https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Spearra&feedformat=atom
AoPS Wiki - User contributions [en]
2024-03-29T02:14:31Z
User contributions
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_19&diff=100583
1990 AHSME Problems/Problem 19
2019-01-18T17:34:57Z
<p>Spearra: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
For how many integers <math>N</math> between <math>1</math> and <math>1990</math> is the improper fraction <math>\frac{N^2+7}{N+4}</math> <math>\underline{not}</math> in lowest terms?<br />
<br />
<math>\text{(A) } 0\quad<br />
\text{(B) } 86\quad<br />
\text{(C) } 90\quad<br />
\text{(D) } 104\quad<br />
\text{(E) } 105</math><br />
<br />
== Solution ==<br />
What we want to know is for how many <math>n</math> is <cmath>\gcd(n^2+7, n+4) > 1.</cmath> We start by setting <cmath>n+4 \equiv 0 \mod m</cmath> for some arbitrary <math>m</math>. This shows that <math>m</math> evenly divides <math>n+4</math>. Next we want to see under which conditions <math>m</math> also divides <math>n^2 + 7</math>. We know from the previous statement that <cmath>n \equiv -4 \mod m</cmath> and thus <cmath>n^2 \equiv (-4)^2 \equiv 16 \mod m.</cmath> Next we simply add <math>7</math> to get <cmath>n^2 + 7 \equiv 23 \mod m.</cmath> However, we also want <cmath>n^2 + 7 \equiv 0 \mod m</cmath> which leads to <cmath>n^2 + 7\equiv 23 \equiv 0 \mod m</cmath> from the previous statement. Since from that statement <math>23</math> divides <math>m</math> evenly, <math>m</math> must be of the form <math>23x</math>, for some arbitrary integer <math>x</math>. After this, we can set <cmath>n+4=23x</cmath> and <cmath>n=23x-4.</cmath> Finally, we must find the largest <math>x</math> such that <cmath>23x-4<1990.</cmath> This is a simple linear inequality for which the answer is <math>x=86</math>, or <math>\fbox{B}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1990|num-b=18|num-a=20}} <br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_17&diff=99393
1993 AHSME Problems/Problem 17
2018-12-12T07:42:55Z
<p>Spearra: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<asy><br />
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle, black+linewidth(.75));<br />
draw((0,-1)--(0,1), black+linewidth(.75));<br />
draw((-1,0)--(1,0), black+linewidth(.75));<br />
draw((-1,-1/sqrt(3))--(1,1/sqrt(3)), black+linewidth(.75));<br />
draw((-1,1/sqrt(3))--(1,-1/sqrt(3)), black+linewidth(.75));<br />
draw((-1/sqrt(3),-1)--(1/sqrt(3),1), black+linewidth(.75));<br />
draw((1/sqrt(3),-1)--(-1/sqrt(3),1), black+linewidth(.75));<br />
</asy><br />
<br />
Amy painted a dartboard over a square clock face using the "hour positions" as boundaries.[See figure.] If <math>t</math> is the area of one of the eight triangular regions such as that between 12 o'clock and 1 o'clock, and <math>q</math> is the area of one of the four corner quadrilaterals such as that between 1 o'clock and 2 o'clock, then <math>\frac{q}{t}=</math><br />
<br />
<math>\text{(A) } 2\sqrt{3}-2\quad<br />
\text{(B) } \frac{3}{2}\quad<br />
\text{(C) } \frac{\sqrt{5}+1}{2}\quad<br />
\text{(D) } \sqrt{3}\quad<br />
\text{(E) } 2</math><br />
<br />
== Solution ==<br />
Assume the length of the side of the square is 2, WLOG. This means the side of one t section is 1. As the lines are at clock face positions, each section has a <math>\tfrac{360}{12} = 30</math> degree angle from the center. So each section t is a <math>30-60-90</math> triangle with a long leg of 1. Therefore, the short leg is <math>\tfrac{1}{\sqrt3}</math>.<br />
<br />
<br />
This makes the area of each <math>t = \tfrac{1}{2}\cdot b \cdot h = \tfrac{1}{2}\cdot 1 \cdot \tfrac{1}{\sqrt3} = \tfrac{1}{2\sqrt3}</math><br />
<br />
<br />
The total area comprises <math>4q+8t</math>, so <cmath>4q+(8\cdot \tfrac{1}{2\sqrt3}) = 2^2=4</cmath><br />
<br />
<cmath>4q + \tfrac{4}{\sqrt3} = 4</cmath><br />
<br />
<cmath>4q = 4 - \tfrac{4}{\sqrt3}</cmath><br />
<br />
<cmath>q = 1 - \tfrac{1}{\sqrt3} = \tfrac{\sqrt3 - 1 }{\sqrt3}</cmath><br />
<br />
<br />
<cmath>\frac{q}{t} = \frac{\tfrac{\sqrt3 - 1 }{\sqrt3}}{\tfrac{1}{2\sqrt3}} = 2\cdot (\sqrt3 - 1) = \boxed{2\sqrt3-2}</cmath><br />
<br />
<br />
<br />
<math>\fbox{A}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1993|num-b=18|num-a=19}} <br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_17&diff=99392
1993 AHSME Problems/Problem 17
2018-12-12T07:40:32Z
<p>Spearra: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<asy><br />
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle, black+linewidth(.75));<br />
draw((0,-1)--(0,1), black+linewidth(.75));<br />
draw((-1,0)--(1,0), black+linewidth(.75));<br />
draw((-1,-1/sqrt(3))--(1,1/sqrt(3)), black+linewidth(.75));<br />
draw((-1,1/sqrt(3))--(1,-1/sqrt(3)), black+linewidth(.75));<br />
draw((-1/sqrt(3),-1)--(1/sqrt(3),1), black+linewidth(.75));<br />
draw((1/sqrt(3),-1)--(-1/sqrt(3),1), black+linewidth(.75));<br />
