https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Grn+trtle&feedformat=atom AoPS Wiki - User contributions [en] 2021-05-16T00:42:19Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_20&diff=31050 2004 AMC 10B Problems/Problem 20 2009-03-30T22:19:10Z <p>Grn trtle: </p> <hr /> <div>== Problem ==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;, respectively. If &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt; intersect at &lt;math&gt;T&lt;/math&gt; so that &lt;math&gt;AT/DT=3&lt;/math&gt; and &lt;math&gt;BT/ET=4&lt;/math&gt;, what is &lt;math&gt;CD/BD&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12} &lt;/math&gt;<br /> <br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,2*WNW);<br /> &lt;/asy&gt;<br /> <br /> == Solution (Mass points) ==<br /> <br /> The presence of only ratios in the problem essentially cries out for mass points.<br /> <br /> As per the problem, we assign a mass of &lt;math&gt;1&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;, and a mass of &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;D&lt;/math&gt;. Then, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;T&lt;/math&gt;, &lt;math&gt;T&lt;/math&gt; has a mass of &lt;math&gt;4&lt;/math&gt;.<br /> <br /> Now, were we to assign a mass of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;4&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;, we'd have &lt;math&gt;5T&lt;/math&gt;. Scaling this down by &lt;math&gt;4/5&lt;/math&gt; (to get &lt;math&gt;4T&lt;/math&gt;, which puts &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; in terms of the masses of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;), we assign a mass of &lt;math&gt;\frac{4}{5}&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;\frac{16}{5}&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;.<br /> <br /> Now, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;E&lt;/math&gt;, we must give &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;\frac{16}{5}-1=\frac{11}{5}&lt;/math&gt;.<br /> <br /> Finally, the ratio of &lt;math&gt;CD&lt;/math&gt; to &lt;math&gt;BD&lt;/math&gt; is given by the ratio of the mass of &lt;math&gt;B&lt;/math&gt; to the mass of &lt;math&gt;C&lt;/math&gt;, which is &lt;math&gt;\frac{4}{5}\cdot\frac{5}{11}=\boxed{\textbf{(D)}\ \frac{4}{11}}&lt;/math&gt;.<br /> <br /> == Solution (Coordinates) ==<br /> <br /> Affine transformations preserve ratios of distances, and for any pair of triangles there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any &lt;math&gt;\triangle ABC&lt;/math&gt;, and we just need to compute it for any single triangle.<br /> <br /> We can choose the points &lt;math&gt;A=(-3,0)&lt;/math&gt;, &lt;math&gt;B=(0,4)&lt;/math&gt;, and &lt;math&gt;D=(1,0)&lt;/math&gt;. This way we will have &lt;math&gt;T=(0,0)&lt;/math&gt;, and &lt;math&gt;E=(0,-1)&lt;/math&gt;. The situation is <br /> shown in the picture below:<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,NW);<br /> label(&quot;$3$&quot;,A--T,N);<br /> label(&quot;$4$&quot;,B--T,W);<br /> label(&quot;$1$&quot;,D--T,N);<br /> label(&quot;$1$&quot;,E--T,W);<br /> <br /> &lt;/asy&gt;<br /> <br /> The point &lt;math&gt;C&lt;/math&gt; is the intersection of the lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt;. The points on the first line have the form &lt;math&gt;(t,4-4t)&lt;/math&gt;, the points on the second line have the form &lt;math&gt;(t,-1-t/3)&lt;/math&gt;. Solving for &lt;math&gt;t&lt;/math&gt; we get &lt;math&gt;t=15/11&lt;/math&gt;, hence &lt;math&gt;C=(15/11,-16/11)&lt;/math&gt;.