are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br />
<br />
<math>\textbf{(A)}\ 312 \qquad<br />
\textbf{(B)}\ 343 \qquad<br />
\textbf{(C)}\ 625 \qquad<br />
\textbf{(D)}\ 729 \qquad<br />
\textbf{(E)}\ 1680</math><br />
<br />
==Solution 1==<br />
Let us call the six sides of our cube <math>a,b,c,d,e,</math> and <math>f</math> (where <math>a</math> is opposite <math>d</math>, <math>c</math> is opposite <math>e</math>, and <math>b</math> is opposite <math>f</math>.<br />
Thus, for the eight vertices, we have the following products: <math>abc</math>,<math>abe</math>,<math>bcd</math>,<math>bde</math>,<math>acf</math>,<math>cdf</math>,<math>cef</math>, and <math>def</math>.<br />
Let us find the sum of these products:<br />
<math>abc+abe+bcd+bde+acf+cdf+aef+def</math><br />
We notice <math>b</math> is a factor of the first four terms, and <math>f</math> is factor is the last four terms.<br />
<math>b(ac+ae+cd+de)+f(ac+ae+cd+de)</math><br />
Now, we can factor even more:<br />
<math>(b+f)(ac+ae+cd+de)</math><br />
<math>(b+f)(a(c+e)+d(c+e))</math><br />
<math>(b+f)(a+d)(c+e)</math><br />
We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is <math>(7+2)</math>,<math>(6+3)</math>, and <math>(5+4)</math> all factors.<br />
<math> (7+2)(6+3)(5+4)</math><br />
<math>9</math> <math> *</math> <math> 9</math> <math> *</math> <math>9</math><br />
<math> 729 </math><br />
Thus our answer is <math>\textbf{(D)}\ 729</math>.<br />
<br />
==Solution 2==<br />
We first find the factorization <math>(b+f)(a+d)(c+e)</math> using the method in Solution 1. By using AM-GM, we get, <math>(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3</math>. To maximize, the factorization, we get the answer is <math>\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}</math><br />
<br />
==Solution 3 (Cheap Solution)==<br />
===Warning===<br />
Only use this if you are stuck on the problem or are low on time, and if you don't want to get the correct answer.<br />
<br />
===Solution===<br />
Create a pairing that seems to intuitively seem to be optimal value. So put a number and it's complement(the number that's the difference of 9 and this number). <math>1680</math> is way too high using reasonability after you do this so you put <math>\boxed{\textbf{D}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_2&diff=892451989 AIME Problems/Problem 22017-12-29T16:25:38Z<p>Matrix math: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Ten [[point]]s are marked on a [[circle]]. How many distinct [[convex polygon]]s of three or more sides can be drawn using some (or all) of the ten points as [[vertex | vertices]]?<br />
<br />
== Solution ==<br />
Any [[subset]] of the ten points with three or more members can be made into exactly one such polygon. Thus, we need to count the number of such subsets. There are <math>2^{10} = 1024</math> total subsets of a ten-member [[set]], but of these <math>{10 \choose 0} = 1</math> have 0 members, <math>{10 \choose 1} = 10</math> have 1 member and <math>{10 \choose 2} = 45</math> have 2 members. Thus the answer is <math>1024 - 1 - 10 - 45 = \boxed{968}</math>.<br />
<br />
Note <math>{N \choose 0}+{N \choose 1} + {N \choose 2} + \dots + {N \choose N}</math> is equivalent to <math>2^N</math><br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_10&diff=855591985 AIME Problems/Problem 102017-05-05T13:57:41Z<p>Matrix math: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
How many of the first 1000 [[positive integer]]s can be expressed in the form<br />
<br />
<math>\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor</math>,<br />
<br />
where <math>x</math> is a [[real number]], and <math>\lfloor z \rfloor</math> denotes the greatest [[integer]] less than or equal to <math>z</math>?<br />
__TOC__<br />
== Solution ==<br />
We will be able to reach the same number of integers while <math>x</math> ranges from 0 to 1 as we will when <math>x</math> ranges from <math>n</math> to <math>n + 1</math> for any [[integer]] <math>n</math> (Quick proof: <math>\lfloor 2(n+x)\rfloor + \ldots = \lfloor 2n + 2x\rfloor + \ldots = 2n + \lfloor 2x\rfloor \ldots</math>). Since <math>\lfloor 2\cdot50 \rfloor + \lfloor 4\cdot50 \rfloor + \lfloor 6\cdot50 \rfloor + \lfloor 8\cdot50 \rfloor = 100 + 200 + 300 + 400</math>, the answer must be exactly 50 times the number of integers we will be able to reach as <math>x</math> ranges from 0 to 1, including 1 but excluding 0.<br />
<br />
=== Solution 1 ===<br />
