https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Willwang123&feedformat=atom AoPS Wiki - User contributions [en] 2022-09-27T17:38:42Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_25&diff=47995 2011 AMC 8 Problems/Problem 25 2012-08-20T21:10:45Z <p>Willwang123: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> A circle with radius &lt;math&gt;1&lt;/math&gt; is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br /> <br /> &lt;asy&gt;<br /> filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,mediumgray,black);<br /> filldraw(Circle((0,0),1), mediumgray,black);<br /> filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The area of the smaller square is the one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: &lt;math&gt;2*2*1/2=2.&lt;/math&gt;<br /> <br /> The circle's shaded area is the area of the smaller square subtracted from the area of the circle: &lt;math&gt;\pi - 2.&lt;/math&gt;<br /> <br /> If you draw the diagonals of the smaller square, you will see that the larger square is split &lt;math&gt;4&lt;/math&gt; congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Approximating &lt;math&gt;\pi&lt;/math&gt; to &lt;math&gt;3.14,&lt;/math&gt; the ratio of the circle's shaded area to the area between the two squares is about<br /> <br /> &lt;cmath&gt;\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}&lt;/cmath&gt;<br /> <br /> ===Solution 2===<br /> For the ratio of the circle's shaded area to the area between the squares to be &lt;math&gt;1,&lt;/math&gt; they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately &lt;math&gt;\boxed{\textbf{(A)}\ \frac12}&lt;/math&gt;.<br /> <br /> Note that this solution is not rigorous, because we still should show that the ratio is less than &lt;math&gt;\frac{3}{4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2011|num-b=24|after=Last Problem}}</div> Willwang123 https://artofproblemsolving.com/wiki/index.php?title=Imaginary_number&diff=45301 Imaginary number 2012-03-06T02:34:28Z <p>Willwang123: </p> <hr /> <div>An '''imaginary number''' is a [[complex number]] whose [[real part]] is equal to 0. In the [[complex plane]], these numbers lie on the [[imaginary axis]]. They are sometimes also refered to as ''pure imaginary numbers'', <br /> <br /> The number &lt;math&gt;z&lt;/math&gt; is imaginary if and only if &lt;math&gt;z = i \textrm{Im}(z)&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt; is the [[imaginary unit]] and &lt;math&gt;\textrm{Im}&lt;/math&gt; is the [[imaginary part]] [[function]].<br /> <br /> The imaginary unit is &lt;math&gt;i&lt;/math&gt;, which is &lt;math&gt;\sqrt{-1}&lt;/math&gt;. &lt;math&gt;i&lt;/math&gt; can be treated as a variable with the distinctive property that &lt;math&gt;i^2= -1&lt;/math&gt;. Powers of &lt;math&gt;i&lt;/math&gt; can be equal to only 4 possibilities: &lt;math&gt;-i, i, 1, -1&lt;/math&gt; because the values cycle. This means that &lt;math&gt;i^1=i&lt;/math&gt;, &lt;math&gt;i^2=-1&lt;/math&gt;, &lt;math&gt;i^3=-i&lt;/math&gt;, &lt;math&gt;i^4=1&lt;/math&gt;, &lt;math&gt;i^5=i....&lt;/math&gt;. <br /> <br /> Multiplying imaginary/complex numbers is not so much different from multiplying variables. In fact, the easiest way to multiply imaginary numbers is to treat &lt;math&gt;i&lt;/math&gt; as a variable and then check to see if there is either &lt;math&gt;i^2&lt;/math&gt; or &lt;math&gt;i^4&lt;/math&gt; in which case you switch it its real value accordingly.<br /> <br /> Dividing imaginary/complex numbers is also somewhat different than normal variables. The first thing you want to do in a fraction with complex or imaginary numbers in the numerator or denominator is to make the denominator real. To do this you'll want to multiply both sides by the denominator's [[Complex conjugate]]. This will make the denominator and possibly the numerator real. From there you must simplify the fraction, and you'll either get a complex or real number.