<br /> <br /> The ratio &lt;math&gt;CD/BD&lt;/math&gt; can now be computed simply by observing the &lt;math&gt;x&lt;/math&gt; coordinates of &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\frac 4{11}}<br /> &lt;/cmath&gt;<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=19|num-a=21}}</div> Grn trtle https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_3&diff=28285 2001 AMC 12 Problems/Problem 3 2008-10-29T22:42:35Z <p>Grn trtle: /* Solution */</p> <hr /> <div>== Problem ==<br /> The state income tax where Kristin lives is levied at the rate of &lt;math&gt;p%&lt;/math&gt; of the first<br /> &lt;dollar/&gt;&lt;math&gt;28000&lt;/math&gt; of annual income plus &lt;math&gt;(p + 2)%&lt;/math&gt; of any amount above &lt;dollar/&gt;&lt;math&gt;28000&lt;/math&gt;. Kristin<br /> noticed that the state income tax she paid amounted to &lt;math&gt;(p + 0.25)%&lt;/math&gt; of her<br /> annual income. What was her annual income?<br /> <br /> &lt;math&gt;\text{(A)}\,&lt;/math&gt;&lt;dollar/&gt;&lt;math&gt;28000\qquad \text{(B)}\,&lt;/math&gt;&lt;dollar/&gt;&lt;math&gt;32000\qquad \text{(C)}\,&lt;/math&gt;&lt;dollar/&gt;&lt;math&gt;35000\qquad \text{(D)}\,&lt;/math&gt;&lt;dollar/&gt;&lt;math&gt;42000\qquad \text{(E)}\,&lt;/math&gt;&lt;dollar/&gt;&lt;math&gt;56000&lt;/math&gt;<br /> <br /> == Solution (unverified) ==<br /> Let the income amount be denoted by &lt;math&gt;A&lt;/math&gt;.<br /> <br /> We know that &lt;math&gt;\frac{A(p+.25)}{100}=\frac{28000p}{100}+\frac{(p+2)(A-28000)}{100}&lt;/math&gt;.<br /> <br /> We can now try to solve for &lt;math&gt;A&lt;/math&gt;:<br /> <br /> &lt;math&gt;(p+.25)A=28000p+Ap+2A-28000p-56000&lt;/math&gt;<br /> <br /> &lt;math&gt;.25A=2A-56000&lt;/math&gt;<br /> <br /> &lt;math&gt;A=32000&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2001|num-b=2|num-a=4}}</div> Grn trtle https://artofproblemsolving.com/wiki/index.php?title=1977_Canadian_MO_Problems/Problem_1&diff=27836 1977 Canadian MO Problems/Problem 1 2008-09-06T22:46:15Z <p>Grn trtle: Added another solution that uses simpler algebra.</p> <hr /> <div>== Problem ==<br /> <br /> If &lt;math&gt;f(x)=x^2+x,&lt;/math&gt; prove that the equation &lt;math&gt;4f(a)=f(b)&lt;/math&gt; has no solutions in positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b.&lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> Directly plugging &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; into the function, &lt;math&gt;4a^2+4a=b^2+b.&lt;/math&gt; We now have a quadratic in &lt;math&gt;a.&lt;/math&gt;<br /> <br /> Applying the quadratic formula, &lt;math&gt;a=\frac{-1\pm \sqrt{b^2+b+1}}{2}. &lt;/math&gt;<br /> <br /> In order for both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; to be integers, the [[discriminant]] must be a [[perfect square]]. However, since &lt;math&gt;b^2&lt; b^2+b+1 &lt;(b+1)^2,&lt;/math&gt; the quantity &lt;math&gt;b^2+b+1&lt;/math&gt; cannot be a perfect square when &lt;math&gt;b&lt;/math&gt; is an integer. Hence, when &lt;math&gt;b&lt;/math&gt; is a positive integer, &lt;math&gt;a&lt;/math&gt; cannot be.<br /> <br /> == Alternate Solution ==<br /> Write out the expanded form of &lt;math&gt;4f(a)=f(b)&lt;/math&gt;:<br /> <br /> &lt;math&gt;4a^2+4a=b^2+b&lt;/math&gt;<br /> <br /> Now simply factor it to get: &lt;math&gt;4a(a+1)=b(b+1)&lt;/math&gt;<br /> <br /> Subtract &lt;math&gt;b(b+1)&lt;/math&gt; from both sides: &lt;math&gt;4a(a+1)-b(b+1)=0&lt;/math&gt;<br /> <br /> For the left side to equal &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;4a&lt;/math&gt; or &lt;math&gt;a+1&lt;/math&gt; must be &lt;math&gt;0&lt;/math&gt; AND &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;b+1&lt;/math&gt; must be &lt;math&gt;0&lt;/math&gt;.<br /> <br /> Set each one equal to &lt;math&gt;0&lt;/math&gt; to find the possible solutions:<br /> <br /> &lt;math&gt;4a=0&lt;/math&gt;<br /> <br /> &lt;math&gt;a+1=0&lt;/math&gt;<br /> <br /> &lt;math&gt;b=0&lt;/math&gt;<br /> <br /> &lt;math&gt;b+1=0&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;a&lt;/math&gt; must be &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;. The same applies to &lt;math&gt;b&lt;/math&gt;. None of these solutions are greater than &lt;math&gt;0&lt;/math&gt;.<br /> <br /> &lt;math&gt;\therefore&lt;/math&gt; There are no positive solutions for &lt;math&gt;a&lt;/math&gt; or &lt;math&gt;b&lt;/math&gt;, Q.E.D.