https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Limac&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-18T04:45:12Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_23&diff=36563 2003 AMC 10A Problems/Problem 23 2011-02-04T00:13:55Z <p>Limac: /* Solution 2 */ fix</p> <hr /> <div>== Problem ==<br /> A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure we have &lt;math&gt;3&lt;/math&gt; rows of small congruent equilateral triangles, with &lt;math&gt;5&lt;/math&gt; small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of &lt;math&gt;2003&lt;/math&gt; small equilateral triangles? <br /> <br /> [[Image:2003amc10a23.gif]]<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018 &lt;/math&gt;<br /> <br /> == Solution 1==<br /> There are &lt;math&gt;1+3+5+...+2003=1002^{2}=1004004&lt;/math&gt; small equilateral triangles. <br /> <br /> Each small equilateral triangle needs &lt;math&gt;3&lt;/math&gt; toothpicks to make it. <br /> <br /> But, each toothpick that isn't one of the &lt;math&gt;1002\cdot3=3006&lt;/math&gt; toothpicks on the outside of the large equilateral triangle is a side for &lt;math&gt;2&lt;/math&gt; small equilateral triangles. <br /> <br /> So, the number of toothpicks on the inside of the large equilateral triangle is &lt;math&gt;\frac{10040004\cdot3-3006}{2}=1504503&lt;/math&gt;<br /> <br /> Therefore the total number of toothpicks is &lt;math&gt;1504503+3006=1,507,509 \Rightarrow \mathrm{(C)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We see that the bottom row of &lt;math&gt;2003&lt;/math&gt; small triangles is formed from &lt;math&gt;1002&lt;/math&gt; downward-facing triangles and &lt;math&gt;1001&lt;/math&gt; upward-facing triangles. Since each downward-facing triangle uses three distinct toothpicks, and since the total number of downward-facing triangles is &lt;math&gt;1002+1001+...+1=\frac{1003\cdot1002}{2}=502503&lt;/math&gt;, we have that the total number of toothpicks is &lt;math&gt;3\cdot 502503=1507509&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_4&diff=36354 1984 AIME Problems/Problem 4 2011-01-17T23:59:33Z <p>Limac: /* Solution */ answer was incorrect</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;\displaystyle S&lt;/math&gt; be a list of [[positive integer]]s - not necessarily [[distinct]] - in which the number &lt;math&gt;\displaystyle 68&lt;/math&gt; appears. The [[arithmetic mean]] of the numbers in &lt;math&gt;\displaystyle S&lt;/math&gt; is &lt;math&gt;\displaystyle 56&lt;/math&gt;. However, if &lt;math&gt;\displaystyle 68&lt;/math&gt; is removed, the arithmetic mean of the numbers is &lt;math&gt;\displaystyle 55&lt;/math&gt;. What's the largest number that can appear in &lt;math&gt;\displaystyle S&lt;/math&gt;?<br /> <br /> == Solution ==<br /> Suppose &lt;math&gt;S&lt;/math&gt; has &lt;math&gt;n&lt;/math&gt; members other than 68, and the sum of these members is &lt;math&gt;s&lt;/math&gt;. Then we're given that &lt;math&gt;\frac{s + 68}{n + 1} = 56&lt;/math&gt; and &lt;math&gt;\frac{s}{n} = 55&lt;/math&gt;. Multiplying to clear [[denominator]]s, we have &lt;math&gt;s + 68 = 56n + 56&lt;/math&gt; and &lt;math&gt;s = 55n&lt;/math&gt; so &lt;math&gt;68 = n + 56&lt;/math&gt;, &lt;math&gt;n = 12&lt;/math&gt; and &lt;math&gt;s = 12\cdot 55 = 660&lt;/math&gt;. Because the sum and number of the elements of &lt;math&gt;S&lt;/math&gt; are fixed, if we want to maximize the largest number in &lt;math&gt;S&lt;/math&gt;, we should take all but one member of &lt;math&gt;S&lt;/math&gt; to be as small as possible. Since all members of &lt;math&gt;S&lt;/math&gt; are positive integers, the smallest possible value of a member is 1. Thus the largest possible element is &lt;math&gt;660 - 11 = \boxed{649}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=3|num-a=5}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_9&diff=33857 2007 AIME II Problems/Problem 9 2010-03-13T02:46:20Z <p>Limac: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> [[Rectangle]] &lt;math&gt;ABCD&lt;/math&gt; is given with &lt;math&gt;AB=63&lt;/math&gt; and &lt;math&gt;BC=448.&lt;/math&gt; Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie on &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; respectively, such that &lt;math&gt;AE=CF=84.&lt;/math&gt; The [[inscribed circle]] of [[triangle]] &lt;math&gt;BEF&lt;/math&gt; is [[tangent]] to &lt;math&gt;EF&lt;/math&gt; at point &lt;math&gt;P,&lt;/math&gt; and the inscribed circle of triangle &lt;math&gt;DEF&lt;/math&gt; is tangent to &lt;math&gt;EF&lt;/math&gt; at [[point]] &lt;math&gt;Q.&lt;/math&gt; Find &lt;math&gt;PQ.&lt;/math&gt;<br /> <br /> == Solution ==<br /> [[Image:2007 AIME II-9.png]]<br /> <br /> === Solution 1 ===<br /> Several [[Pythagorean triple]]s exist amongst the numbers given. &lt;math&gt;BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105&lt;/math&gt;. Also, the length of &lt;math&gt;EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287&lt;/math&gt;.<br /> <br /> Use the [[Two Tangent theorem]] on &lt;math&gt;\triangle BEF&lt;/math&gt;. Since both circles are inscribed in congruent triangles, they are congruent; therefore, &lt;math&gt;EP = FQ = \frac{287 - PQ}{2}&lt;/math&gt;. By the Two Tangent theorem, note that &lt;math&gt;EP = EX = \frac{287 - PQ}{2}&lt;/math&gt;, making &lt;math&gt;BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]&lt;/math&gt;. Also, &lt;math&gt;BX = BY&lt;/math&gt;. &lt;math&gt;FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]&lt;/math&gt;. <br /> <br /> Finally, &lt;math&gt;FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}&lt;/math&gt;. Also, &lt;math&gt;FP = FQ + PQ = \frac{287 - PQ}{2} + PQ&lt;/math&gt;. Equating, we see that &lt;math&gt;\frac{805 - PQ}{2} = \frac{287 + PQ}{2}&lt;/math&gt;, so &lt;math&gt;PQ = 259&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> By the [[Two Tangent theorem]], we have that &lt;math&gt;FY = PQ + QF&lt;/math&gt;. Solve for &lt;math&gt;PQ = FY - QF&lt;/math&gt;. Also, &lt;math&gt;QF = EP = EX&lt;/math&gt;, so &lt;math&gt;PQ = FY - EX&lt;/math&gt;. Since &lt;math&gt;BX = BY&lt;/math&gt;, this can become &lt;math&gt;PQ = FY - EX + (BY - BX)&lt;/math&gt;&lt;math&gt; = \left(FY + BY\right) - \left(EX + BX\right) = FB - EB&lt;/math&gt;. Substituting in their values, the answer is &lt;math&gt;364 - 105 = 259&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> Call the incenter of &lt;math&gt;\triangle BEF&lt;/math&gt; &lt;math&gt;O_1&lt;/math&gt; and the incenter of &lt;math&gt;\triangle DFE&lt;/math&gt; &lt;math&gt;O_2&lt;/math&gt;. Draw triangles &lt;math&gt;\triangle O_1PQ,\triangle PQO_2&lt;/math&gt;.<br /> <br /> Drawing &lt;math&gt;BE&lt;/math&gt;, We find that &lt;math&gt;BE = \sqrt {63^2 + 84^2} = 105&lt;/math&gt;. Applying the same thing for &lt;math&gt;F&lt;/math&gt;, we find that &lt;math&gt;FD = 105&lt;/math&gt; as well. Draw a line through &lt;math&gt;E,F&lt;/math&gt; parallel to the sides of the rectangle, to intersect the opposite side at &lt;math&gt;E_1,F_1&lt;/math&gt; respectively. Drawing &lt;math&gt;\triangle EE_1F&lt;/math&gt; and &lt;math&gt;FF_1E&lt;/math&gt;, we can find that &lt;math&gt;EF = \sqrt {63^2 + 280^2} = 287&lt;/math&gt;. We then use Heron's formula to get:<br /> <br /> &lt;cmath&gt;[BEF] = [DEF] = 11 466&lt;/cmath&gt;.<br /> <br /> So the inradius of the triangle-type things is &lt;math&gt;\frac {637}{21}&lt;/math&gt;.<br /> <br /> Now, we just have to find &lt;math&gt;O_1Q = O_2P&lt;/math&gt;, which can be done with simple subtraction, and then we can use the [[Pythagorean Theorem]] to find &lt;math&gt;PQ&lt;/math&gt;. <br /> <br /> {{incomplete|solution}}<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=II|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_11&diff=33628 2006 AMC 10B Problems/Problem 11 2010-02-22T22:08:46Z <p>Limac: /* Solution */</p> <hr /> <div>== Problem ==<br /> What is the tens digit in the sum &lt;math&gt; 7!+8!+9!+...+2006!&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9 &lt;/math&gt;<br /> <br /> == Solution ==<br /> Since &lt;math&gt;10!&lt;/math&gt; is [[divisibility | divisible]] by &lt;math&gt;100&lt;/math&gt;, any [[factorial]] greater than &lt;math&gt;10!&lt;/math&gt; is also divisible by &lt;math&gt;100&lt;/math&gt;. The last two [[digit]]s of all factorials greater than &lt;math&gt;10!&lt;/math&gt; are &lt;math&gt;00&lt;/math&gt;, so the last two digits of &lt;math&gt;10!+11!+...+2006!&lt;/math&gt; is &lt;math&gt;00&lt;/math&gt;. <br /> (*)<br /> <br /> So all that is needed is the tens digit of the sum &lt;math&gt;7!+8!+9!&lt;/math&gt;<br /> <br /> &lt;math&gt;7!+8!+9!=5040+40320+362880=408240&lt;/math&gt;<br /> <br /> So the tens digit is &lt;math&gt;4 \Rightarrow C&lt;/math&gt;<br /> <br /> (*) A slightly faster method would have to take the &lt;math&gt;\pmod {100}&lt;/math&gt; residue of &lt;math&gt;7! + 8! + 9!.&lt;/math&gt; Since &lt;math&gt;7! = 5040,&lt;/math&gt; we can rewrite the sum as &lt;cmath&gt; 5040 + 8\cdot 5040 + 72\cdot 5040 \equiv 40 + 8\cdot 40 + 72\cdot 40 = 40 + 320 + 2880 \equiv 40 \pmod{100}. &lt;/cmath&gt; Since the last two digits of the sum is &lt;math&gt;40&lt;/math&gt;, the tens digit is &lt;math&gt;4.&lt;/math&gt;<br /> <br /> == See Also ==<br /> *[[2006 AMC 10B Problems]]<br /> <br /> *[[2006 AMC 10B Problems/Problem 10|Previous Problem]]<br /> <br /> *[[2006 AMC 10B Problems/Problem 12|Next Problem]]<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_11&diff=33627 2006 AMC 10B Problems/Problem 11 2010-02-22T22:07:56Z <p>Limac: /* Solution */</p> <hr /> <div>== Problem ==<br /> What is the tens digit in the sum &lt;math&gt; 7!+8!+9!+...+2006!&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9 &lt;/math&gt;<br /> <br /> == Solution ==<br /> Since &lt;math&gt;10!&lt;/math&gt; is [[divisibility | divisible]] by &lt;math&gt;100&lt;/math&gt;, any [[factorial]] greater than &lt;math&gt;10!&lt;/math&gt; is also divisible by &lt;math&gt;100&lt;/math&gt;. The last two [[digit]]s of all factorials greater than &lt;math&gt;10!&lt;/math&gt; are &lt;math&gt;00&lt;/math&gt;, so the last two digits of &lt;math&gt;10!+11!+...+2006!&lt;/math&gt; is &lt;math&gt;00&lt;/math&gt;. <br /> (*)<br /> <br /> So all that is needed is the tens digit of the sum &lt;math&gt;7!+8!+9!&lt;/math&gt;<br /> <br /> &lt;math&gt;7!+8!+9!=5040+40320+362880=408240&lt;/math&gt;<br /> <br /> So the tens digit is &lt;math&gt;4 \Rightarrow C&lt;/math&gt;<br /> <br /> (*) A slightly faster method would have to take the &lt;math&gt;\pmod {100}&lt;/math&gt; residue of &lt;math&gt;7! + 8! + 9!.&lt;/math&gt; Since &lt;math&gt;7! = 5040,&lt;/math&gt; we can rewrite the sum as &lt;cmath&gt; 5040 + 8\cdot 5040 + 72\cdot 5040 \equiv 40 + 8\cdot 40 + 72\cdot 40 = 30 + 320 + 2880 \equiv 40 \pmod{100}. &lt;/cmath&gt; Since the last two digits of the sum is &lt;math&gt;40&lt;/math&gt;, the tens digit is &lt;math&gt;4.&lt;/math&gt;<br /> <br /> == See Also ==<br /> *[[2006 AMC 10B Problems]]<br /> <br /> *[[2006 AMC 10B Problems/Problem 10|Previous Problem]]<br /> <br /> *[[2006 AMC 10B Problems/Problem 12|Next Problem]]<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_11&diff=33626 2006 AMC 10B Problems/Problem 11 2010-02-22T22:07:35Z <p>Limac: /* Solution */</p> <hr /> <div>== Problem ==<br /> What is the tens digit in the sum &lt;math&gt; 7!+8!+9!+...+2006!&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9 &lt;/math&gt;<br /> <br /> == Solution ==<br /> Since &lt;math&gt;10!&lt;/math&gt; is [[divisibility | divisible]] by &lt;math&gt;100&lt;/math&gt;, any [[factorial]] greater than &lt;math&gt;10!&lt;/math&gt; is also divisible by &lt;math&gt;100&lt;/math&gt;. The last two [[digit]]s of all factorials greater than &lt;math&gt;10!&lt;/math&gt; are &lt;math&gt;00&lt;/math&gt;, so the last two digits of &lt;math&gt;10!+11!+...+2006!&lt;/math&gt; is &lt;math&gt;00&lt;/math&gt;. <br /> (*)<br /> So all that is needed is the tens digit of the sum &lt;math&gt;7!+8!+9!&lt;/math&gt;<br /> <br /> &lt;math&gt;7!+8!+9!=5040+40320+362880=408240&lt;/math&gt;<br /> <br /> So the tens digit is &lt;math&gt;4 \Rightarrow C&lt;/math&gt;<br /> <br /> (*) A slightly faster method would have to take the &lt;math&gt;\pmod {100}&lt;/math&gt; residue of &lt;math&gt;7! + 8! + 9!.&lt;/math&gt; Since &lt;math&gt;7! = 5040,&lt;/math&gt; we can rewrite the sum as &lt;cmath&gt; 5040 + 8\cdot 5040 + 72\cdot 5040 \equiv 40 + 8\cdot 40 + 72\cdot 40 = 30 + 320 + 2880 \equiv 40 \pmod{100}. &lt;/cmath&gt; Since the last two digits of the sum is &lt;math&gt;40&lt;/math&gt;, the tens digit is &lt;math&gt;4.&lt;/math&gt;<br /> <br /> == See Also ==<br /> *[[2006 AMC 10B Problems]]<br /> <br /> *[[2006 AMC 10B Problems/Problem 10|Previous Problem]]<br /> <br /> *[[2006 AMC 10B Problems/Problem 12|Next Problem]]<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_8&diff=33446 2002 AIME II Problems/Problem 8 2010-02-14T22:40:41Z <p>Limac: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find the least positive integer &lt;math&gt;k&lt;/math&gt; for which the equation &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor=k&lt;/math&gt; has no integer solutions for &lt;math&gt;n&lt;/math&gt;. (The notation &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; means the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> Note that if &lt;math&gt;\frac{2002}n - \frac{2002}{n+1}\leq 1&lt;/math&gt;, then either &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor&lt;/math&gt;,<br /> or &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1&lt;/math&gt;. Either way, we won't skip any natural numbers.