</asy><br />
<br />
Amy painted a dartboard over a square clock face using the "hour positions" as boundaries.[See figure.] If <math>t</math> is the area of one of the eight triangular regions such as that between 12 o'clock and 1 o'clock, and <math>q</math> is the area of one of the four corner quadrilaterals such as that between 1 o'clock and 2 o'clock, then <math>\frac{q}{t}=</math><br />
<br />
<math>\text{(A) } 2\sqrt{3}-2\quad<br />
\text{(B) } \frac{3}{2}\quad<br />
\text{(C) } \frac{\sqrt{5}+1}{2}\quad<br />
\text{(D) } \sqrt{3}\quad<br />
\text{(E) } 2</math><br />
<br />
== Solution ==<br />
Assume the length of the side of the square is 2, WLOG. This means the side of one t section is 1. As the lines are at clock face positions, each section has a <math>\tfrac{360}{12} = 30</math> degree angle from the center. So each section t is a <math>30-60-90</math> triangle with a long leg of 1. Therefore, the short leg is <math>\tfrac{1}{\sqrt3}</math>.<br />
<br />
This makes the area of each <math>t = \tfrac{1}{2}\cdot b \cdot h = \tfrac{1}{2}\cdot 1 \cdot \tfrac{1}{\sqrt3} = \tfrac{1}{2\sqrt3}</math><br />
<br />
The total area comprises <math>4q+8t</math>, so <math>4q+(8\cdot \tfrac{1}{2\sqrt3}) = 2^2=4</math><br />
<br />
<math>4q + \tfrac{4}{\sqrt3} = 4</math><br />
<br />
<math>4q = 4 - \tfrac{4}{\sqrt3}</math><br />
<br />
<math>q = 1 - \tfrac{1}{\sqrt3} = \tfrac{\sqrt3 - 1 }{\sqrt3}</math><br />
<br />
<br />
<math>\frac{q}{t} = \frac{\tfrac{\sqrt3 - 1 }{\sqrt3}}{\tfrac{1}{2\sqrt3}} = 2\cdot (\sqrt3 - 1) = \boxed{2\sqrt3-2}</math><br />
<br />
<br />
<br />
<math>\fbox{A}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1993|num-b=18|num-a=19}} <br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_17&diff=99391
1993 AHSME Problems/Problem 17
2018-12-12T07:34:53Z
<p>Spearra: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<asy><br />
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle, black+linewidth(.75));<br />
draw((0,-1)--(0,1), black+linewidth(.75));<br />
draw((-1,0)--(1,0), black+linewidth(.75));<br />
draw((-1,-1/sqrt(3))--(1,1/sqrt(3)), black+linewidth(.75));<br />
draw((-1,1/sqrt(3))--(1,-1/sqrt(3)), black+linewidth(.75));<br />
draw((-1/sqrt(3),-1)--(1/sqrt(3),1), black+linewidth(.75));<br />
draw((1/sqrt(3),-1)--(-1/sqrt(3),1), black+linewidth(.75));<br />
</asy><br />
<br />
Amy painted a dartboard over a square clock face using the "hour positions" as boundaries.[See figure.] If <math>t</math> is the area of one of the eight triangular regions such as that between 12 o'clock and 1 o'clock, and <math>q</math> is the area of one of the four corner quadrilaterals such as that between 1 o'clock and 2 o'clock, then <math>\frac{q}{t}=</math><br />
<br />
<math>\text{(A) } 2\sqrt{3}-2\quad<br />
\text{(B) } \frac{3}{2}\quad<br />
\text{(C) } \frac{\sqrt{5}+1}{2}\quad<br />
\text{(D) } \sqrt{3}\quad<br />
\text{(E) } 2</math><br />
<br />
== Solution ==<br />
Assume the length of the side of the square is 4, WLOG. This means the side of one t section is 2. As the lines are at clock face positions, each section has a <math>\tfrac{360}{12} = 30</math> degree angle from the center. So each section t is a <math>30-60-90</math> triangle with a long leg of 2. Therefore, the short leg is <math>\tfrac{2}{\sqrt3}</math>.<br />
<br />
This makes the area of each <math>t = \tfrac{1}{2}\cdot b \cdot h = \tfrac{1}{2}\cdot 2 \cdot \tfrac{2}{\sqrt3} = \tfrac{2}{\sqrt3}</math><br />
<br />
The total area comprises <math>4q+8t</math>, so <math>4q+(8\cdot \tfrac{2}{\sqrt3}) = 4^2=16</math><br />
<br />
<math>4q + \tfrac{16}{\sqrt3} = 16</math><br />
<br />
<math>4q = 16 - \tfrac{16}{\sqrt3}</math><br />
<br />
<math>q = 4 - \tfrac{4}{\sqrt3} = \tfrac{4\sqrt3 - 4 }{\sqrt3}</math><br />
<br />
<br />
<math>\frac{q}{t} = \frac{\tfrac{4\sqrt3 - 4 }{\sqrt3}}{\tfrac{2}{\sqrt3}} = \frac{4\sqrt3 - 4}{2} = \boxed{2\sqrt3-2}</math><br />
<br />
<br />
<br />
<math>\fbox{A}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1993|num-b=18|num-a=19}} <br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=1976_AHSME_Problems&diff=98466
1976 AHSME Problems
2018-11-02T06:40:16Z
<p>Spearra: /* Problem 30 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
If one minus the reciprocal of <math>(1-x)</math> equals the reciprocal of <math>(1-x)</math>, then <math>x</math> equals<br />
<br />
<math>\textbf{(A) }-2\qquad<br />
\textbf{(B) }-1\qquad<br />
\textbf{(C) }1/2\qquad<br />
\textbf{(D) }2\qquad <br />
\textbf{(E) }3 </math> <br />
<br />
[[1976 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
For how many real numbers <math>x</math> is <math>\sqrt{-(x+1)^2}</math> a real number?<br />
<br />
<math>\textbf{(A) }\text{none}\qquad<br />
\textbf{(B) }\text{one}\qquad<br />
\textbf{(C) }\text{two}\qquad\\<br />
\textbf{(D) }\text{a finite number greater than two}\qquad<br />
\textbf{(E) }\infty </math> <br />