<br /> <br /> {{alternate solutions}}<br /> <br /> {{Old CanadaMO box|before=First question|num-a=2|year=1977}}<br /> <br /> [[Category:Olympiad Algebra Problems]]</div> Grn trtle https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems&diff=27812 2000 AMC 12 Problems 2008-09-06T06:42:08Z <p>Grn trtle: /* Problem 21 */</p> <hr /> <div>== Problem 1 ==<br /> <br /> In the year &lt;math&gt;2001&lt;/math&gt;, the United States will host the International Mathematical Olympiad. Let &lt;math&gt;I,M,&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; be distinct positive integers such that the product &lt;math&gt;I \cdot M \cdot O = 2001 &lt;/math&gt;. What is the largest possible value of the sum &lt;math&gt;I + M + O&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 23 } \qquad \mathrm{(B) \ 55 } \qquad \mathrm{(C) \ 99 } \qquad \mathrm{(D) \ 111 } \qquad \mathrm{(E) \ 671 } &lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> &lt;math&gt;2000(2000^{2000}) =&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ 2000^{2001} } \qquad \mathrm{(B) \ 4000^{2000} } \qquad \mathrm{(C) \ 2000^{4000} } \qquad \mathrm{(D) \ 4,000,000^{2000} } \qquad \mathrm{(E) \ 2000^{4,000,000} } &lt;/math&gt;<br /> <br /> <br /> [[2000 AMC 12 Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> Each day, Jenny ate &lt;math&gt;20\%&lt;/math&gt; of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, &lt;math&gt;32&lt;/math&gt; remained. How many jellybeans were in the jar originally?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 40 } \qquad \mathrm{(B) \ 50 } \qquad \mathrm{(C) \ 55 } \qquad \mathrm{(D) \ 60 } \qquad \mathrm{(E) \ 75 } &lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> The Fibonacci sequence &lt;math&gt;1,1,2,3,5,8,13,21,\ldots &lt;/math&gt; starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 4 } \qquad \mathrm{(C) \ 6 } \qquad \mathrm{(D) \ 7 } \qquad \mathrm{(E) \ 9 } &lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> If &lt;math&gt;|x - 2| = p,&lt;/math&gt; where &lt;math&gt;x &lt; 2,&lt;/math&gt; then &lt;math&gt;x - p =&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ -2 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 2-2p } \qquad \mathrm{(D) \ 2p-2 } \qquad \mathrm{(E) \ |2p-2| } &lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> Two different prime numbers between &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } &lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> How many positive integers &lt;math&gt;b&lt;/math&gt; have the property that &lt;math&gt;\log_{b} 729&lt;/math&gt; is a positive integer?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } &lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; non-overlapping squares. If the pattern continued, how many non-overlapping squares would there be in figure &lt;math&gt;100&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text {(A)}10401 \qquad \text {(B)}19801 \qquad \text {(C)} 20201 \qquad \text {(D)} 39801 \qquad \text {(E)}40801&lt;/math&gt;<br /> <br /> [[Image:2000 AHSME number 8.png]]<br /> <br /> [[2000 AMC 12 Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71,76,80,82, and 91. What was the last score Mrs. Walters entered?<br /> <br /> &lt;math&gt;\text{(A)} \ 71 \qquad \text{(B)} \ 76 \qquad \text{(C)} \ 80 \qquad \text{(D)} \ 82 \qquad \text{(E)} \ 91&lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The point &lt;math&gt;P = (1,2,3)&lt;/math&gt; is reflected in the &lt;math&gt;xy&lt;/math&gt;-plane, then its image &lt;math&gt;Q&lt;/math&gt; is rotated by &lt;math&gt;180^\circ&lt;/math&gt; about the &lt;math&gt;x&lt;/math&gt;-axis to produce &lt;math&gt;R&lt;/math&gt;, and finally, &lt;math&gt;R&lt;/math&gt; is translated by 5 units in the positive-&lt;math&gt;y&lt;/math&gt; direction to produce &lt;math&gt;S&lt;/math&gt;. What are the coordinates of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \text {(A) } (1,7, - 3) \qquad \text {(B) } ( - 1,7, - 3) \qquad \text {(C) } ( - 1, - 2,8) \qquad \text {(D) } ( - 1,3,3) \qquad \text {(E) } (1,3,3)<br /> &lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Two non-zero real