<br /> <br /> The smallest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;\frac{2002}n - \frac{2002}{n+1} &gt; 1&lt;/math&gt; is &lt;math&gt;n=45&lt;/math&gt;. (The inequality simplifies to &lt;math&gt;n(n+1)&gt;2002&lt;/math&gt;, which is easy to solve by trial, as the solution is obviously &lt;math&gt;\simeq \sqrt{2002}&lt;/math&gt;.)<br /> <br /> We can now compute:<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{45}\right\rfloor=44 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{44}\right\rfloor=45 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{43}\right\rfloor=46 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{42}\right\rfloor=47 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{41}\right\rfloor=48 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{40}\right\rfloor=50 &lt;/cmath&gt;<br /> <br /> From the observation above (and the fact that &lt;math&gt;\left\lfloor\frac{2002}{2002}\right\rfloor=1&lt;/math&gt;) we know that all integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;44&lt;/math&gt; will be achieved for some values of &lt;math&gt;n&lt;/math&gt;. Similarly, for &lt;math&gt;n&lt;40&lt;/math&gt; we obviously have &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor &gt; 50&lt;/math&gt;.<br /> <br /> Hence the least positive integer &lt;math&gt;k&lt;/math&gt; for which the equation &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor=k&lt;/math&gt; has no integer solutions for &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\boxed{049}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Rewriting the given information and simplifying it a bit, we have<br /> &lt;cmath&gt; \begin{align*} <br /> k \le \frac{2002}{n} &lt; k+1 &amp;\implies \frac{1}{k} \ge \frac{n}{2002} &gt; \frac{1}{k+1}. \\ &amp;\implies \frac{2002}{k} \ge n &gt; \frac{2002}{k+1}. <br /> \end{align*} &lt;/cmath&gt;<br /> <br /> Now note that in order for there to be no integer solutions to &lt;math&gt;n,&lt;/math&gt; we must have &lt;math&gt;\left\lfloor \frac{2002}{k} \right\rfloor = \left\lfloor \frac{2002}{k+1} \right\rfloor.&lt;/math&gt; We seek the smallest such &lt;math&gt;k.&lt;/math&gt; A bit of experimentation yields that &lt;math&gt;k=49&lt;/math&gt; is the smallest solution, as for &lt;math&gt;k=49,&lt;/math&gt; it is true that &lt;math&gt;\left\lfloor \frac{2002}{49} \right\rfloor = \left\lfloor \frac{2002}{50} \right\rfloor = 40.&lt;/math&gt; Furthermore, &lt;math&gt;k=49&lt;/math&gt; is the smallest such case. (If unsure, we could check if the result holds for &lt;math&gt;k=48,&lt;/math&gt; and as it turns out, it doesn't.) Therefore, the answer is &lt;math&gt;\boxed{049}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2002|n=II|num-b=7|num-a=9}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_20&diff=33337 2003 AMC 10A Problems/Problem 20 2010-02-08T04:07:35Z <p>Limac: /* Problem20 */</p> <hr /> <div>== Problem20 ==<br /> A base-10 three digit number &lt;math&gt;n&lt;/math&gt; is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of &lt;math&gt;n&lt;/math&gt; are both three-digit numerals? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7 &lt;/math&gt;<br /> <br /> == Solution ==<br /> To be a three digit number in base-10: <br /> <br /> &lt;math&gt;10^{2} \leq n \leq 10^{3}-1&lt;/math&gt;<br /> <br /> &lt;math&gt;100 \leq n \leq 999&lt;/math&gt; <br /> <br /> Thus there are &lt;math&gt;900&lt;/math&gt; three-digit numbers in base-10<br /> <br /> To be a three-digit number in base-9: <br /> <br /> &lt;math&gt;9^{2} \leq n \leq 9^{3}-1&lt;/math&gt;<br /> <br /> &lt;math&gt;81 \leq n \leq 728&lt;/math&gt;<br /> <br /> To be a three-digit number in base-11: <br /> <br /> &lt;math&gt;11^{2} \leq n \leq 11^{3}-1&lt;/math&gt;<br /> <br /> &lt;math&gt;121 \leq n \leq 1330&lt;/math&gt;<br /> <br /> So, &lt;math&gt;121 \leq n \leq 728&lt;/math&gt; <br /> <br /> Thus, there are &lt;math&gt;608&lt;/math&gt; base-10 three-digit numbers that are three digit numbers in base-9 and base-11. <br /> <br /> Therefore the desired probability is &lt;math&gt;\frac{608}{900}\approx 0.7 \Rightarrow E&lt;/math&gt;. <br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_25&diff=33328 2009 AMC 10B Problems/Problem 25 2010-02-07T15:58:36Z <p>Limac: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #25]] and [[2009 AMC 12B Problems|2009 AMC 12B #17]]}}<br /> <br /> == Problem ==<br /> Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?<br /> <br /> &lt;math&gt;\mathrm{(A)}\frac 18\qquad<br /> \mathrm{(B)}\frac {3}{16}\qquad<br /> \mathrm{(C)}\frac 14\qquad<br /> \mathrm{(D)}\frac 38\qquad<br /> \mathrm{(E)}\frac 12&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> There are two possible stripe orientations for each of the six faces of the cube, so there are &lt;math&gt;2^6 = 64&lt;/math&gt; possible stripe combinations. There are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripe orientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in either of two different ways. So a total of &lt;math&gt;3 \cdot 2 \cdot 2 = 12&lt;/math&gt; stripe combinations on the cube result in a continuous stripe around the cube. The required probability is &lt;math&gt;\frac {12}{64} = \boxed{\frac {3}{16}}&lt;/math&gt;.<br /> <br /> Here's another way similar to this: <br /> <br /> So there are &lt;math&gt;2^6&lt;/math&gt; choices for the stripes as mentioned above. Now, let's just consider the &quot;view point&quot; of one of the faces. We can choose any of the 2 orientation for the stripe (it can go from up to down, or from right to left). Once that orientation is chosen, each of the other faces that contribute to that loop only have 1 choice, which is to go in the direction of the loop. That gives us a total count of 2 possibilities for any one of the faces. Since there are six faces, and this argument is valid for all of them, we conclude that there are 2(6) = 12 total ways to have the stripe. Therefore, the probability is 12/64 = 3/16.<br /> <br /> === Solution 2 ===<br /> Without loss of generality, orient the cube so that the stripe on the top face goes from front to back. There are two mutually exclusive ways for there to be an encircling stripe: either the front, bottom and back faces are painted to complete an encircling stripe with the top face's stripe or the front, right, back and left faces are painted to form an encircling stripe. The probability of the first case is &lt;math&gt;(\frac 12)^3 = \frac 18&lt;/math&gt;, and the probability of the second case is &lt;math&gt;(\frac 12)^4 = \frac {1}{16}&lt;/math&gt;. The cases are disjoint, so the probabilities sum &lt;math&gt;\frac 18 + \frac {1}{16} = \boxed {\frac {3}{16}}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> There are three possible orientations of an encircling stripe. For any one of these to appear, the stripes on the four faces through which the continuous stripe is to pass must be properly aligned. The probability of each such stripe alignment is &lt;math&gt;(\frac 12)^4 = \frac {1}{16}&lt;/math&gt;. Since there are three such possibilities and they are disjoint, the total probability is &lt;math&gt;3 \cdot \frac {1}{16} = \boxed{\frac {3}{16}}&lt;/math&gt;. The answer is &lt;math&gt;\mathrm{(B)}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Consider a vertex on the cube and the three faces that are adjacent to that vertex. If no two stripes on those three faces are aligned, then there is no stripe encircling the cube. The probability that the stripes aren't aligned is &lt;math&gt;\frac{1}{4}&lt;/math&gt;, since for each alignment of one stripe, there is one and only one way to align the other two stripes out of four total possibilities. therefore there is a probability of &lt;math&gt;\frac{3}{4}&lt;/math&gt; that two stripes are aligned.<br /> <br /> Now consider the opposing vertex and the three sides adjacent to it. Given the two connected stripes next to our first vertex, we have two more that must be connected to make a continuous stripe. There is a probability of &lt;math&gt;\left(\frac{1}{2}\right)^2=\frac{1}{4}&lt;/math&gt; that they are aligned, so there is a probability of &lt;math&gt;\frac{3}{4}\cdot \frac{1}{4}=\frac{3}{16}&lt;/math&gt; that there is a continuous stripe.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2009|ab=B|num-b=24|after=Last question}}<br /> {{AMC12 box|year=2009|ab=B|num-b=16|num-a=18}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_22&diff=33323 2003 AMC 10A Problems/Problem 22 2010-02-06T14:20:18Z <p>Limac: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, we have &lt;math&gt;AB=8&lt;/math&gt;, &lt;math&gt;BC=9&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;BC&lt;/math&gt; with &lt;math&gt;BH=6&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;AD&lt;/math&gt; with &lt;math&gt;DE=4&lt;/math&gt;, line &lt;math&gt;EC&lt;/math&gt; intersects line &lt;math&gt;AH&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; is on line &lt;math&gt;AD&lt;/math&gt; with &lt;math&gt;GF \perp AF&lt;/math&gt;. Find the length of &lt;math&gt;GF&lt;/math&gt;. <br /> <br /> [[Image:2003amc10a22.gif]]<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30 &lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;math&gt;\angle GCH = \angle ABH&lt;/math&gt; (Opposite angles are equal).<br /> <br /> &lt;math&gt;\angle F = \angle B&lt;/math&gt; (Both are 90 degrees).<br /> <br /> &lt;math&gt;\angle BHA = \angle HAD&lt;/math&gt; (Alt. Interior Angles are congruent).<br /> <br /> Therefore &lt;math&gt;Triangles\: GFA&lt;/math&gt; and &lt;math&gt;ABH&lt;/math&gt; are similar.<br /> &lt;math&gt;GCH&lt;/math&gt; and &lt;math&gt;GEA&lt;/math&gt; are also similar.<br /> <br /> &lt;math&gt;DA&lt;/math&gt; is 9, therefore &lt;math&gt;EA&lt;/math&gt; must equal 5. Similarly, &lt;math&gt;CH&lt;/math&gt; must equal 3. <br /> <br /> Because &lt;math&gt;GCH&lt;/math&gt; and &lt;math&gt;GEA&lt;/math&gt; are similar, the ratio of &lt;math&gt;CH\; =\; 3&lt;/math&gt; and &lt;math&gt;EA\; =\; 5&lt;/math&gt;, must also hold true for &lt;math&gt;GH&lt;/math&gt; and &lt;math&gt;HA&lt;/math&gt;. &lt;math&gt;\frac{GH}{GA} = \frac{3}{5}&lt;/math&gt;, so &lt;math&gt;HA&lt;/math&gt; is &lt;math&gt;\frac{2}{5}&lt;/math&gt; of &lt;math&gt;GA&lt;/math&gt;. By Pythagorean theorem, &lt;math&gt;(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10&lt;/math&gt;.<br /> <br /> &lt;math&gt;HA\: =\: 10 =\: \frac{2}{5}*(GA)&lt;/math&gt;. <br /> <br /> &lt;math&gt;GA\: =\: 25.&lt;/math&gt;<br /> <br /> So &lt;math&gt;\frac{GA}{HA}\: =\: \frac{GF}{BA}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac{25}{10}\: =\: \frac{GF}{8}&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;GF\: = \boxed{20} = \boxed{\mathrm{(B)}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Since &lt;math&gt;ABCD&lt;/math&gt; is a rectangle, &lt;math&gt;CD=AB=8&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;ABCD&lt;/math&gt; is a rectangle and &lt;math&gt;GF \perp AF&lt;/math&gt;, &lt;math&gt;\angle GFE = \angle CDE = \angle ABC = 90^\circ &lt;/math&gt;. <br /> <br /> Since &lt;math&gt;ABCD&lt;/math&gt; is a rectangle, &lt;math&gt;AD || BC&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;AH&lt;/math&gt; is a [[transversal]], and &lt;math&gt;\angle GAF = \angle AHB&lt;/math&gt;. <br /> <br /> This is sufficient to prove that &lt;math&gt; GFE \approx CDE&lt;/math&gt; and &lt;math&gt; GFA \approx ABH&lt;/math&gt;.<br /> <br /> Using ratios: <br /> <br /> &lt;math&gt;\frac{GF}{FE}=\frac{CD}{DE}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{GF}{FD+4}=\frac{8}{4}=2&lt;/math&gt;<br /> <br /> &lt;math&gt;GF=2 \cdot (FD+4)=2 \cdot FD+8&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{GF}{FA}=\frac{AB}{BH}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{GF}{FD+9}=\frac{8}{6}=\frac{4}{3}&lt;/math&gt;<br /> <br /> &lt;math&gt;GF=\frac{4}{3} \cdot (FD+9)=\frac{4}{3} \cdot FD+12&lt;/math&gt;<br /> <br /> Since &lt;math&gt;GF&lt;/math&gt; can't have 2 different lengths, both expressions for &lt;math&gt;GF&lt;/math&gt; must be equal. <br /> <br /> &lt;math&gt;2 \cdot FD+8=\frac{4}{3} \cdot FD+12&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3} \cdot FD=4&lt;/math&gt;<br /> <br /> &lt;math&gt;FD=6&lt;/math&gt;<br /> <br /> &lt;math&gt;GF=2 \cdot FD+8=2\cdot6+8=20 \Rightarrow B&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> Since &lt;math&gt;ABCD&lt;/math&gt; is a rectangle, &lt;math&gt;CD=3&lt;/math&gt;, &lt;math&gt;EA=5&lt;/math&gt;, and &lt;math&gt;CD=8&lt;/math&gt;. From the [[Pythagorean Theorem]], &lt;math&gt;CE^2=CD^2+DE^2=80\Rightarrow CE=4\sqrt{5}&lt;/math&gt;.<br /> ==== Lemma ====<br /> Statement: &lt;math&gt;GCH \approx GEA&lt;/math&gt;<br /> <br /> Proof: &lt;math&gt;\angle CGH=\angle EGA&lt;/math&gt;, obviously.<br /> <br /> &lt;math&gt;\begin{eqnarray}<br /> \angle HCE=180^{\circ}-\angle CHG\\<br /> \angle DCE=\angle CHG-90^{\circ}\\<br /> \angle CEED=180-\angle CHG\\<br /> \angle GEA=\angle GCH<br /> \end{eqnarray}&lt;/math&gt;<br /> <br /> Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.