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<hr />
<div>== Problem 1==<br />
<br />
When simplified, <math>(-\frac{1}{125})^{-2/3}</math> becomes:<br />
<br />
<math>\textbf{(A)}\ \frac{1}{25} \qquad<br />
\textbf{(B)}\ -\frac{1}{25} \qquad<br />
\textbf{(C)}\ 25\qquad<br />
\textbf{(D)}\ -25\qquad<br />
\textbf{(E)}\ 25\sqrt{-1}</math> <br />
<br />
[[1961 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2==<br />
<br />
An automobile travels <math>a/6</math> feet in <math>r</math> seconds. If this rate is maintained for <math>3</math> minutes, how many yards does it travel in <math>3</math> minutes?<br />
<br />
<math>\textbf{(A)}\ \frac{a}{1080r}\qquad<br />
\textbf{(B)}\ \frac{30r}{a}\qquad<br />
\textbf{(C)}\ \frac{30a}{r}\qquad<br />
\textbf{(D)}\ \frac{10r}{a}\qquad<br />
\textbf{(E)}\ \frac{10a}{r} </math><br />
<br />
[[1961 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3==<br />
<br />
If the graphs of <math>2y+x+3=0</math> and <math>3y+ax+2=0</math> are to meet at right angles, the value of <math>a</math> is:<br />
<br />
<math>\textbf{(A)}\ \pm \frac{2}{3} \qquad<br />
\textbf{(B)}\ -\frac{2}{3}\qquad<br />
\textbf{(C)}\ -\frac{3}{2} \qquad<br />
\textbf{(D)}\ 6\qquad<br />
\textbf{(E)}\ -6 </math><br />
<br />
[[1961 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4==<br />
<br />
Let the set consisting of the squares of the positive integers be called <math>u</math>; thus <math>u</math> is the set <math>1, 4, 9, 16 \ldots</math>. <br />
If a certain operation on one or more members of the set always yields a member of the set, <br />
we say that the set is closed under that operation. Then <math>u</math> is closed under:<br />
<br />
<math>\textbf{(A)}\ \text{Addition}\qquad<br />
\textbf{(B)}\ \text{Multiplication} \qquad<br />
\textbf{(C)}\ \text{Division} \qquad\\<br />
\textbf{(D)}\ \text{Extraction of a positive integral root}\qquad<br />
\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1961 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5==<br />
<br />
Let <math>S=(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1</math>. Then <math>S</math> equals:<br />
<br />
<math>\textbf{(A)}\ (x-2)^4 \qquad<br />
\textbf{(B)}\ (x-1)^4 \qquad<br />
\textbf{(C)}\ x^4 \qquad<br />
\textbf{(D)}\ (x+1)^4 \qquad<br />
\textbf{(E)}\ x^4+1 </math><br />
<br />
[[1961 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6==<br />
<br />
When simplified, <math>\log{8} \div \log{\frac{1}{8}}</math> becomes:<br />
<br />
<math>\textbf{(A)}\ 6\log{2} \qquad<br />
\textbf{(B)}\ \log{2} \qquad<br />
\textbf{(C)}\ 1 \qquad<br />
\textbf{(D)}\ 0\qquad<br />
\textbf{(E)}\ -1 </math> <br />
<br />
[[1961 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7==<br />
<br />
When simplified, the third term in the expansion of <math>(\frac{a}{\sqrt{x}}-\frac{\sqrt{x}}{a^2})^6</math> is:<br />
<br />
<math>\textbf{(A)}\ \frac{15}{x}\qquad<br />
\textbf{(B)}\ -\frac{15}{x}\qquad<br />
\textbf{(C)}\ -\frac{6x^2}{a^9} \qquad<br />
\textbf{(D)}\ \frac{20}{a^3}\qquad<br />
\textbf{(E)}\ -\frac{20}{a^3}</math><br />
<br />
[[1961 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8==<br />
<br />
Let the two base angles of a triangle be <math>A</math> and <math>B</math>, with <math>B</math> larger than <math>A</math>. <br />
The altitude to the base divides the vertex angle <math>C</math> into two parts, <math>C_1</math> and <math>C_2</math>, with <math>C_2</math> adjacent to side <math>a</math>. Then:<br />
<br />
<math>\textbf{(A)}\ C_1+C_2=A+B \qquad<br />
\textbf{(B)}\ C_1-C_2=B-A \qquad\\<br />
\textbf{(C)}\ C_1-C_2=A-B \qquad<br />
\textbf{(D)}\ C_1+C_2=B-A\qquad<br />
\textbf{(E)}\ C_1-C_2=A+B</math><br />
<br />
[[1961 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9==<br />
<br />
Let <math>r</math> be the result of doubling both the base and exponent of <math>a^b</math>, and <math>b</math> does not equal to <math>0</math>. <br />
If <math>r</math> equals the product of <math>a^b</math> by <math>x^b</math>, then <math>x</math> equals:<br />
<br />
<math>\textbf{(A)}\ a \qquad<br />
\textbf{(B)}\ 2a \qquad<br />
\textbf{(C)}\ 4a \qquad<br />
\textbf{(D)}\ 2\qquad<br />
\textbf{(E)}\ 4 </math> <br />
<br />
[[1961 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10==<br />
<br />
Each side of <math>\triangle ABC</math> is <math>12</math> units. <math>D</math> is the foot of the perpendicular dropped from <math>A</math> on <math>BC</math>, <br />
and <math>E</math> is the midpoint of <math>AD</math>. The length of <math>BE</math>, in the same unit, is:<br />
<br />
<math>\textbf{(A)}\ \sqrt{18} \qquad<br />
\textbf{(B)}\ \sqrt{28} \qquad<br />
\textbf{(C)}\ 6 \qquad<br />
\textbf{(D)}\ \sqrt{63} \qquad<br />
\textbf{(E)}\ \sqrt{98}</math> <br />
<br />
[[1961 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11==<br />
<br />
Two tangents are drawn to a circle from an exterior point <math>A</math>; they touch the circle at points <math>B</math> and <math>C</math> respectively. <br />
A third tangent intersects segment <math>AB</math> in <math>P</math> and <math>AC</math> in <math>R</math>, and touches the circle at <math>Q</math>. If <math>AB=20</math>, then the perimeter of <math>\triangle APR</math> is<br />
<br />
<math>\textbf{(A)}\ 42\qquad<br />
\textbf{(B)}\ 40.5 \qquad<br />
\textbf{(C)}\ 40\qquad<br />
\textbf{(D)}\ 39\frac{7}{8} \qquad<br />
\textbf{(E)}\ \text{not determined by the given information}</math><br />
<br />
[[1961 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12==<br />
<br />
The first three terms of a geometric progression are <math>\sqrt{2}, \sqrt[3]{2}, \sqrt[6]{2}</math>. Find the fourth term.<br />
<br />
<math>\textbf{(A)}\ 1\qquad<br />
\textbf{(B)}\ \sqrt[7]{2}\qquad<br />
\textbf{(C)}\ \sqrt[8]{2}\qquad<br />
\textbf{(D)}\ \sqrt[9]{2}\qquad<br />
\textbf{(E)}\ \sqrt[10]{2}</math> <br />
<br />
[[1961 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13==<br />
<br />
The symbol <math>|a|</math> means <math>a</math> is a positive number or zero, and <math>-a</math> if <math>a</math> is a negative number. <br />
For all real values of <math>t</math> the expression <math>\sqrt{t^4+t^2}</math> is equal to?<br />
<br />
<math>\textbf{(A)}\ t^3\qquad<br />
\textbf{(B)}\ t^2+t\qquad<br />
\textbf{(C)}\ |t^2+t|\qquad<br />
\textbf{(D)}\ t\sqrt{t^2+1}\qquad<br />
\textbf{(E)}\ |t|\sqrt{1+t^2}</math><br />
<br />
[[1961 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14==<br />
<br />
A rhombus is given with one diagonal twice the length of the other diagonal. <br />
Express the side of the rhombus is terms of <math>K</math>, where <math>K</math> is the area of the rhombus in square inches.<br />
<br />
<math>\textbf{(A)}\ \sqrt{K}\qquad<br />
\textbf{(B)}\ \frac{1}{2}\sqrt{2K}\qquad<br />
\textbf{(C)}\ \frac{1}{3}\sqrt{3K}\qquad<br />
\textbf{(D)}\ \frac{1}{4}\sqrt{4K}\qquad<br />
\textbf{(E)}\ \text{None of these are correct}</math> <br />
<br />
[[1961 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15==<br />
<br />
If <math>x</math> men working <math>x</math> hours a day for <math>x</math> days produce <math>x</math> articles, then the number of articles <br />
(not necessarily an integer) produced by <math>y</math> men working <math>y</math> hours a day for <math>y</math> days is:<br />
<br />
<math>\textbf{(A)}\ \frac{x^3}{y^2}\qquad<br />
\textbf{(B)}\ \frac{y^3}{x^2}\qquad<br />
\textbf{(C)}\ \frac{x^2}{y^3}\qquad<br />
\textbf{(D)}\ \frac{y^2}{x^3}\qquad<br />
\textbf{(E)}\ y</math><br />
<br />
[[1961 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16==<br />
<br />
An altitude <math>h</math> of a triangle is increased by a length <math>m</math>. How much must be taken from the corresponding base <math>b</math> <br />
so that the area of the new triangle is one-half that of the original triangle? <br />
<br />
<math>\textbf{(A)}\ \frac{bm}{h+m}\qquad<br />
\textbf{(B)}\ \frac{bh}{2h+2m}\qquad<br />
\textbf{(C)}\ \frac{b(2m+h)}{m+h}\qquad<br />
\textbf{(D)}\ \frac{b(m+h)}{2m+h}\qquad<br />
\textbf{(E)}\ \frac{b(2m+h)}{2(h+m)}</math> <br />
<br />
[[1961 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17==<br />
<br />
In the base ten number system the number <math>526</math> means <math>5 \times 10^2+2 \times 10 + 6</math>. <br />
In the Land of Mathesis, however, numbers are written in the base <math>r</math>. <br />
Jones purchases an automobile there for <math>440</math> monetary units (abbreviated m.u). <br />
He gives the salesman a <math>1000</math> m.u bill, and receives, in change, <math>340</math> m.u. The base <math>r</math> is:<br />
<br />
<math>\textbf{(A)}\ 2\qquad<br />
\textbf{(B)}\ 5\qquad<br />
\textbf{(C)}\ 7\qquad<br />
\textbf{(D)}\ 8\qquad<br />
\textbf{(E)}\ 12 </math><br />
<br />
[[1961 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18==<br />
<br />
The yearly changes in the population census of a town for four consecutive years are, <br />
