https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Toiletpaperninjastars&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T13:31:00ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II&diff=1234432016 AIME II2020-06-01T22:52:17Z<p>Toiletpaperninjastars: Adding AIME box for navigation</p>
<hr />
<div>'''2016 [[American Invitational Mathematics Examination|AIME]] II''' problems and solutions. The test was held on Wednesday, March 16, 2016. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br />
<br />
* [[2016 AIME II Problems|Entire Test]]<br />
* [[2016 AIME II Answer Key|Answer Key]]<br />
** [[2016 AIME II Problems/Problem 1|Problem 1]] <br />
** [[2016 AIME II Problems/Problem 2|Problem 2]]<br />
** [[2016 AIME II Problems/Problem 3|Problem 3]]<br />
** [[2016 AIME II Problems/Problem 4|Problem 4]]<br />
** [[2016 AIME II Problems/Problem 5|Problem 5]]<br />
** [[2016 AIME II Problems/Problem 6|Problem 6]]<br />
** [[2016 AIME II Problems/Problem 7|Problem 7]]<br />
** [[2016 AIME II Problems/Problem 8|Problem 8]]<br />
** [[2016 AIME II Problems/Problem 9|Problem 9]]<br />
** [[2016 AIME II Problems/Problem 10|Problem 10]]<br />
** [[2016 AIME II Problems/Problem 11|Problem 11]]<br />
** [[2016 AIME II Problems/Problem 12|Problem 12]]<br />
** [[2016 AIME II Problems/Problem 13|Problem 13]]<br />
** [[2016 AIME II Problems/Problem 14|Problem 14]]<br />
** [[2016 AIME II Problems/Problem 15|Problem 15]]<br />
<br />
{{AIME box|year=2016|n=II|before=[[2016 AIME I]]|after=[[2017 AIME I]]}}</div>Toiletpaperninjastarshttps://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I&diff=1234422016 AIME I2020-06-01T22:51:32Z<p>Toiletpaperninjastars: Adding AIME box for navigation</p>
<hr />
<div>'''2016 [[American Invitational Mathematics Examination|AIME]] I''' problems and solutions. The test was held on Thursday, March 3, 2016. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br />
<br />
* [[2016 AIME I Problems|Entire Test]]<br />
* [[2016 AIME I Answer Key|Answer Key]]<br />
** [[2016 AIME I Problems/Problem 1|Problem 1]] <br />
** [[2016 AIME I Problems/Problem 2|Problem 2]]<br />
** [[2016 AIME I Problems/Problem 3|Problem 3]]<br />
** [[2016 AIME I Problems/Problem 4|Problem 4]]<br />
** [[2016 AIME I Problems/Problem 5|Problem 5]]<br />
** [[2016 AIME I Problems/Problem 6|Problem 6]]<br />
** [[2016 AIME I Problems/Problem 7|Problem 7]]<br />
** [[2016 AIME I Problems/Problem 8|Problem 8]]<br />
** [[2016 AIME I Problems/Problem 9|Problem 9]]<br />
** [[2016 AIME I Problems/Problem 10|Problem 10]]<br />
** [[2016 AIME I Problems/Problem 11|Problem 11]]<br />
** [[2016 AIME I Problems/Problem 12|Problem 12]]<br />
** [[2016 AIME I Problems/Problem 13|Problem 13]]<br />
** [[2016 AIME I Problems/Problem 14|Problem 14]]<br />
** [[2016 AIME I Problems/Problem 15|Problem 15]]<br />
<br />
{{AIME box|year=2016|n=I|before=[[2015 AIME II]]|after=[[2016 AIME II]]}}</div>Toiletpaperninjastarshttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II&diff=1234412015 AIME II2020-06-01T22:50:55Z<p>Toiletpaperninjastars: Adding AIME box for navigation</p>
<hr />
<div>'''2015 [[American Invitational Mathematics Examination|AIME]] II''' problems and solutions. The test was held on Wednesday, March 25, 2015. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br />
<br />
* [[2015 AIME II Problems|Entire Test]]<br />
* [[2015 AIME II Answer Key|Answer Key]]<br />
** [[2015 AIME II Problems/Problem 1|Problem 1]] <br />
** [[2015 AIME II Problems/Problem 2|Problem 2]]<br />
** [[2015 AIME II Problems/Problem 3|Problem 3]]<br />
** [[2015 AIME II Problems/Problem 4|Problem 4]]<br />
** [[2015 AIME II Problems/Problem 5|Problem 5]]<br />
** [[2015 AIME II Problems/Problem 6|Problem 6]]<br />
** [[2015 AIME II Problems/Problem 7|Problem 7]]<br />
** [[2015 AIME II Problems/Problem 8|Problem 8]]<br />
** [[2015 AIME II Problems/Problem 9|Problem 9]]<br />
** [[2015 AIME II Problems/Problem 10|Problem 10]]<br />
** [[2015 AIME II Problems/Problem 11|Problem 11]]<br />
** [[2015 AIME II Problems/Problem 12|Problem 12]]<br />
** [[2015 AIME II Problems/Problem 13|Problem 13]]<br />
** [[2015 AIME II Problems/Problem 14|Problem 14]]<br />
** [[2015 AIME II Problems/Problem 15|Problem 15]]<br />
<br />