<br /> <br /> <br /> <br /> Let &lt;math&gt;GC=x&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{eqnarray}<br /> \dfrac{x}{3}=\dfrac{x+4\sqrt{5}}{5}\\<br /> 5x=3x+12\sqrt{5}\\<br /> 2x=12\sqrt{5}\\<br /> x=6\sqrt{5}<br /> \end{eqnarray}&lt;/cmath&gt;<br /> <br /> Also, &lt;math&gt;\triangle GFE\approx \triangle CDE&lt;/math&gt;, therefore<br /> <br /> &lt;cmath&gt;\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}&lt;/cmath&gt;<br /> <br /> We can multiply both sides by &lt;math&gt;\sqrt{5}&lt;/math&gt; to get that &lt;math&gt;GF&lt;/math&gt; is twice of 10, or &lt;math&gt;20\Rightarrow \mathrm{(B)}&lt;/math&gt;<br /> <br /> === Solution 4 ===<br /> We extend BC such that it intersects GF at X. Since ABCD is a rectangle, it follows that CD=8, therefore, XF=8. Let GX=y. From the similarity of triangles GCH and GEA, we have the ratio 3:5 (as CH=9-6=3, and EA=9-4=5). GX and GF are the altitudes of GCH and GEA, respectively. Thus, y:y+8 = 3:5, from which we have y=12, thus GF=y+8=12+8=20. B.<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=21|num-a=23}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_22&diff=33322 2003 AMC 10A Problems/Problem 22 2010-02-06T14:19:01Z <p>Limac: New solution</p> <hr /> <div>== Problem ==<br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, we have &lt;math&gt;AB=8&lt;/math&gt;, &lt;math&gt;BC=9&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;BC&lt;/math&gt; with &lt;math&gt;BH=6&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;AD&lt;/math&gt; with &lt;math&gt;DE=4&lt;/math&gt;, line &lt;math&gt;EC&lt;/math&gt; intersects line &lt;math&gt;AH&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; is on line &lt;math&gt;AD&lt;/math&gt; with &lt;math&gt;GF \perp AF&lt;/math&gt;. Find the length of &lt;math&gt;GF&lt;/math&gt;. <br /> <br /> [[Image:2003amc10a22.gif]]<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30 &lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;math&gt;\angle GCH = \angle ABH&lt;/math&gt; (Opposite angles are equal).<br /> <br /> &lt;math&gt;\angle F = \angle B&lt;/math&gt; (Both are 90 degrees).<br /> <br /> &lt;math&gt;\angle BHA = \angle HAD&lt;/math&gt; (Alt. Interior Angles are congruent).<br /> <br /> Therefore &lt;math&gt;Triangles\: GFA&lt;/math&gt; and &lt;math&gt;ABH&lt;/math&gt; are similar.<br /> &lt;math&gt;GCH&lt;/math&gt; and &lt;math&gt;GEA&lt;/math&gt; are also similar.<br /> <br /> &lt;math&gt;DA&lt;/math&gt; is 9, therefore &lt;math&gt;EA&lt;/math&gt; must equal 5. Similarly, &lt;math&gt;CH&lt;/math&gt; must equal 3. <br /> <br /> Because &lt;math&gt;GCH&lt;/math&gt; and &lt;math&gt;GEA&lt;/math&gt; are similar, the ratio of &lt;math&gt;CH\; =\; 3&lt;/math&gt; and &lt;math&gt;EA\; =\; 5&lt;/math&gt;, must also hold true for &lt;math&gt;GH&lt;/math&gt; and &lt;math&gt;HA&lt;/math&gt;. &lt;math&gt;\frac{GH}{GA} = \frac{3}{5}&lt;/math&gt;, so &lt;math&gt;HA&lt;/math&gt; is &lt;math&gt;\frac{2}{5}&lt;/math&gt; of &lt;math&gt;GA&lt;/math&gt;. By Pythagorean theorem, &lt;math&gt;(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10&lt;/math&gt;.<br /> <br /> &lt;math&gt;HA\: =\: 10 =\: \frac{2}{5}*(GA)&lt;/math&gt;. <br /> <br /> &lt;math&gt;GA\: =\: 25.&lt;/math&gt;<br /> <br /> So &lt;math&gt;\frac{GA}{HA}\: =\: \frac{GF}{BA}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac{25}{10}\: =\: \frac{GF}{8}&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;GF\: = \boxed{20} = \boxed{\mathrm{(B)}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Since &lt;math&gt;ABCD&lt;/math&gt; is a rectangle, &lt;math&gt;CD=AB=8&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;ABCD&lt;/math&gt; is a rectangle and &lt;math&gt;GF \perp AF&lt;/math&gt;, &lt;math&gt;\angle GFE = \angle CDE = \angle ABC = 90^\circ &lt;/math&gt;. <br /> <br /> Since &lt;math&gt;ABCD&lt;/math&gt; is a rectangle, &lt;math&gt;AD || BC&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;AH&lt;/math&gt; is a [[transversal]], and &lt;math&gt;\angle GAF = \angle AHB&lt;/math&gt;. <br /> <br /> This is sufficient to prove that &lt;math&gt; GFE \approx CDE&lt;/math&gt; and &lt;math&gt; GFA \approx ABH&lt;/math&gt;.<br /> <br /> Using ratios: <br /> <br /> &lt;math&gt;\frac{GF}{FE}=\frac{CD}{DE}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{GF}{FD+4}=\frac{8}{4}=2&lt;/math&gt;<br /> <br /> &lt;math&gt;GF=2 \cdot (FD+4)=2 \cdot FD+8&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{GF}{FA}=\frac{AB}{BH}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{GF}{FD+9}=\frac{8}{6}=\frac{4}{3}&lt;/math&gt;<br /> <br /> &lt;math&gt;GF=\frac{4}{3} \cdot (FD+9)=\frac{4}{3} \cdot FD+12&lt;/math&gt;<br /> <br /> Since &lt;math&gt;GF&lt;/math&gt; can't have 2 different lengths, both expressions for &lt;math&gt;GF&lt;/math&gt; must be equal. <br /> <br /> &lt;math&gt;2 \cdot FD+8=\frac{4}{3} \cdot FD+12&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3} \cdot FD=4&lt;/math&gt;<br /> <br /> &lt;math&gt;FD=6&lt;/math&gt;<br /> <br /> &lt;math&gt;GF=2 \cdot FD+8=2\cdot6+8=20 \Rightarrow B&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> Since &lt;math&gt;ABCD&lt;/math&gt; is a rectangle, &lt;math&gt;CD=3&lt;/math&gt;, &lt;math&gt;EA=5&lt;/math&gt;, and &lt;math&gt;CD=8&lt;/math&gt;. From the [[Pythagorean Theorem]], &lt;math&gt;CE^2=CD^2+DE^2=80\Rightarrow CE=4\sqrt{5}&lt;/math&gt;.<br /> ==== Lemma ====<br /> Statement: &lt;math&gt;GCH \approx GEA&lt;/math&gt;<br /> <br /> Proof: &lt;math&gt;\angle CGH=\angle EGA&lt;/math&gt;, obviously.<br /> <br /> &lt;math&gt;\begin{eqnarray}<br /> \angle HCE=180^{\circ}-\angle CHG\\<br /> \angle DCE=\angle CHG-90^{\circ}\\<br /> \angle CEED=180-\angle CHG\\<br /> \angle GEA=\angle GCH<br /> \end{eqnarray}&lt;/math&gt;<br /> <br /> Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.<br /> <br /> <br /> <br /> Let &lt;math&gt;GC=x&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{eqnarray}<br /> \dfrac{x}{3}=\dfrac{x+4\sqrt{5}}{5}\\<br /> 5x=3x+12\sqrt{5}\\<br /> 2x=12\sqrt{5}\\<br /> x=6\sqrt{5}<br /> \end{eqnarray}&lt;/cmath&gt;<br /> <br /> Also, &lt;math&gt;\triangle GFE\approx \triangle CDE&lt;/math&gt;, therefore<br /> <br /> &lt;cmath&gt;\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}&lt;/cmath&gt;<br /> <br /> We can multiply both sides by &lt;math&gt;\sqrt{5}&lt;/math&gt; to get that &lt;math&gt;GF&lt;/math&gt; is twice of 10, or &lt;math&gt;20\Rightarrow \mathrm{(B)}&lt;/math&gt;<br /> <br /> === Solution 4 ===<br /> We extend BC such that it intersects GF at X. Since ABCD is a rectangle, it follows that CD=8, therefore, XF=8. Let GX=y. From the similarity of triangles GCH and GEA, we have the ratio 3:5. GX and GF are the altitudes of GCH and GEA, respectively. Thus, y:y+8 = 3:5, from which we have y=12, thus GF=y+8=12+8=20. B.<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=21|num-a=23}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_1&diff=33315 2005 AMC 10A Problems/Problem 1 2010-02-06T00:55:33Z <p>Limac: /* Problem */</p> <hr /> <div>==Problem==<br /> While eating out, Mike and Joe each tipped their server &lt;math&gt;2&lt;/math&gt; dollars. Mike tipped &lt;math&gt;10\%&lt;/math&gt; of his bill and Joe tipped &lt;math&gt;20\%&lt;/math&gt; of his bill. What was the difference, in dollars between their bills? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 20 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; be Mike's bill and &lt;math&gt;j&lt;/math&gt; be Joe's bill. <br /> <br /> &lt;math&gt;\frac{10}{100}m=2&lt;/math&gt;<br /> <br /> &lt;math&gt;m=20&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{20}{100}j=2&lt;/math&gt;<br /> <br /> &lt;math&gt;j=10&lt;/math&gt;<br /> <br /> So the desired difference is &lt;math&gt;m-j=20-10=10 \Rightarrow D&lt;/math&gt;<br /> <br /> ==See Also==<br /> *[[2005 AMC 10A Problems]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 2|Next Problem]]<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_6&diff=33247 1993 AIME Problems/Problem 6 2010-01-30T02:01:06Z <p>Limac: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> What is the smallest [[positive]] [[integer]] than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Denote the first of each of the series of consecutive integers as &lt;math&gt;a,\ b,\ c&lt;/math&gt;. Therefore, &lt;math&gt;n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;. Simplifying, &lt;math&gt;9a = 10b + 9 = 11c + 19&lt;/math&gt;. The relationship between &lt;math&gt;a,\ b&lt;/math&gt; suggests that &lt;math&gt;b&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. Also, &lt;math&gt;10b -10 = 10(b-1) = 11c&lt;/math&gt;, so &lt;math&gt;b-1&lt;/math&gt; is divisible by &lt;math&gt;11&lt;/math&gt;. We find that the least possible value of &lt;math&gt;b = 45&lt;/math&gt;, so the answer is &lt;math&gt;10(45) + 45 = 495&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Let the desired integer be &lt;math&gt;n&lt;/math&gt;. From the information given, it can be determined that, for positive integers &lt;math&gt;a, \ b, \ c&lt;/math&gt;:<br /> <br /> &lt;math&gt;n = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;<br /> <br /> This can be rewritten as the following congruences:<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{9}&lt;/math&gt; <br /> <br /> &lt;math&gt;n \equiv 5 \pmod{10}&lt;/math&gt;<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{11}&lt;/math&gt;<br /> <br /> Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is &lt;math&gt;\boxed{495}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the desired integer. From the given information, we have<br /> &lt;cmath&gt; 9x &amp;= a \\ 11y &amp;= a \\ 10z + 5 &amp;= a, &lt;/cmath&gt; here, &lt;math&gt;x,&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have &lt;math&gt;z&lt;/math&gt; as the 4th term of the sequence. Since, &lt;math&gt;a&lt;/math&gt; is a multiple of &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;11,&lt;/math&gt; it is also a multiple of &lt;math&gt;\lcm[9,11]=99.&lt;/math&gt; Hence, &lt;math&gt;a=99m,&lt;/math&gt; for some &lt;math&gt;m.&lt;/math&gt; So, we have &lt;math&gt;10z + 5 = 99m.&lt;/math&gt; It follows that &lt;math&gt;99(5) = 495&lt;/math&gt; is the smallest integer that can be represented in such a way.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=5|num-a=7}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_6&diff=33246 1993 AIME Problems/Problem 6 2010-01-30T02:00:37Z <p>Limac: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> What is the smallest [[positive]] [[integer]] than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Denote the first of each of the series of consecutive integers as &lt;math&gt;a,\ b,\ c&lt;/math&gt;. Therefore, &lt;math&gt;n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;. Simplifying, &lt;math&gt;9a = 10b + 9 = 11c + 19&lt;/math&gt;. The relationship between &lt;math&gt;a,\ b&lt;/math&gt; suggests that &lt;math&gt;b&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. Also, &lt;math&gt;10b -10 = 10(b-1) = 11c&lt;/math&gt;, so &lt;math&gt;b-1&lt;/math&gt; is divisible by &lt;math&gt;11&lt;/math&gt;. We find that the least possible value of &lt;math&gt;b = 45&lt;/math&gt;, so the answer is &lt;math&gt;10(45) + 45 = 495&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Let the desired integer be &lt;math&gt;n&lt;/math&gt;. From the information given, it can be determined that, for positive integers &lt;math&gt;a, \ b, \ c&lt;/math&gt;:<br /> <br /> &lt;math&gt;n = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;<br /> <br /> This can be rewritten as the following congruences:<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{9}&lt;/math&gt; <br /> <br /> &lt;math&gt;n \equiv 5 \pmod{10}&lt;/math&gt;<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{11}&lt;/math&gt;<br /> <br /> Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is &lt;math&gt;\boxed{495}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the desired integer. From the given information, we have<br /> &lt;cmath&gt; 9x &amp;= a \\ 11y &amp;= a \\ 10z + 5 &amp;= a, &lt;/cmath&gt; here, &lt;math&gt;x,&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have &lt;math&gt;z&lt;/math&gt; as the 4th term of the sequence. Since, &lt;math&gt;s&lt;/math&gt; is a multiple of &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;11,&lt;/math&gt; it is also a multiple of &lt;math&gt;\lcm[9,11]=99.&lt;/math&gt; Hence, &lt;math&gt;a=99m,&lt;/math&gt; for some &lt;math&gt;m.&lt;/math&gt; So, we have &lt;math&gt;10z + 5 = 99m.&lt;/math&gt; It follows that &lt;math&gt;99(5) = 495&lt;/math&gt; is the smallest integer that can be represented in such a way.