respectively, 25% increase, 25% increase, 25% decrease, 25% decrease. <br />
The net change over the four years, to the nearest percent, is:<br />
<br />
<math>\textbf{(A)}\ -12 \qquad<br />
\textbf{(B)}\ -1 \qquad<br />
\textbf{(C)}\ 0 \qquad<br />
\textbf{(D)}\ 1\qquad<br />
\textbf{(E)}\ 12 </math><br />
<br />
[[1961 AHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19==<br />
<br />
Consider the graphs of <math>y=2\log{x}</math> and <math>y=\log{2x}</math>. We may say that:<br />
<br />
<math>\textbf{(A)}\ \text{They do not intersect}\qquad \\<br />
\textbf{(B)}\ \text{They intersect at 1 point only}\qquad \\<br />
\textbf{(C)}\ \text{They intersect at 2 points only} \qquad \\<br />
\textbf{(D)}\ \text{They intersect at a finite number of points but greater than 2} \qquad \\<br />
\textbf{(E)}\ \text{They coincide} </math> <br />
<br />
[[1961 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20==<br />
<br />
The set of points satisfying the pair of inequalities <math>y>2x</math> and <math>y>4-x</math> is contained entirely in quadrants:<br />
<br />
<math>\textbf{(A)}\ \text{I and II}\qquad<br />
\textbf{(B)}\ \text{II and III}\qquad<br />
\textbf{(C)}\ \text{I and III}\qquad<br />
\textbf{(D)}\ \text{III and IV}\qquad<br />
\textbf{(E)}\ \text{I and IV}</math> <br />
<br />
[[1961 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21==<br />
<br />
Medians <math>AD</math> and <math>CE</math> of <math>\triangle ABC</math> intersect in <math>M</math>. The midpoint of <math>AE</math> is <math>N</math>. <br />
Let the area of <math>\triangle MNE</math> be <math>k</math> times the area of <math>\triangle ABC</math>. Then <math>k</math> equals:<br />
<br />
<math>\textbf{(A)}\ \frac{1}{6}\qquad<br />
\textbf{(B)}\ \frac{1}{8}\qquad<br />
\textbf{(C)}\ \frac{1}{9}\qquad<br />
\textbf{(D)}\ \frac{1}{12}\qquad<br />
\textbf{(E)}\ \frac{1}{16}</math> <br />
<br />
[[1961 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22==<br />
<br />
If <math>3x^3-9x^2+kx-12</math> is divisible by <math>x-3</math>, then it is also divisible by:<br />
<br />
<math>\textbf{(A)}\ 3x^2-x+4\qquad<br />
\textbf{(B)}\ 3x^2-4\qquad<br />
\textbf{(C)}\ 3x^2+4\qquad<br />
\textbf{(D)}\ 3x-4 \qquad<br />
\textbf{(E)}\ 3x+4 </math> <br />
<br />
[[1961 AHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23==<br />
<br />
Points <math>P</math> and <math>Q</math> are both in the line segment <math>AB</math> and on the same side of its midpoint. <math>P</math> divides <math>AB</math> in the ratio <math>2:3</math>, <br />
and <math>Q</math> divides <math>AB</math> in the ratio <math>3:4</math>. If <math>PQ=2</math>, then the length of <math>AB</math> is:<br />
<br />
<math>\textbf{(A)}\ 60\qquad<br />
\textbf{(B)}\ 70\qquad<br />
\textbf{(C)}\ 75\qquad<br />
\textbf{(D)}\ 80\qquad<br />
\textbf{(E)}\ 85 </math><br />
<br />
[[1961 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24==<br />
<br />
Thirty-one books are arranged from left to right in order of increasing prices. <br />
The price of each book differs by <math>\textdollar{2}</math> from that of each adjacent book. <br />
For the price of the book at the extreme right a customer can buy the middle book and the adjacent one. Then:<br />
<br />
<math>\textbf{(A)}\ \text{The adjacent book referred to is at the left of the middle book}\qquad \\<br />
\textbf{(B)}\ \text{The middle book sells for \textdollar 36} \qquad \\<br />
\textbf{(C)}\ \text{The cheapest book sells for \textdollar4} \qquad \\<br />
\textbf{(D)}\ \text{The most expensive book sells for \textdollar64 } \qquad \\<br />
\textbf{(E)}\ \text{None of these is correct } </math> <br />
<br />
[[1961 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25==<br />
<br />
<math>\triangle ABC</math> is isosceles with base <math>AC</math>. Points <math>P</math> and <math>Q</math> are respectively in <math>CB</math> and <math>AB</math> and such that <math>AC=AP=PQ=QB</math>. <br />
The number of degrees in <math>\angle B</math> is:<br />
<br />
<math>\textbf{(A)}\ 25\frac{5}{7}\qquad<br />
\textbf{(B)}\ 26\frac{1}{3}\qquad<br />
\textbf{(C)}\ 30\qquad<br />
\textbf{(D)}\ 40\qquad<br />
\textbf{(E)}\ \text{Not determined by the information given}</math><br />
<br />
[[1961 AHSME Problems/Problem 25|Solution]]<br />
<br />
== Problem 26==<br />
<br />
For a given arithmetic series the sum of the first <math>50</math> terms is <math>200</math>, and the sum of the next <math>50</math> terms is <math>2700</math>. <br />
The first term in the series is:<br />
<br />
<math>\textbf{(A)}\ -1221 \qquad<br />
\textbf{(B)}\ -21.5 \qquad<br />
\textbf{(C)}\ -20.5 \qquad<br />
\textbf{(D)}\ 3 \qquad<br />
\textbf{(E)}\ 3.5 </math> <br />
<br />
[[1961 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27==<br />
<br />
Given two equiangular polygons <math>P_1</math> and <math>P_2</math> with different numbers of sides; <br />
each angle of <math>P_1</math> is <math>x</math> degrees and each angle of <math>P_2</math> is <math>kx</math> degrees, <br />
where <math>k</math> is an integer greater than <math>1</math>. <br />
The number of possibilities for the pair <math>(x, k)</math> is:<br />
<br />
<math>\textbf{(A)}\ \infty\qquad<br />
\textbf{(B)}\ \text{finite, but greater than 2}\qquad<br />
\textbf{(C)}\ 2\qquad<br />
\textbf{(D)}\ 1\qquad<br />
\textbf{(E)}\ 0 </math><br />
<br />
[[1961 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28==<br />
<br />
If <math>2137^{753}</math> is multiplied out, the units' digit in the final product in the final product is:<br />
<br />
<math>\textbf{(A)}\ 1\qquad<br />
\textbf{(B)}\ 3\qquad<br />
\textbf{(C)}\ 5\qquad<br />
\textbf{(D)}\ 7\qquad<br />
\textbf{(E)}\ 9</math> <br />
<br />
[[1961 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29==<br />
<br />
Let the roots of <math>ax^2+bx+c=0</math> be <math>r</math> and <math>s</math>. The equation with roots <math>ar+b</math> and <math>as+b</math> is:<br />
<br />
<math>\textbf{(A)}\ x^2-bx-ac=0\qquad<br />
\textbf{(B)}\ x^2-bx+ac=0 \qquad\\<br />
\textbf{(C)}\ x^2+3bx+ca+2b^2=0 \qquad<br />
\textbf{(D)}\ x^2+3bx-ca+2b^2=0 \qquad\\<br />
\textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0</math> <br />
<br />
[[1961 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30==<br />
<br />
If <math>\log_{10}2=a</math> and <math>\log_{10}3=b</math>, then <math>\log_{5}12=</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{a+b}{a+1}\qquad<br />
\textbf{(B)}\ \frac{2a+b}{a+1}\qquad<br />
\textbf{(C)}\ \frac{a+2b}{1+a}\qquad<br />
\textbf{(D)}\ \frac{2a+b}{1-a}\qquad<br />
\textbf{(E)}\ \frac{a+2b}{1-a}</math> <br />
<br />
[[1961 AHSME Problems/Problem 30|Solution]]<br />
<br />
== Problem 31==<br />
<br />
In <math>\triangle ABC</math> the ratio <math>AC:CB</math> is <math>3:4</math>. The bisector of the exterior angle at <math>C</math> intersects <math>BA</math> extended at <math>P</math> <br />
(<math>A</math> is between <math>P</math> and <math>B</math>). The ratio <math>PA:AB</math> is:<br />
<br />
<math>\textbf{(A)}\ 1:3 \qquad<br />
\textbf{(B)}\ 3:4 \qquad<br />
\textbf{(C)}\ 4:3 \qquad<br />
\textbf{(D)}\ 3:1 \qquad<br />
\textbf{(E)}\ 7:1 </math><br />
<br />
[[1961 AHSME Problems/Problem 31|Solution]]<br />
<br />
== Problem 32==<br />
<br />
A regular polygon of <math>n</math> sides is inscribed in a circle of radius <math>R</math>. The area of the polygon is <math>3R^2</math>. Then <math>n</math> equals: <br />
<br />
<math>\textbf{(A)}\ 8\qquad<br />
\textbf{(B)}\ 10\qquad<br />
\textbf{(C)}\ 12\qquad<br />
\textbf{(D)}\ 15\qquad<br />
\textbf{(E)}\ 18 </math> <br />
<br />
[[1961 AHSME Problems/Problem 32|Solution]]<br />
<br />
== Problem 33==<br />
<br />
The number of solutions of <math>2^{2x}-3^{2y}=55</math>, in which <math>x</math> and <math>y</math> are integers, is:<br />
<br />
<math>\textbf{(A)}\ 0\qquad<br />
\textbf{(B)}\ 1\qquad<br />
\textbf{(C)}\ 2\qquad<br />
\textbf{(D)}\ 3\qquad<br />
\textbf{(E)}\ \text{More than three, but finite}</math> <br />
<br />
[[1961 AHSME Problems/Problem 33|Solution]]<br />
<br />
== Problem 34==<br />
<br />
Let S be the set of values assumed by the fraction <math>\frac{2x+3}{x+2}</math>.<br />
When <math>x</math> is any member of the interval <math>x \ge 0</math>. If there exists a number <math>M</math> such that no number of the set <math>S</math> is greater than <math>M</math>, <br />
then <math>M</math> is an upper bound of <math>S</math>. If there exists a number <math>m</math> such that such that no number of the set <math>S</math> is less than <math>m</math>, <br />
then <math>m</math> is a lower bound of <math>S</math>. We may then say:<br />
<br />
<math>\textbf{(A)}\ \text{m is in S, but M is not in S}\qquad\\<br />
\textbf{(B)}\ \text{M is in S, but m is not in S}\qquad\\<br />
\textbf{(C)}\ \text{Both m and M are in S}\qquad\\<br />
\textbf{(D)}\ \text{Neither m nor M are in S}\qquad\\<br />
\textbf{(E)}\ \text{M does not exist either in or outside S} </math> <br />