<br />
[[1976 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
The sum of the distances from one vertex of a square with sides of length <math>2</math> to the midpoints of each of the sides of the square is<br />
<br />
<math>\textbf{(A) }2\sqrt{5}\qquad<br />
\textbf{(B) }2+\sqrt{3}\qquad<br />
\textbf{(C) }2+2\sqrt{3}\qquad<br />
\textbf{(D) }2+\sqrt{5}\qquad<br />
\textbf{(E) }2+2\sqrt{5}</math> <br />
<br />
[[1976 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let a geometric progression with n terms have first term one, common ratio <math>r</math> and sum <math>s</math>, where <math>r</math> and <math>s</math> are not zero. <br />
The sum of the geometric progression formed by replacing each term of the original progression by its reciprocal is<br />
<br />
<math>\textbf{(A) }\frac{1}{s}\qquad<br />
\textbf{(B) }\frac{1}{r^ns}\qquad<br />
\textbf{(C) }\frac{s}{r^{n-1}}\qquad<br />
\textbf{(D) }\frac{r^n}{s}\qquad<br />
\textbf{(E) } \frac{r^{n-1}}{s}</math> <br />
<br />
[[1976 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
How many integers greater than <math>10</math> and less than <math>100</math>, written in base-<math>10</math> notation, <br />
are increased by <math>9</math> when their digits are reversed?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad<br />
\textbf{(B)}\ 1 \qquad<br />
\textbf{(C)}\ 8 \qquad<br />
\textbf{(D)}\ 9 \qquad<br />
\textbf{(E)}\ 10 </math> <br />
<br />
[[1976 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
If <math>c</math> is a real number and the negative of one of the solutions of <math>x^2-3x+c=0</math> is a solution of <math>x^2+3x-c=0</math>, then the solutions of <math>x^2-3x+c=0</math> are<br />
<br />
<math>\textbf{(A) }1,~2\qquad<br />
\textbf{(B) }-1,~-2\qquad<br />
\textbf{(C) }0,~3\qquad<br />
\textbf{(D) }0,~-3\qquad <br />
\textbf{(E) }\frac{3}{2},~\frac{3}{2} </math> <br />
<br />
<br />
[[1976 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
If <math>x</math> is a real number, then the quantity <math>(1-|x|)(1+x)</math> is positive if and only if<br />
<br />
<math>\textbf{(A) }|x|<1\qquad<br />
\textbf{(B) }|x|>1\qquad<br />
\textbf{(C) }x<-1\text{ or }-1<x<1\qquad\\<br />
\textbf{(D) }x<1\qquad <br />
\textbf{(E) }x<-1 </math> <br />
<br />
[[1976 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
A point in the plane, both of whose rectangular coordinates are integers with absolute values less than or equal to four, <br />
is chosen at random, with all such points having an equal probability of being chosen. What is the probability that the distance <br />
from the point to the origin is at most two units?<br />
<br />
<math>\textbf{(A) }\frac{13}{81}\qquad<br />
\textbf{(B) }\frac{15}{81}\qquad<br />
\textbf{(C) }\frac{13}{64}\qquad<br />
\textbf{(D) }\frac{\pi}{16}\qquad<br />
\textbf{(E) }\text{the square of a rational number}</math> <br />
<br />
[[1976 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
In triangle <math>ABC</math>, <math>D</math> is the midpoint of <math>AB</math>; <math>E</math> is the midpoint of <math>DB</math>; and <math>F</math> is the midpoint of <math>BC</math>. <br />
If the area of <math>\triangle ABC</math> is <math>96</math>, then the area of <math>\triangle AEF</math> is<br />
<br />
<math>\textbf{(A) }16\qquad<br />
\textbf{(B) }24\qquad<br />
\textbf{(C) }32\qquad<br />
\textbf{(D) }36\qquad <br />
\textbf{(E) }48 </math><br />
<br />
[[1976 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
If <math>m,~n,~p</math>, and <math>q</math> are real numbers and <math>f(x)=mx+n</math> and <math>g(x)=px+q</math>, then the equation <math>f(g(x))=g(f(x))</math> has a solution<br />
<br />
<math>\textbf{(A) }\text{for all choices of }m,~n,~p, \text{ and } q\qquad\\<br />
\textbf{(B) }\text{if and only if }m=p\text{ and }n=q\qquad\\<br />
\textbf{(C) }\text{if and only if }mq-np=0\qquad\\<br />
\textbf{(D) }\text{if and only if }n(1-p)-q(1-m)=0\qquad\\<br />
\textbf{(E) }\text{if and only if }(1-n)(1-p)-(1-q)(1-m)=0 </math> <br />
<br />
[[1976 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
Which of the following statements is (are) equivalent to the statement <br />
"If the pink elephant on planet alpha has purple eyes, then the wild pig on planet beta does not have a long nose"?<br />
<br />
<math>\begin{array}{l}\textbf{I. }\text{"If the wild pig on planet beta has a long nose, then the pink elephant on planet alpha has purple eyes."}\\<br />
\textbf{II. }\text{"If the pink elephant on planet alpha does not have purple eyes, then the wild pig on planet beta does not have a long nose."}\\<br />
\textbf{III. }\text{"If the wild pig on planet beta has a long nose, then the pink elephant on planet alpha does not have purple eyes."}\\<br />
\textbf{IV. }\text{"The pink elephant on planet alpha does not have purple eyes, or the wild pig on planet beta does not have a long nose."}\end{array}</math><br />