{{AIME box|year=2015|n=II|before=[[2015 AIME I]]|after=[[2016 AIME I]]}}</div>Toiletpaperninjastarshttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I&diff=1234402015 AIME I2020-06-01T22:50:20Z<p>Toiletpaperninjastars: Adding AIME box for navigation</p>
<hr />
<div>'''2015 [[American Invitational Mathematics Examination|AIME]] I''' problems and solutions. The test was held on Thursday, March 19, 2015. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br />
<br />
* [[2015 AIME I Problems|Entire Test]]<br />
* [[2015 AIME I Answer Key|Answer Key]]<br />
** [[2015 AIME I Problems/Problem 1|Problem 1]] <br />
** [[2015 AIME I Problems/Problem 2|Problem 2]]<br />
** [[2015 AIME I Problems/Problem 3|Problem 3]]<br />
** [[2015 AIME I Problems/Problem 4|Problem 4]]<br />
** [[2015 AIME I Problems/Problem 5|Problem 5]]<br />
** [[2015 AIME I Problems/Problem 6|Problem 6]]<br />
** [[2015 AIME I Problems/Problem 7|Problem 7]]<br />
** [[2015 AIME I Problems/Problem 8|Problem 8]]<br />
** [[2015 AIME I Problems/Problem 9|Problem 9]]<br />
** [[2015 AIME I Problems/Problem 10|Problem 10]]<br />
** [[2015 AIME I Problems/Problem 11|Problem 11]]<br />
** [[2015 AIME I Problems/Problem 12|Problem 12]]<br />
** [[2015 AIME I Problems/Problem 13|Problem 13]]<br />
** [[2015 AIME I Problems/Problem 14|Problem 14]]<br />
** [[2015 AIME I Problems/Problem 15|Problem 15]]<br />
<br />
{{AIME box|year=2015|n=I|before=[[2014 AIME II]]|after=[[2015 AIME II]]}}</div>Toiletpaperninjastarshttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II&diff=1234392014 AIME II2020-06-01T22:49:07Z<p>Toiletpaperninjastars: Adding the aime box at the bottom for navigation</p>
<hr />
<div>'''2014 [[American Invitational Mathematics Examination|AIME]] II''' problems and solutions. The test was held on March 26, 2014. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br />
<br />
* [[2014 AIME II Problems|Entire Test]]<br />
* [[2014 AIME II Answer Key|Answer Key]]<br />
** [[2014 AIME II Problems/Problem 1|Problem 1]] <br />
** [[2014 AIME II Problems/Problem 2|Problem 2]]<br />
** [[2014 AIME II Problems/Problem 3|Problem 3]]<br />
** [[2014 AIME II Problems/Problem 4|Problem 4]]<br />
** [[2014 AIME II Problems/Problem 5|Problem 5]]<br />
** [[2014 AIME II Problems/Problem 6|Problem 6]]<br />
** [[2014 AIME II Problems/Problem 7|Problem 7]]<br />
** [[2014 AIME II Problems/Problem 8|Problem 8]]<br />
** [[2014 AIME II Problems/Problem 9|Problem 9]]<br />
** [[2014 AIME II Problems/Problem 10|Problem 10]]<br />
** [[2014 AIME II Problems/Problem 11|Problem 11]]<br />
** [[2014 AIME II Problems/Problem 12|Problem 12]]<br />
** [[2014 AIME II Problems/Problem 13|Problem 13]]<br />
** [[2014 AIME II Problems/Problem 14|Problem 14]]<br />
** [[2014 AIME II Problems/Problem 15|Problem 15]]<br />
<br />
{{AIME box|year=2014|n=II|before=[[2014 AIME I]]|after=[[2015 AIME I]]}}</div>Toiletpaperninjastarshttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems&diff=976222004 AIME I Problems2018-09-02T01:39:03Z<p>Toiletpaperninjastars: /* Problem 12 */</p>
<hr />
<div>{{AIME Problems|year=2004|n=I}}<br />
<br />
== Problem 1 ==<br />
The digits of a positive integer <math> n </math> are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when <math> n </math> is divided by 37?<br />
<br />
[[2004 AIME I Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Set <math> A </math> consists of <math> m </math> consecutive integers whose sum is <math> 2m, </math> and set <math> B </math> consists of <math> 2m </math> consecutive integers whose sum is <math> m. </math> The absolute value of the difference between the greatest element of <math> A </math> and the greatest element of <math> B </math> is 99. Find <math> m. </math><br />
<br />