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=5|num-a=7}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_6&diff=33245 1993 AIME Problems/Problem 6 2010-01-30T02:00:10Z <p>Limac: Added a new solution</p> <hr /> <div>== Problem ==<br /> What is the smallest [[positive]] [[integer]] than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Denote the first of each of the series of consecutive integers as &lt;math&gt;a,\ b,\ c&lt;/math&gt;. Therefore, &lt;math&gt;n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;. Simplifying, &lt;math&gt;9a = 10b + 9 = 11c + 19&lt;/math&gt;. The relationship between &lt;math&gt;a,\ b&lt;/math&gt; suggests that &lt;math&gt;b&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. Also, &lt;math&gt;10b -10 = 10(b-1) = 11c&lt;/math&gt;, so &lt;math&gt;b-1&lt;/math&gt; is divisible by &lt;math&gt;11&lt;/math&gt;. We find that the least possible value of &lt;math&gt;b = 45&lt;/math&gt;, so the answer is &lt;math&gt;10(45) + 45 = 495&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Let the desired integer be &lt;math&gt;n&lt;/math&gt;. From the information given, it can be determined that, for positive integers &lt;math&gt;a, \ b, \ c&lt;/math&gt;:<br /> <br /> &lt;math&gt;n = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;<br /> <br /> This can be rewritten as the following congruences:<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{9}&lt;/math&gt; <br /> <br /> &lt;math&gt;n \equiv 5 \pmod{10}&lt;/math&gt;<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{11}&lt;/math&gt;<br /> <br /> Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is &lt;math&gt;\boxed{495}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the desired integer. From the given information, we have<br /> &lt;cmath&gt; 9x = a \\ 11y = a \\ 10z + 5 = a, &lt;/cmath&gt; here, &lt;math&gt;x,&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have &lt;math&gt;z&lt;/math&gt; as the 4th term of the sequence. Since, &lt;math&gt;s&lt;/math&gt; is a multiple of &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;11,&lt;/math&gt; it is also a multiple of &lt;math&gt;\lcm[9,11]=99.&lt;/math&gt; Hence, &lt;math&gt;a=99m,&lt;/math&gt; for some &lt;math&gt;m.&lt;/math&gt; So, we have &lt;math&gt;10z + 5 = 99m.&lt;/math&gt; It follows that &lt;math&gt;99(5) = 495&lt;/math&gt; is the smallest integer that can be represented in such a way.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=5|num-a=7}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_23&diff=33144 2002 AMC 12B Problems/Problem 23 2010-01-04T03:09:03Z <p>Limac: /* Alternate Solution */</p> <hr /> <div>== Problem ==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, we have &lt;math&gt;AB = 1&lt;/math&gt; and &lt;math&gt;AC = 2&lt;/math&gt;. Side &lt;math&gt;\overline{BC}&lt;/math&gt; and the [[Median of a triangle|median]] from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt; have the same length. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{1+\sqrt{2}}{2}<br /> \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2<br /> \qquad\mathrm{(C)}\ \sqrt{2}<br /> \qquad\mathrm{(D)}\ \frac 32<br /> \qquad\mathrm{(E)}\ \sqrt{3}&lt;/math&gt;<br /> == Solution ==<br /> [[Image:2002_12B_AMC-23.png]]<br /> <br /> Let &lt;math&gt;D&lt;/math&gt; be the foot of the median from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt;, and we let &lt;math&gt;AD = BC = 2a&lt;/math&gt;. Then by the [[Law of Cosines]] on &lt;math&gt;\triangle ABD, \triangle ACD&lt;/math&gt;, we have <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 1^2 &amp;= a^2 + (2a)^2 - 2(a)(2a)\cos ADB \\<br /> 2^2 &amp;= a^2 + (2a)^2 - 2(a)(2a)\cos ADC <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\cos ADC = \cos (180 - ADB) = -\cos ADB&lt;/math&gt;, we can add these two equations and get <br /> <br /> &lt;cmath&gt;5 = 10a^2&lt;/cmath&gt;<br /> <br /> Hence &lt;math&gt;a = \frac{1}{\sqrt{2}}&lt;/math&gt; and &lt;math&gt;BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> == Alternate Solution ==<br /> <br /> From [[Stewart's Theorem]], we have &lt;math&gt;(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.&lt;/math&gt; Simplifying, we get &lt;math&gt;(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2002|ab=B|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_23&diff=33143 2002 AMC 12B Problems/Problem 23 2010-01-04T03:07:26Z <p>Limac: A new solution.</p> <hr /> <div>== Problem ==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, we have &lt;math&gt;AB = 1&lt;/math&gt; and &lt;math&gt;AC = 2&lt;/math&gt;. Side &lt;math&gt;\overline{BC}&lt;/math&gt; and the [[Median of a triangle|median]] from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt; have the same length. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{1+\sqrt{2}}{2}<br /> \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2<br /> \qquad\mathrm{(C)}\ \sqrt{2}<br /> \qquad\mathrm{(D)}\ \frac 32<br /> \qquad\mathrm{(E)}\ \sqrt{3}&lt;/math&gt;<br /> == Solution ==<br /> [[Image:2002_12B_AMC-23.png]]<br /> <br /> Let &lt;math&gt;D&lt;/math&gt; be the foot of the median from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt;, and we let &lt;math&gt;AD = BC = 2a&lt;/math&gt;. Then by the [[Law of Cosines]] on &lt;math&gt;\triangle ABD, \triangle ACD&lt;/math&gt;, we have <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 1^2 &amp;= a^2 + (2a)^2 - 2(a)(2a)\cos ADB \\<br /> 2^2 &amp;= a^2 + (2a)^2 - 2(a)(2a)\cos ADC <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\cos ADC = \cos (180 - ADB) = -\cos ADB&lt;/math&gt;, we can add these two equations and get <br /> <br /> &lt;cmath&gt;5 = 10a^2&lt;/cmath&gt;<br /> <br /> Hence &lt;math&gt;a = \frac{1}{\sqrt{2}}&lt;/math&gt; and &lt;math&gt;BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> == Alternate Solution ==<br /> <br /> From [[Stewart's Theorem]], we have &lt;math&gt;(2)(1/2a)(2) + (1)(1/2a)(1) = (a)(a)(a) + (1/2a)a(1/2a).&lt;/math&gt; Simplifying, we get &lt;math&gt;(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2002|ab=B|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_10&diff=33077 2005 AMC 10A Problems/Problem 10 2009-12-21T20:57:07Z <p>Limac: /* Alternate Solution */</p> <hr /> <div>==Problem==<br /> There are two values of &lt;math&gt;a&lt;/math&gt; for which the equation &lt;math&gt; 4x^2 + ax + 8x + 9 = 0 &lt;/math&gt; has only one solution for &lt;math&gt;x&lt;/math&gt;. What is the sum of those values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 &lt;/math&gt;<br /> <br /> ==Solution==<br /> A [[quadratic equation]] has exactly one [[root]] if and only if it is a [[perfect square]]. So set<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = (mx + n)^2&lt;/math&gt;<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2&lt;/math&gt;<br /> <br /> Two [[polynomial]]s are equal only if their [[coefficient]]s are equal, so we must have<br /> <br /> &lt;math&gt;m^2 = 4, n^2 = 9&lt;/math&gt;<br /> <br /> &lt;math&gt;m = \pm 2, n = \pm 3&lt;/math&gt;<br /> <br /> &lt;math&gt;a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12&lt;/math&gt;<br /> <br /> &lt;math&gt;a = 4&lt;/math&gt; or &lt;math&gt;a = -20&lt;/math&gt;.<br /> <br /> So the desired sum is &lt;math&gt; (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} &lt;/math&gt;<br /> <br /> <br /> <br /> Alternatively, note that whatever the two values of &lt;math&gt;a&lt;/math&gt; are, they must lead to equations of the form &lt;math&gt;px^2 + qx + r =0&lt;/math&gt; and &lt;math&gt;px^2 - qx + r = 0&lt;/math&gt;. So the two choices of &lt;math&gt;a&lt;/math&gt; must make &lt;math&gt;a_1 + 8 = q&lt;/math&gt; and &lt;math&gt;a_2 + 8 = -q&lt;/math&gt; so &lt;math&gt;a_1 + a_2 + 16 = 0&lt;/math&gt; and &lt;math&gt;a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Alternate Solution==<br /> <br /> Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have <br /> &lt;cmath&gt; (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 144. &lt;/cmath&gt; We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the &lt;math&gt;\pm&lt;/math&gt; sign when added). So we must have <br /> &lt;cmath&gt; \frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. &lt;/cmath&gt;<br /> Therefore, we have &lt;math&gt;(-16)(2)/2 = -16 \implies \boxed{A}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> *[[2005 AMC 10A Problems]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 9|Previous Problem]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 11|Next Problem]]<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_10&diff=33076 2005 AMC 10A Problems/Problem 10 2009-12-21T20:56:43Z <p>Limac: /* Alternate Solution */ Another way to do this problem</p> <hr /> <div>==Problem==<br /> There are two values of &lt;math&gt;a&lt;/math&gt; for which the equation &lt;math&gt; 4x^2 + ax + 8x + 9 = 0 &lt;/math&gt; has only one solution for &lt;math&gt;x&lt;/math&gt;. What is the sum of those values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 &lt;/math&gt;<br /> <br /> ==Solution==<br /> A [[quadratic equation]] has exactly one [[root]] if and only if it is a [[perfect square]]. So set<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = (mx + n)^2&lt;/math&gt;<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2&lt;/math&gt;<br /> <br /> Two [[polynomial]]s are equal only if their [[coefficient]]s are equal, so we must have<br /> <br /> &lt;math&gt;m^2 = 4, n^2 = 9&lt;/math&gt;<br /> <br /> &lt;math&gt;m = \pm 2, n = \pm 3&lt;/math&gt;<br /> <br /> &lt;math&gt;a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12&lt;/math&gt;<br /> <br /> &lt;math&gt;a = 4&lt;/math&gt; or &lt;math&gt;a = -20&lt;/math&gt;.<br /> <br /> So the desired sum is &lt;math&gt; (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} &lt;/math&gt;<br /> <br /> <br /> <br /> Alternatively, note that whatever the two values of &lt;math&gt;a&lt;/math&gt; are, they must lead to equations of the form &lt;math&gt;px^2 + qx + r =0&lt;/math&gt; and &lt;math&gt;px^2 - qx + r = 0&lt;/math&gt;. So the two choices of &lt;math&gt;a&lt;/math&gt; must make &lt;math&gt;a_1 + 8 = q&lt;/math&gt; and &lt;math&gt;a_2 + 8 = -q&lt;/math&gt; so &lt;math&gt;a_1 + a_2 + 16 = 0&lt;/math&gt; and &lt;math&gt;a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Alternate Solution==<br /> <br /> Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have <br /> &lt;cmath&gt; (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 144. &lt;/cmath&gt; We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the &lt;math&gt;\pm&lt;/math&gt; sign when added). So we must have <br /> &lt;cmath&gt; \frac{-16 \pm \sqrt{\text{something}}}{2} + \frac{-16 \pm \sqrt{\text{something}}}{2}. &lt;/cmath&gt;<br /> Therefore, we have &lt;math&gt;(-16)(2)/2 = -16 \implies \boxed{A}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> *[[2005 AMC 10A Problems]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 9|Previous Problem]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 11|Next Problem]]<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_10&diff=33075 2005 AMC 10A Problems/Problem 10 2009-12-21T20:55:31Z <p>Limac: /* Solution */</p> <hr /> <div>==Problem==<br /> There are two values of &lt;math&gt;a&lt;/math&gt; for which the equation &lt;math&gt; 4x^2 + ax + 8x + 9 = 0 &lt;/math&gt; has only one solution for &lt;math&gt;x&lt;/math&gt;. What is the sum of those values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 &lt;/math&gt;<br /> <br /> ==Solution==<br /> A [[quadratic equation]] has exactly one [[root]] if and only if it is a [[perfect square]]. So set<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = (mx + n)^2&lt;/math&gt;<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2&lt;/math&gt;<br /> <br /> Two [[polynomial]]s are equal only if their [[coefficient]]s are equal, so we must have<br /> <br /> &lt;math&gt;m^2 = 4, n^2 = 9&lt;/math&gt;<br /> <br /> &lt;math&gt;m = \pm 2, n = \pm 3&lt;/math&gt;<br /> <br /> &lt;math&gt;a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12&lt;/math&gt;<br /> <br /> &lt;math&gt;a = 4&lt;/math&gt; or &lt;math&gt;a = -20&lt;/math&gt;.<br /> <br /> So the desired sum is &lt;math&gt; (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} &lt;/math&gt;<br /> <br /> <br /> <br /> Alternatively, note that whatever the two values of &lt;math&gt;a&lt;/math&gt; are, they must lead to equations of the form &lt;math&gt;px^2 + qx + r =0&lt;/math&gt; and &lt;math&gt;px^2 - qx + r = 0&lt;/math&gt;. So the two choices of &lt;math&gt;a&lt;/math&gt; must make &lt;math&gt;a_1 + 8 = q&lt;/math&gt; and &lt;math&gt;a_2 + 8 = -q&lt;/math&gt; so &lt;math&gt;a_1 + a_2 + 16 = 0&lt;/math&gt; and &lt;math&gt;a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Alternate Solution==<br /> <br /> Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have <br /> &lt;cmath&gt; (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 144. &lt;/cmath&gt; We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the &lt;math&gt;\pm&lt;/math&gt; sign when added). So we must have <br /> &lt;cmath&gt; \frac{-16 \pm \sqrt{\text{something}}{2} + \frac{-16 \pm \sqrt{something}}{2}. &lt;/cmath&gt;<br /> Therefore, we have &lt;math&gt;(-16)(2)/2 = -16 \implies \boxed{A}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> *[[2005 AMC 10A Problems]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 9|Previous Problem]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 11|Next Problem]]<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_1&diff=33074 2003 AMC 12A Problems/Problem 1 2009-12-21T18:40:06Z <p>Limac: /* Problem */</p> <hr /> <div>== Problem ==<br /> What is the Difference between the sum of the first &lt;math&gt;2003&lt;/math&gt; even counting numbers and the sum of the first &lt;math&gt;2003&lt;/math&gt; odd counting numbers? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 &lt;/math&gt;<br /> <br /> == Solution ==<br /> The first &lt;math&gt;2003&lt;/math&gt; even counting numbers are &lt;math&gt;2,4,6,...,4006&lt;/math&gt;. <br /> <br /> The first &lt;math&gt;2003&lt;/math&gt; odd counting numbers are &lt;math&gt;1,3,5,...,4005&lt;/math&gt;. <br /> <br /> Thus, the problem is asking for the value of &lt;math&gt;(2+4+6+...+4006)-(1+3+5+...