<br />
<math> \textbf{(A) }\textbf{I. }\text{and }\textbf{II. }\text{only}\qquad<br />
\textbf{(B) }\textbf{III. }\text{and }\textbf{IV. }\text{only}\qquad<br />
\textbf{(C) }\textbf{II. }\text{and }\textbf{IV. }\text{only}\qquad<br />
\textbf{(D) }\textbf{II. }\text{and }\textbf{III. }\text{only}\qquad<br />
\textbf{(E) }\textbf{III. }\text{only} </math> <br />
<br />
[[1976 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
A supermarket has <math>128</math> crates of apples. Each crate contains at least <math>120</math> apples and at most <math>144</math> apples. <br />
What is the largest integer <math>n</math> such that there must be at least <math>n</math> crates containing the same number of apples?<br />
<br />
<math>\textbf{(A) }4\qquad<br />
\textbf{(B) }5\qquad<br />
\textbf{(C) }6\qquad<br />
\textbf{(D) }24\qquad <br />
\textbf{(E) }25 </math><br />
<br />
[[1976 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
If <math>x</math> cows give <math>x+1</math> cans of milk in <math>x+2</math> days, how many days will it take <math>x+3</math> cows to give <math>x+5</math> cans of milk?<br />
<br />
<math>\textbf{(A) }\frac{x(x+2)(x+5)}{(x+1)(x+3)}\qquad<br />
\textbf{(B) }\frac{x(x+1)(x+5)}{(x+2)(x+3)}\qquad\\<br />
\textbf{(C) }\frac{(x+1)(x+3)(x+5)}{x(x+2)}\qquad<br />
\textbf{(D) }\frac{(x+1)(x+3)}{x(x+2)(x+5)}\qquad \\<br />
\textbf{(E) }\text{none of these}</math><br />
<br />
[[1976 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
The measures of the interior angles of a convex polygon are in arithmetic progression. <br />
If the smallest angle is <math>100^\circ</math>, and the largest is <math>140^\circ</math>, then the number of sides the polygon has is <br />
<br />
<math>\textbf{(A) }6\qquad<br />
\textbf{(B) }8\qquad<br />
\textbf{(C) }10\qquad<br />
\textbf{(D) }11\qquad <br />
\textbf{(E) }12 </math> <br />
<br />
[[1976 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
If <math>r</math> is the remainder when each of the numbers <math>1059,~1417</math>, and <math>2312</math> is divided by <math>d</math>, where <math>d</math> is an integer greater than <math>1</math>, then <math>d-r</math> equals <br />
<br />
<math>\textbf{(A) }1\qquad<br />
\textbf{(B) }15\qquad<br />
\textbf{(C) }179\qquad<br />
\textbf{(D) }d-15\qquad <br />
\textbf{(E) }d-1 </math><br />
<br />
[[1976 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
In triangles <math>ABC</math> and <math>DEF</math>, lengths <math>AC,~BC,~DF</math>, and <math>EF</math> are all equal. Length <math>AB</math> is twice the length of the altitude of <math>\triangle DEF</math><br />
from <math>F</math> to <math>DE</math>. Which of the following statements is (are) true?<br />
<br />
<math>\begin{array}{l}\textbf{I. }\angle ACB \text{ and }\angle DFE\text{ must be complementary.}\\<br />
\textbf{II. }\angle ACB \text{ and }\angle DFE\text{ must be supplementary.}\\<br />
\textbf{III. }\text{The area of }\triangle ABC\text{ must equal the area of }\triangle DEF.\\<br />
\textbf{IV. }\text{The area of }\triangle ABC\text{ must equal twice the area of }\triangle DEF.\end{array}</math><br />
<br />
<math>\textbf{(A) }\textbf{II. }\text{only}\qquad<br />
\textbf{(B) }\textbf{III. }\text{only}\qquad<br />
\textbf{(C) }\textbf{IV. }\text{only}\qquad<br />
\textbf{(D) }\text{I. }\text{and }\textbf{III. }\text{only}\qquad <br />
\textbf{(E) }\textbf{II. }\text{and }\textbf{III. }\text{only}</math> <br />
<br />
[[1976 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
If <math>\theta</math> is an acute angle, and <math>\sin 2\theta=a</math>, then <math>\sin\theta+\cos\theta</math> equals<br />
<br />
<math>\textbf{(A) }\sqrt{a+1}\qquad<br />
\textbf{(B) }(\sqrt{2}-1)a+1\qquad<br />
\textbf{(C) }\sqrt{a+1}-\sqrt{a^2-a}\qquad\\<br />
\textbf{(D) }\sqrt{a+1}+\sqrt{a^2-a}\qquad <br />
\textbf{(E) }\sqrt{a+1}+a^2-a </math><br />
<br />
[[1976 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
<asy><br />
//size(100);//local<br />
size(200);<br />
real r1=2;<br />
pair<br />
O=(0,0),<br />
D=(.5,.5*sqrt(3)),<br />
C=(D.x+.5*3,D.y),<br />
B,<br />
B_prime=endpoint(arc(D, 3, 0,-2));<br />
B=B_prime;<br />
path<br />
c1=circle(O, r1);<br />
pair C=midpoint(D--B_prime);<br />
path arc2=arc(B_prime, 6/2, 158.25,250);<br />
draw(c1);<br />
draw(O--D);<br />
draw(D--C);<br />
draw(C--B_prime);<br />
pair A=beginpoint(arc2);<br />
draw(B_prime--A);<br />
//dot(O^^D^^C^^A);<br />
//dot(B_prime);<br />
label("\scriptsize{$O$}",O,.6dir(D--O));<br />
label("\scriptsize{$C$}",C,.5dir(-55));<br />
label("\scriptsize{$D$}", D,.2NW);<br />
//label("\scriptsize{$B$}",B,S);<br />
label("\scriptsize{$B$}", B_prime, .5*dir(D--B_prime));<br />
label("\scriptsize{$A$}",A,.5dir(NE));<br />
label("\tiny{2}", O--D, .45*LeftSide);<br />
label("\tiny{3}", D--C, .45*LeftSide);<br />
label("\tiny{6}", B_prime--A, .45*RightSide);<br />