<br /> <br /> It is a common mistake to say that &lt;math&gt;\sqrt{-1}&lt;/math&gt; is &lt;math&gt;i&lt;/math&gt;, rather than &lt;math&gt;\pm i&lt;/math&gt;, which allows for many mathematical fallacies to be made such as &lt;math&gt;1=2&lt;/math&gt;. Another mistake that is made often is that &lt;math&gt;\sqrt{-4}= 4i&lt;/math&gt;, but is really &lt;math&gt;\sqrt{-4}=2i&lt;/math&gt; since it is under a radical.<br /> <br /> ==See Also==<br /> <br /> <br /> {{stub}}<br /> <br /> [[Category:Complex numbers]]</div> Willwang123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_7&diff=38287 2011 AIME II Problems/Problem 7 2011-04-30T18:57:59Z <p>Willwang123: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal tot he number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let ''m'' be the maximum number of red marbles for which such an arrangement is possible, and let N be the number of ways he can arrange the ''m''+5 marbles to satisfy the requirement. Find the remainder when N is divided by 1000.<br /> <br /> ==Solution==<br /> <br /> We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 &quot;not the same colors&quot; and 0 &quot;same colors.&quot; Now, for every red marble we add, we will add one &quot;same color&quot; pair and keep all 10 &quot;not the same color&quot; pairs. It follows that we can add 10 more red marbles for a total of 16 = ''m''. We can place those ten marbles in any of 6 &quot;boxes&quot;: To the left of the first green marble, to the right of the first but left of the second, etc. up until to the right of the last. This is a stars-and-bars problem, the solution of which can be found as &lt;math&gt;\binom{n+k}{k}&lt;/math&gt; where n is the number of stars and k is the number of bars. There are 10 stars (The unassigned Rs, since each &quot;box&quot; must contain at least one, are not counted here) and 5 &quot;bars,&quot; the green marbles. So the answer is &lt;math&gt;\binom{15}{5} = 3003&lt;/math&gt;, take the remainder when divided by 1000 to get the answer: &lt;math&gt;\framebox[1.3\width]{003.}&lt;/math&gt;</div> Willwang123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_7&diff=38286 2011 AIME II Problems/Problem 7 2011-04-30T18:57:41Z <p>Willwang123: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal tot he number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let ''m'' be the maximum number of red marbles for which such an arrangement is possible, and let N be the number of ways he can arrange the ''m''+5 marbles to satisfy the requirement. Find the remainder when N is divided by 1000.<br /> <br /> ==Solution==<br /> <br /> We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 &quot;not the same colors&quot; and 0 &quot;same colors.&quot; Now, for every red marble we add, we will add one &quot;same color&quot; pair and keep all 10 &quot;not the same color&quot; pairs. It follows that we can add 10 more red marbles for a total of 16 = ''m''. We can place those ten marbles in any of 6 &quot;boxes&quot;: To the left of the first green marble, to the right of the first but left of the second, etc. up until to the right of the last. This is a stars-and-bars problem, the solution of which can be found as &lt;math&gt;\binom{n+k}{k}&lt;/math&gt; where n is the number of stars and k is the number of bars. There are 10 stars (The unassigned Rs, since each &quot;box&quot; must contain at least one, are not counted here) and 5 &quot;bars,&quot; the green marbles. So the answer is &lt;math&gt;\binom{15}{5} = 3003, take the remainder when divided by 1000 to get the answer: &lt;/math&gt;\framebox[1.3\width]{003.}$</div> Willwang123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_6&diff=38285 2011 AIME II Problems/Problem 6 2011-04-30T18:55:00Z <p>Willwang123: /* Problem 6 */</p> <hr /> <div>==Problem 6==<br /> <br /> Define an ordered quadruple &lt;math&gt;(a, b, c, d)&lt;/math&gt; as interesting if &lt;math&gt;1 \le a&lt;b&lt;c&lt;d \le 10&lt;/math&gt;, and &lt;math&gt;a+d&gt;b+c&lt;/math&gt;. How many interesting ordered quadruples are there?