Noting that all of the numbers are even, we can reduce this to any real number <math>x</math> between <math>0</math> to <math>\frac 12</math>, as this will be equivalent to <math>\frac n2</math> to <math>\frac {n+1}2</math> for any integer <math>n</math> (same reasoning as above). So now we only need to test every 10 numbers; and our answer will be 100 times the number of integers we can reach between 1 and 10. <br />
<br />
We can now approach this by directly searching for the integers (this solution) or brute forcing all of the cases (next solution):<br />
<br />
We can match up the greatest integer functions with one of the partitions of the integer. If we let <math>x = \frac 12</math> then we get the solution <math>10</math>; now consider when <math>x < \frac 12</math>: <math>\lfloor 2x \rfloor = 0</math>, <math>\lfloor 4x \rfloor \le 1</math>, <math>\lfloor 6x \rfloor \le 2</math>, <math>\lfloor 8x \rfloor \le 3</math>. But according to this the maximum we can get is <math>1+2+3 = 6</math>, so we only need to try the first 6 numbers.<br />
<br />
*<math>1</math>: Easily possible, for example try plugging in <math>x =\frac 18</math>. <br />
*<math>2</math>: Also simple, for example using <math>\frac 16</math>.<br />
*<math>3</math>: The partition must either be <math>1+1+1</math> or <math>1+2</math>. If <math>\lfloor 4x \rfloor = 1</math>, then <math>x \ge \frac 14</math>, but then <math>\lfloor 8x \rfloor \ge 2</math>; not possible; and vice versa to show that the latter partition doesn't work. So we cannot obtain <math>3</math>.<br />
*<math>4</math>: We can partition as <math>1+1+2</math>, and from the previous case we see that <math>\frac 14</math> works.<br />
*<math>5</math>: We can partition as <math>1+2+2</math>, from which we find that <math>\frac 13</math> works.<br />
*<math>6</math>: We can partition as <math>1+2+3</math>, from which we find that <math>\frac 38</math> works.<br />
<br />
Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers <math>1,2,4,5,6,10</math>; hence our solution is <math>6 \cdot 100 = \boxed{600}</math>.<br />
<br />
=== Solution 2 ===<br />
As we change the value of <math>x</math>, the value of our [[expression]] changes only when <math>x</math> crosses [[rational number]] of the form <math>\frac{m}{n}</math>, where <math>n</math> is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form <math>\frac{m}{\textrm{lcm}(2, 4, 6, 8)} = \frac{m}{24}</math>. This gives us 24 calculations to make; we summarize the results here:<br />
<br />
<math>\frac{1}{24}, \frac{2}{24} \to 0</math><br />
<br />
<math>\frac{3}{24} \to 1</math><br />
<br />
<math>\frac{4}{24}, \frac{5}{24} \to 2</math><br />
<br />
<math>\frac{6}{24}, \frac{7}{24} \to 4</math><br />
<br />
<math>\frac{8}{24} \to 5</math><br />
<br />
<math>\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6</math><br />
<br />
<math>\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10</math><br />
<br />
<math>\frac{15}{24} \to 11</math><br />
<br />
<math>\frac{16}{24},\frac{17}{24} \to 12</math><br />
<br />
<math>\frac{18}{24}, \frac{19}{24} \to 14</math><br />
<br />
<math>\frac{20}{24}\to 15</math><br />
<br />
<math>\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16</math><br />
<br />
<math>\frac{24}{24} \to 20</math><br />
<br />
Thus, we hit 12 of the first 20 integers and so we hit <math>50 \cdot 12 = \boxed{600}</math> of the first <math>1000</math>.<br />
<br />
===Solution 3===<br />
<br />
Recall from Hermite's Identity that <math>\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor</math>. Then we can rewrite <math>\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 4\lfloor x\rfloor + \left\lfloor x + \frac18\right\rfloor + \left\lfloor x + \frac16\right\rfloor + 2\left\lfloor x + \frac14\right\rfloor + \left\lfloor x + \frac13\right\rfloor</math><br />
<math>+ \left\lfloor x + \frac38\right\rfloor + 4\left\lfloor x + \frac12\right\rfloor + \left\lfloor x + \frac58\right\rfloor + \left\lfloor x + \frac23\right\rfloor + 2\left\lfloor x + \frac34\right\rfloor + \left\lfloor x + \frac56\right\rfloor + \left\lfloor x + \frac78\right\rfloor</math>. There are <math>12</math> terms here (we don't actually have to write all of it out; we can just see where there will be duplicates and subtract accordingly from <math>20</math>). Starting from every integer <math>x</math>, we can keep adding to achieve one higher value for each of these terms, but after raising the last term, we will have raised the whole sum by <math>20</math> while only achieving <math>12</math> of those <math>20</math> values. We can conveniently shift the <math>1000</math> (since it can be achieved) to the position of the <math>0</math> so that there are only complete cycles of <math>20</math>, and the answer is <math>\frac {12}{20}\cdot1000 = \boxed{600}</math>.<br />