[[2004 AIME I Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A convex polyhedron <math> P </math> has 26 vertices, 60 edges, and 36 faces, 24 of which are triangular, and 12 of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does <math> P </math> have?<br />
<br />
[[2004 AIME I Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A square has sides of length 2. Set <math> S </math> is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set <math> S </math> enclose a region whose area to the nearest hundredth is <math> k. </math> Find <math> 100k. </math><br />
<br />
[[2004 AIME I Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. What is <math> m+n </math>?<br />
<br />
[[2004 AIME I Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
An integer is called snakelike if its decimal representation <math> a_1a_2a_3\cdots a_k </math> satisfies <math> a_i<a_{i+1} </math> if <math> i </math> is odd and <math> a_i>a_{i+1} </math> if <math> i </math> is even. How many snakelike integers between 1000 and 9999 have four distinct digits?<br />
<br />
[[2004 AIME I Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Let <math> C </math> be the coefficient of <math> x^2 </math> in the expansion of the product <math> (1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x). </math> Find <math> |C|. </math><br />
<br />
[[2004 AIME I Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Define a regular <math> n </math>-pointed star to be the union of <math> n </math> line segments <math> P_1P_2, P_2P_3,\ldots, P_nP_1 </math> such that<br />
<br />
* the points <math> P_1, P_2,\ldots, P_n </math> are coplanar and no three of them are collinear,<br />
* each of the <math> n </math> line segments intersects at least one of the other line segments at a point other than an endpoint,<br />
* all of the angles at <math> P_1, P_2,\ldots, P_n </math> are congruent,<br />
* all of the <math> n </math> line segments <math> P_2P_3,\ldots, P_nP_1 </math> are congruent, and<br />
* the path <math> P_1P_2, P_2P_3,\ldots, P_nP_1 </math> turns counterclockwise at an angle of less than 180 degrees at each vertex.<br />
<br />
There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?<br />
<br />
[[2004 AIME I Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Let <math> ABC </math> be a triangle with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 rectangle. A segment is drawn to divide triangle <math> ABC </math> into a triangle <math> U_1 </math> and a trapezoid <math> V_1 </math> and another segment is drawn to divide rectangle <math> DEFG </math> into a triangle <math> U_2 </math> and a trapezoid <math> V_2 </math> such that <math> U_1 </math> is similar to <math> U_2 </math> and <math> V_1 </math> is similar to <math> V_2. </math> The minimum value of the area of <math> U_1 </math> can be written in the form <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math><br />
<br />
[[2004 AIME I Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A circle of radius 1 is randomly placed in a 15-by-36 rectangle <math> ABCD </math> so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal <math> AC </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m + n. </math><br />
<br />
[[2004 AIME I Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid <math> C </math> and a frustum-shaped solid <math> F, </math> in such a way that the ratio between the areas of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the volumes of <math> C </math> and <math> F </math> are both equal to <math> k. </math> Given that <math> k=m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math><br />
<br />
[[2004 AIME I Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let <math> S </math> be the set of ordered pairs <math> (x, y) </math> such that <math> 0 < x \le 1, 0<y\le 1, </math> and <math> \left \lfloor{\log_2{\left(\frac 1x\right)}}\right \rfloor </math> and <math> \left \lfloor{\log_5{\left(\frac 1y\right)}}\right \rfloor </math> are both even. Given that the area of the graph of <math> S </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math> The notation <math> \left \lfloor{z}\right \rfloor</math> denotes the greatest integer that is less than or equal to <math> z. </math><br />