label("\tiny{3}", waypoint(C--B_prime,.1), .45*N);<br />
//Credit to Klaus-Anton for the diagram</asy><br />
<br />
In the adjoining figure, <math>AB</math> is tangent at <math>A</math> to the circle with center <math>O</math>; point <math>D</math> is interior to the circle; <br />
and <math>DB</math> intersects the circle at <math>C</math>. If <math>BC=DC=3</math>, <math>OD=2</math>, and <math>AB=6</math>, then the radius of the circle is<br />
<br />
<math>\textbf{(A) }3+\sqrt{3}\qquad<br />
\textbf{(B) }15/\pi\qquad<br />
\textbf{(C) }9/2\qquad<br />
\textbf{(D) }2\sqrt{6}\qquad <br />
\textbf{(E) }\sqrt{22}</math><br />
<br />
[[1976 AHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
A polynomial <math>p(x)</math> has remainder three when divided by <math>x-1</math> and remainder five when divided by <math>x-3</math>. <br />
The remainder when <math>p(x)</math> is divided by <math>(x-1)(x-3)</math> is<br />
<br />
<math>\textbf{(A) }x-2\qquad<br />
\textbf{(B) }x+2\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }8\qquad <br />
\textbf{(E) }15 </math><br />
<br />
[[1976 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
Let <math>a,~b</math>, and <math>x</math> be positive real numbers distinct from one. Then <math>4(\log_ax)^2+3(\log_bx)^2=8(\log_ax)(\log_bx)</math><br />
<br />
<math>\textbf{(A) }\text{for all values of }a,~b,\text{ and }x\qquad\\<br />
\textbf{(B) }\text{if and only if }a=b^2\qquad\\<br />
\textbf{(C) }\text{if and only if }b=a^2\qquad\\<br />
\textbf{(D) }\text{if and only if }x=ab\qquad\\<br />
\textbf{(E) }\text{for none of these} </math> <br />
<br />
[[1976 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
What is the smallest positive odd integer <math>n</math> such that the product <math>2^{1/7}2^{3/7}\cdots2^{(2n+1)/7}</math> is greater than <math>1000</math>? <br />
(In the product the denominators of the exponents are all sevens, and the numerators are the successive odd integers from <math>1</math> to <math>2n+1</math>.)<br />
<br />
<math>\textbf{(A) }7\qquad<br />
\textbf{(B) }9\qquad<br />
\textbf{(C) }11\qquad<br />
\textbf{(D) }17\qquad <br />
\textbf{(E) }19 </math><br />
<br />
[[1976 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
Given an equilateral triangle with side of length <math>s</math>, consider the locus of all points <math>\mathit{P}</math> in the plane of the <br />
triangle such that the sum of the squares of the distances from <math>\mathit{P}</math> to the vertices of the triangle is a fixed number <math>a</math>. This locus<br />
<br />
<math>\textbf{(A) }\text{is a circle if }a>s^2\qquad\\<br />
\textbf{(B) }\text{contains only three points if }a=2s^2\text{ and is a circle if }a>2s^2\qquad\\<br />
\textbf{(C) }\text{is a circle with positive radius only if }s^2<a<2s^2\qquad\\<br />
\textbf{(D) }\text{contains only a finite number of points for any value of }a\qquad\\<br />
\textbf{(E) }\text{is none of these} </math> <br />
<br />
[[1976 AHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
For integers <math>k</math> and <math>n</math> such that <math>1\le k<n</math>, let <math>C^n_k=\frac{n!}{k!(n-k)!}</math>. Then <math>\left(\frac{n-2k-1}{k+1}\right)C^n_k</math> is an integer<br />
<br />
<math>\textbf{(A) }\text{for all }k\text{ and }n\qquad \\<br />
\textbf{(B) }\text{for all even values of }k\text{ and }n,\text{ but not for all }k\text{ and }n\qquad \\<br />
\textbf{(C) }\text{for all odd values of }k\text{ and }n,\text{ but not for all }k\text{ and }n\qquad \\<br />
\textbf{(D) }\text{if }k=1\text{ or }n-1,\text{ but not for all odd values }k\text{ and }n\qquad \\<br />
\textbf{(E) }\text{if }n\text{ is divisible by }k,\text{ but not for all even values }k\text{ and }n </math> <br />
<br />
[[1976 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
<asy><br />
size(150);<br />
pair A=(0,0),B=(1,0),C=(0,1),D=(-1,0),E=(0,.5),F=(sqrt(2)/2,.25);<br />
draw(circle(A,1)^^D--B);<br />
draw(circle(E,.5)^^circle( F ,.25));<br />
label("$A$", D, W);<br />
label("$K$", A, S);<br />
label("$B$", B, dir(0));<br />
label("$L$", E, N);<br />
label("$M$",shift(-.05,.05)*F);<br />
//Credit to Klaus-Anton for the diagram<br />
</asy><br />
<br />
In the adjoining figure, circle <math>\mathit{K}</math> has diameter <math>\mathit{AB}</math>; circle <math>\mathit{L}</math> is tangent to circle <math>\mathit{K}</math> <br />
and to <math>\mathit{AB}</math> at the center of circle <math>\mathit{K}</math>; and circle <math>\mathit{M}</math> tangent to circle <math>\mathit{K}</math>, <br />
to circle <math>\mathit{L}</math> and <math>\mathit{AB}</math>. The ratio of the area of circle <math>\mathit{K}</math> to the area of circle <math>\mathit{M}</math> is<br />
<br />
<math>\textbf{(A) }12\qquad<br />
\textbf{(B) }14\qquad<br />
\textbf{(C) }16\qquad<br />
\textbf{(D) }18\qquad <br />
\textbf{(E) }\text{not an integer} </math> <br />
<br />