<br />
===Solution 4===<br />
<br />
Imagine that we increase <math>x</math> from <math>0</math> to <math>1</math>. At the beginning, the value of our expression is <math>0</math>, at the end it is <math>2+4+6+8=20</math>. How many integers between <math>1</math> and <math>20</math> did we skip? We skip some integers precisely at those points where at least two of <math>2x</math>, <math>4x</math>, <math>6x</math>, and <math>8x</math> become integers at the same time.<br />
<br />
Obviously, for <math>x=1/2</math> and <math>x=1</math> all four values become integers at the same time, hence we skip three integers at each of these locations. Additionally, for <math>x=1/4</math> and <math>x=3/4</math> the values <math>4x</math> and <math>8x</math> become integers at the same time, hence we skip one integer at each of the locations. <br />
<br />
Therefore for <math>x\in(0,1]</math> we skip a total of <math>3+3+1+1=8</math> integers. As in Solution 2, we conclude that we hit <math>12</math> of the integers from <math>1</math> to <math>20</math>, and so we hit <math>50 \cdot 12 = \boxed{600}</math> of the first <math>1000</math>.<br />
<br />
== See also ==<br />
* [[Floor function]]<br />
{{AIME box|year=1985|num-b=9|num-a=11}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_10&diff=855581985 AIME Problems/Problem 102017-05-05T13:57:04Z<p>Matrix math: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
How many of the first 1000 [[positive integer]]s can be expressed in the form<br />
<br />
<math>\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor</math>,<br />
<br />
where <math>x</math> is a [[real number]], and <math>\lfloor z \rfloor</math> denotes the greatest [[integer]] less than or equal to <math>z</math>?<br />
__TOC__<br />
== Solution ==<br />
We will be able to reach the same number of integers while <math>x</math> ranges from 0 to 1 as we will when <math>x</math> ranges from <math>n</math> to <math>n + 1</math> for any [[integer]] <math>n</math> (Quick proof: <math>\lfloor 2(n+x)\rfloor + \ldots = \lfloor 2n + 2x\rfloor + \ldots = 2n + \lfloor 2x\rfloor \ldots</math>). Since <math>\lfloor 2\cdot50 \rfloor + \lfloor 4\cdot50 \rfloor + \lfloor 6\cdot50 \rfloor + \lfloor 8\cdot50 \rfloor = 100 + 200 + 300 + 400</math>, the answer must be exactly 50 times the number of integers we will be able to reach as <math>x</math> ranges from 0 to 1, including 1 but excluding 0.<br />
<br />
=== Solution 1 ===<br />
Noting that all of the numbers are even, we can reduce this to any real number <math>x</math> between <math>0</math> to <math>\frac 12</math>, as this will be equivalent to <math>\frac n2</math> to <math>\frac {n+1}2</math> for any integer <math>n</math> (same reasoning as above). So now we only need to test every 10 numbers; and our answer will be 100 times the number of integers we can reach between 1 and 10. <br />
<br />
We can now approach this by directly searching for the integers (this solution) or brute forcing all of the cases (next solution):<br />
<br />
We can match up the greatest integer functions with one of the partitions of the integer. If we let <math>x = \frac 12</math> then we get the solution <math>10</math>; now consider when <math>x < \frac 12</math>: <math>\lfloor 2x \rfloor = 0</math>, <math>\lfloor 4x \rfloor \le 1</math>, <math>\lfloor 6x \rfloor \le 2</math>, <math>\lfloor 8x \rfloor \le 3</math>. But according to this the maximum we can get is <math>1+2+3 = 6</math>, so we only need to try the first 6 numbers.<br />
<br />
*<math>1</math>: Easily possible, for example try plugging in <math>x =\frac 18</math>. <br />
*<math>2</math>: Also simple, for example using <math>\frac 16</math>.<br />
*<math>3</math>: The partition must either be <math>1+1+1</math> or <math>1+2</math>. If <math>\lfloor 4x \rfloor = 1</math>, then <math>x \ge \frac 14</math>, but then <math>\lfloor 8x \rfloor \ge 2</math>; not possible; and vice versa to show that the latter partition doesn't work. So we cannot obtain <math>3</math>.<br />
*<math>4</math>: We can partition as <math>1+1+2</math>, and from the previous case we see that <math>\frac 14</math> works.<br />
*<math>5</math>: We can partition as <math>1+2+2</math>, from which we find that <math>\frac 13</math> works.<br />
*<math>6</math>: We can partition as <math>1+2+3</math>, from which we find that <math>\frac 38</math> works.<br />
<br />
Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers <math>1,2,4,5,6,10</math>; hence our solution is <math>6 \cdot 100 = 600</math>.<br />
<br />