<br />
[[2004 AIME I Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
The polynomial <math> P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17} </math> has 34 complex roots of the form <math> z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34, </math> with <math> 0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{34} < 1 </math> and <math> r_k>0. </math> Given that <math> a_1 + a_2 + a_3 + a_4 + a_5 = m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math><br />
<br />
[[2004 AIME I Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A unicorn is tethered by a 20-foot silver rope to the base of a magician's cylindrical tower whose radius is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from the nearest point on the tower, and the length of the rope that is touching the tower is <math> \frac{a-\sqrt{b}}c </math> feet, where <math> a, b, </math> and <math> c </math> are positive integers, and <math> c </math> is prime. Find <math> a+b+c. </math><br />
<br />
[[2004 AIME I Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
For all positive integers <math>x</math>, let<br />
<cmath><br />
f(x)=\begin{cases}1 &\mbox{if x = 1}\\ \frac x{10} &\mbox{if x is divisible by 10}\\ x+1 &\mbox{otherwise}\end{cases}<br />
</cmath><br />
and define a sequence as follows: <math>x_1=x</math> and <math>x_{n+1}=f(x_n)</math> for all positive integers <math>n</math>. Let <math>d(x)</math> be the smallest <math>n</math> such that <math>x_n=1</math>. (For example, <math>d(100)=3</math> and <math>d(87)=7</math>.) Let <math>m</math> be the number of positive integers <math>x</math> such that <math>d(x)=20</math>. Find the sum of the distinct prime factors of <math>m</math>.<br />
<br />
[[2004 AIME I Problems/Problem 15|Solution]]<br />
<br />
== See Also ==<br />
* [[2004 AIME I]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Toiletpaperninjastarshttps://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems&diff=972591982 AHSME Problems2018-08-16T01:43:03Z<p>Toiletpaperninjastars: </p>
<hr />
<div>== Problem 1 ==<br />
<br />
When the polynomial <math>x^3-2</math> is divided by the polynomial <math>x^2-2</math>, the remainder is <br />
<br />
<math>\text{(A)} \ 2 \qquad <br />
\text{(B)} \ -2 \qquad <br />
\text{(C)} \ -2x-2 \qquad <br />
\text{(D)} \ 2x+2 \qquad <br />
\text{(E)} \ 2x-2</math> <br />
<br />
[[1982 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
If a number eight times as large as <math>x</math> is increased by two, then one fourth of the result equals <br />
<br />
<math>\text{(A)} \ 2x + \frac{1}{2} \qquad <br />
\text{(B)} \ x + \frac{1}{2} \qquad <br />
\text{(C)} \ 2x+2 \qquad <br />
\text{(D)}\ 4\qquad<br />
\text{(E)}\ 8 </math> <br />
<br />
[[1982 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Evaluate <math>(x^x)^{(x^x)}</math> at <math>x = 2</math>. <br />
<br />
<math>\text{(A)} \ 16 \qquad <br />
\text{(B)} \ 64 \qquad <br />
\text{(C)} \ 256 \qquad <br />
\text{(D)} \ 1024 \qquad <br />
\text{(E)} \ 65,536 </math> <br />
<br />
[[1982 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
The perimeter of a semicircular region, measured in centimeters, is numerically equal to its area, <br />
measured in square centimeters. The radius of the semicircle, measured in centimeters, is <br />
<br />
<math>\text{(A)} \pi \qquad <br />
\text{(B)} \frac{2}{\pi} \qquad <br />
\text{(C)} 1 \qquad <br />
\text{(D)}\frac{1}{2}\qquad<br />
\text{(E)}\frac{4}{\pi}+2 </math> <br />
<br />
<br />
<br />
[[1982 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
Two positive numbers <math>x</math> and <math>y</math> are in the ratio <math>a: b</math> where <math>0 < a < b</math>. If <math>x+y = c</math>, then the smaller of <math>x</math> and <math>y</math> is <br />
<br />
<math>\text{(A)} \ \frac{ac}{b} \qquad <br />
\text{(B)} \ \frac{bc-ac}{b} \qquad <br />
\text{(C)} \ \frac{ac}{a+b} \qquad <br />
\text{(D)}\ \frac{bc}{a+b}\qquad<br />
\text{(E)}\ \frac{ac}{b-a} </math> <br />
<br />
[[1982 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