[[1976 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
For a sequence <math>u_1,u_2\dots</math>, define <math>\Delta^1(u_n)=u_{n+1}-u_n</math> and, for all integer <br />
<math>k>1, \Delta^k(u_n)=\Delta^1(\Delta^{k-1}(u_n))</math>. <br />
If <math>u_n=n^3+n</math>, then <math>\Delta^k(u_n)=0</math> for all <math>n</math><br />
<br />
<math>\textbf{(A) }\text{if }k=1\qquad \\<br />
\textbf{(B) }\text{if }k=2,\text{ but not if }k=1\qquad \\<br />
\textbf{(C) }\text{if }k=3,\text{ but not if }k=2\qquad \\<br />
\textbf{(D) }\text{if }k=4,\text{ but not if }k=3\qquad\\<br />
\textbf{(E) }\text{for no value of }k </math><br />
<br />
[[1976 AHSME Problems/Problem 25|Solution]]<br />
<br />
== Problem 26 ==<br />
<br />
<asy><br />
size(150);<br />
dotfactor=4;<br />
draw(circle((0,0),4));<br />
draw(circle((10,-6),3));<br />
pair O,A,P,Q;<br />
O = (0,0);<br />
A = (10,-6);<br />
P = (-.55, -4.12);<br />
Q = (10.7, -2.86);<br />
dot("$O$", O, NE);<br />
dot("$O'$", A, SW);<br />
dot("$P$", P, SW);<br />
dot("$Q$", Q, NE);<br />
draw((2*sqrt(2),2*sqrt(2))--(10 + 3*sqrt(2)/2, -6 + 3*sqrt(2)/2)--cycle);<br />
draw((-1.68*sqrt(2),-2.302*sqrt(2))--(10 - 2.6*sqrt(2)/2, -6 - 3.4*sqrt(2)/2)--cycle);<br />
draw(P--Q--cycle);<br />
//Credit to happiface for the diagram<br />
</asy><br />
<br />
In the adjoining figure, every point of circle <math>\mathit{O'}</math> is exterior to circle <math>\mathit{O}</math>. <br />
Let <math>\mathit{P}</math> and <math>\mathit{Q}</math> be the points of intersection of an internal common tangent with the two external common tangents. <br />
Then the length of <math>PQ</math> is<br />
<br />
<math>\textbf{(A) }\text{the average of the lengths of the internal and external common tangents}\qquad\\<br />
\textbf{(B) }\text{equal to the length of an external common tangent if and only if circles }\mathit{O}\text{ and }\mathit{O'} \text{have equal radii}\\<br />
\textbf{(C) }\text{always equal to the length of an external common tangent}\qquad\\<br />
\textbf{(D) }\text{greater than the length of an external common tangent}\qquad\\<br />
\textbf{(E) }\text{the geometric mean of the lengths of the internal and external common tangents} </math> <br />
<br />
[[1976 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27 ==<br />
<br />
If <math>N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}</math>, then <math>N</math> equals<br />
<br />
<math>\textbf{(A) }1\qquad<br />
\textbf{(B) }2\sqrt{2}-1\qquad<br />
\textbf{(C) }\frac{\sqrt{5}}{2}\qquad<br />
\textbf{(D) }\sqrt{\frac{5}{2}}\qquad<br />
\textbf{(E) }\text{none of these} </math> <br />
<br />
[[1976 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28 ==<br />
<br />
Lines <math>\mathit{L}_1,\mathit{L}_2,\dots,\mathit{L}_{100}</math> are distinct. All lines <math>\mathit{L}_{4n}, n</math> a positive integer, are parallel to each other. <br />
All lines <math>\mathit{L}_{4n-3}</math>, <math>n</math> a positive integer, pass through a given point <math>\mathit{A}</math>. <br />
The maximum number of points of intersection of pairs of lines from the complete set <math>\{\mathit{L}_1,\mathit{L}_2,\dots,\mathit{L}_{100}\}</math> is<br />
<br />
<math>\textbf{(A) }4350\qquad<br />
\textbf{(B) }4351\qquad<br />
\textbf{(C) }4900\qquad<br />
\textbf{(D) }4901\qquad <br />
\textbf{(E) }9851</math><br />
<br />
[[1976 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29 ==<br />
<br />
Ann and Barbara were comparing their ages and found that Barbara is as old as Ann was when Barbara was as old as <br />
Ann had been when Barbara was half as old as Ann is. If the sum of their present ages is <math>44</math> years, then Ann's age is<br />
<br />
<math>\textbf{(A) }22\qquad<br />
\textbf{(B) }24\qquad<br />
\textbf{(C) }25\qquad<br />
\textbf{(D) }26\qquad <br />
\textbf{(E) }28 </math> <br />
<br />
[[1976 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30 ==<br />
<br />
How many distinct ordered triples <math>(x,y,z)</math> satisfy the equations <br />
<cmath>x+2y+4z=12</cmath><br />
<cmath>xy+4yz+2xz=22</cmath><br />
<cmath>xyz=6</cmath><br />
<br />
<br />
<math>\textbf{(A) }\text{none}\qquad<br />
\textbf{(B) }1\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }4\qquad<br />
\textbf{(E) }6 </math><br />
<br />
[[1976 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=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=Elementary_symmetric_sum&diff=98285
Elementary symmetric sum
2018-10-26T01:53:00Z
<p>Spearra: Corrected invalid wikipedia link to Sums of Powers</p>
<hr />
<div>An '''elementary symmetric sum''' is a type of [[summation]].<br />
<br />
== Definition ==<br />
The <math>k</math>-th '''elementary symmetric sum''' of a [[set]] of <math>n</math> numbers is the sum of all products of <math>k</math> of those numbers (<math>1 \leq k \leq n</math>). For example, if <math>n = 4</math>, and our set of numbers is <math>\{a, b, c, d\}</math>, then:<br />