=== Solution 2 ===<br />
As we change the value of <math>x</math>, the value of our [[expression]] changes only when <math>x</math> crosses [[rational number]] of the form <math>\frac{m}{n}</math>, where <math>n</math> is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form <math>\frac{m}{\textrm{lcm}(2, 4, 6, 8)} = \frac{m}{24}</math>. This gives us 24 calculations to make; we summarize the results here:<br />
<br />
<math>\frac{1}{24}, \frac{2}{24} \to 0</math><br />
<br />
<math>\frac{3}{24} \to 1</math><br />
<br />
<math>\frac{4}{24}, \frac{5}{24} \to 2</math><br />
<br />
<math>\frac{6}{24}, \frac{7}{24} \to 4</math><br />
<br />
<math>\frac{8}{24} \to 5</math><br />
<br />
<math>\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6</math><br />
<br />
<math>\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10</math><br />
<br />
<math>\frac{15}{24} \to 11</math><br />
<br />
<math>\frac{16}{24},\frac{17}{24} \to 12</math><br />
<br />
<math>\frac{18}{24}, \frac{19}{24} \to 14</math><br />
<br />
<math>\frac{20}{24}\to 15</math><br />
<br />
<math>\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16</math><br />
<br />
<math>\frac{24}{24} \to 20</math><br />
<br />
Thus, we hit 12 of the first 20 integers and so we hit <math>50 \cdot 12 = \boxed{600}</math> of the first <math>1000</math>.<br />
<br />
===Solution 3===<br />
<br />
Recall from Hermite's Identity that <math>\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor</math>. Then we can rewrite <math>\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 4\lfloor x\rfloor + \left\lfloor x + \frac18\right\rfloor + \left\lfloor x + \frac16\right\rfloor + 2\left\lfloor x + \frac14\right\rfloor + \left\lfloor x + \frac13\right\rfloor</math><br />
<math>+ \left\lfloor x + \frac38\right\rfloor + 4\left\lfloor x + \frac12\right\rfloor + \left\lfloor x + \frac58\right\rfloor + \left\lfloor x + \frac23\right\rfloor + 2\left\lfloor x + \frac34\right\rfloor + \left\lfloor x + \frac56\right\rfloor + \left\lfloor x + \frac78\right\rfloor</math>. There are <math>12</math> terms here (we don't actually have to write all of it out; we can just see where there will be duplicates and subtract accordingly from <math>20</math>). Starting from every integer <math>x</math>, we can keep adding to achieve one higher value for each of these terms, but after raising the last term, we will have raised the whole sum by <math>20</math> while only achieving <math>12</math> of those <math>20</math> values. We can conveniently shift the <math>1000</math> (since it can be achieved) to the position of the <math>0</math> so that there are only complete cycles of <math>20</math>, and the answer is <math>\frac {12}{20}\cdot1000 = \boxed{600}</math>.<br />
<br />
===Solution 4===<br />
<br />
Imagine that we increase <math>x</math> from <math>0</math> to <math>1</math>. At the beginning, the value of our expression is <math>0</math>, at the end it is <math>2+4+6+8=20</math>. How many integers between <math>1</math> and <math>20</math> did we skip? We skip some integers precisely at those points where at least two of <math>2x</math>, <math>4x</math>, <math>6x</math>, and <math>8x</math> become integers at the same time.<br />
<br />
Obviously, for <math>x=1/2</math> and <math>x=1</math> all four values become integers at the same time, hence we skip three integers at each of these locations. Additionally, for <math>x=1/4</math> and <math>x=3/4</math> the values <math>4x</math> and <math>8x</math> become integers at the same time, hence we skip one integer at each of the locations. <br />
<br />
Therefore for <math>x\in(0,1]</math> we skip a total of <math>3+3+1+1=8</math> integers. As in Solution 2, we conclude that we hit <math>12</math> of the integers from <math>1</math> to <math>20</math>, and so we hit <math>50 \cdot 12 = \boxed{600}</math> of the first <math>1000</math>.<br />
<br />
== See also ==<br />
* [[Floor function]]<br />
{{AIME box|year=1985|num-b=9|num-a=11}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_10&diff=855571985 AIME Problems/Problem 102017-05-05T13:56:29Z<p>Matrix math: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
How many of the first 1000 [[positive integer]]s can be expressed in the form<br />
<br />
<math>\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor</math>,<br />
<br />
where <math>x</math> is a [[real number]], and <math>\lfloor z \rfloor</math> denotes the greatest [[integer]] less than or equal to <math>z</math>?<br />
__TOC__<br />
== Solution ==<br />