<br /> <br /> ==Solution==<br /> Rearranging the inequality we get &lt;math&gt;d-c &gt; b-a&lt;/math&gt;. Let &lt;math&gt;e = 11&lt;/math&gt;, then &lt;math&gt;(a, b-a, c-b, d-c, e-d)&lt;/math&gt; is a partition of 11 into 5 positive integers or equivalently:<br /> &lt;math&gt;(a-1, b-a-1, c-b-1, d-c-1, e-d-1)&lt;/math&gt; is a partition of 6 into 5 non-negative integer parts. Via a standard balls and urns argument, the number of ways to partition 6 into 5 non-negative parts is &lt;math&gt;\binom{6+4}4 = \binom{10}4 = 210&lt;/math&gt;. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry there as many partitions where the fourth is less than the second. So, if &lt;math&gt;N&lt;/math&gt; is the number of partitions where the second element is equal to the fourth, our answer is &lt;math&gt;(210-N)/2&lt;/math&gt;.<br /> <br /> We find &lt;math&gt;N&lt;/math&gt; as a sum of 4 cases:<br /> * two parts equal to zero, &lt;math&gt;\binom82 = 28&lt;/math&gt; ways,<br /> * two parts equal to one, &lt;math&gt;\binom62 = 15&lt;/math&gt; ways,<br /> * two parts equal to two, &lt;math&gt;\binom42 = 6&lt;/math&gt; ways,<br /> * two parts equal to three, &lt;math&gt;\binom22 = 1&lt;/math&gt; way.<br /> Therefore, &lt;math&gt;N = 28 + 15 + 6 + 1 = 50&lt;/math&gt; and our answer is &lt;math&gt;(210 - 50)/2 = \fbox{80.}&lt;/math&gt;</div> Willwang123 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems&diff=37046 1983 AIME Problems 2011-02-21T19:07:08Z <p>Willwang123: /* Problem 2 */</p> <hr /> <div>{{AIME Problems|year=1983}}<br /> <br /> == Problem 1 ==<br /> Let &lt;math&gt;x&lt;/math&gt;,&lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt;, and let &lt;math&gt;w&lt;/math&gt; be a positive number such that &lt;math&gt;\log_xw=24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt;, and &lt;math&gt;\log_{xyz}w=12&lt;/math&gt;. Find &lt;math&gt;\log_zw&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Let &lt;math&gt;f(x)=|x-p|+|x-15|+|x-p-15|&lt;/math&gt;, where &lt;math&gt;p \leq x \leq 15&lt;/math&gt;. Determine the minimum value taken by &lt;math&gt;f(x)&lt;/math&gt; by &lt;math&gt;x&lt;/math&gt; in the interval &lt;math&gt;0 &lt; x \leq 15&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> What is the product of the real roots of the equation &lt;math&gt;x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}&lt;/math&gt;?<br /> <br /> [[1983 AIME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is &lt;math&gt;\sqrt{50}&lt;/math&gt; cm, the length of &lt;math&gt;AB&lt;/math&gt; is 6 cm, and that of &lt;math&gt;BC&lt;/math&gt; is 2 cm. The angle &lt;math&gt;ABC&lt;/math&gt; is a right angle. Find the square of the distance (in centimeters) from &lt;math&gt;B&lt;/math&gt; to the center of the circle.<br /> <br /> &lt;asy&gt;<br /> size(150); defaultpen(linewidth(0.65)+fontsize(11));<br /> real r=10;<br /> pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;<br /> path P=circle(O,r);<br /> C=intersectionpoint(B--(B.x+r,B.y),P);<br /> draw(P);<br /> draw(C--B--O--A--B);<br /> dot(O); dot(A); dot(B); dot(C);<br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$A$&quot;,A,NE);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C\$&quot;,C,SE);<br /> &lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Suppose that the sum of the squares of two complex numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;7&lt;/math&gt; and the sum of the cubes is &lt;math&gt;10&lt;/math&gt;. What is the largest real value that &lt;math&gt;x + y&lt;/math&gt; can have?<br /> <br /> [[1983 AIME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Let &lt;math&gt;a_n&lt;/math&gt; equal &lt;math&gt;6^{n}+8^{n}&lt;/math&gt;. Determine the remainder upon dividing &lt;math&gt;a_ {83}&lt;/math&gt; by &lt;math&gt;49&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let &lt;math&gt;P&lt;/math&gt; be the probability that at least two of the three had been sitting next to each other. If &lt;math&gt;P&lt;/math&gt; is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?