<br />
[[1961 AHSME Problems/Problem 34|Solution]]<br />
<br />
== Problem 35==<br />
<br />
The number <math>695</math> is to be written with a factorial base of numeration, that is, <math>695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!</math> <br />
where <math>a_1, a_2, a_3 ... a_n</math> are integers such that <math>0 \le a_k \le k,</math> and <math>n!</math> means <math>n(n-1)(n-2)...2 \times 1</math>. Find <math>a_4</math> <br />
<br />
<math>\textbf{(A)}\ 0\qquad<br />
\textbf{(B)}\ 1\qquad<br />
\textbf{(C)}\ 2\qquad<br />
\textbf{(D)}\ 3\qquad<br />
\textbf{(E)}\ 4 </math><br />
<br />
[[1961 AHSME Problems/Problem 35|Solution]]<br />
<br />
== Problem 36==<br />
<br />
In <math>\triangle ABC</math> the median from <math>A</math> is given perpendicular to the median from <math>B</math>. If <math>BC=7</math> and <math>AC=6</math>, find the length of <math>AB</math>. <br />
<br />
<math>\textbf{(A)}\ 4\qquad<br />
\textbf{(B)}\ \sqrt{17} \qquad<br />
\textbf{(C)}\ 4.25\qquad<br />
\textbf{(D)}\ 2\sqrt{5} \qquad<br />
\textbf{(E)}\ 4.5 </math> <br />
<br />
[[1961 AHSME Problems/Problem 36|Solution]]<br />
<br />
== Problem 37==<br />
<br />
In racing over a distance <math>d</math> at uniform speed, <math>A</math> can beat <math>B</math> by <math>20</math> yards, <math>B</math> can beat <math>C</math> by <math>10</math> yards, <br />
and <math>A</math> can beat <math>C</math> by <math>28</math> yards. Then <math>d</math>, in yards, equals:<br />
<br />
<math>\textbf{(A)}\ \text{Not determined by the given information}\qquad<br />
\textbf{(B)}\ 58\qquad<br />
\textbf{(C)}\ 100\qquad<br />
\textbf{(D)}\ 116\qquad<br />
\textbf{(E)}\ 120</math> <br />
<br />
[[1961 AHSME Problems/Problem 37|Solution]]<br />
<br />
== Problem 38==<br />
<br />
<math>\triangle ABC</math> is inscribed in a semicircle of radius <math>r</math> so that its base <math>AB</math> coincides with diameter <math>AB</math>. <br />
Point <math>C</math> does not coincide with either <math>A</math> or <math>B</math>. Let <math>s=AC+BC</math>. Then, for all permissible positions of <math>C</math>:<br />
<br />
<math>\textbf{(A)}\ s^2\le8r^2\qquad<br />
\textbf{(B)}\ s^2=8r^2 \qquad<br />
\textbf{(C)}\ s^2 \ge 8r^2 \qquad\\<br />
\textbf{(D)}\ s^2\le4r^2 \qquad<br />
\textbf{(E)}\ x^2=4r^2 </math> <br />
<br />
[[1961 AHSME Problems/Problem 38|Solution]]<br />
<br />
== Problem 39==<br />
<br />
Any five points are taken inside or on a square with side length <math>1</math>. Let a be the smallest possible number with the <br />
property that it is always possible to select one pair of points from these five such that the distance between them <br />
is equal to or less than <math>a</math>. Then <math>a</math> is:<br />
<br />
<math>\textbf{(A)}\ \sqrt{3}/3\qquad<br />
\textbf{(B)}\ \sqrt{2}/2\qquad<br />
\textbf{(C)}\ 2\sqrt{2}/3\qquad<br />
\textbf{(D)}\ 1 \qquad<br />
\textbf{(E)}\ \sqrt{2}</math><br />
<br />
[[1961 AHSME Problems/Problem 39|Solution]]<br />
<br />
== Problem 40==<br />
<br />
Find the minimum value of <math>\sqrt{x^2+y^2}</math> if <math>5x+12y=60</math>. <br />
<br />
<math>\textbf{(A)}\ \frac{60}{13}\qquad<br />
\textbf{(B)}\ \frac{13}{5}\qquad<br />
\textbf{(C)}\ \frac{13}{12}\qquad<br />
\textbf{(D)}\ 1\qquad<br />
\textbf{(E)}\ 0 </math><br />
<br />
[[1961 AHSME Problems/Problem 40|Solution]]<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME box|year=1961|before=[[1960 AHSME]]|after=[[1962 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>INomOnCountdownhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_20&diff=910762015 AMC 12B Problems/Problem 202018-02-12T00:54:44Z<p>INomOnCountdown: </p>
<hr />
<div>==Problem==<br />
For every positive integer <math>n</math>, let <math>\text{mod}_5 (n)</math> be the remainder obtained when <math>n</math> is divided by 5. Define a function <math>f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}</math> recursively as follows:<br />
<br />
<cmath>f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text{,}\\<br />
f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\<br />
f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4.<br />
\end{cases}</cmath> <br />
<br />
What is <math>f(2015,2)</math>?<br />
<br />
<math>\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4</math><br />
<br />
==Solution==<br />
Simply draw a table of values of <math>f(i,j)</math> for the first few values of <math>i</math>:<br />
<br />
<cmath>\begin{array}{|c || c | c | c | c | c |}<br />
\hline<br />
i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline<br />
0 & 1 & 2 & 3 & 4 & 0\\ \hline<br />
1 & 2 & 3 & 4 & 0 & 1\\ \hline<br />
2 & 3 & 0 & 2 & 4 & 1\\ \hline<br />
3 & 0 & 3 & 4 & 1 & 0\\ \hline<br />
4 & 3 & 1 & 3 & 1 & 3\\ \hline<br />
5 & 1 & 1 & 1 & 1 & 1\\ \hline<br />
\end{array}</cmath><br />
<br />
Now we claim that for <math>i \ge 5</math>, <math>f(i,j) = 1</math> for all values <math>0 \le j \le 4</math>. We will prove this by induction on <math>i</math> and <math>j</math>. The base cases for <math>i = 5</math>, have already been proven.<br />
<br />
For our inductive step, we must show that for all valid values of <math>j</math>, <math>f(i, j) = 1</math> if for all valid values of <math>j</math>, <math>f(i - 1, j) = 1</math>.<br />
<br />
We prove this itself by induction on <math>j</math>. For the base case, <math>j=0</math>, <math>f(i, 0) = f(i-1, 1) = 1</math>. For the inductive step, we need <math>f(i, j) = 1</math> if <math>f(i, j-1) = 1</math>. Then, <math>f(i, j) = f(i-1, f(i, j-1)).</math> <math>f(i, j-1) = 1</math> by our inductive hypothesis from our inner induction and <math>f(i-1, 1) = 1</math> from our outer inductive hypothesis. Thus, <math>f(i, j) = 1</math>, completing the proof.<br />
<br />
It is now clear that for <math>i \ge 5</math>, <math>f(i,j) = 1</math> for all values <math>0 \le j \le 4</math>.<br />
<br />
Thus, <math>f(2015,2) = \boxed{\textbf{(B)} \; 1}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=21|num-b=19}}<br />
{{MAA Notice}}</div>INomOnCountdownhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_20&diff=910752015 AMC 12B Problems/Problem 202018-02-12T00:54:29Z<p>INomOnCountdown: Adding a proof to the solution.</p>
<hr />
<div>==Problem==<br />
For every positive integer <math>n</math>, let <math>\text{mod}_5 (n)</math> be the remainder obtained when <math>n</math> is divided by 5. Define a function <math>f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}</math> recursively as follows:<br />
<br />
<cmath>f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text{,}\\<br />
f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\<br />
f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4.<br />
\end{cases}</cmath> <br />
<br />
What is <math>f(2015,2)</math>?<br />
<br />
<math>\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4</math><br />
<br />
==Solution #1==<br />
Simply draw a table of values of <math>f(i,j)</math> for the first few values of <math>i</math>:<br />
<br />
<cmath>\begin{array}{|c || c | c | c | c | c |}<br />
\hline<br />
i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline<br />
0 & 1 & 2 & 3 & 4 & 0\\ \hline<br />
1 & 2 & 3 & 4 & 0 & 1\\ \hline<br />
2 & 3 & 0 & 2 & 4 & 1\\ \hline<br />
3 & 0 & 3 & 4 & 1 & 0\\ \hline<br />
4 & 3 & 1 & 3 & 1 & 3\\ \hline<br />
5 & 1 & 1 & 1 & 1 & 1\\ \hline<br />
\end{array}</cmath><br />
<br />
Now we claim that for <math>i \ge 5</math>, <math>f(i,j) = 1</math> for all values <math>0 \le j \le 4</math>. We will prove this by induction on <math>i</math> and <math>j</math>. The base cases for <math>i = 5</math>, have already been proven.<br />
<br />
For our inductive step, we must show that for all valid values of <math>j</math>, <math>f(i, j) = 1</math> if for all valid values of <math>j</math>, <math>f(i - 1, j) = 1</math>.<br />
<br />
We prove this itself by induction on <math>j</math>. For the base case, <math>j=0</math>, <math>f(i, 0) = f(i-1, 1) = 1</math>. For the inductive step, we need <math>f(i, j) = 1</math> if <math>f(i, j-1) = 1</math>. Then, <math>f(i, j) = f(i-1, f(i, j-1)).</math> <math>f(i, j-1) = 1</math> by our inductive hypothesis from our inner induction and <math>f(i-1, 1) = 1</math> from our outer inductive hypothesis. Thus, <math>f(i, j) = 1</math>, completing the proof.<br />
<br />
It is now clear that for <math>i \ge 5</math>, <math>f(i,j) = 1</math> for all values <math>0 \le j \le 4</math>.<br />
<br />
Thus, <math>f(2015,2) = \boxed{\textbf{(B)} \; 1}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=21|num-b=19}}<br />
{{MAA Notice}}</div>INomOnCountdownhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_21&diff=910742015 AMC 12B Problems/Problem 212018-02-12T00:37:42Z<p>INomOnCountdown: Added Solution #3.</p>
<hr />
<div>==Problem==<br />
Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let <math>s</math> denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of <math>s</math>?<br />