+4005)&lt;/math&gt;. <br /> <br /> &lt;math&gt;(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005) &lt;/math&gt; <br /> <br /> &lt;math&gt;= 1+1+1+...+1 = 2003 \Rightarrow D&lt;/math&gt;<br /> <br /> == See Also ==<br /> *[[2003 AMC 12A Problems]]<br /> <br /> *[[2003 AMC 12A/Problem 2|Next Problem]]<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=British_Flag_Theorem&diff=32890 British Flag Theorem 2009-10-30T23:55:47Z <p>Limac: </p> <hr /> <div>The '''British flag theorem''' says that if a point P is chosen inside [[rectangle]] ABCD then &lt;math&gt;AP^{2}+PC^{2}=BP^{2}+DP^{2}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> pair A,B,C,D,P;<br /> A=(0,0);<br /> B=(200,0);<br /> C=(200,150);<br /> D=(0,150);<br /> P=(124,85);<br /> draw(A--B--C--D--cycle);<br /> label(&quot;A&quot;,A,(-1,0));<br /> dot(A);<br /> label(&quot;B&quot;,B,(0,-1));<br /> dot(B);<br /> label(&quot;C&quot;,C,(1,0));<br /> dot(C);<br /> label(&quot;D&quot;,D,(0,1));<br /> dot(D);<br /> dot(P);<br /> label(&quot;P&quot;,P,(1,1));<br /> draw((0,85)--(200,85));<br /> draw((124,0)--(124,150));<br /> label(&quot;$w$&quot;,(124,0),(0,-1));<br /> label(&quot;$x$&quot;,(200,85),(1,0));<br /> label(&quot;$y$&quot;,(124,150),(0,1));<br /> label(&quot;$z$&quot;,(0,85),(-1,0));<br /> dot((124,0));<br /> dot((200,85));<br /> dot((124,150));<br /> dot((0,85));<br /> &lt;/asy&gt;<br /> <br /> The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case.<br /> <br /> == Proof ==<br /> <br /> In Figure 1, by the [[Pythagorean theorem]], we have:<br /> <br /> * &lt;math&gt;AP^{2} = Aw^{2} + Az^{2}&lt;/math&gt;<br /> * &lt;math&gt;PC^{2} = wB^{2} + zD^{2}&lt;/math&gt;<br /> * &lt;math&gt;BP^{2} = wB^{2} + Az^{2}&lt;/math&gt;<br /> * &lt;math&gt;PD^{2} = zD^{2} + Aw^{2}&lt;/math&gt;<br /> <br /> Therefore:<br /> <br /> *&lt;math&gt;AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} =\nolinebreak BP^{2} +\nolinebreak PD^{2}&lt;/math&gt;<br /> <br /> <br /> [[Category:geometry]]<br /> <br /> [[Category:Theorems]]<br /> <br /> {{stub}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems&diff=32154 2008 AMC 10A Problems 2009-06-20T19:09:00Z <p>Limac: </p> <hr /> <div>==Problem 1==<br /> A bakery owner turns on his doughnut machine at &lt;math&gt;\text{8:30}\ {\small\text{AM}}&lt;/math&gt;. At &lt;math&gt;\text{11:10}\ {\small\text{AM}}&lt;/math&gt; the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \text{1:50}\ {\small\text{PM}}\qquad\mathrm{(B)}\ \text{3:00}\ {\small\text{PM}}\qquad\mathrm{(C)}\ \text{3:30}\ {\small\text{PM}}\qquad\mathrm{(D)}\ \text{4:30}\ {\small\text{PM}}\qquad\mathrm{(E)}\ \text{5:50}\ {\small\text{PM}}&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is &lt;math&gt;2:1&lt;/math&gt;. The ratio of the rectangle's length to its width is &lt;math&gt;2:1&lt;/math&gt;. What percent of the rectangle's area is inside the square?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12.5\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 75\qquad\mathrm{(E)}\ 87.5&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> For the positive integer &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;&lt;n&gt;&lt;/math&gt; denote the sum of all the positive divisors of &lt;math&gt;n&lt;/math&gt; with the exception of &lt;math&gt;n&lt;/math&gt; itself. For example, &lt;math&gt;&lt;4&gt;=1+2=3&lt;/math&gt; and &lt;math&gt;&lt;12&gt;=1+2+3+4+6=16&lt;/math&gt;. What is &lt;math&gt;&lt;&lt;&lt;6&gt;&gt;&gt;&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 24\qquad\mathrm{(D)}\ 32\qquad\mathrm{(E)}\ 36&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Suppose that &lt;math&gt;\tfrac{2}{3}&lt;/math&gt; of &lt;math&gt;10&lt;/math&gt; bananas are worth as much as &lt;math&gt;8&lt;/math&gt; oranges. How many oranges are worth as much as &lt;math&gt;\tfrac{1}{2}&lt;/math&gt; of &lt;math&gt;5&lt;/math&gt; bananas?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ \frac{5}{2}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{7}{2}\qquad\mathrm{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Which of the following is equal to the product<br /> &lt;cmath&gt;\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdot\cdots\cdot\frac{4n+4}{4n}\cdot\cdots\cdot\frac{2008}{2004}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 251\qquad\mathrm{(B)}\ 502\qquad\mathrm{(C)}\ 1004\qquad\mathrm{(D)}\ 2008\qquad\mathrm{(E)}\ 4016&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of 3 kilometers per hour, bikes at a rate of 20 kilometers per hour, and runs at a rate of 10 kilometers per hour. Which of the following is closest to the triathlete's average speed, in kilometers per hour, for the entire race?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 5\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The fraction<br /> &lt;cmath&gt;\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}&lt;/cmath&gt;<br /> simplifies to which of the following?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Heather compares the price of a new computer at two different stores. Store &lt;math&gt;A&lt;/math&gt; offers &lt;math&gt;15\%&lt;/math&gt; off the sticker price followed by a &lt;math&gt;\&lt;/math&gt;&lt;math&gt;90&lt;/math&gt; rebate, and store &lt;math&gt;B&lt;/math&gt; offers &lt;math&gt;25\%&lt;/math&gt; off the same sticker price with no rebate. Heather saves &lt;math&gt;\&lt;/math&gt;&lt;math&gt;15&lt;/math&gt; by buying the computer at store &lt;math&gt;A&lt;/math&gt; instead of store &lt;math&gt;B&lt;/math&gt;. What is the sticker price of the computer, in dollars?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Suppose that<br /> &lt;cmath&gt;\frac{2x}{3}-\frac{x}{6}&lt;/cmath&gt;<br /> is an integer. Which of the following statements must be true about &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \text{It is negative.}\qquad\mathrm{(B)}\ \text{It is even, but not necessarily a multiple of 3.}\\\qquad\mathrm{(C)}\ \text{It is a multiple of 3, but not necessarily even.}\\\qquad\mathrm{(D)}\ \text{It is a multiple of 6, but not necessarily a multiple of 12.}\\\qquad\mathrm{(E)}\ \text{It is a multiple of 12.}&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Each of the sides of a square &lt;math&gt;S_1&lt;/math&gt; with area &lt;math&gt;16&lt;/math&gt; is bisected, and a smaller square &lt;math&gt;S_2&lt;/math&gt; is constructed using the bisection points as vertices. The same process is carried out on &lt;math&gt;S_2&lt;/math&gt; to construct an even smaller square &lt;math&gt;S_3&lt;/math&gt;. What is the area of &lt;math&gt;S_3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing toward the shore at a constant rate of 4 miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> In a collection of red, blue, and green marbles, there are &lt;math&gt;25\%&lt;/math&gt; more red marbles than blue marbles, and there are &lt;math&gt;60\%&lt;/math&gt; more green marbles than red marbles. Suppose that there are &lt;math&gt;r&lt;/math&gt; red marbles. What is the total number of marbles in the collection?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2.85r\qquad\mathrm{(B)}\ 3r\qquad\mathrm{(C)}\ 3.4r\qquad\mathrm{(D)}\ 3.85r\qquad\mathrm{(E)}\ 4.25r&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Doug can paint a room in &lt;math&gt;5&lt;/math&gt; hours. Dave can paint the same room in &lt;math&gt;7&lt;/math&gt; hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let &lt;math&gt;t&lt;/math&gt; be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by &lt;math&gt;t&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t+1\right)=1\qquad\mathrm{(B)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t+1=1\qquad\mathrm{(C)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t=1\\\mathrm{(D)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1\qquad\mathrm{(E)}\ \left(5+7\right)t=1&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Older television screens have an aspect ratio of &lt;math&gt;4: 3&lt;/math&gt;. That is, the ratio of the width to the height is &lt;math&gt;4: 3&lt;/math&gt;. The aspect ratio of many movies is not &lt;math&gt;4: 3&lt;/math&gt;, so they are sometimes shown on a television screen by &quot;letterboxing&quot; - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of &lt;math&gt;2: 1&lt;/math&gt; and is shown on an older television screen with a &lt;math&gt;27&lt;/math&gt;-inch diagonal. What is the height, in inches, of each darkened strip?<br /> &lt;asy&gt;unitsize(1mm);<br /> filldraw((0,0)--(21.6,0)--(21.6,2.7)--(0,2.7)--cycle,grey,black);<br /> filldraw((0,13.5)--(21.6,13.5)--(21.6,16.2)--(0,16.2)--cycle,grey,black);<br /> draw((0,0)--(21.6,0)--(21.6,16.2)--(0,16.2)--cycle);&lt;/asy&gt;<br /> &lt;math&gt;\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 2.25\qquad\mathrm{(C)}\ 2.5\qquad\mathrm{(D)}\ 2.7\qquad\mathrm{(E)}\ 3&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; lie on a [[circle]] centered at &lt;math&gt;O&lt;/math&gt;, and &lt;math&gt;\angle AOB = 60^\circ&lt;/math&gt;. A second circle is internally [[tangent]] to the first and tangent to both &lt;math&gt;\overline{OA}&lt;/math&gt; and &lt;math&gt;\overline{OB}&lt;/math&gt;. What is the ratio of the area of the smaller circle to that of the larger circle?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \frac{1}{9}\qquad\mathrm{(C)}\ \frac{1}{8}\qquad\mathrm{(D)}\ \frac{1}{6}\qquad\mathrm{(E)}\ \frac{1}{4}&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> An equilateral triangle has side length 6. What is the area of the region containing all points that are outside the triangle but not more than 3 units from a point of the triangle?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\right)^2\pi\\\mathrm{(E)}\ 9\left(\sqrt{3}{+1\right)^2\pi&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> A right triangle has perimeter 32 and area 20. What is the length of its hypotenuse?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{57}{4}\qquad\mathrm{(B)}\ \frac{59}{4}\qquad\mathrm{(C)}\ \frac{61}{4}\qquad\mathrm{(D)}\ \frac{63}{4}\qquad\mathrm{(E)}\ \frac{65}{4}&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> Rectangle &lt;math&gt;PQRS&lt;/math&gt; lies in a plane with &lt;math&gt;PQ=RS=2&lt;/math&gt; and &lt;math&gt;QR=SP=6&lt;/math&gt;. The rectangle is rotated &lt;math&gt;90^\circ&lt;/math&gt; clockwise about &lt;math&gt;R&lt;/math&gt;, then rotated &lt;math&gt;90^\circ&lt;/math&gt; clockwise about the point &lt;math&gt;S&lt;/math&gt; moved to after the first rotation. What is the length of the path traveled by point &lt;math&gt;P&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \left(2\sqrt{3}+\sqrt{5}\right)\pi\qquad\mathrm{(B)}\ 6\pi\qquad\mathrm{(C)}\ \left(3+\sqrt{10}\right)\pi\qquad\mathrm{(D)}\ \left(\sqrt{3}+2\sqrt{5}\right)\pi\\\mathrm{(E)}\ 2\sqrt{10}\pi&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Trapezoid &lt;math&gt;ABCD&lt;/math&gt; has bases &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; and diagonals intersecting at &lt;math&gt;K&lt;/math&gt;. Suppose that &lt;math&gt;AB = 9&lt;/math&gt;, &lt;math&gt;DC = 12&lt;/math&gt;, and the area of &lt;math&gt;\triangle AKD&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt;. What is the area of trapezoid &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A cube with side length &lt;math&gt;1&lt;/math&gt; is sliced by a plane that passes through two diagonally opposite vertices &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; and the midpoints &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; of two opposite edges not containing &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;C&lt;/math&gt;, as shown. What is the area of quadrilateral &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> unitsize(3cm);<br /> defaultpen(fontsize(8)+linewidth(0.7));<br /> currentprojection=obliqueX;<br /> <br /> draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype(&quot;4 4&quot;));<br /> draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1));<br /> draw((0.5,1,0)--(0,1,0)--(0,1,1));<br /> dot((0.5,0,0));<br /> label(&quot;$A$&quot;,(0.5,0,0),WSW);<br /> dot((0,1,1));<br /> label(&quot;$C$&quot;,(0,1,1),NE);<br /> dot((0.5,1,0.5));<br /> label(&quot;$D$&quot;,(0.5,1,0.5),ESE);<br /> dot((0,0,0.5));<br /> label(&quot;$B$&quot;,(0,0,0.5),NW);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{\sqrt{6}}{2}\qquad\mathrm{(B)}\ \frac{5}{4}\qquad\mathrm{(C)}\ \sqrt{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be 6. To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts 1. If it comes up tails, he takes half of the previous term and subtracts 1. What is the probability that the fourth term in Jacob's sequence is an integer?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{1}{6}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{1}{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Two subsets of the set &lt;math&gt;S=\lbrace a,b,c,d,e\rbrace&lt;/math&gt; are to be chosen so that their union is &lt;math&gt;S&lt;/math&gt; and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> Let &lt;math&gt;k={2008}^{2}+{2}^{2008}&lt;/math&gt;. What is the units digit of &lt;math&gt;k^2+2^k&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> A round table has radius &lt;math&gt;4&lt;/math&gt;. Six rectangular place mats are placed on the table. Each place mat has width &lt;math&gt;1&lt;/math&gt; and length &lt;math&gt;x&lt;/math&gt; as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length &lt;math&gt;x&lt;/math&gt;. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;asy&gt;unitsize(4mm);<br /> defaultpen(linewidth(.8)+fontsize(8));<br /> draw(Circle((0,0),4));<br /> path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br /> draw(mat);<br /> draw(rotate(60)*mat);<br /> draw(rotate(120)*mat);<br /> draw(rotate(180)*mat);<br /> draw(rotate(240)*mat);<br /> draw(rotate(300)*mat);<br /> label(&quot;$$x$$&quot;,(-1.55,2.1),E);<br /> label(&quot;$$1$$&quot;,(-0.5,3.8),S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> [[2008 AMC 10A Problems/Problem 25|Solution]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_5&diff=31752 2008 AMC 10B Problems/Problem 5 2009-05-19T20:28:12Z <p>Limac: </p> <hr /> <div>==Problem==<br /> For [[real number]]s &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, define &lt;math&gt;a * b=(a-b)^2&lt;/math&gt;. What is &lt;math&gt;(x-y)^2*(y-x)^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ x^2+y^2\qquad\mathrm{(C)}\ 2x^2\qquad\mathrm{(D)}\ 2y^2\qquad\mathrm{(E)}\ 4xy&lt;/math&gt;<br /> <br /> ==Solution==<br /> Since &lt;math&gt;(-a)^2 = a^2&lt;/math&gt;, it follows that &lt;math&gt;(x-y)^2 = (y-x)^2&lt;/math&gt;, and <br /> &lt;cmath&gt;(x-y)^2 * (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.&lt;/cmath&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2008|ab=B|num-b=4|num-a=6}}<br /> <br /> [[Category:Articles with dollar signs]]<br /> [[Category:Introductory Algebra Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_5&diff=31751 2008 AMC 10B Problems/Problem 5 2009-05-19T20:27:46Z <p>Limac: </p> <hr /> <div>==Problem==<br /> For [[real number]]s &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, define &lt;math&gt;a * b=(a-b)^2&lt;/math&gt;. What is &lt;math&gt;(x-y)^2 (y-x)^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ x^2+y^2\qquad\mathrm{(C)}\ 2x^2\qquad\mathrm{(D)}\ 2y^2\qquad\mathrm{(E)}\ 4xy&lt;/math&gt;<br /> <br /> ==Solution==<br /> Since &lt;math&gt;(-a)^2 = a^2&lt;/math&gt;, it follows that &lt;math&gt;(x-y)^2 = (y-x)^2&lt;/math&gt;, and <br /> &lt;cmath&gt;(x-y)^2 * (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.&lt;/cmath&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2008|ab=B|num-b=4|num-a=6}}<br /> <br /> [[Category:Articles with dollar signs]]<br /> [[Category:Introductory Algebra Problems]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=AMC_10&diff=31240 AMC 10 2009-04-12T15:52:08Z <p>Limac: </p> <hr /> <div>The '''American Mathematics Contest 10''' ('''AMC 10'''), along with the [[AMC 12]], is one of the first exams in the series of exams used to challenge bright students, grades 10 and below, on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO).<br /> <br /> High scoring AMC 10 and AMC 12 students are invited to take the [[American Invitational Mathematics Examination]] (AIME).<br /> <br /> The AMC 10 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC.<br /> <br /> <br /> == Format ==<br /> <br /> The AMC 10 is a 25 question, 75 minute multiple choice test. Problems generally increase in difficulty as the exam progresses. Calculators were permitted; however, as of 2008, calculators are not allowed any more.<br /> <br /> The AMC 10 is scored in a way that penalizes guesses. Correct answers are worth 6 points, incorrect questions are worth 0 points, and unanswered answers are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, unanswered questions were awarded 2.5 points. In 2006 and 2007, unanswered questions were awarded 2 points. Students that score over 120 points or are in the top 1% of the AMC 10 contest are invited to take the [[AIME]].[http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html]<br /> <br /> == Curriculum ==<br /> The AMC 10 tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solved by students without any background in calculus.<br /> <br /> <br /> == Resources ==<br /> === Links ===<br /> * [http://www.unl.edu/amc/ AMC homepage], their [http://www.unl.edu/amc/e-exams/e5-amc10/amc10.shtml AMC 10 page], and [http://www.unl.edu/amc/mathclub/index.html practice problems]<br /> * The [[AoPS]] [http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AMC10.php AMC 10 guide] (Warning: Scoring system is out of date) <br /> * [http://www.artofproblemsolving.com/Forum/index.php?f=133 AMC Forum] for discussion of the AMC and problems from AMC exams.<br /> * [[AMC 10 Problems and Solutions | Past AMC 10 exams]].<br /> <br /> === Recommended reading ===<br /> * [http://www.artofproblemsolving.com/Books/AoPS_B_CP_AMC.php Problem and solution books for past AMC exams].<br /> * Introduction to Counting &amp; Probability by Dr. [[David Patrick]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3 Information]<br /> * Introduction to Algebra by [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=200 Information]<br /> * Introduction to Geometry by [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9 Information]<br /> * Introduction to Number Theory by [[Mathew Crawford]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=10 information]<br /> * The Art of Problem Solving Volume I by [[Sandor Lehoczky]] and [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=1 Information].<br /> <br /> === AMC Preparation Classes ===<br /> '''These should be taken if a student is having trouble or wants a more clear and confirmed understanding of concepts.'''<br /> * [[AoPS]] hosts an [http://www.artofproblemsolving.com/Classes/AoPS_C_About.php online school] teaching introductory classes in topics covered by the AMC 10 as well as an AMC 10 preparation class.<br /> * [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AMC 10 and AMC 12. [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jam Schedule]<br /> * [[EPGY]] offers an AMC 10 preparation class.<br /> <br /> == See also ==<br /> * [[Mathematics competitions]]<br /> * [[ARML]]<br /> * [[Mathematics summer programs]]<br /> <br /> <br /> <br /> [[Category:Mathematics competitions]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=AMC_10&diff=31239 AMC 10 2009-04-12T15:51:25Z <p>Limac: </p> <hr /> <div>The '''American Mathematics Contest 10''' ('''AMC 10'''), along with the [[AMC 12]], is one of the first exams in the series of exams used to challenge bright students, grades 10 and below, on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO).<br /> <br /> High scoring AMC 10 and AMC 12 students are invited to take the [[American Invitational Mathematics Examination]] (AIME).<br /> <br /> The AMC 10 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC.<br /> <br /> <br /> == Format ==<br /> <br /> The AMC 10 is a 25 question, 75 minute multiple choice test. Problems generally increase in difficulty as the exam progresses. Calculators were permitted; however, as of 2008, calculators are not allowed any more.<br /> <br /> The AMC 10 is scored in a way that penalizes guesses. Correct answers are worth 6 points, incorrect questions are worth 0 points, and unanswered answers are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, unanswered questions were awarded 2.5 points. In 2006 and 2007, unanswered questions were awarded 2 points. Students that score over 120 points or in the top 1% of the AMC 10 contest are invited to take the [[AIME]].[http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html]<br /> <br /> == Curriculum ==<br /> The AMC 10 tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solved by students without any background in calculus.<br /> <br /> <br /> == Resources ==<br /> === Links ===<br /> * [http://www.unl.edu/amc/ AMC homepage], their [http://www.unl.edu/amc/e-exams/e5-amc10/amc10.shtml AMC 10 page], and [http://www.unl.edu/amc/mathclub/index.html practice problems]<br /> * The [[AoPS]] [http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AMC10.php AMC 10 guide] (Warning: Scoring system is out of date) <br /> * [http://www.artofproblemsolving.com/Forum/index.php?f=133 AMC Forum] for discussion of the AMC and problems from AMC exams.<br /> * [[AMC 10 Problems and Solutions | Past AMC 10 exams]].<br /> <br /> === Recommended reading ===<br /> * [http://www.artofproblemsolving.com/Books/AoPS_B_CP_AMC.php Problem and solution books for past AMC exams].<br /> * Introduction to Counting &amp; Probability by Dr. [[David Patrick]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3 Information]<br /> * Introduction to Algebra by [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=200 Information]<br /> * Introduction to Geometry by [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9 Information]<br /> * Introduction to Number Theory by [[Mathew Crawford]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=10 information]<br /> * The Art of Problem Solving Volume I by [[Sandor Lehoczky]] and [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=1 Information].<br /> <br /> === AMC Preparation Classes ===<br /> '''These should be taken if a student is having trouble or wants a more clear and confirmed understanding of concepts.'''<br /> * [[AoPS]] hosts an [http://www.artofproblemsolving.com/Classes/AoPS_C_About.php online school] teaching introductory classes in topics covered by the AMC 10 as well as an AMC 10 preparation class.<br /> * [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AMC 10 and AMC 12. [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jam Schedule]<br /> * [[EPGY]] offers an AMC 10 preparation class.<br /> <br /> == See also ==<br /> * [[Mathematics competitions]]<br /> * [[ARML]]<br /> * [[Mathematics summer programs]]<br /> <br /> <br /> <br /> [[Category:Mathematics competitions]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems&diff=31237 2009 AMC 10B Problems 2009-04-12T15:31:21Z <p>Limac: </p> <hr /> <div>== Problem 1 ==<br /> <br /> Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars, How many bagels did she buy?<br /> <br /> &lt;math&gt;<br /> \text{(A) } 1<br /> \qquad<br /> \text{(B) } 2<br /> \qquad<br /> \text{(C) } 3<br /> \qquad<br /> \text{(D) } 4<br /> \qquad<br /> \text{(E) } 5<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Which of the following is equal to &lt;math&gt;\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac 14<br /> \qquad<br /> \text{(B) } \frac 13<br /> \qquad<br /> \text{(C) } \frac 12<br /> \qquad<br /> \text{(D) } \frac 23<br /> \qquad<br /> \text{(E) } \frac 34<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> Paula the painter had just enough paint for &lt;math&gt;30&lt;/math&gt; identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for &lt;math&gt;25&lt;/math&gt; rooms. How many cans of paint did she use for the &lt;math&gt;25&lt;/math&gt; rooms?<br /> <br /> &lt;math&gt;<br /> \text{(A) } 10<br /> \qquad<br /> \text{(B) } 12<br /> \qquad<br /> \text{(C) } 15<br /> \qquad<br /> \text{(D) } 18<br /> \qquad<br /> \text{(E) } 25<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths &lt;math&gt;15&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; meters. What fraction of the yard is occupied by the flower beds?<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(2mm);<br /> defaultpen(linewidth(.8pt));<br /> <br /> fill((0,0)--(0,5)--(5,5)--cycle,gray);<br /> fill((25,0)--(25,5)--(20,5)--cycle,gray);<br /> draw((0,0)--(0,5)--(25,5)--(25,0)--cycle);<br /> draw((0,0)--(5,5));<br /> draw((20,5)--(25,0));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac {1}{8}<br /> \qquad<br /> \text{(B) } \frac {1}{6}<br /> \qquad<br /> \text{(C) } \frac {1}{5}<br /> \qquad<br /> \text{(D) } \frac {1}{4}<br /> \qquad<br /> \text{(E) } \frac {1}{3}<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Twenty percent less than 60 is one-third more than what number?<br /> <br /> &lt;math&gt;<br /> \text{(A) } 16<br /> \qquad<br /> \text{(B) } 30<br /> \qquad<br /> \text{(C) } 32<br /> \qquad<br /> \text{(D) } 36<br /> \qquad<br /> \text{(E) } 48<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?<br /> <br /> &lt;math&gt;<br /> \text{(A) } 10<br /> \qquad<br /> \text{(B) } 12<br /> \qquad<br /> \text{(C) } 16<br /> \qquad<br /> \text{(D) } 18<br /> \qquad<br /> \text{(E) } 24<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> By inserting parentheses, it is possible to give the expression<br /> &lt;cmath&gt;2\times3 + 4\times5&lt;/cmath&gt;<br /> several values. How many different values can be obtained?