<br /> <br /> [[1983 AIME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> What is the largest 2-digit prime factor of the integer &lt;math&gt;{200\choose 100}&lt;/math&gt;?<br /> <br /> [[1983 AIME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Find the minimum value of &lt;math&gt;\frac{9x^2\sin^2 x + 4}{x\sin x}&lt;/math&gt; for &lt;math&gt;0 &lt; x &lt; \pi&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The numbers &lt;math&gt;1447&lt;/math&gt;, &lt;math&gt;1005&lt;/math&gt;, and &lt;math&gt;1231&lt;/math&gt; have something in common. Each is a four-digit number beginning with &lt;math&gt;1&lt;/math&gt; that has exactly two identical digits. How many such numbers are there?<br /> <br /> [[1983 AIME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The solid shown has a square base of side length &lt;math&gt;s&lt;/math&gt;. The upper edge is parallel to the base and has length &lt;math&gt;2s&lt;/math&gt;. All edges have length &lt;math&gt;s&lt;/math&gt;. Given that &lt;math&gt;s=6\sqrt{2}&lt;/math&gt;, what is the volume of the solid?<br /> <br /> &lt;asy&gt;<br /> size(170);import three; pathpen = black+linewidth(0.65); pointpen = black;<br /> currentprojection = perspective(30,-20,10);real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> D(A--B--C--D--A--E--D); D(B--F--C); D(E--F);<br /> MP(&quot;A&quot;,A);MP(&quot;B&quot;,B);MP(&quot;C&quot;,C);MP(&quot;D&quot;,D);MP(&quot;E&quot;,E,N);MP(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The length of diameter &lt;math&gt;AB&lt;/math&gt; is a two digit integer. Reversing the digits gives the length of a perpendicular chord &lt;math&gt;CD&lt;/math&gt;. The distance from their intersection point &lt;math&gt;H&lt;/math&gt; to the center &lt;math&gt;O&lt;/math&gt; is a positive rational number. Determine the length of &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> &lt;asy&gt;pointpen=black; pathpen=black+linewidth(0.65);<br /> pair O=(0,0),A=(-65/2,0),B=(65/2,0);<br /> pair H=(-((65/2)^2-28^2)^.5,0),C=(H.x,28),D=(H.x,-28);<br /> D(CP(O,A));D(MP(&quot;A&quot;,A,W)--MP(&quot;B&quot;,B,E));D(MP(&quot;C&quot;,C,N)--MP(&quot;D&quot;,D));<br /> dot(MP(&quot;H&quot;,H,SE));dot(MP(&quot;O&quot;,O,SE));&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> For &lt;math&gt;\{1, 2, 3, \ldots, n\}&lt;/math&gt; and each of its non-empty subsets, an alternating sum is defined as follows. Arrange the number in the subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example, the alternating sum for &lt;math&gt;\{1, 2, 3, 6,9\}&lt;/math&gt; is &lt;math&gt;9-6+3-2+1=6&lt;/math&gt; and for &lt;math&gt;\{5\}&lt;/math&gt; it is simply &lt;math&gt;5&lt;/math&gt;. Find the sum of all such alternating sums for &lt;math&gt;n=7&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> In the adjoining figure, two circles with radii &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are drawn with their centers &lt;math&gt;12&lt;/math&gt; units apart. At &lt;math&gt;P&lt;/math&gt;, one of the points of intersection, a line is drawn in such a way that the chords &lt;math&gt;QP&lt;/math&gt; and &lt;math&gt;PR&lt;/math&gt; have equal length. (&lt;math&gt;P&lt;/math&gt; is the midpoint of &lt;math&gt;QR&lt;/math&gt;) Find the square of the length of &lt;math&gt;QP&lt;/math&gt;. <br /> <br /> [[Image:1983_AIME-14.png]]<br /> <br /> [[1983 AIME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> The adjoining figure shows two intersecting chords in a circle, with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the sine of the minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this fraction is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> <br /> [[Image:1983_AIME-15.png]]<br /> <br /> [[1983 AIME Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[1983 AIME]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:AIME Problems]]</div> Willwang123