<br />
<math>\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15</math><br />
<br />
==Solution 1==<br />
We can translate this wordy problem into this simple equation:<br />
<br />
<cmath>\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil</cmath><br />
<br />
We will proceed to solve this equation via casework.<br />
<br />
Case 1: <math>\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}</math><br />
<br />
Our equation becomes <math>\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}</math>, where <math>j \in \{0,1,2,3,4\}</math> Using the fact that <math>s</math> is an integer, we quickly find that <math>j=1</math> and <math>j=4</math> yield <math>s=64</math> and <math>s=66</math>, respectively.<br />
<br />
<br><br />
<br />
Case 2: <math>\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}+\frac{1}{2}</math><br />
<br />
Our equation becomes <math>\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}</math>, where <math>j \in \{0,1,2,3,4\}</math> Using the fact that <math>s</math> is an integer, we quickly find that <math>j=2</math> yields <math>s=63</math>.<br />
<br />
Summing up we get <math>63+64+66=193</math>. The sum of the digits is <math>\boxed{\textbf{(D)}\; 13}</math>.<br />
<br />
==Solution 2==<br />
It can easily be seen that the problem can be expressed by the equation:<br />
<cmath>\left\lceil \frac{s}{2} \right\rceil - \left\lceil \frac{s}{5} \right\rceil = 19</cmath><br />
<br />
However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:<br />
<br />
<cmath>\frac{s+a}{2} - \frac{s+b}{5} = 19</cmath> Where <math>a \in \{0,1\}</math> and <math>b \in \{0,1,2,3,4\}</math> Multiplying both sides by ten and simplifying, we get:<br />
<cmath>5s+5a-2s-2b=190</cmath><br />
<cmath>3s = 190+2b-5a</cmath><br />
<cmath>s = 63 + \frac{1+2b-5a}{3}</cmath><br />
<br />
Because s must be an integer, we need to find the values of <math>a</math> and <math>b</math> such that <math>2b-5a \equiv 2 \mod 3</math>. We solve using casework.<br />
<br />
Case 1: <math>a = 0</math><br />
<br />
If <math>a = 0</math>, we have <math>2b \equiv 2 \mod 3</math>. We can easily see that <math>b = 1</math> or <math>b = 4</math>, which when plugged into our original equation lead to <math>s = 64</math> and <math>s=66</math> respectively.<br />
<br />
<br><br />
<br />
Case 2: <math>a = 1</math><br />
<br />
If <math>a = 1</math>, we have <math>2b-5 \equiv 2 \mod 3</math>, which can be rewritten as <math>2b \equiv 1 \mod 3 </math>. We can again easily see that <math>b = 2</math> is the only solution, which when plugged into our original equation lead to <math>s = 63</math>.<br />
<br />
Adding these together we get <math>64+66+63=193</math>. The sum of the digits is <math>\boxed{\textbf{(D)}\; 13}</math>.<br />
<br />
==Solution 3==<br />
<br />
As before, we write the equation:<br />
<br />
<cmath>\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil.</cmath><br />
<br />
To get a ballpark estimate of where <math>s</math> might lie, we remove the ceiling functions to find:<br />
<br />
<cmath>\frac{s}{2} - 19 = \frac{s}{5}.</cmath><br />
<br />
This gives <math>\frac{3s}{10} = 19</math>, and thus values for <math>s</math> will be around <math>\frac{190}{3} = 63.\overline3</math>.<br />
<br />
Now, to establish some bounds around this estimated working value, we note that if <math>s=60</math>, Cozy takes 30 steps while Dash takes 12, a difference of 18. If <math>s=70</math>, Cozy takes 35 steps while Dash takes 14, a difference of 21. When <math>s</math> increases from a multiple of ten, the difference will never decrease beyond what it is at the multiple of ten, and likewise, when it decreases, it never becomes greater than at the multiple of ten, so any working values of <math>s</math> will be between <math>60</math> and <math>70</math>.<br />
<br />
Then, by inspection, <math>s=63, 64,</math> or <math>66</math>, so <math>\sum s = 193 \implies \boxed{\textbf{(D)}\; 13}.</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=22|num-b=20}}<br />
{{MAA Notice}}</div>INomOnCountdownhttps://artofproblemsolving.com/wiki/index.php?title=Joining_an_ARML_team&diff=85149Joining an ARML team2017-04-02T18:56:56Z<p>INomOnCountdown: /* Texas */</p>
<hr />
<div>Team selection for the [[American Regions Mathematics League]] varies from team to team.<br />
<br />
==Arizona==<br />
<br />
The Arizona ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==California==<br />
<br />
===Northern California===<br />
<br />
Practice meetings are held at UC Davis, Mathematical Sciences Building, room 2112, on Thursdays at 7pm. For more information, e-mail Eric Brattain-Morrin at [mailto:eric.brattain@gmail.com eric.brattain@gmail.com] and visit [http://www.math.ucdavis.edu/~exploration/arml/contact.php the team website].<br />
<br />
===San Diego===<br />
<br />
The SD team is entering just its third year of participation, so a permanent process of selecting team members has yet to be decided upon. The team is organized by and practices at the San Diego Math Circle (SDMC), and most of the students on last year's team were regular attendees at SDMC. Also, since the 2007 team contained no seniors, the organizers for the 2008 team are not planning on extending invites to new students unless scores on the AMC exams or the San Diego Math Olympiad (SDMO) are particularly high. <br />
<br />
A student who wishes to attend practice should try to take the SDMO. If a student wishes to speak with one of the coaches for the team, they may do so by contacting AoPS member 'Generating'.<br />
<br />
* [http://www.sdmathcircle.org/Welcome.php San Diego Math Circle website]<br />
<br />
===San Francisco Bay Area===<br />
<br />
The San Francisco Bay Area ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml]. Teams are coordinated by Tim Sanders, AoPS user farmertim. Coaching and practices are run by Moor Xu, AoPS user annoyingpi. One can contact them for further information, although nearly all questions are answered by the link provided above. <br />
In SFBA, anyone is welcome to sign up for an ARML team. The number of teams that will be going is determined by the amount of interest, and it very rare that someone who signs up will be excluded. Only the top three teams are selected by merit, and the other teams are chosen by geographical placement. The three top teams usually all place in the top 20, often even in the top 15 or 10.<br />
<br />
===Southern California===<br />
<br />
The Southern California team is open to residents of the following Southern California counties: Santa Barbara, Ventura, Los Angeles, Orange, Kern, San Bernardino, and Riverside. The organization fields three or four teams (45 or 60 students) and competes at the western ARML site in Las Vegas.<br />
<br />
Practices are held throughout the school year, approximately once a month, on the campus of California State University, Long Beach. Practices are normally held on Saturday afternoons. In addition, there is a Santa Barbara area group that meets and practices in Goleta and becomes part of the Southern California team.<br />
<br />
Team selection is based on all of the following criteria:<br />
<br />
* Attendence at practice sessions.<br />
* Performance on problems at practice sessions.<br />
* Performance at ARML itself in previous years.<br />
* Performance on AMC and AIME, including current and previous years.<br />
* Performance at CSULB Math Day at the Beach, a contest held in March.<br />
<br />
The coach is Dr. Kent Merryfield, a professor at CSULB. His AoPS user name is Kent Merryfield. Please contact him for further information.<br />
<br />
* [http://www.csulb.edu/depts/math/?q=node/20 Math Day at the Beach website.]<br />
* [http://www.csulb.edu/depts/math/?q=node/32 SoCal ARML website.]<br />
* [[Southern California ARML | SoCal ARML wiki page.]]<br />
<br />
== Connecticut ==<br />
Connecticut team selection is based on performance in math leagues across the state, from which a few upperclassmen can automatically qualify, and a runoff which is held in early March. To participate in the runoff, either have the math team coach at your school contact the league director in February to get the date or, if you don't participate in a league, contact Daniel Bochicchio, the current coach, at [mailto:dbochicchio@eosmith.org dbochicchio@eosmith.org]. Connecticut sends three teams.<br />
<br />
== Florida ==<br />
Florida ARML sends three teams to ARML each year. The selection criteria for the Florida ARML team takes into consideration several factors:<br />
<br />
*[[AMC]] and [[AIME]] performance<br />
*Past [[AMC]], [[AIME]], and [[USAMO]] scores<br />
*Past [[ARML]] performance<br />
*FAMAT-designated competitions<br />
*An annual statewide tryout test <br />
<br />
Florida ARML is organized by the Florida Student Association of Mathematics. Email [mailto:eliross2@aol.com eliross2@aol.com] for more information.<br />
<br />
== Georgia ==<br />