The sum of all but one of the interior angles of a convex polygon equals <math>2570^\circ</math>. The remaining angle is <br />
<br />
<math>\text{(A)} \ 90^\circ \qquad <br />
\text{(B)} \ 105^\circ \qquad <br />
\text{(C)} \ 120^\circ \qquad <br />
\text{(D)}\ 130^\circ\qquad<br />
\text{(E)}\ 144^\circ </math><br />
<br />
[[1982 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
If the operation <math>x \star y</math> is defined by <math>x \star y = (x+1)(y+1) - 1</math>, then which one of the following is FALSE? <br />
<br />
<math>\text{(A)} \ x \star y = y\star x \text{ for all real } x,y. \\<br />
\text{(B)} \ x \star (y + z) = ( x \star y ) + (x \star z) \text{ for all real } x,y, \text{ and } z.\\<br />
\text{(C)} \ (x-1) \star (x+1) = (x \star x) - 1 \text{ for all real } x. \\<br />
\text{(D)} \ x \star 0 = x \text{ for all real } x. \\<br />
\text{(E)} \ x \star (y \star z) = (x \star y) \star z \text{ for all real } x,y, \text{ and } z. </math> <br />
<br />
[[1982 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
By definition, <math>r! = r(r - 1) \cdots 1</math> and <math>\binom{j}{k} = \frac {j!}{k!(j - k)!}</math>, where <math>r,j,k</math> are positive integers and <math>k < j</math>. <br />
If <math>\binom{n}{1}, \binom{n}{2}, \binom{n}{3}</math> form an arithmetic progression with <math>n > 3</math>, then <math>n</math> equals <br />
<br />
<math>\textbf{(A)}\ 5\qquad <br />
\textbf{(B)}\ 7\qquad <br />
\textbf{(C)}\ 9\qquad <br />
\textbf{(D)}\ 11\qquad <br />
\textbf{(E)}\ 12</math> <br />
<br />
[[1982 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
A vertical line divides the triangle with vertices <math>(0,0), (1,1)</math>, and <math>(9,1)</math> in the <math>xy\text{-plane}</math> into two regions of equal area. <br />
The equation of the line is <math>x=</math> <br />
<br />
<math>\text {(A)} 2.5 \qquad <br />
\text {(B)} 3.0 \qquad <br />
\text {(C)} 3.5 \qquad <br />
\text {(D)} 4.0\qquad <br />
\text {(E)} 4.5 </math> <br />
<br />
[[1982 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
In the adjoining diagram, <math>BO</math> bisects <math>\angle CBA</math>, <math>CO</math> bisects <math>\angle ACB</math>, and <math>MN</math> is parallel to <math>BC</math>. <br />
If <math>AB=12, BC=24</math>, and <math>AC=18</math>, then the perimeter of <math>\triangle AMN</math> is<br />
<br />
<asy><br />
size(200);<br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18))[0], O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*dir(180)), N=intersectionpoint(A--C, O--O+40*dir(0));<br />
draw(B--M--O--B--C--O--N--C^^N--A--M);<br />
label("$A$", A, dir(90));<br />
label("$B$", B, dir(O--B));<br />
label("$C$", C, dir(O--C));<br />
label("$M$", M, dir(90)*dir(B--A));<br />
label("$N$", N, dir(90)*dir(A--C));<br />
label("$O$", O, dir(90));</asy><br />
<br />
<math>\text {(A)} 30 \qquad <br />
\text {(B)} 33 \qquad <br />
\text {(C)} 36 \qquad <br />
\text {(D)} 39 \qquad <br />
\text {(E)} 42 </math> <br />
<br />
[[1982 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
How many integers with four different digits are there between <math>1,000</math> and <math>9,999</math> such that the absolute value of <br />
the difference between the first digit and the last digit is <math>2</math>? <br />
<br />
<math>\text {(A)} 672 \qquad <br />
\text {(B)} 784 \qquad <br />
\text {(C)} 840 \qquad <br />
\text {(D)} 896 \qquad <br />
\text {(E)} 1008</math> <br />
<br />
[[1982 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
Let <math>f(x) = ax^7+bx^3+cx-5</math>, where <math>a,b</math> and <math>c</math> are constants. If <math>f(-7) = 7</math>, the <math>f(7)</math> equals <br />
<br />
<math>\text {(A)} -17 \qquad <br />
\text {(B)} -7 \qquad<br />
\text {(C)} 14 \qquad <br />
\text {(D)} 21\qquad <br />
\text {(E)} \text{not uniquely determined}</math> <br />
<br />
[[1982 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
If <math>a>1, b>1</math>, and <math>p=\frac{\log_b(\log_ba)}{\log_ba}</math>, then <math>a^p</math> equals <br />
<br />