<br /> <br /> &lt;math&gt;<br /> \text{(A) } 2<br /> \qquad<br /> \text{(B) } 3<br /> \qquad<br /> \text{(C) } 4<br /> \qquad<br /> \text{(D) } 5<br /> \qquad<br /> \text{(E) } 6<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> In a certain year the price of gasoline rose by &lt;math&gt;20\%&lt;/math&gt; during January, fell by &lt;math&gt;20\%&lt;/math&gt; during February, rose by &lt;math&gt;25\%&lt;/math&gt; during March, and fell by &lt;math&gt;x\%&lt;/math&gt; during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is &lt;math&gt;x&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) } 12<br /> \qquad<br /> \text{(B) } 17<br /> \qquad<br /> \text{(C) } 20<br /> \qquad<br /> \text{(D) } 25<br /> \qquad<br /> \text{(E) } 35<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> Segment &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; intersect at &lt;math&gt;C&lt;/math&gt;, as shown, &lt;math&gt;AB=BC=CD=CE&lt;/math&gt;, and &lt;math&gt;\angle A = \frac 52 \angle B&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle D&lt;/math&gt;? <br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> dotfactor=4;<br /> <br /> pair C=(0,0), Ep=dir(35), D=dir(-35), B=dir(145);<br /> pair A=intersectionpoints(Circle(B,1),C--(-1*Ep));<br /> pair[] ds={A,B,C,D,Ep};<br /> <br /> dot(ds);<br /> draw(A--Ep--D--B--cycle);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,NW);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$E$&quot;,Ep,E);<br /> label(&quot;$D$&quot;,D,E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) } 52.5<br /> \qquad<br /> \text{(B) } 55<br /> \qquad<br /> \text{(C) } 57.7<br /> \qquad<br /> \text{(D) } 60<br /> \qquad<br /> \text{(E) } 62.5<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> A flagpole is originally &lt;math&gt;5&lt;/math&gt; meters tall. A hurricane snaps the flagpole at a point &lt;math&gt;x&lt;/math&gt; meters above the ground so that the upper part, still attached to the stump, touches the ground &lt;math&gt;1&lt;/math&gt; meter away from the base. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \text{(A) } 2.0<br /> \qquad<br /> \text{(B) } 2.1<br /> \qquad<br /> \text{(C) } 2.2<br /> \qquad<br /> \text{(D) } 2.3<br /> \qquad<br /> \text{(E) } 2.4<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> How many &lt;math&gt;7&lt;/math&gt;-digit palindromes (numbers that read the same backward as forward) can be formed using the digits &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \text{(A) } 6<br /> \qquad<br /> \text{(B) } 12<br /> \qquad<br /> \text{(C) } 24<br /> \qquad<br /> \text{(D) } 36<br /> \qquad<br /> \text{(E) } 48<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> Distinct points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt; lie on a line, with &lt;math&gt;AB=BC=CD=1&lt;/math&gt;. Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie on a second line, parallel to the first, with &lt;math&gt;EF=1&lt;/math&gt;. A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?<br /> <br /> &lt;math&gt;<br /> \text{(A) } 3<br /> \qquad<br /> \text{(B) } 4<br /> \qquad<br /> \text{(C) } 5<br /> \qquad<br /> \text{(D) } 6<br /> \qquad<br /> \text{(E) } 7<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> As shown below, convex pentagon &lt;math&gt;ABCDE&lt;/math&gt; has sides &lt;math&gt;AB=3&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;CD=6&lt;/math&gt;, &lt;math&gt;DE=3&lt;/math&gt;, and &lt;math&gt;EA=7&lt;/math&gt;. The pentagon is originally positioned in the plane with vertex &lt;math&gt;A&lt;/math&gt; at the origin and vertex &lt;math&gt;B&lt;/math&gt; on the positive &lt;math&gt;x&lt;/math&gt;-axis. The pentagon is then rolled clockwise to the right along the &lt;math&gt;x&lt;/math&gt;-axis. Which side will touch the point &lt;math&gt;x=2009&lt;/math&gt; on the &lt;math&gt;x&lt;/math&gt;-axis? <br /> <br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), Ep=7*dir(105), B=3*dir(0);<br /> pair D=Ep+B;<br /> pair C=intersectionpoints(Circle(D,6),Circle(B,4));<br /> pair[] ds={A,B,C,D,Ep};<br /> <br /> dot(ds);<br /> draw(B--C--D--Ep--A);<br /> draw((6,6)..(8,4)..(8,3),EndArrow(3));<br /> xaxis(&quot;$x$&quot;,-8,14,EndArrow(3));<br /> <br /> label(&quot;$E$&quot;,Ep,NW);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$(0,0)=A$&quot;,A,SW);<br /> <br /> label(&quot;$3$&quot;,midpoint(A--B),N);<br /> label(&quot;$4$&quot;,midpoint(B--C),NW);<br /> label(&quot;$6$&quot;,midpoint(C--D),NE);<br /> label(&quot;$3$&quot;,midpoint(D--Ep),S);<br /> label(&quot;$7$&quot;,midpoint(Ep--A),W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) } \overline{AB}<br /> \qquad<br /> \text{(B) } \overline{BC}<br /> \qquad<br /> \text{(C) } \overline{CD}<br /> \qquad<br /> \text{(D) } \overline{DE}<br /> \qquad<br /> \text{(E) } \overline{EA}<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> On Monday, Millie puts a quart of seeds, &lt;math&gt;25\%&lt;/math&gt; of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only &lt;math&gt;25\%&lt;/math&gt; of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \text{Tuesday}<br /> \qquad<br /> \text{(B) } \text{Wednesday}<br /> \qquad<br /> \text{(C) } \text{Thursday}<br /> \qquad<br /> \text{(D) } \text{Friday}<br /> \qquad<br /> \text{(E) } \text{Saturday}<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> When a bucket is two-thirds full of water, the bucket and water weigh &lt;math&gt;a&lt;/math&gt; kilograms. When the bucket is one-half full of water the total weight is &lt;math&gt;b&lt;/math&gt; kilograms. In terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, what is the total weight in kilograms when the bucket is full of water?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac23a + \frac13b<br /> \qquad<br /> \text{(B) } \frac32a - \frac12b<br /> \qquad<br /> \text{(C) } \frac32a + b<br /> \qquad<br /> \text{(D) } \frac32a + 2b<br /> \qquad<br /> \text{(E) } 3a - 2b<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> Points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; lie on a circle centered at &lt;math&gt;O&lt;/math&gt;, each of &lt;math&gt;\overline{BA}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; are tangent to the circle, and &lt;math&gt;\triangle ABC&lt;/math&gt; is equilateral. The circle intersects &lt;math&gt;\overline{BO}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;. What is &lt;math&gt;\frac{BD}{BO}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac {\sqrt2}{3}<br /> \qquad<br /> \text{(B) } \frac {1}{2}<br /> \qquad<br /> \text{(C) } \frac {\sqrt3}{3}<br /> \qquad<br /> \text{(D) } \frac {\sqrt2}{2}<br /> \qquad<br /> \text{(E) } \frac {\sqrt3}{2}<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from &lt;math&gt;(a,0)&lt;/math&gt; to &lt;math&gt;(3,3)&lt;/math&gt;, divides the entire region into two regions of equal area. What is &lt;math&gt;a&lt;/math&gt;?<br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> <br /> fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray);<br /> <br /> xaxis(&quot;$x$&quot;,-0.5,4,EndArrow(HookHead,4));<br /> yaxis(&quot;$y$&quot;,-0.5,4,EndArrow(4));<br /> <br /> draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0));<br /> draw((1,0)--(1,2)--(3,2));<br /> draw((2/3,0)--(3,3));<br /> <br /> label(&quot;$(a,0)$&quot;,(2/3,0),S);<br /> label(&quot;$(3,3)$&quot;,(3,3),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac 12<br /> \qquad<br /> \text{(B) } \frac 35<br /> \qquad<br /> \text{(C) } \frac 23<br /> \qquad<br /> \text{(D) } \frac 34<br /> \qquad<br /> \text{(E) } \frac 45<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=8&lt;/math&gt; and &lt;math&gt;BC=6&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;AB&lt;/math&gt; with &lt;math&gt;\overline{ME}\perp\overline{AC}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AME&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac{65}{8}<br /> \qquad<br /> \text{(B) } \frac{25}{3}<br /> \qquad<br /> \text{(C) } 9<br /> \qquad<br /> \text{(D) } \frac{75}{8}<br /> \qquad<br /> \text{(E) } \frac{85}{8}<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> A particular &lt;math&gt;12&lt;/math&gt;-hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a &lt;math&gt;1&lt;/math&gt;, it mistakenly displays a &lt;math&gt;9&lt;/math&gt;. For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac 12<br /> \qquad<br /> \text{(B) } \frac 58<br /> \qquad<br /> \text{(C) } \frac 34<br /> \qquad<br /> \text{(D) } \frac 56<br /> \qquad<br /> \text{(E) } \frac 9{10}<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has a right angle at &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;AB=1&lt;/math&gt;, and &lt;math&gt;BC=2&lt;/math&gt;. The bisector of &lt;math&gt;\angle BAC&lt;/math&gt; meets &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;. What is &lt;math&gt;BD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,1), B=(0,0), C=(2,0);<br /> pair D=extension(A,bisectorpoint(B,A,C),B,C);<br /> pair[] ds={A,B,C,D};<br /> <br /> dot(ds);<br /> draw(A--B--C--A--D);<br /> <br /> label(&quot;$1$&quot;,midpoint(A--B),W);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$D$&quot;,D,S);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$A$&quot;,A,NW);<br /> draw(rightanglemark(C,B,A,2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac {\sqrt3 - 1}{2}<br /> \qquad<br /> \text{(B) } \frac {\sqrt5 - 1}{2}<br /> \qquad<br /> \text{(C) } \frac {\sqrt5 + 1}{2}<br /> \qquad<br /> \text{(D) } \frac {\sqrt6 + \sqrt2}{2}<br /> \qquad<br /> \text{(E) } 2\sqrt 3 - 1<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> What is the remainder when &lt;math&gt;3^0 + 3^1 + 3^2 + \cdots + 3^{2009}&lt;/math&gt; is divided by 8?<br /> <br /> &lt;math&gt;<br /> \text{(A) } 0<br /> \qquad<br /> \text{(B) } 1<br /> \qquad<br /> \text{(C) } 2<br /> \qquad<br /> \text{(D) } 4<br /> \qquad<br /> \text{(E) } 6<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> A cubical cake with edge length &lt;math&gt;2&lt;/math&gt; inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where &lt;math&gt;M&lt;/math&gt; is the midpoint of a top edge. The piece whose top is triangle &lt;math&gt;B&lt;/math&gt; contains &lt;math&gt;c&lt;/math&gt; cubic inches of cake and &lt;math&gt;s&lt;/math&gt; square inches of icing. What is &lt;math&gt;c+s&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> <br /> draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle);<br /> draw((1,1)--(-1,0));<br /> pair P=foot((1,-1),(1,1),(-1,0));<br /> draw((1,-1)--P);<br /> draw(rightanglemark((-1,0),P,(1,-1),4));<br /> <br /> label(&quot;$M$&quot;,(-1,0),W);<br /> label(&quot;$C$&quot;,(-0.1,-0.3));<br /> label(&quot;$A$&quot;,(-0.4,0.7));<br /> label(&quot;$B$&quot;,(0.7,0.4));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac{24}{5}<br /> \qquad<br /> \text{(B) } \frac{32}{5}<br /> \qquad<br /> \text{(C) } 8+\sqrt5<br /> \qquad<br /> \text{(D) } 5+\frac{16\sqrt5}{5}<br /> \qquad<br /> \text{(E) } 10+5\sqrt5<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac 1{16}<br /> \qquad<br /> \text{(B) } \frac 18<br /> \qquad<br /> \text{(C) } \frac 3{16}<br /> \qquad<br /> \text{(D) } \frac 14<br /> \qquad<br /> \text{(E) } \frac 5{16}<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with &lt;math&gt;9&lt;/math&gt; trapezoids, let &lt;math&gt;x&lt;/math&gt; be the angle measure in degrees of the larger interior angle of the trapezoid. What is &lt;math&gt;x&lt;/math&gt;? <br /> <br /> &lt;asy&gt;<br /> unitsize(4mm);<br /> defaultpen(linewidth(.8pt));<br /> int i;<br /> real r=5, R=6;<br /> <br /> path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0);<br /> for(i=0; i&lt;9; ++i)<br /> {<br /> draw(rotate(20*i)*t);<br /> }<br /> draw((-r,0)--(R+1,0));<br /> draw((-R,0)--(-R-1,0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) } 100<br /> \qquad<br /> \text{(B) } 102<br /> \qquad<br /> \text{(C) } 104<br /> \qquad<br /> \text{(D) } 106<br /> \qquad<br /> \text{(E) } 108<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac 18<br /> \qquad<br /> \text{(B) } \frac {3}{16}<br /> \qquad<br /> \text{(C) } \frac 14<br /> \qquad<br /> \text{(D) } \frac 38<br /> \qquad<br /> \text{(E) } \frac 12<br /> &lt;/math&gt;<br /> <br /> [[2009 AMC 10B Problems/Problem 25|Solution]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_4&diff=31236 2009 AMC 10B Problems/Problem 4 2009-04-12T15:29:48Z <p>Limac: </p> <hr /> <div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #4]] and [[2009 AMC 12B Problems|2009 AMC 12B #4]]}}<br /> <br /> == Problem ==<br /> A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths &lt;math&gt;15&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; meters. What fraction of the yard is occupied by the flower beds?<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(2mm);<br /> defaultpen(linewidth(.8pt));<br /> <br /> fill((0,0)--(0,5)--(5,5)--cycle,gray);<br /> fill((25,0)--(25,5)--(20,5)--cycle,gray);<br /> draw((0,0)--(0,5)--(25,5)--(25,0)--cycle);<br /> draw((0,0)--(5,5));<br /> draw((20,5)--(25,0));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\frac {1}{8}\qquad<br /> \mathrm{(B)}\frac {1}{6}\qquad<br /> \mathrm{(C)}\frac {1}{5}\qquad<br /> \mathrm{(D)}\frac {1}{4}\qquad<br /> \mathrm{(E)}\frac {1}{3}&lt;/math&gt;<br /> <br /> == Solution ==<br /> Each triangle has leg length &lt;math&gt;\frac 12 \cdot (25 - 15) = 5&lt;/math&gt; meters and area &lt;math&gt;\frac 12 \cdot 5^2 = \frac {25}{2}&lt;/math&gt; square meters. Thus the flower beds have a total area of 25 square meters. The entire yard has length 25 m and width 5 m, so its area is 125 square meters. The fraction of the yard occupied by the flower beds is &lt;math&gt;\frac {25}{125} = \boxed{\frac15}&lt;/math&gt;. The answer is &lt;math&gt;\mathrm{(C)}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2009|ab=B|num-b=3|num-a=5}}<br /> {{AMC12 box|year=2009|ab=B|num-b=3|num-a=5}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems&diff=31198 2000 AMC 8 Problems 2009-04-10T14:37:10Z <p>Limac: </p> <hr /> <div>==Problem 1==<br /> Aunt Anna is &lt;math&gt;42&lt;/math&gt; years old. Caitlin is &lt;math&gt;5&lt;/math&gt; years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 15<br /> \qquad<br /> \mathrm{(B)}\ 16<br /> \qquad<br /> \mathrm{(C)}\ 17<br /> \qquad<br /> \mathrm{(D)}\ 21<br /> \qquad<br /> \mathrm{(E)}\ 37<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of these numbers is less than its reciprocal?