<br />
[[Georgia ARML]] has sent at least two teams to ARML each year since 1991. Beginning in 2008, we took advantage of the opening of the new Southern Site to add a third team (the "C" team). In 2011, we added a fourth team (the "D" team). In the past, the A team competed in the A division and the B team in the B division, but recently, the A team, the B team and the C team compete in the A division. We also field a few alternates as well. Students interested in participation should do well in [[Georgia mathematics competitions | local tournaments]] and the [[AMC]] and [[AIME]]. In addition to schools invited to the annual [[GCTM State Math Tournament | varsity state tournament]], the Georgia ARML coaches invite other individuals that are under serious consideration for the ARML team. The coaches select the team members during the state tournament based on [[AMC | USAMO index]], performance in local tournaments, and score at the state tournament. It is vital that any student under consideration has what we call "verified ability". This can be described as having good results from more than one contest or tournament. "Good results" can be described as placing higher than someone who was on the ARML team last year -- that is, doing better than someone who we know is good. <br />
<br />
The coaches attempt to select the best 30 students in the state (regardless of age or grade) to comprise the A and B teams -- those selected usually consist of USAMO qualifiers, the top 12 at the state tournament, and team veterans. (Of course, these groups are represented by overlapping Venn diagrams.) It is rare that a first-time senior is selected, although it does happen. Then the coaches select the best 15 young students (10th grade and below) to comprise a C team -- a team specifically comprised of up-and-comers who the coaches hope will be A and B team material in the future. Finally, the coaches allow anyone not selected, but who is interested in competing at ARML, to form a team of "walk-ons" which becomes the D team. It is rare for a student from the walk-ons to be so good that they move up to the A or B teams, but it has happened.<br />
<br />
The team usually practices on Sundays from the state tournament until the trip to ARML. The specific compositions of the A, B, C, and D teams are not usually determined until immediately before ARML. A member's team placement (on A, B, C, or D) depends on the person's performance against other team members in practice and the coaches's discretion.<br />
<br />
== Maine ==<br />
<br />
The two Maine ARML Teams consist of approximately the top 30 scorers on 5 [[MAML]] (Maine Association of Math Leagues) Meets. Training includes the problem set "Pete's Fabulous 42."<br />
<br />
* [http://www.maml.net MAML Website]<br />
<br />
==Maryland==<br />
<br />
===Montgomery (Montgomery County, MD)===<br />
<br />
Montgomery County typically sends four teams of high-schoolers and one team of middle-schoolers to ARML, with all 5 teams (Montgomery A, B C, D, and Junior) teams competing in division A. <br />
<br />
Top scorers from the Montgomery County high school math league are invited to ARML practices after the regular season, which ends around February. Top scorers from the middle school league are also invited. The Montgomery County ARML contact is Ms. Zenia Yang at the Yang Academy. Interested students from Montgomery County Public Schools who did not participate in the math leagues can ask their high school math team coaches to contact Ms. Yang and attend the weekly ARML practices to try out for the teams.<br />
<br />
Practices are usually held at Montgomery Blair High School on Thursdays from 6:30pm-9:30pm. Team selection is done by individual scores at practice and at the discretion of the coaches.<br />
<br />
===Chesapeake ARML===<br />
<br />
The Chesapeake ARML team draws on all areas and schools in Maryland not served by the Montgomery teams or by the public high schools in Howard or Baltimore Counties. Selection is based on AMC/AIME/USAMO scores, UMD Math Contest scores, participation on Chesapeake teams for PUMaC, HMMT, JHMT, Purple Comet, ARML Power Contest, and similar, and a series of practices done in April and May. For more information, contact the Chesapeake head coach, Catherine Asaro, at asaro@sff.net or here on AoPS in the Maryland Forum. Web site at http://chesapeakemath.catherineasaro.net<br />
<br />
===Howard County===<br />
The Howard County ARML teams draws the twelve public high schools in Howard County, MD.<br />
<br />
===Baltimore County===<br />
The Baltimore County ARML teams draws on the public high schools in Baltimore County, MD.<br />
<br />
==Michigan==<br />
<br />
Michigan sends three teams to Iowa. The teams are collectively known as the Michigan All-Stars. In general participants must place in the top 100 at MMPC, a local competition, to be in the running for the team. All members of the top 100 receive and invitation to be on the team. There are 3 practices that decide who is on which of the three teams. Although MMPC automatically qualifies for team selection, it is possible to get on a team by contacting coach Dave Friday. The three teams are the Reals, Naturals, and Primes. Two compete in A and 1 in B... Michigan was also awarded the best shirt design at ARML Iowa 2015...<br />
<br />
==Minnesota==<br />
<br />
Minnesota sends two teams to ARML each year, with the Gold and Maroon teams usually competing in divisions A and B, respectively.<br />
<br />
Roughly 35 students are invited to ARML practices, which take place on three consecutive Saturdays in May. There is no practice Memorial Day weekend. Invitations to the ARML team are extended to the top 10 state scorers on the AMC12, the top 10 regular-season scorers on the Minnesota High School Math League, and the top 10 scorers on the Invitational Event at the statewide math league tournament (held in March).<br />
<br />
Since these lists tend to overlap quite a bit, invitations are usually given to students "further down" these lists until enough invites have been given to fill two 15-person teams. <br />
<br />
In addition, an extra 5 or so younger students (typically in grades 8 through 10) are invited to be ARML "students in training". They may or may not go to ARML, but often serve as alternates (in the event that other students cannot attend). The expectation is that a student in training will learn from the practices, and the following year will be on one of the two teams. Since the creation of the AMC8 and AMC10 exams, the top scorers from these exams have typically been invited to ARML practices, either as team members or students in training.<br />
<br />
The selection of the Gold and Maroon teams is determined by students' performance at practices, and is not announced until the night before the competition.<br />
<br />
Usually at least one student in training is invited to go to ARML. This is to prevent a last-minute no-show (due to illness or emergency, for example) from crippling one of the teams.<br />
<br />
Invitations to participate on the MN team are usually sent out shortly after the MN State High School Math League statewide tournament in March. <br />
<br />
* [http://www.macalester.edu/mathleague/index.htm Minnesota State High School Math League site]<br />
<br />
==Missouri==<br />
<br />
Missouri sends two teams, Red and Blue, and a few alternates for a total of 36-37 students to the Iowa site ARML competition.<br />
<br />
Open practices are held usually once a month during the school year in two places, Springfield and St. Louis. Attendance is encouraged but not required. There are usually one or two "all day practices" a year at Missouri University of Science and Technology.<br />
<br />
Applications are based on AMC/AIME scores, GPML scores, other math competitions, and past ARML experience. Most if not all applicants are selected for the team.<br />
<br />
Since ARML is NOT a strictly a “state” team competition and because neither Kansas nor southern Illinois have ARML teams, ARML has given the “Missouri” team permission to include students from Kansas and from western Illinois (eg Edwardsville, etc.) on the MO ARML team. (eg. the 2008 team fielded 6 from KS, and 1 from IL)<br />
<br />
[http://math.missouristate.edu/MissouriARML.htm MO ARML homepage]<br />
<br />
==Nevada==<br />
<br />
The Nevada ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==New Jersey==<br />
New Jersey sends four different ARML teams. Also, many NJ students are on the Lehigh Valley teams.<br />
<br />
=== BCA/AAST ===<br />
AAST (a school in Bergen County Academies) sent 6 teams in 2012, all named after Greek Gods. Although there is a testing process to determine who's on which team, anyone in the AAST Math Team is permitted to participate in ARML. [https://bcts.bergen.org/index.php/math-team-home website]<br />
<br />
=== Central Jersey ===<br />