We will be able to reach the same number of integers while <math>x</math> ranges from 0 to 1 as we will when <math>x</math> ranges from <math>n</math> to <math>n + 1</math> for any [[integer]] <math>n</math> (Quick proof: <math>\lfloor 2(n+x)\rfloor + \ldots = \lfloor 2n + 2x\rfloor + \ldots = 2n + \lfloor 2x\rfloor \ldots</math>). Since <math>\lfloor 2\cdot50 \rfloor + \lfloor 4\cdot50 \rfloor + \lfloor 6\cdot50 \rfloor + \lfloor 8\cdot50 \rfloor = 100 + 200 + 300 + 400</math>, the answer must be exactly 50 times the number of integers we will be able to reach as <math>x</math> ranges from 0 to 1, including 1 but excluding 0.<br />
<br />
=== Solution 1 ===<br />
Noting that all of the numbers are even, we can reduce this to any real number <math>x</math> between <math>0</math> to <math>\frac 12</math>, as this will be equivalent to <math>\frac n2</math> to <math>\frac {n+1}2</math> for any integer <math>n</math> (same reasoning as above). So now we only need to test every 10 numbers; and our answer will be 100 times the number of integers we can reach between 1 and 10. <br />
<br />
We can now approach this by directly searching for the integers (this solution) or brute forcing all of the cases (next solution):<br />
<br />
We can match up the greatest integer functions with one of the partitions of the integer. If we let <math>x = \frac 12</math> then we get the solution <math>10</math>; now consider when <math>x < \frac 12</math>: <math>\lfloor 2x \rfloor = 0</math>, <math>\lfloor 4x \rfloor \le 1</math>, <math>\lfloor 6x \rfloor \le 2</math>, <math>\lfloor 8x \rfloor \le 3</math>. But according to this the maximum we can get is <math>1+2+3 = 6</math>, so we only need to try the first 6 numbers.<br />
<br />
*<math>1</math>: Easily possible, for example try plugging in <math>x =\frac 18</math>. <br />
*<math>2</math>: Also simple, for example using <math>\frac 16</math>.<br />
*<math>3</math>: The partition must either be <math>1+1+1</math> or <math>1+2</math>. If <math>\lfloor 4x \rfloor = 1</math>, then <math>x \ge \frac 14</math>, but then <math>\lfloor 8x \rfloor \ge 2</math>; not possible; and vice versa to show that the latter partition doesn't work. So we cannot obtain <math>3</math>.<br />
*<math>4</math>: We can partition as <math>1+1+2</math>, and from the previous case we see that <math>\frac 14</math> works.<br />
*<math>5</math>: We can partition as <math>1+2+2</math>, from which we find that <math>\frac 13</math> works.<br />
*<math>6</math>: We can partition as <math>1+2+3</math>, from which we find that <math>\frac 38</math> works.<br />
<br />
Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers <math>1,2,4,5,6,10</math>; hence our solution is <math>6 \cdot 100 = 600</math>.<br />
<br />
=== Solution 2 ===<br />
As we change the value of <math>x</math>, the value of our [[expression]] changes only when <math>x</math> crosses [[rational number]] of the form <math>\frac{m}{n}</math>, where <math>n</math> is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form <math>\frac{m}{\textrm{lcm}(2, 4, 6, 8)} = \frac{m}{24}</math>. This gives us 24 calculations to make; we summarize the results here:<br />
<br />
<math>\frac{1}{24}, \frac{2}{24} \to 0</math><br />
<br />
<math>\frac{3}{24} \to 1</math><br />
<br />
<math>\frac{4}{24}, \frac{5}{24} \to 2</math><br />
<br />
<math>\frac{6}{24}, \frac{7}{24} \to 4</math><br />
<br />
<math>\frac{8}{24} \to 5</math><br />
<br />
<math>\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6</math><br />
<br />
<math>\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10</math><br />
<br />
<math>\frac{15}{24} \to 11</math><br />
<br />
<math>\frac{16}{24},\frac{17}{24} \to 12</math><br />
<br />
<math>\frac{18}{24}, \frac{19}{24} \to 14</math><br />
<br />
<math>\frac{20}{24}\to 15</math><br />
<br />
<math>\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16</math><br />
<br />
<math>\frac{24}{24} \to 20</math><br />
<br />
Thus, we hit 12 of the first 20 integers and so we hit <math>50 \cdot 12 = \boxed{600}</math> of the first <math>1000</math>.<br />
<br />
===Solution 3===<br />
<br />
Recall from Hermite's Identity that <math>\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor</math>. Then we can rewrite <math>\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 4\lfloor x\rfloor + \left\lfloor x + \frac18\right\rfloor + \left\lfloor x + \frac16\right\rfloor + 2\left\lfloor x + \frac14\right\rfloor + \left\lfloor x + \frac13\right\rfloor</math><br />