<math>\text {(A)} 1 \qquad <br />
\text {(B)} b \qquad <br />
\text {(C)} \log_ab \qquad <br />
\text {(D)} \log_ba \qquad <br />
\text {(E)} a^{\log_ba} </math> <br />
<br />
[[1982 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
In the adjoining figure, points <math>B</math> and <math>C</math> lie on line segment <math>AD</math>, and <math>AB, BC</math>, and <math>CD</math> are diameters of circle <br />
<math>O, N</math>, and <math>P</math>, respectively. Circles <math>O, N</math>, and <math>P</math> all have radius <math>15</math> and the line <math>AG</math> is tangent to circle <math>P</math> at <math>G</math>. <br />
If <math>AG</math> intersects circle <math>N</math> at points <math>E</math> and <math>F</math>, then chord <math>EF</math> has length<br />
<br />
<asy><br />
size(250);<br />
defaultpen(fontsize(10));<br />
pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1];<br />
draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7));<br />
dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P);<br />
label("$A$", A, W);<br />
label("$B$", B, SE);<br />
label("$C$", C, NE);<br />
label("$D$", D, dir(0));<br />
label("$P$", P, S);<br />
label("$N$", N, S);<br />
label("$O$", O, S);<br />
label("$E$", E, dir(120));<br />
label("$F$", F, NE);<br />
label("$G$", G, dir(100));</asy><br />
<br />
<math>\text {(A)} 20 \qquad <br />
\text {(B)} 15\sqrt{2} \qquad <br />
\text {(C)} 24 \qquad <br />
\text{(D)} 25 \qquad <br />
\text {(E)} \text{none of these}</math> <br />
<br />
[[1982 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
Let <math>[z]</math> denote the greatest integer not exceeding <math>z</math>. Let <math>x</math> and <math>y</math> satisfy the simultaneous equations <br />
<br />
<cmath>\begin{align*} y&=2[x]+3 \\ y&=3[x-2]+5. \end{align*}</cmath><br />
<br />
If <math>x</math> is not an integer, then <math>x+y</math> is <br />
<br />
<math>\text {(A) } \text{ an integer} \qquad <br />
\text {(B) } \text{ between 4 and 5} \qquad <br />
\text{(C) }\text{ between -4 and 4}\qquad\\<br />
\text{(D) }\text{ between 15 and 16}\qquad<br />
\text{(E) } 16.5 </math> <br />
<br />
[[1982 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
A wooden cube has edges of length <math>3</math> meters. Square holes, of side one meter, centered in each face are cut through to the opposite face. <br />
The edges of the holes are parallel to the edges of the cube. The entire surface area including the inside, in square meters, is <br />
<br />
<math>\text {(A)} 54 \qquad <br />
\text {(B)} 72 \qquad <br />
\text {(C)} 76 \qquad <br />
\text {(D)} 84\qquad <br />
\text {(E)} 86 </math> <br />
<br />
[[1982 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
How many real numbers <math>x</math> satisfy the equation <math>3^{2x+2}-3^{x+3}-3^{x}+3=0</math>? <br />
<br />
<math>\text {(A)} 0 \qquad <br />
\text {(B)} 1 \qquad <br />
\text {(C)} 2 \qquad <br />
\text {(D)} 3 \qquad <br />
\text {(E)} 4 </math> <br />
<br />
[[1982 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
In the adjoining figure of a rectangular solid, <math>\angle DHG=45^\circ</math> and <math>\angle FHB=60^\circ</math>. Find the cosine of <math>\angle BHD</math>.<br />
<br />
<asy><br />
import three;defaultpen(linewidth(0.7)+fontsize(10));<br />
currentprojection=orthographic(1/3+1/10,1-1/10,1/3);<br />
real r=sqrt(3);<br />
triple A=(0,0,r), B=(0,r,r), C=(1,r,r), D=(1,0,r), E=O, F=(0,r,0), G=(1,0,0), H=(1,r,0);<br />
draw(D--G--H--D--A--B--C--D--B--F--H--B^^C--H);<br />
draw(A--E^^G--E^^F--E, linetype("4 4"));<br />
label("$A$", A, N);<br />
label("$B$", B, dir(0));<br />
label("$C$", C, N);<br />
label("$D$", D, W);<br />
label("$E$", E, NW);<br />
label("$F$", F, S);<br />
label("$G$", G, W);<br />
label("$H$", H, S);<br />
triple H45=(1,r-0.15,0.1), H60=(1-0.05, r, 0.07);<br />
label("$45^\circ$", H45, dir(125), fontsize(8));<br />
label("$60^\circ$", H60, dir(25), fontsize(8));</asy><br />
<br />
<math>\text {(A)} \frac{\sqrt{3}}{6} \qquad <br />
\text {(B)} \frac{\sqrt{2}}{6} \qquad <br />