numbers, &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b,&lt;/math&gt; satisfy &lt;math&gt;ab = a - b&lt;/math&gt;. Which of the following is a possible value of &lt;math&gt;\frac {a}{b} + \frac {b}{a} - ab&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)} \ - 2 \qquad \text{(B)} \ \frac { - 1}{2} \qquad \text{(C)} \ \frac {1}{3} \qquad \text{(D)} \ \frac {1}{2} \qquad \text{(E)} \ 2&lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> Let A, M, and C be nonnegative integers such that &lt;math&gt;A + M + C=12&lt;/math&gt;. What is the maximum value of &lt;math&gt;A \cdot M \cdot C&lt;/math&gt;+&lt;math&gt;A \cdot M&lt;/math&gt;+&lt;math&gt;M \cdot C&lt;/math&gt;+&lt;math&gt;A\cdot C&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } &lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?<br /> <br /> &lt;math&gt;\text {(A)}\ 3 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 6 \qquad \text {(E)}\ 7&lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> When the [[mean]], [[median]], and [[mode]] of the list<br /> <br /> &lt;cmath&gt;10,2,5,2,4,2,x&lt;/cmath&gt;<br /> <br /> are arranged in increasing order, they form a non-constant [[arithmetic progression]]. What is the sum of all possible real values of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text {(A)}\ 3 \qquad \text {(B)}\ 6 \qquad \text {(C)}\ 9 \qquad \text {(D)}\ 17 \qquad \text {(E)}\ 20&lt;/math&gt;<br /> [[2000 AMC 12 Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Let &lt;math&gt;f&lt;/math&gt; be a [[function]] for which &lt;math&gt;f(x/3) = x^2 + x + 1&lt;/math&gt;. Find the sum of all values of &lt;math&gt;z&lt;/math&gt; for which &lt;math&gt;f(3z) = 7&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3&lt;/math&gt;<br /> [[2000 AMC 12 Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> [[2000 AMC 12 Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> [[2000 AMC 12 Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> [[2000 AMC 12 Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> [[2000 AMC 12 Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> [[2000 AMC 12 Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> Through a point on the [[hypotenuse]] of a [[right triangle]], lines are drawn [[parallel]] to the legs of the triangle so that the triangle is divided into a [[square]] and two smaller right triangles. The area of one of the two small right triangles is &lt;math&gt;m&lt;/math&gt; times the area of the square. The [[ratio]] of the area of the other small right triangle to the area of the square is <br /> <br /> &lt;math&gt;\text {(A)}\ \frac{1}{2m+1} \qquad \text {(B)}\ m \qquad \text {(C)}\ 1-m \qquad \text {(D)}\ \frac{1}{4m} \qquad \text {(E)}\ \frac{1}{8m^2}&lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> [[2000 AMC 12 Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> Professor Gamble buys a lottery ticket, which requires that he pick six different integers from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;46&lt;/math&gt;, inclusive. He chooses his numbers so that the sum of the base-ten logarithms of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property— the sum of the base-ten logarithms is an integer. What is the probability that Professor Gamble holds the winning ticket?<br /> <br /> &lt;math&gt;\text {(A)}\ 1/5 \qquad \text {(B)}\ 1/4 \qquad \text {(C)}\ 1/3 \qquad \text {(D)}\ 1/2 \qquad \text {(E)}\ 1 &lt;/math&gt;<br /> <br /> [[2000 AMC 12 Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> [[2000 AMC 12 Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> [[2000 AMC 12 Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> * [[AMC 12]]<br /> * [[AMC 12 Problems and Solutions]]<br /> * [[2000 AMC 12]]<br /> * [[Mathematics competition resources]]</div> Grn trtle https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_2&diff=27700 2007 AMC 12A Problems/Problem 2 2008-09-02T01:03:52Z <p>Grn trtle: </p> <hr /> <div>== Problem ==<br /> An aquarium has a [[rectangular prism|rectangular base]] that measures 100 cm by 40 cm and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base that measures 40 cm by 20 cm and a height of 10 cm is placed in the aquarium. By how many centimeters does the water rise?