<math>+ \left\lfloor x + \frac38\right\rfloor + 4\left\lfloor x + \frac12\right\rfloor + \left\lfloor x + \frac58\right\rfloor + \left\lfloor x + \frac23\right\rfloor + 2\left\lfloor x + \frac34\right\rfloor + \left\lfloor x + \frac56\right\rfloor + \left\lfloor x + \frac78\right\rfloor</math>. There are <math>12</math> terms here (we don't actually have to write all of it out; we can just see where there will be duplicates and subtract accordingly from <math>20</math>). Starting from every integer <math>x</math>, we can keep adding to achieve one higher value for each of these terms, but after raising the last term, we will have raised the whole sum by <math>20</math> while only achieving <math>12</math> of those <math>20</math> values. We can conveniently shift the <math>1000</math> (since it can be achieved) to the position of the <math>0</math> so that there are only complete cycles of <math>20</math>, and the answer is <math>\frac {12}{20}\cdot1000 = \boxed{600}</math>.<br />
<br />
===Solution 4===<br />
<br />
Imagine that we increase <math>x</math> from <math>0</math> to <math>1</math>. At the beginning, the value of our expression is <math>0</math>, at the end it is <math>2+4+6+8=20</math>. How many integers between <math>1</math> and <math>20</math> did we skip? We skip some integers precisely at those points where at least two of <math>2x</math>, <math>4x</math>, <math>6x</math>, and <math>8x</math> become integers at the same time.<br />
<br />
Obviously, for <math>x=1/2</math> and <math>x=1</math> all four values become integers at the same time, hence we skip three integers at each of these locations. Additionally, for <math>x=1/4</math> and <math>x=3/4</math> the values <math>4x</math> and <math>8x</math> become integers at the same time, hence we skip one integer at each of the locations. <br />
<br />
Therefore for <math>x\in(0,1]</math> we skip a total of <math>3+3+1+1=8</math> integers. As in Solution 2, we conclude that we hit <math>12</math> of the integers from <math>1</math> to <math>20</math>, and so we hit <math>50 \cdot 12 = 600</math> of the first <math>1000</math>.<br />
<br />
== See also ==<br />
* [[Floor function]]<br />
{{AIME box|year=1985|num-b=9|num-a=11}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_16&diff=855562003 AMC 12A Problems/Problem 162017-05-05T13:50:55Z<p>Matrix math: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>A point P is chosen at random in the interior of equilateral triangle <math>ABC</math>. What is the probability that <math>\triangle ABP</math> has a greater area than each of <math>\triangle ACP</math> and <math>\triangle BCP</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
== Solution==<br />
<asy><br />
draw((0,10)--(8.660254037844385792,-5)--(-8.660254037844385792,-5)--cycle);<br />
dot((0,0));<br />
label("$P$",(0,0),N);<br />
</asy><br />
===Solution 1===<br />
<br />
After we pick point <math>P</math>, we realize that <math>ABC</math> is symmetric for this purpose, and so the probability that <math>ACP</math> is the greatest area, or <math>ABP</math> or <math>BCP</math>, are all the same. Since they add to <math>1</math>, the probability that <math>ACP</math> has the greatest area is <math>\boxed{\mathrm{(C)}\ \dfrac{1}{3}}</math><br />
<br />
===Solution 2===<br />
<br />
We will use geometric probability. Let us take point P, and draw the perpendiculars to BC, CA, and AB, and call the feet of these perpendiculars D, E, and F respectively. The area of triangle ACP is simply <math>\frac{1}{2} * AC * PF</math>. Similarly we can find the area of triangles BCP and ABP. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose P, PD + PE + PF = the height of the triangle. Setting the area of triangle ABP greater than ACP and BCP, we want PF to be the largest of PF, PD, and PE. We then realize that PF = PD = PE when P is the incenter of ABC. Let us call the incenter of the triangle Q. If we want PF to be the largest of the three, by testing points we realize that P must be in the interior of quadrilateral QDCE. So our probability (using geometric probability) is the area of QDCE divided by the area of ABC. We will now show that the three quadrilaterals, QDCE, QEAF, and QFBD are congruent. As the definition of point Q yields, QF = QD = QE. Since ABC is equilateral, Q is also the circumcenter of ABC, so QA = QB = QC. By the Pythagorean Theorem, BD = DC = CE = EA = AF = FB. Also, angles BDQ, BFQ, CEQ, CDQ, AFQ, and AEQ are all equal to 90 degrees. Angles DBF, FAE, ECD are all equal to 60 degrees, so it is now clear that QDCE, QEAF, QFBD are all congruent. Summing up these areas gives us the area of ABC. QDCE contributes to a third of that area so <math>\frac{[QDCE]}{[ABC]}=\frac{1}{3}</math> (C).<br />
<br />
==See Also==<br />