\text {(C)} \frac{\sqrt{6}}{3} \qquad <br />
\text{(D)}\frac{\sqrt{6}}{4}\qquad<br />
\text{(E)}\frac{\sqrt{6}-\sqrt{2}}{4} </math><br />
<br />
[[1982 AHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
Let <math>f(x)=|x-2|+|x-4|-|2x-6|</math> for <math>2 \leq x\leq 8</math>. The sum of the largest and smallest values of <math>f(x)</math> is <br />
<br />
<math>\text {(A)} 1 \qquad <br />
\text {(B)} 2 \qquad <br />
\text {(C)} 4 \qquad <br />
\text {(D)} 6 \qquad <br />
\text {(E)}\text{none of these} </math> <br />
<br />
[[1982 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
The number of pairs of positive integers <math>(x,y)</math> which satisfy the equation <math>x^2+y^2=x^3</math> is <br />
<br />
<math>\text {(A)} 0 \qquad <br />
\text {(B)} 1 \qquad <br />
\text {(C)} 2 \qquad <br />
\text {(D)} \text{not finite} \qquad <br />
\text {(E)} \text{none of these} </math> <br />
<br />
[[1982 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
In the adjoining figure, the triangle <math>ABC</math> is a right triangle with <math>\angle BCA=90^\circ</math>. Median <math>CM</math> is perpendicular to median <math>BN</math>, <br />
and side <math>BC=s</math>. The length of <math>BN</math> is<br />
<br />
<asy><br />
size(200);<br />
defaultpen(linewidth(0.7)+fontsize(10));real r=54.72;<br />
pair B=origin, C=dir(r), A=intersectionpoint(B--(9,0), C--C+4*dir(r-90)), M=midpoint(B--A), N=midpoint(A--C), P=intersectionpoint(B--N, C--M);<br />
draw(M--C--A--B--C^^B--N);<br />
pair point=P;<br />
markscalefactor=0.005;<br />
draw(rightanglemark(C,P,B));<br />
label("$A$", A, dir(point--A));<br />
label("$B$", B, dir(point--B));<br />
label("$C$", C, dir(point--C));<br />
label("$M$", M, S);<br />
label("$N$", N, dir(C--A)*dir(90));<br />
label("$s$", B--C, NW);</asy><br />
<br />
<math>\text {(A)} s\sqrt 2 \qquad <br />
\text {(B)} \frac 32s\sqrt2 \qquad <br />
\text {(C)} 2s\sqrt2 \qquad <br />
\text{(D)}\frac{1}{2}s\sqrt5\qquad<br />
\text{(E)}\frac{1}{2}s\sqrt6</math> <br />
<br />
[[1982 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
In a narrow alley of width <math>w</math> a ladder of length a is placed with its foot at point P between the walls. <br />
Resting against one wall at <math>Q</math>, the distance k above the ground makes a <math>45^\circ</math> angle with the ground. <br />
Resting against the other wall at <math>R</math>, a distance h above the ground, the ladder makes a <math>75^\circ</math> angle with the ground. <br />
The width <math>w</math> is equal to <br />
<br />
<math> \text{(A)}a\qquad<br />
\text{(B)}RQ\qquad<br />
\text{(C)}k\qquad<br />
\text{(D)}\frac{h+k}{2}\qquad<br />
\text{(E)}h </math> <br />
<br />
[[1982 AHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle. <br />
The cosine of the smallest angle is <br />
<br />
<math> \text{(A)}\frac{3}{4}\qquad<br />
\text{(B)}\frac{7}{10}\qquad<br />
\text{(C)}\frac{2}{3}\qquad<br />
\text{(D)}\frac{9}{14}\qquad<br />
\text{(E)}\text{none of these} </math><br />
<br />
[[1982 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If <math>AG=2, GF=13, FC=1</math>, <br />
and <math>HJ=7</math>, then <math>DE</math> equals <br />
<br />
<asy><br />
defaultpen(fontsize(10));<br />
real r=sqrt(22);<br />
pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J);<br />
dot(A^^B^^C^^D^^E^^F^^G^^H^^J);<br />
draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7));<br />
label("$A$", A, N);<br />
label("$B$", B, dir(210));<br />
label("$C$", C, dir(330));<br />
label("$D$", D, SW);<br />
label("$E$", E, SE);<br />
label("$F$", F, dir(170));<br />
label("$G$", G, dir(250));<br />
label("$H$", H, SE);<br />
label("$J$", J, dir(0));<br />
label("2", A--G, dir(30));<br />
label("13", F--G, dir(180+30));<br />
label("1", F--C, dir(30));<br />
label("7", H--J, dir(-30));</asy><br />
<br />
<math>\text {(A)} 2\sqrt{22} \qquad <br />
\text {(B)} 7\sqrt{3} \qquad <br />
\text {(C)} 9 \qquad <br />
\text {(D)} 10 \qquad <br />
\text {(E)} 13</math> <br />
<br />
[[1982 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