<br />
1st Symmetric Sum = <math>S_1 = a+b+c+d</math><br />
<br />
2nd Symmetric Sum = <math>S_2 = ab+ac+ad+bc+bd+cd</math><br />
<br />
3rd Symmetric Sum = <math>S_3 = abc+abd+acd+bcd</math><br />
<br />
4th Symmetric Sum = <math>S_4 = abcd</math><br />
<br />
==Notation==<br />
The first elementary symmetric sum of <math>f(x)</math> is often written <math>\sum_{sym}f(x)</math>. The <math>n</math>th can be written <math>\sum_{sym}^{n}f(x)</math><br />
<br />
== Uses ==<br />
Any [[symmetric sum]] can be written as a [[polynomial]] of the elementary symmetric sum functions. For example, <math>x^3 + y^3 + z^3 = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - xz) + 3xyz = S_1^3 - 3S_1S_2 + 3S_3</math>. This is often used to solve systems of equations involving [https://en.wikipedia.org/wiki/Sums_of_powers sums of powers], combined with Vieta's formulas.<br />
<br />
Elementary symmetric sums show up in [[Vieta's formulas]]. In a monic polynomial, the coefficient of the <math>x^1</math> term is <math>e_1</math>, and the coefficient of the <math>x^k</math> term is <math>e_k</math>, where the symmetric sums are taken over the roots of the polynomial.<br />
<br />
==See Also==<br />
*[[Symmetric sum]]<br />
*[[Cyclic sum]]<br />
<br />
[[Category:Elementary algebra]]<br />
[[Category:Definition]]</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_14&diff=98254
1978 AHSME Problems/Problem 14
2018-10-24T04:42:51Z
<p>Spearra: Created page with " Assuming the solutions to the equation are n and m, by Vieta's formulas, <math>n_n + m_n = 18_n</math>. <math>n_n = 10_n</math>, so <math>10_n + m_n = 18_n</math>. <cm..."</p>
<hr />
<div><br />
<br />
Assuming the solutions to the equation are n and m, by Vieta's formulas, <math>n_n + m_n = 18_n</math>. <br />
<br />
<math>n_n = 10_n</math>, so <math>10_n + m_n = 18_n</math>. <br />
<br />
<cmath>m_n = 8_n</cmath>.<br />
<br />
Also by Vieta's formulas, <math>n_n \cdot m_n = b_n</math>. <br />
<cmath>10_n \cdot 8_n = \boxed{80_n}</cmath>.<br />
<br />
The answer is (C) <math>80</math></div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_11&diff=92256
2018 AMC 10B Problems/Problem 11
2018-02-23T02:04:30Z
<p>Spearra: /* Solution 3 */</p>
<hr />
<div>Which of the following expressions is never a prime number when <math>p</math> is a prime number?<br />
<br />
<math>\textbf{(A) } p^2+16 \qquad \textbf{(B) } p^2+24 \qquad \textbf{(C) } p^2+26 \qquad \textbf{(D) } p^2+46 \qquad \textbf{(E) } p^2+96</math><br />
<br />
==Solution 1==<br />
<br />
Because squares of a non-multiple of 3 is always <math>1\mod 3</math>, the only expression is always a multiple of <math>3</math> is <math>\boxed{\textbf{(C) } p^2+26} </math>. This is excluding when <math>p=0\mod3</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite.<br />
<br />
==Solution 2 (Bad Solution)==<br />
<br />
We proceed with guess and check:<br />
<math>5^2+16=41 \qquad<br />
7^2+24=73 \qquad<br />
5^2+46=71 \qquad<br />
19^2+96=457</math>.<br />
Clearly only <math>\boxed{(\textbf{C})}</math> is our only option left.<br />
(franchester)<br />
<br />
==Solution 3==<br />
From Fermat's Little Theorom, <math>p^2 \equiv 1\pmod 3</math> if <math>p</math> is coprime with <math>3</math>.<br />
So for any <math>n\equiv2\pmod3</math>, <math>p^2+n \equiv 0\pmod 3</math> - divisible by 3, so not a prime. <br />
The only choice <math>\equiv2\pmod3</math> is <math>\boxed{(\textbf{C})}</math><br />
<br />
==Solution 4==<br />
Primes can only be <math>1</math> or <math>-1\mod 6</math>. Therefore, the square of a prime can only be <math>1\mod 6</math>. <math>p^2+26</math> then must be <math>3\mod 6</math>, so it is always divisible by <math>3</math>. Therefore, the answer is <math>\boxed{\text{(C)}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_11&diff=92255
2018 AMC 10B Problems/Problem 11
2018-02-23T02:04:03Z
<p>Spearra: /* Solution 3 */</p>
<hr />
<div>Which of the following expressions is never a prime number when <math>p</math> is a prime number?<br />
<br />
<math>\textbf{(A) } p^2+16 \qquad \textbf{(B) } p^2+24 \qquad \textbf{(C) } p^2+26 \qquad \textbf{(D) } p^2+46 \qquad \textbf{(E) } p^2+96</math><br />
<br />
==Solution 1==<br />
<br />
Because squares of a non-multiple of 3 is always <math>1\mod 3</math>, the only expression is always a multiple of <math>3</math> is <math>\boxed{\textbf{(C) } p^2+26} </math>. This is excluding when <math>p=0\mod3</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite.<br />
<br />
==Solution 2 (Bad Solution)==<br />
<br />
We proceed with guess and check:<br />
<math>5^2+16=41 \qquad<br />
7^2+24=73 \qquad<br />
5^2+46=71 \qquad<br />
19^2+96=457</math>.<br />
Clearly only <math>\boxed{(\textbf{C})}</math> is our only option left.<br />
(franchester)<br />
<br />
==Solution 3==<br />
From Fermat's Little Theorom, <math>p^2 \equiv 1\pmod 3</math> if <math>p</math> is co-prime with <math>3</math>.<br />