{{AMC12 box|year=2003|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_5&diff=854441989 AIME Problems/Problem 52017-04-28T01:46:34Z<p>Matrix math: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
When a certain biased coin is flipped five times, the [[probability]] of getting heads exactly once is not equal to <math>0</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>.<br />
<br />
== Solution ==<br />
===Solution 1 ===<br />
Denote the probability of getting a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>. After canceling out terms, we get <math>1 - h = 2h</math>, so <math>h = \frac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>.<br />
<br />
<br />
=== Solution 2 ===<br />
Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math>. To find the probability getting <math>3</math> heads, we need to find <math>{5\choose3}\dfrac{(h)^3(t)^2}{(h + t)^5} =10\cdot\dfrac{(h)^3(2h)^2}{(h + 2h)^5}</math> (recall that <math>h</math> cannot be <math>0</math>). The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_5&diff=854301989 AIME Problems/Problem 52017-04-25T14:43:33Z<p>Matrix math: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
When a certain biased coin is flipped five times, the [[probability]] of getting heads exactly once is not equal to <math>0</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>.<br />
<br />
== Solution ==<br />
===Solution 1 ===<br />
Denote the probability of getting a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>. After canceling out terms, we get <math>1 - h = 2h</math>, so <math>h = \frac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>.<br />
<br />
<br />
=== Solution 2 ===<br />
Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math>. To find the probability getting <math>3</math> heads, we need to find <math>{5\choose3}\frac{(h)^3(t)^2}{(h + t)^5} =10\cdot\frac{(h)^3(2h)^2}{(h + 2h)^5}</math> (recall that <math>h</math> cannot be <math>0</math>). The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_5&diff=854291989 AIME Problems/Problem 52017-04-25T14:42:48Z<p>Matrix math: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
When a certain biased coin is flipped five times, the [[probability]] of getting heads exactly once is not equal to <math>0</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>.<br />
<br />
== Solution ==<br />
===Solution 1 ===<br />
Denote the probability of getting a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>. After canceling out terms, we get <math>1 - h = 2h</math>, so <math>h = \frac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>.<br />
<br />
<br />
=== Solution 2 ===<br />
Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math>. To find the probability getting 3 heads, we need to find <math>{5\choose3}\frac{(h)^3(t)^2}{(h + t)^5} =10\cdot\frac{(h)^3(2h)^2}{(h + 2h)^5}</math> (recall that <math>h</math> cannot be <math>0</math>). The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_13&diff=854241990 AIME Problems/Problem 132017-04-24T13:58:27Z<p>Matrix math: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}</math>. Given that <math>9^{4000}_{}</math> has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of <math>T_{}^{}</math> have 9 as their leftmost digit?<br />
<br />
== Solution ==<br />
When a number is multiplied by <math>9</math>, it gains a digit unless the new number starts with a 9. <br />
<br />
Since <math>9^{4000}</math> has 3816 digits more than <math>9^1</math>, <math>4000 - 3816 = \boxed{184}</math> numbers have 9 as their leftmost digits.<br />
<br />
== See also ==<br />
{{AIME box|year=1990|num-b=12|num-a=14}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_5&diff=854231989 AIME Problems/Problem 52017-04-24T07:01:43Z<p>Matrix math: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
When a certain biased coin is flipped five times, the [[probability]] of getting heads exactly once is not equal to <math>0</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>.<br />
<br />
== Solution ==<br />
===Solution 1 ===<br />
Denote the probability of getting a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>. After canceling out terms, we get <math>1 - h = 2h</math>, so <math>h = \frac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>.<br />
<br />
<br />
=== Solution 2 ===<br />
Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math>. To find the probability getting 3 heads, we need to find <math>{5\choose3}\frac{(h)^3(t)^2}{(h + t)^5} =10\cdot\frac{(h)^3(2h)^2}{(h + 2h)^5}</math> (recall that <math>h</math> cannot be 0). The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_5&diff=854031989 AIME Problems/Problem 52017-04-23T16:46:21Z<p>Matrix math: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