The adjacent map is part of a city: the small rectangles are rocks, and the paths in between are streets. <br />
Each morning, a student walks from intersection <math>A</math> to intersection <math>B</math>, always walking along streets shown, <br />
and always going east or south. For variety, at each intersection where he has a choice, he chooses with <br />
probability <math>\frac{1}{2}</math> whether to go east or south. Find the probability that through any given morning, he goes through <math>C</math>. <br />
<br />
<asy><br />
defaultpen(linewidth(0.7)+fontsize(8));<br />
size(250);<br />
path p=origin--(5,0)--(5,3)--(0,3)--cycle;<br />
path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle;<br />
int i,j;<br />
for(i=0; i<5; i=i+1) {<br />
for(j=0; j<6; j=j+1) {<br />
draw(shift(6*i, 4*j)*p);<br />
}}<br />
clip((4,2)--(25,2)--(25,21)--(4,21)--cycle);<br />
fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black);<br />
label("A", (6,19), SE);<br />
label("B", (23,4), NW);<br />
label("C", (23,8), NW);<br />
draw((26,11.5)--(30,11.5), Arrows(5));<br />
draw((28,9.5)--(28,13.5), Arrows(5));<br />
label("N", (28,13.5), N);<br />
label("W", (26,11.5), W);<br />
label("E", (30,11.5), E);<br />
label("S", (28,9.5), S);</asy><br />
<br />
<math> \text{(A)}\frac{11}{32}\qquad<br />
\text{(B)}\frac{1}{2}\qquad<br />
\text{(C)}\frac{4}{7}\qquad<br />
\text{(D)}\frac{21}{32}\qquad<br />
\text{(E)}\frac{3}{4} </math> <br />
<br />
[[1982 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 26 ==<br />
<br />
If the base <math>8</math> representation of a perfect square is <math>ab3c</math>, where <math>a\ne 0</math>, then <math>c</math> equals <br />
<br />
<math>\text{(A)} 0\qquad <br />
\text{(B)}1 \qquad <br />
\text{(C)} 3\qquad <br />
\text{(D)} 4\qquad <br />
\text{(E)} \text{not uniquely determined} </math> <br />
<br />
[[1982 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 27 ==<br />
<br />
Suppose <math>z=a+bi</math> is a solution of the polynomial equation <math>c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0=0</math>, where <math>c_0, c_1, c_2, c_3, a</math>, and <math>b</math> <br />
are real constants and <math>i^2=-1</math>. Which of the following must also be a solution? <br />
<br />
<math>\text{(A)} -a-bi\qquad <br />
\text{(B)} a-bi\qquad <br />
\text{(C)} -a+bi\qquad <br />
\text{(D)}b+ai \qquad <br />
\text{(E)} \text{none of these} </math> <br />
<br />
[[1982 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28 ==<br />
<br />
A set of consecutive positive integers beginning with <math>1</math> is written on a blackboard. <br />
One number is erased. The average (arithmetic mean) of the remaining numbers is <math>35\frac{7}{17}</math>. What number was erased? <br />
<br />
<math>\text{(A)} 6\qquad <br />
\text{(B)}7 \qquad <br />
\text{(C)}8 \qquad <br />
\text{(D)} 9\qquad <br />
\text{(E)}\text{cannot be determined} </math> <br />
<br />
[[1982 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29 ==<br />
<br />
Let <math>x,y</math>, and <math>z</math> be three positive real numbers whose sum is <math>1</math>. If no one of these numbers is more than twice any other, <br />
then the minimum possible value of the product <math>xyz</math> is <br />
<br />
<math> \textbf{(A)}\ \frac{1}{32}\qquad<br />
\textbf{(B)}\ \frac{1}{36}\qquad<br />
\textbf{(C)}\ \frac{4}{125}\qquad<br />
\textbf{(D)}\ \frac{1}{127}\qquad<br />
\textbf{(E)}\ \text{none of these} </math> <br />
<br />
[[1982 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30 ==<br />
<br />
Find the units digit of the decimal expansion of <br />
<br />
<math>(15 + \sqrt{220})^{19} + (15 + \sqrt{220})^{82}</math>. <br />
<br />
<math>\textbf{(A)}\ 0\qquad <br />
\textbf{(B)}\ 2\qquad <br />
\textbf{(C)}\ 5\qquad <br />
\textbf{(D)}\ 9\qquad <br />
\textbf{(E)}\ \text{none of these}</math> <br />
<br />
[[1982 AHSME Problems/Problem 30|Solution]]<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME box|year=1982|before=[[1981 AHSME]]|after=[[1983 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>Toiletpaperninjastars