Qualification for the Central Jersey ARML team requires that<br />
*the student has participated (and chosen as a representative for the school) in at least 4 of the Central Jersey Math League meets,<br />
*and the average score meets a certain minimum which varies per year (this year, it is 3.0).<br />
<br />
Afterwards, interested students have to attend training sessions which take place at Highland Park High School or Hillsborough.<br />
<br />
More information can be found at [http://uzza.us/cjml/arml.html CJML website].<br />
<br />
=== WW-P ===<br />
<br />
WW-P, short for West Windsor-Plainsboro, is a team founded in 2012. Currently, it consists of students only from the WW-P school district, but students in close areas may be welcome as well. The selection process is a series of individual tests, and other experience is taken into consideration if needed. Two teams were sent in 2013; the A1 team ranked 3rd nationally. Also, they got 1st nationally in the B-Division two times in a row. More teams may be formed if more students join the program. <br />
<br />
If you would like more information, please email southmathclub@gmail.com.<br />
<br />
=== Raritan Valley ===<br />
<br />
[http://www.raritanvalleymathteam.org/ website]<br />
<br />
==New Mexico==<br />
<br />
The New Mexico ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==New York City==<br />
For information about the New York City Math Team, please visit [http://www.nycmathteam.org the NYC Math Team homepage].<br />
<br />
==North Carolina==<br />
<br />
North Carolina sends two 15-person ARML teams to compete at Penn State. The 2006 NC "A" team placed 1st in Division A. Invitations to spring team practice sessions are extended based on performances at the [http://courses.ncssm.edu/goebel/statecon/state.htm NC State Math Contest], AMC, AIME, Duke Math Meet and other math competitions. The top 15 or so finishers at the NC State Math Contest (Comprehensive) and all USAMO qualifiers are among those typically invited to practice sessions. Performance at these practice sessions and math contests determine team assignments. <br />
<br />
[http://courses.ncssm.edu/goebel/statecon/Inform/sites.htm NC State Math Contest qualifying events] are held throughout the state in February and March. Archie Benton, John Noland and several others coach and prepare the team with e-mail problems and the spring practice sessions held at the NC School of Science and Mathematics in Durham.<br />
<br />
==Ohio==<br />
<br />
Invitation is based on OCTM (a state-wide competition) and AMC scores. Also, at the OHMIO (second level of OCTM) the offer to join is extended to anyone who is interested. Team placement is based on a combination of OCTM scores, AMC scores, and how well the person does in practices. Ohio normally sends two teams, but is sending three this year because enough students were interested. Also, starting this year, the Ohio A team is competing in division A. The other two teams are competing in division B.<br />
<br />
The first practice is Sunday, April 22, and the second is Saturday, May 19.<br />
<br />
==Ontario==<br />
<br />
Ontario typically sends three teams of fifteen students. Team selections involve a team selection test and team practice performance scores. Practice sessions start in March and occur once a month until the actual competition. The practices are full day, 10AM - 4PM, on Saturdays. <br />
<br />
More information can be found at: http://www.torontomathcircles.ca/team-selection.html<br />
<br />
==Pennsylvania==<br />
===Pittsburgh===<br />
The Western PA region typically sends one team to ARML each year. In addition, weekly practices are held at Carnegie Mellon University throughout the academic year; these are open to any interested high school students in the Pittsburgh area (well-prepared middle school students are also welcome). <br />
<br />
A schedule for these practices, and other relevant details, can be found on the webpage of Misha Lavrov, the head coach: http://math.cmu.edu/~mlavrov/.<br />
<br />
===Lehigh Valley===<br />
For information about the Lehigh Valley ARML teams, please visit [http://www.lehigh.edu/~dmd1/arml.html the Lehigh Valley ARML homepage] and [http://www.lehigh.edu/~dmd1/logistics.html the Lehigh Valley ARML logistics page.].<br />
<br />
==South Carolina==<br />
<br />
To join the SC All-State Team, one must take a preliminary exam administered through their school. For more information, please contact [mailto:coach@scall-statemathteam.com coach@scall-statemathteam.com].<br />
<br />
The preliminary exam is composed of 25 questions (non multiple choice), and is usually composed of easy to mid range AMC-12 level questions. From this exam, approximately 50-60 (in 2006 it was 49) of the top scorers from the state are selected into the South Carolina All State Mathematics Team. The qualifying floor this year was 11 out of the 25 questions. After an individual is accepted into the SC All State Team, he or she is invited to one or two ARML practices which are usually composed of individual tests, team tests, and a power round test.<br />
<br />
* [[South Carolina ARML | SC ARML wiki page]]<br />
<br />
==Texas==<br />
<br />
Texas sends three teams each year to the Iowa ARML site. Interested students can submit a form on the team website to indicate interest. Dr. Edward Early, the head coach, forms every year's team by inviting any returning students who participated in the competition last year and who qualified for AIME again. The remaining slots are filled with newcomers. Most new invitations are sent out to students in 7th-10th grade, and selection is primarily based on performance on MATHCOUNTS and the AMC 10/12.<br />
<br />
The team's website is located at http://edwarde.create.stedwards.edu/ARML.html.<br />
<br />
==West Virginia==<br />
<br />
The West Virginia team is selected using the top 15 winners in West Virginia State Math Field Day. Winners 16-30 are used as potential alternates for the team. West Virginia State Math Field Day uses a similar format as the ARML, having an Individual Exam, Individual Short Answer Section, Team Questions, a Team Power Question, and 2 sets of relays of 5 each (there are 10 members in each team).<br />
<br />
==See Other==<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=820893#820893 Newer discussion on AoPS message boards.]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40434 Older discussion on AoPS message boards.]<br />
* [[ARML]]<br />
[[Category:ARML]]</div>INomOnCountdownhttps://artofproblemsolving.com/wiki/index.php?title=Joining_an_ARML_team&diff=85148Joining an ARML team2017-04-02T18:56:30Z<p>INomOnCountdown: /* Texas */</p>
<hr />
<div>Team selection for the [[American Regions Mathematics League]] varies from team to team.<br />
<br />
==Arizona==<br />
<br />
The Arizona ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==California==<br />
<br />
===Northern California===<br />
<br />
Practice meetings are held at UC Davis, Mathematical Sciences Building, room 2112, on Thursdays at 7pm. For more information, e-mail Eric Brattain-Morrin at [mailto:eric.brattain@gmail.com eric.brattain@gmail.com] and visit [http://www.math.ucdavis.edu/~exploration/arml/contact.php the team website].<br />
<br />
===San Diego===<br />
<br />
The SD team is entering just its third year of participation, so a permanent process of selecting team members has yet to be decided upon. The team is organized by and practices at the San Diego Math Circle (SDMC), and most of the students on last year's team were regular attendees at SDMC. Also, since the 2007 team contained no seniors, the organizers for the 2008 team are not planning on extending invites to new students unless scores on the AMC exams or the San Diego Math Olympiad (SDMO) are particularly high. <br />
<br />
A student who wishes to attend practice should try to take the SDMO. If a student wishes to speak with one of the coaches for the team, they may do so by contacting AoPS member 'Generating'.<br />
<br />
* [http://www.sdmathcircle.org/Welcome.php San Diego Math Circle website]<br />
<br />
===San Francisco Bay Area===<br />
<br />
The San Francisco Bay Area ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml]. Teams are coordinated by Tim Sanders, AoPS user farmertim. Coaching and practices are run by Moor Xu, AoPS user annoyingpi. One can contact them for further information, although nearly all questions are answered by the link provided above. <br />
In SFBA, anyone is welcome to sign up for an ARML team. The number of teams that will be going is determined by the amount of interest, and it very rare that someone who signs up will be excluded. Only the top three teams are selected by merit, and the other teams are chosen by geographical placement. The three top teams usually all place in the top 20, often even in the top 15 or 10.<br />
<br />
===Southern California===<br />
<br />