When a certain biased coin is flipped five times, the [[probability]] of getting heads exactly once is not equal to <math>0</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>.<br />
<br />
== Solution ==<br />
===Solution 1 ===<br />
Denote the probability of getting a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>. After canceling out terms, we get <math>1 - h = 2h</math>, so <math>h = \frac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>.<br />
<br />
<br />
=== Solution 2 ===<br />
Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math> (recall that <math>h</math> cannot be 0). To find the probability getting 3 heads, we need to find <math>{5\choose3}\frac{(h)^3(t)^2}{(h + t)^5} =10\cdot\frac{(h)^3(2h)^2}{(h + 2h)^5}</math>. The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_5&diff=854021989 AIME Problems/Problem 52017-04-23T16:42:26Z<p>Matrix math: /* Solution */</p>
<hr />
<div>== Problem ==<br />
When a certain biased coin is flipped five times, the [[probability]] of getting heads exactly once is not equal to <math>0</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>.<br />
<br />
== Solution ==<br />
===Solution 1 ===<br />
Denote the probability of getting a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>. After canceling out terms, we get <math>1 - h = 2h</math>, so <math>h = \frac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>.<br />
<br />
<br />
=== Solution 2 ===<br />
Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math>. To find the probability getting 3 heads, we need to find <math>{5\choose3}\frac{(h)^3(t)^2}{(h + t)^5} =10\cdot\frac{(h)^3(2h)^2}{(h + 2h)^5}</math>. The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Matrix mathhttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_1&diff=854011989 AIME Problems/Problem 12017-04-23T15:47:52Z<p>Matrix math: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Compute <math>\sqrt{(31)(30)(29)(28)+1}</math>.<br />
<br />
== Solution ==<br />
=== Solution 1===<br />
Notice <math>{31*28 = 868}</math> and <math>{30*29 =870}</math>. So now our expression is <math>\sqrt{(870)(868) + 1}</math>. Setting 870 equal to <math>y</math>, we get <math>\sqrt{(y-1)^{2}}</math> which then equals <math>{(y-1)}</math>. So since <math>{y = 870}</math>, <math>{y-1}=869</math>, our answer is <math>\boxed{869}</math>.<br />
<br />
=== Solution 2===<br />
Note that the four numbers to multiply are symmetric with the center at <math>29.5</math>. <br />
Multiply the symmetric pairs to get <math>31\cdot 28=868</math> and <math>30\cdot 29=870</math>.<br />
<math>\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}</math>.<br />
<br />
=== Solution 3===<br />
The last digit under the radical is <math>1</math>, so the square root must either end in <math>1</math> or <math>9</math>, since <math>x^2 = 1\pmod {10}</math> means <math>x = \pm 1</math>. Additionally, the number must be near <math>29 \cdot 30 = 870</math>, narrowing the reasonable choices to <math>869</math> and <math>871</math>.<br />
<br />
Continuing the logic, the next-to-last digit under the radical is the same as the last digit of <math>28 \cdot 29 \cdot 3 \cdot 31</math>, which is <math>6</math>. Quick computation shows that <math>869^2</math> ends in <math>61</math>, while <math>871^2</math> ends in <math>41</math>. Thus, the answer is <math>\boxed{869}</math>.<br />
<br />
===Solution 4===<br />
Similar to Solution 1 above, call the consecutive integers <math>\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)</math> to make use of symmetry. Note that <math>n</math> itself is not an integer - in this case, <math>n = 29.5</math>. The expression becomes <math>\sqrt{\left(n-\frac{3}{2}\right)\left(n + \frac{3}{2}\right)\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) + 1}</math>. Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives <math>\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}</math>. The inside is a perfect square trinomial, since <math>b^2 = 4ac</math>. It's equal to <math>\sqrt{\left(n^2 - \frac{5}{4}\right)^2}</math>, which simplifies to <math>n^2 - \frac{5}{4}</math>. You can plug in the value of <math>n</math> from there, or further simplify to <math>\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1</math>, which is easier to compute. Either way, plugging in <math>n=29.5</math> gives <math>\boxed{869}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1989|before=First Question|num-a=2}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Matrix math