So for any <math>n\equiv2\pmod3</math>, <math>p^2+n \equiv 0\pmod 3</math> - divisible by 3, so not a prime. <br />
The only choice <math>\equiv2\pmod3</math> is <math>\boxed{(\textbf{C})}</math><br />
<br />
==Solution 4==<br />
Primes can only be <math>1</math> or <math>-1\mod 6</math>. Therefore, the square of a prime can only be <math>1\mod 6</math>. <math>p^2+26</math> then must be <math>3\mod 6</math>, so it is always divisible by <math>3</math>. Therefore, the answer is <math>\boxed{\text{(C)}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_11&diff=92190
2018 AMC 10B Problems/Problem 11
2018-02-22T00:15:39Z
<p>Spearra: </p>
<hr />
<div>Which of the following expressions is never a prime number when <math>p</math> is a prime number?<br />
<br />
<math>\textbf{(A) } p^2+16 \qquad \textbf{(B) } p^2+24 \qquad \textbf{(C) } p^2+26 \qquad \textbf{(D) } p^2+46 \qquad \textbf{(E) } p^2+96</math><br />
<br />
==Solution 1==<br />
<br />
Because squares of a non-multiple of 3 is always <math>1\mod 3</math>, the only expression is always a multiple of <math>3</math> is <math>\boxed{\textbf{(C) } p^2+26} </math>. This is excluding when <math>p=0\mod3</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite.<br />
<br />
==Solution 2 (Bad Solution)==<br />
<br />
We proceed with guess and check:<br />
<math>5^2+16=41 \qquad<br />
7^2+24=73 \qquad<br />
5^2+46=71 \qquad<br />
19^2+96=457</math>.<br />
Clearly only <math>\boxed{(\textbf{C})}</math> is our only option left.<br />
(franchester)<br />
<br />
==Solution 3==<br />
From Fermat's Little Theorom, <math>p^2 \equiv 1\pmod 3</math>.<br />
So for any <math>n\equiv2\pmod3</math>, <math>p^2+n \equiv 0\pmod 3</math> - divisible by 3, so not a prime. <br />
The only choice <math>\equiv2\pmod3</math> is <math>\boxed{(\textbf{C})}</math><br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_13&diff=90718
2018 AMC 10A Problems/Problem 13
2018-02-09T01:53:13Z
<p>Spearra: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
<br />
A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point <math>A</math> falls on point <math>B</math>. What is the length in inches of the crease?<br />
<asy><br />
draw((0,0)--(4,0)--(4,3)--(0,0));<br />
label("$A$", (0,0), SW);<br />
label("$B$", (4,3), NE);<br />
label("$C$", (4,0), SE);<br />
label("$4$", (2,0), S);<br />
label("$3$", (4,1.5), E);<br />
label("$5$", (2,1.5), NW);<br />
fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br />
</asy><br />
<math>\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 </math><br />
<br />
==Solution 1==<br />
<br />
First, we need to realize that the crease line is just the perpendicular bisector of side <math>AB</math>, the hypotenuse of right triangle <math>\triangle ABC</math>. Call the midpoint of <math>AB</math> point <math>D</math>. Draw this line and call the intersection point with <math>AC</math> as <math>E</math>. Now, <math>\triangle ABC</math> is similar to <math>\triangle ADE</math> by <math>AA</math> similarity. Setting up the ratios, we find that<br />
<cmath>\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.</cmath><br />
Thus, our answer is <math>\boxed{D}</math>.<br />
<br />
~Nivek<br />
<br />
==Solution 2 (if you are already out of time)==<br />
<br />
Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and measure. It will be <math>\boxed{D} \frac{15}{8}</math> inches in length.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2018|ab=A|num-b=12|num-a=14}}<br />
{{AMC12 box|year=2018|ab=A|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=88724
AMC historical results
2017-11-30T15:39:27Z
<p>Spearra: /* AMC 10A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 56.99<br />
*AIME floor: 96<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: <br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: <br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100 (Top 5% 106.5)<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=88723
AMC historical results
2017-11-30T15:39:03Z
<p>Spearra: /* AMC 10B */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 56.99<br />
*AIME floor: 96<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: <br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: <br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100 (Top 5% 106.5)<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>
Spearra
https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=88722
AMC historical results
2017-11-30T15:38:39Z
<p>Spearra: /* AMC 10A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 56.99<br />
*AIME floor: 96<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: <br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: <br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 117<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100 (Top 5% 106.5)<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>
Spearra