The Southern California team is open to residents of the following Southern California counties: Santa Barbara, Ventura, Los Angeles, Orange, Kern, San Bernardino, and Riverside. The organization fields three or four teams (45 or 60 students) and competes at the western ARML site in Las Vegas.<br />
<br />
Practices are held throughout the school year, approximately once a month, on the campus of California State University, Long Beach. Practices are normally held on Saturday afternoons. In addition, there is a Santa Barbara area group that meets and practices in Goleta and becomes part of the Southern California team.<br />
<br />
Team selection is based on all of the following criteria:<br />
<br />
* Attendence at practice sessions.<br />
* Performance on problems at practice sessions.<br />
* Performance at ARML itself in previous years.<br />
* Performance on AMC and AIME, including current and previous years.<br />
* Performance at CSULB Math Day at the Beach, a contest held in March.<br />
<br />
The coach is Dr. Kent Merryfield, a professor at CSULB. His AoPS user name is Kent Merryfield. Please contact him for further information.<br />
<br />
* [http://www.csulb.edu/depts/math/?q=node/20 Math Day at the Beach website.]<br />
* [http://www.csulb.edu/depts/math/?q=node/32 SoCal ARML website.]<br />
* [[Southern California ARML | SoCal ARML wiki page.]]<br />
<br />
== Connecticut ==<br />
Connecticut team selection is based on performance in math leagues across the state, from which a few upperclassmen can automatically qualify, and a runoff which is held in early March. To participate in the runoff, either have the math team coach at your school contact the league director in February to get the date or, if you don't participate in a league, contact Daniel Bochicchio, the current coach, at [mailto:dbochicchio@eosmith.org dbochicchio@eosmith.org]. Connecticut sends three teams.<br />
<br />
== Florida ==<br />
Florida ARML sends three teams to ARML each year. The selection criteria for the Florida ARML team takes into consideration several factors:<br />
<br />
*[[AMC]] and [[AIME]] performance<br />
*Past [[AMC]], [[AIME]], and [[USAMO]] scores<br />
*Past [[ARML]] performance<br />
*FAMAT-designated competitions<br />
*An annual statewide tryout test <br />
<br />
Florida ARML is organized by the Florida Student Association of Mathematics. Email [mailto:eliross2@aol.com eliross2@aol.com] for more information.<br />
<br />
== Georgia ==<br />
<br />
[[Georgia ARML]] has sent at least two teams to ARML each year since 1991. Beginning in 2008, we took advantage of the opening of the new Southern Site to add a third team (the "C" team). In 2011, we added a fourth team (the "D" team). In the past, the A team competed in the A division and the B team in the B division, but recently, the A team, the B team and the C team compete in the A division. We also field a few alternates as well. Students interested in participation should do well in [[Georgia mathematics competitions | local tournaments]] and the [[AMC]] and [[AIME]]. In addition to schools invited to the annual [[GCTM State Math Tournament | varsity state tournament]], the Georgia ARML coaches invite other individuals that are under serious consideration for the ARML team. The coaches select the team members during the state tournament based on [[AMC | USAMO index]], performance in local tournaments, and score at the state tournament. It is vital that any student under consideration has what we call "verified ability". This can be described as having good results from more than one contest or tournament. "Good results" can be described as placing higher than someone who was on the ARML team last year -- that is, doing better than someone who we know is good. <br />
<br />
The coaches attempt to select the best 30 students in the state (regardless of age or grade) to comprise the A and B teams -- those selected usually consist of USAMO qualifiers, the top 12 at the state tournament, and team veterans. (Of course, these groups are represented by overlapping Venn diagrams.) It is rare that a first-time senior is selected, although it does happen. Then the coaches select the best 15 young students (10th grade and below) to comprise a C team -- a team specifically comprised of up-and-comers who the coaches hope will be A and B team material in the future. Finally, the coaches allow anyone not selected, but who is interested in competing at ARML, to form a team of "walk-ons" which becomes the D team. It is rare for a student from the walk-ons to be so good that they move up to the A or B teams, but it has happened.<br />
<br />
The team usually practices on Sundays from the state tournament until the trip to ARML. The specific compositions of the A, B, C, and D teams are not usually determined until immediately before ARML. A member's team placement (on A, B, C, or D) depends on the person's performance against other team members in practice and the coaches's discretion.<br />
<br />
== Maine ==<br />
<br />
The two Maine ARML Teams consist of approximately the top 30 scorers on 5 [[MAML]] (Maine Association of Math Leagues) Meets. Training includes the problem set "Pete's Fabulous 42."<br />
<br />
* [http://www.maml.net MAML Website]<br />
<br />
==Maryland==<br />
<br />
===Montgomery (Montgomery County, MD)===<br />
<br />
Montgomery County typically sends four teams of high-schoolers and one team of middle-schoolers to ARML, with all 5 teams (Montgomery A, B C, D, and Junior) teams competing in division A. <br />
<br />
Top scorers from the Montgomery County high school math league are invited to ARML practices after the regular season, which ends around February. Top scorers from the middle school league are also invited. The Montgomery County ARML contact is Ms. Zenia Yang at the Yang Academy. Interested students from Montgomery County Public Schools who did not participate in the math leagues can ask their high school math team coaches to contact Ms. Yang and attend the weekly ARML practices to try out for the teams.<br />
<br />
Practices are usually held at Montgomery Blair High School on Thursdays from 6:30pm-9:30pm. Team selection is done by individual scores at practice and at the discretion of the coaches.<br />
<br />
===Chesapeake ARML===<br />
<br />
The Chesapeake ARML team draws on all areas and schools in Maryland not served by the Montgomery teams or by the public high schools in Howard or Baltimore Counties. Selection is based on AMC/AIME/USAMO scores, UMD Math Contest scores, participation on Chesapeake teams for PUMaC, HMMT, JHMT, Purple Comet, ARML Power Contest, and similar, and a series of practices done in April and May. For more information, contact the Chesapeake head coach, Catherine Asaro, at asaro@sff.net or here on AoPS in the Maryland Forum. Web site at http://chesapeakemath.catherineasaro.net<br />
<br />
===Howard County===<br />
The Howard County ARML teams draws the twelve public high schools in Howard County, MD.<br />
<br />
===Baltimore County===<br />
The Baltimore County ARML teams draws on the public high schools in Baltimore County, MD.<br />
<br />
==Michigan==<br />
<br />
Michigan sends three teams to Iowa. The teams are collectively known as the Michigan All-Stars. In general participants must place in the top 100 at MMPC, a local competition, to be in the running for the team. All members of the top 100 receive and invitation to be on the team. There are 3 practices that decide who is on which of the three teams. Although MMPC automatically qualifies for team selection, it is possible to get on a team by contacting coach Dave Friday. The three teams are the Reals, Naturals, and Primes. Two compete in A and 1 in B... Michigan was also awarded the best shirt design at ARML Iowa 2015...<br />
<br />
==Minnesota==<br />
<br />
Minnesota sends two teams to ARML each year, with the Gold and Maroon teams usually competing in divisions A and B, respectively.<br />
<br />
Roughly 35 students are invited to ARML practices, which take place on three consecutive Saturdays in May. There is no practice Memorial Day weekend. Invitations to the ARML team are extended to the top 10 state scorers on the AMC12, the top 10 regular-season scorers on the Minnesota High School Math League, and the top 10 scorers on the Invitational Event at the statewide math league tournament (held in March).<br />
<br />
Since these lists tend to overlap quite a bit, invitations are usually given to students "further down" these lists until enough invites have been given to fill two 15-person teams. <br />
<br />
In addition, an extra 5 or so younger students (typically in grades 8 through 10) are invited to be ARML "students in training". They may or may not go to ARML, but often serve as alternates (in the event that other students cannot attend). The expectation is that a student in training will learn from the practices, and the following year will be on one of the two teams. Since the creation of the AMC8 and AMC10 exams, the top scorers from these exams have typically been invited to ARML practices, either as team members or students in training.<br />