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ -2<br /> \qquad<br /> \mathrm{(B)}\ -1<br /> \qquad<br /> \mathrm{(C)}\ 0<br /> \qquad<br /> \mathrm{(D)}\ 1<br /> \qquad<br /> \mathrm{(E)}\ 2<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> How many whole numbers lie in the interval between &lt;math&gt;\frac{5}{3}&lt;/math&gt; and &lt;math&gt;2\pi?&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 2<br /> \qquad<br /> \mathrm{(B)}\ 3<br /> \qquad<br /> \mathrm{(C)}\ 4<br /> \qquad<br /> \mathrm{(D)}\ 5<br /> \qquad<br /> \mathrm{(E)}\ infinitely\ many<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> In &lt;math&gt;1960&lt;/math&gt; only &lt;math&gt;5\%&lt;/math&gt; of the working adults in Carlin City worked at home. By &lt;math&gt;1970&lt;/math&gt; the &quot;at-home&quot; work force had increased to &lt;math&gt;8\%&lt;/math&gt;. In &lt;math&gt;1980&lt;/math&gt; there were approximately &lt;math&gt;15\%&lt;/math&gt; working at home, and in &lt;math&gt;1990&lt;/math&gt; there were &lt;math&gt;30\%&lt;/math&gt;. The graph that best illustrates this is:<br /> <br /> [[2000 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Each principal of Lincoln High School serves exactly one &lt;math&gt;3&lt;/math&gt;-year term. What is the maximum number of principals this school could have during an &lt;math&gt;8&lt;/math&gt;-year period?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 2<br /> \qquad<br /> \mathrm{(B)}\ 3<br /> \qquad<br /> \mathrm{(C)}\ 4<br /> \qquad<br /> \mathrm{(D)}\ 5<br /> \qquad<br /> \mathrm{(E)}\ 8<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Figure &lt;math&gt;ABCD&lt;/math&gt; is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded &lt;math&gt;L&lt;/math&gt;-shaped region is<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 7<br /> \qquad<br /> \mathrm{(B)}\ 10<br /> \qquad<br /> \mathrm{(C)}\ 12.5<br /> \qquad<br /> \mathrm{(D)}\ 14<br /> \qquad<br /> \mathrm{(E)}\ 15<br /> &lt;/math&gt;<br /> <br /> <br /> [[2000 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> What is the minimum possible product of three different numbers of the set &lt;math&gt;\{-8.-6,-4,0,3,5,7\}?&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ -336<br /> \qquad<br /> \mathrm{(B)}\ -280<br /> \qquad<br /> \mathrm{(C)}\ -210<br /> \qquad<br /> \mathrm{(D)}\ -192<br /> \qquad<br /> \mathrm{(E)}\ 0<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Three dice with faces numbered &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;6&lt;/math&gt; are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots &lt;math&gt;NOT&lt;/math&gt; visible in this view is<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 21<br /> \qquad<br /> \mathrm{(B)}\ 22<br /> \qquad<br /> \mathrm{(C)}\ 31<br /> \qquad<br /> \mathrm{(D)}\ 41<br /> \qquad<br /> \mathrm{(E)}\ 53<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Three-digit powers of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; are used in this &lt;math&gt;cross-number&lt;/math&gt; puzzle. What is the only possible digit for the outlined square?<br /> <br /> &lt;math&gt;ACROSS\ DOWN&lt;/math&gt;<br /> <br /> &lt;math&gt;2)\ 2^m \qquad\ 1)\ 5^n&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 0<br /> \qquad<br /> \mathrm{(B)}\ 2<br /> \qquad<br /> \mathrm{(C)}\ 4<br /> \qquad<br /> \mathrm{(D)}\ 6<br /> \qquad<br /> \mathrm{(E)}\ 8<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Ara and Shea were once the same height. Since then Shea has grown &lt;math&gt;20\%&lt;/math&gt; while Ara has grow half as many inches as Shea. Shea is now &lt;math&gt;60&lt;/math&gt; inches tall. How tall, in inches, is Ara now?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 48<br /> \qquad<br /> \mathrm{(B)}\ 51<br /> \qquad<br /> \mathrm{(C)}\ 52<br /> \qquad<br /> \mathrm{(D)}\ 54<br /> \qquad<br /> \mathrm{(E)}\ 55<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> The number &lt;math&gt;64&lt;/math&gt; has the property that it is divisible by its units digit. How many whole numbers between &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;50&lt;/math&gt; have this property?<br /> <br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 15<br /> \qquad<br /> \mathrm{(B)}\ 16<br /> \qquad<br /> \mathrm{(C)}\ 17<br /> \qquad<br /> \mathrm{(D)}\ 18<br /> \qquad<br /> \mathrm{(E)}\ 20<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> ==Problem 13==<br /> ==Problem 14==<br /> What is the units digit of &lt;math&gt;19^{19} + 99^{99}?&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 0<br /> \qquad<br /> \mathrm{(B)}\ 1<br /> \qquad<br /> \mathrm{(C)}\ 2<br /> \qquad<br /> \mathrm{(D)}\ 8<br /> \qquad<br /> \mathrm{(E)}\ 9<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> ==Problem 16==<br /> In order for Mateen to walk a kilometer &lt;math&gt;(1000m)&lt;/math&gt; in his rectangular backyard, he must walk the length &lt;math&gt;25&lt;/math&gt; times or walk its perimeter &lt;math&gt;10&lt;/math&gt; times. What is the area of Mateen's backyard in square meters?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 40<br /> \qquad<br /> \mathrm{(B)}\ 200<br /> \qquad<br /> \mathrm{(C)}\ 400<br /> \qquad<br /> \mathrm{(D)}\ 500<br /> \qquad<br /> \mathrm{(E)}\ 1000<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 16|Solution]]</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems&diff=31197 2000 AMC 8 Problems 2009-04-10T13:54:45Z <p>Limac: New page: ==Problem 1== Aunt Anna is &lt;math&gt;42&lt;/math&gt; years old. Caitlin is &lt;math&gt;5&lt;/math&gt; years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin? &lt;math&gt; \mathrm{(A)...</p> <hr /> <div>==Problem 1==<br /> Aunt Anna is &lt;math&gt;42&lt;/math&gt; years old. Caitlin is &lt;math&gt;5&lt;/math&gt; years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 15<br /> \qquad<br /> \mathrm{(B)}\ 16<br /> \qquad<br /> \mathrm{(C)}\ 17<br /> \qquad<br /> \mathrm{(D)}\ 21<br /> \qquad<br /> \mathrm{(E)}\ 37<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of these numbers is less than its reciprocal?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ -2<br /> \qquad<br /> \mathrm{(B)}\ -1<br /> \qquad<br /> \mathrm{(C)}\ 0<br /> \qquad<br /> \mathrm{(D)}\ 1<br /> \qquad<br /> \mathrm{(E)}\ 2<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> How many whole numbers lie in the interval between &lt;math&gt;\frac{5}{3}&lt;/math&gt; and &lt;math&gt;2\pi?&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 2<br /> \qquad<br /> \mathrm{(B)}\ 3<br /> \qquad<br /> \mathrm{(C)}\ 4<br /> \qquad<br /> \mathrm{(D)}\ 5<br /> \qquad<br /> \mathrm{(E)}&lt;/math&gt; infinitely many<br /> <br /> [[2000 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> In &lt;math&gt;1960&lt;/math&gt; only &lt;math&gt;5%&lt;/math&gt; of the working adults in Carlin City worked at home. By &lt;math&gt;1970&lt;/math&gt; the &quot;at-home&quot; work force had increased to &lt;math&gt;8%&lt;/math&gt;. In &lt;math&gt;1980&lt;/math&gt; there were approximately &lt;math&gt;15%&lt;/math&gt; working at home, and in &lt;math&gt;1990&lt;/math&gt; there were &lt;math&gt;30%&lt;/math&gt;. The graph that best illustrates this is:<br /> <br /> &lt;math&gt;<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8 year period?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 2<br /> \qquad<br /> \mathrm{(B)}\ 3<br /> \qquad<br /> \mathrm{(C)}\ 4<br /> \qquad<br /> \mathrm{(D)}\ 5<br /> \qquad<br /> \mathrm{(E)}\ 8<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Figure &lt;math&gt;ABCD&lt;/math&gt; is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded L-shaped region is<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 7<br /> \qquad<br /> \mathrm{(B)}\ 10<br /> \qquad<br /> \mathrm{(C)}\ 12.5<br /> \qquad<br /> \mathrm{(D)}\ 14<br /> \qquad<br /> \mathrm{(E)}\ 15<br /> &lt;/math&gt;<br /> <br /> <br /> [[2000 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> What is the minimum possible product of three different numbers of the set &lt;math&gt;{-8.-6,-4,0,3,5,7}?&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ -336<br /> \qquad<br /> \mathrm{(B)}\ -280<br /> \qquad<br /> \mathrm{(C)}\ -210<br /> \qquad<br /> \mathrm{(D)}\ -192<br /> \qquad<br /> \mathrm{(E)}\ 0<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Three dice with faces numbered &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;6&lt;/math&gt; are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots &lt;math&gt;NOT&lt;/math&gt; visible in the view is<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 21<br /> \qquad<br /> \mathrm{(B)}\ 22<br /> \qquad<br /> \mathrm{(C)}\ 31<br /> \qquad<br /> \mathrm{(D)}\ 41<br /> \qquad<br /> \mathrm{(E)}\ 53<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 9==</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_12&diff=30922 2009 AIME I Problems/Problem 12 2009-03-21T20:57:23Z <p>Limac: </p> <hr /> <div>== Problem ==<br /> In right &lt;math&gt;\triangle ABC&lt;/math&gt; with hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;AC = 12&lt;/math&gt;, &lt;math&gt;BC = 35&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt; is the altitude to &lt;math&gt;\overline{AB}&lt;/math&gt;. Let &lt;math&gt;\omega&lt;/math&gt; be the circle having &lt;math&gt;\overline{CD}&lt;/math&gt; as a diameter. Let &lt;math&gt;I&lt;/math&gt; be a point outside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\overline{AI}&lt;/math&gt; and &lt;math&gt;\overline{BI}&lt;/math&gt; are both tangent to circle &lt;math&gt;\omega&lt;/math&gt;. The ratio of the perimeter of &lt;math&gt;\triangle ABI&lt;/math&gt; to the length &lt;math&gt;AB&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac {m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=11|num-a=13}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_12&diff=30921 2009 AIME I Problems/Problem 12 2009-03-21T20:57:12Z <p>Limac: </p> <hr /> <div>== Problem ==<br /> In right &lt;math&gt;\triangle ABC&lt;/math&gt; with hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;AC = 12&lt;/math&gt;, &lt;math&gt;BC = 35&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt; is the altitude to &lt;math&gt;\overline{AB}&lt;/math&gt;. Let &lt;math&gt;\omega&lt;/math&gt; be the circle having &lt;math&gt;\overline{CD}&lt;/math&gt; as a diameter. Let &lt;math&gt;I&lt;/math&gt; be a point outside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\overline{AI}&lt;/math&gt; and &lt;math&gt;\overline{BI}&lt;/math&gt; are both tangent to circle &lt;math&gt;\omega&lt;/math&gt;. The ratio of the perimeter of &lt;math&gt;\triangle ABI&lt;/math&gt; to the length &lt;math&gt;AB&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac {m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> dkfj;lsdaf<br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=11|num-a=13}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=30916 2009 AMC 10A Problems/Problem 5 2009-03-21T14:29:49Z <p>Limac: </p> <hr /> <div>==Problem==<br /> What is the sum of the digits of the square of 111,111,111 ?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution==<br /> Using the standard multiplication algorithm, &lt;math&gt;111111111^2=12345678987654321&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> Or <br /> <br /> Add up all the ones(thus deriving the sum of the number) of &lt;math&gt;111,111,111&lt;/math&gt; gives us &lt;math&gt;1+1+1+\dots+1 = 9&lt;/math&gt; Thus, &lt;math&gt;(9)^2 = 81&lt;/math&gt;<br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}</div> Limac https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_6&diff=30915 2009 AMC 10A Problems/Problem 6 2009-03-21T14:20:06Z <p>Limac: Edited to show how solution was dervied.</p> <hr /> <div>== Problem ==<br /> <br /> A circle of radius &lt;math&gt;2&lt;/math&gt; is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded? <br /> <br /> &lt;asy&gt;<br /> unitsize(6mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> dotfactor=4;<br /> <br /> filldraw(Arc((0,0),4,0,180)--cycle,gray,black);<br /> filldraw(Circle((0,2),2),white,black);<br /> dot((0,2));<br /> draw((0,2)--((0,2)+2*dir(60)));<br /> label(&quot;$2$&quot;,midpoint((0,2)--((0,2)+2*dir(60))),SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ \frac{1}{2}<br /> \qquad<br /> \mathrm{(B)}\ \frac{\pi}{6}<br /> \qquad<br /> \mathrm{(C)}\ \frac{2}{\pi}<br /> \qquad<br /> \mathrm{(D)}\ \frac{2}{3}<br /> \qquad<br /> \mathrm{(E)}\ \frac{3}{\pi}<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> Area of the circle inscribed inside the semicircle &lt;math&gt;= \pi r^2 \Rightarrow \pi(2^2) = 4 \pi .&lt;/math&gt;<br /> Area of the larger circle (semicircle's area x 2)&lt;math&gt;= \pi r^2 \Rightarrow \pi(4^2)= 16 \pi&lt;/math&gt; (4, or the diameter of the inscribed circle is the same thing as the radius of the semicircle). Thus, the area of the semicircle is &lt;math&gt;\frac{1}{2}(16 \pi) \Rightarrow 8 \pi .&lt;/math&gt;<br /> Part of the semicircle that is unshaded is &lt;math&gt;\frac{4 \pi}{8 \pi} = \frac{1}{2}&lt;/math&gt; Therefore, the shaded part is &lt;math&gt;1 - \frac{1}{2} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Thus the answer is &lt;math&gt;\frac{1}{2}\Rightarrow \fbox{A}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=5|num-a=7}}</div> Limac