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0.5\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 1.5\qquad \mathrm{(D)}\ 2\qquad \mathrm{(E)}\ 2.5&lt;/math&gt;<br /> <br /> == Solution ==<br /> The water has volume &lt;math&gt;100 \cdot 40 \cdot 40=160000&lt;/math&gt;. The brick has volume 8000. The water and the brick combined have a volume of 168000. The water rises &lt;math&gt;\frac{168000}{4000}-\frac{160000}{4000}=42-40=2&lt;/math&gt; cm &lt;math&gt;\Rightarrow\fbox{D}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2007|ab=A|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Grn trtle https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_1&diff=27699 2007 AMC 12A Problems/Problem 1 2008-09-02T00:55:50Z <p>Grn trtle: </p> <hr /> <div>{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #1]] and [[2007 AMC 10A Problems/Problem 1|2007 AMC 10A #1]]}}<br /> == Problem ==<br /> One ticket to a show costs &lt;math&gt;\&lt;/math&gt;&lt;math&gt;20&lt;/math&gt; at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;math&gt;P&lt;/math&gt; = the amount Pam spent<br /> &lt;math&gt;S&lt;/math&gt; = the amount Susan spent<br /> <br /> * &lt;math&gt;P=5 \cdot (20 \cdot .7) = 70&lt;/math&gt;<br /> * &lt;math&gt;S=4 \cdot (20 \cdot .75) = 60&lt;/math&gt;<br /> <br /> Pam pays 10 more dollars than Susan &lt;math&gt;\Rightarrow\fbox{C}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2007|ab=A|before=First question|num-a=2}}<br /> {{AMC10 box|year=2007|ab=A|before=First question|num-a=2}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Grn trtle https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_10&diff=27694 2007 AMC 12B Problems/Problem 10 2008-09-01T08:43:29Z <p>Grn trtle: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Some boys and girls are having a car wash to raise money for a class trip to China. Initially &lt;math&gt;40&lt;/math&gt;% of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then &lt;math&gt;30&lt;/math&gt;% of the group are girls. How many girls were initially in the group?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4&lt;/math&gt; &lt;math&gt;\mathrm {(B)} 6&lt;/math&gt; &lt;math&gt;\mathrm {(C)} 8&lt;/math&gt; &lt;math&gt;\mathrm {(D)} 10&lt;/math&gt; &lt;math&gt;\mathrm {(E)} 12&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> First, determine the total number of people in the group.<br /> <br /> &lt;math&gt;2=(40/100-30/100)t&lt;/math&gt;<br /> <br /> &lt;math&gt;t=200/10=20&lt;/math&gt;<br /> <br /> Now find the original number of girls:<br /> <br /> &lt;math&gt;40/100t = 800/100 = 8&lt;/math&gt;<br /> <br /> So, there are 8 girls, &lt;math&gt;\Rightarrow \fbox{C}&lt;/math&gt;</div> Grn trtle https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_10&diff=27693 2007 AMC 12B Problems/Problem 10 2008-09-01T08:41:11Z <p>Grn trtle: </p> <hr /> <div>==Problem==<br /> <br /> Some boys and girls are having a car wash to raise money for a class trip to China. Initially &lt;math&gt;40&lt;/math&gt;% of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then &lt;math&gt;30&lt;/math&gt;% of the group are girls. How many girls were initially in the group?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4&lt;/math&gt; &lt;math&gt;\mathrm {(B)} 6&lt;/math&gt; &lt;math&gt;\mathrm {(C)} 8&lt;/math&gt; &lt;math&gt;\mathrm {(D)} 10&lt;/math&gt; &lt;math&gt;\mathrm {(E)} 12&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> First, determine the total number of people in the group.<br /> <br /> &lt;math&gt;2=10/100t&lt;/math&gt;<br /> <br /> &lt;math&gt;t=200/10=20&lt;/math&gt;<br /> <br /> Now find the original number of girls:<br /> <br /> &lt;math&gt;40/100t = 800/100 = 8&lt;/math&gt;<br /> <br /> So, there are 8 girls, &lt;math&gt;\Rightarrow \fbox{C}&lt;/math&gt;</div> Grn trtle