<br />
The selection of the Gold and Maroon teams is determined by students' performance at practices, and is not announced until the night before the competition.<br />
<br />
Usually at least one student in training is invited to go to ARML. This is to prevent a last-minute no-show (due to illness or emergency, for example) from crippling one of the teams.<br />
<br />
Invitations to participate on the MN team are usually sent out shortly after the MN State High School Math League statewide tournament in March. <br />
<br />
* [http://www.macalester.edu/mathleague/index.htm Minnesota State High School Math League site]<br />
<br />
==Missouri==<br />
<br />
Missouri sends two teams, Red and Blue, and a few alternates for a total of 36-37 students to the Iowa site ARML competition.<br />
<br />
Open practices are held usually once a month during the school year in two places, Springfield and St. Louis. Attendance is encouraged but not required. There are usually one or two "all day practices" a year at Missouri University of Science and Technology.<br />
<br />
Applications are based on AMC/AIME scores, GPML scores, other math competitions, and past ARML experience. Most if not all applicants are selected for the team.<br />
<br />
Since ARML is NOT a strictly a “state” team competition and because neither Kansas nor southern Illinois have ARML teams, ARML has given the “Missouri” team permission to include students from Kansas and from western Illinois (eg Edwardsville, etc.) on the MO ARML team. (eg. the 2008 team fielded 6 from KS, and 1 from IL)<br />
<br />
[http://math.missouristate.edu/MissouriARML.htm MO ARML homepage]<br />
<br />
==Nevada==<br />
<br />
The Nevada ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==New Jersey==<br />
New Jersey sends four different ARML teams. Also, many NJ students are on the Lehigh Valley teams.<br />
<br />
=== BCA/AAST ===<br />
AAST (a school in Bergen County Academies) sent 6 teams in 2012, all named after Greek Gods. Although there is a testing process to determine who's on which team, anyone in the AAST Math Team is permitted to participate in ARML. [https://bcts.bergen.org/index.php/math-team-home website]<br />
<br />
=== Central Jersey ===<br />
Qualification for the Central Jersey ARML team requires that<br />
*the student has participated (and chosen as a representative for the school) in at least 4 of the Central Jersey Math League meets,<br />
*and the average score meets a certain minimum which varies per year (this year, it is 3.0).<br />
<br />
Afterwards, interested students have to attend training sessions which take place at Highland Park High School or Hillsborough.<br />
<br />
More information can be found at [http://uzza.us/cjml/arml.html CJML website].<br />
<br />
=== WW-P ===<br />
<br />
WW-P, short for West Windsor-Plainsboro, is a team founded in 2012. Currently, it consists of students only from the WW-P school district, but students in close areas may be welcome as well. The selection process is a series of individual tests, and other experience is taken into consideration if needed. Two teams were sent in 2013; the A1 team ranked 3rd nationally. Also, they got 1st nationally in the B-Division two times in a row. More teams may be formed if more students join the program. <br />
<br />
If you would like more information, please email southmathclub@gmail.com.<br />
<br />
=== Raritan Valley ===<br />
<br />
[http://www.raritanvalleymathteam.org/ website]<br />
<br />
==New Mexico==<br />
<br />
The New Mexico ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==New York City==<br />
For information about the New York City Math Team, please visit [http://www.nycmathteam.org the NYC Math Team homepage].<br />
<br />
==North Carolina==<br />
<br />
North Carolina sends two 15-person ARML teams to compete at Penn State. The 2006 NC "A" team placed 1st in Division A. Invitations to spring team practice sessions are extended based on performances at the [http://courses.ncssm.edu/goebel/statecon/state.htm NC State Math Contest], AMC, AIME, Duke Math Meet and other math competitions. The top 15 or so finishers at the NC State Math Contest (Comprehensive) and all USAMO qualifiers are among those typically invited to practice sessions. Performance at these practice sessions and math contests determine team assignments. <br />
<br />
[http://courses.ncssm.edu/goebel/statecon/Inform/sites.htm NC State Math Contest qualifying events] are held throughout the state in February and March. Archie Benton, John Noland and several others coach and prepare the team with e-mail problems and the spring practice sessions held at the NC School of Science and Mathematics in Durham.<br />
<br />
==Ohio==<br />
<br />
Invitation is based on OCTM (a state-wide competition) and AMC scores. Also, at the OHMIO (second level of OCTM) the offer to join is extended to anyone who is interested. Team placement is based on a combination of OCTM scores, AMC scores, and how well the person does in practices. Ohio normally sends two teams, but is sending three this year because enough students were interested. Also, starting this year, the Ohio A team is competing in division A. The other two teams are competing in division B.<br />
<br />
The first practice is Sunday, April 22, and the second is Saturday, May 19.<br />
<br />
==Ontario==<br />
<br />
Ontario typically sends three teams of fifteen students. Team selections involve a team selection test and team practice performance scores. Practice sessions start in March and occur once a month until the actual competition. The practices are full day, 10AM - 4PM, on Saturdays. <br />
<br />
More information can be found at: http://www.torontomathcircles.ca/team-selection.html<br />
<br />
==Pennsylvania==<br />
===Pittsburgh===<br />
The Western PA region typically sends one team to ARML each year. In addition, weekly practices are held at Carnegie Mellon University throughout the academic year; these are open to any interested high school students in the Pittsburgh area (well-prepared middle school students are also welcome). <br />
<br />
A schedule for these practices, and other relevant details, can be found on the webpage of Misha Lavrov, the head coach: http://math.cmu.edu/~mlavrov/.<br />
<br />
===Lehigh Valley===<br />
For information about the Lehigh Valley ARML teams, please visit [http://www.lehigh.edu/~dmd1/arml.html the Lehigh Valley ARML homepage] and [http://www.lehigh.edu/~dmd1/logistics.html the Lehigh Valley ARML logistics page.].<br />
<br />
==South Carolina==<br />
<br />
To join the SC All-State Team, one must take a preliminary exam administered through their school. For more information, please contact [mailto:coach@scall-statemathteam.com coach@scall-statemathteam.com].<br />
<br />
The preliminary exam is composed of 25 questions (non multiple choice), and is usually composed of easy to mid range AMC-12 level questions. From this exam, approximately 50-60 (in 2006 it was 49) of the top scorers from the state are selected into the South Carolina All State Mathematics Team. The qualifying floor this year was 11 out of the 25 questions. After an individual is accepted into the SC All State Team, he or she is invited to one or two ARML practices which are usually composed of individual tests, team tests, and a power round test.<br />
<br />
* [[South Carolina ARML | SC ARML wiki page]]<br />
<br />
==Texas==<br />
<br />
Texas sends three teams each year to the Iowa ARML site. Interested students submit a form to indicate interest. Dr. Edward Early, the head coach, forms every year's team by inviting any returning students who participated in the competition last year and who qualified for AIME again. The remaining slots are filled with newcomers. Most new invitations are sent out to students in 7th-10th grade, and selection is primarily based on performance on MATHCOUNTS and the AMC 10/12.<br />
<br />
The team's website is located at http://edwarde.create.stedwards.edu/ARML.html.<br />
<br />
==West Virginia==<br />
<br />
The West Virginia team is selected using the top 15 winners in West Virginia State Math Field Day. Winners 16-30 are used as potential alternates for the team. West Virginia State Math Field Day uses a similar format as the ARML, having an Individual Exam, Individual Short Answer Section, Team Questions, a Team Power Question, and 2 sets of relays of 5 each (there are 10 members in each team).<br />
<br />
==See Other==<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=820893#820893 Newer discussion on AoPS message boards.]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40434 Older discussion on AoPS message boards.]<br />
* [[ARML]]<br />
[[Category:ARML]]</div>INomOnCountdown