Difference between revisions of "1966 AHSME Problems"
m (→Problem 5) |
m (→Problem 22) |
||
(11 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
+ | {{AHSC 40 Problems | ||
+ | |year = 1966 | ||
+ | }} | ||
== Problem 1 == | == Problem 1 == | ||
Given that the ratio of <math>3x - 4</math> to <math>y + 15</math> is constant, and <math>y = 3</math> when <math>x = 2</math>, then, when <math>y = 12</math>, <math>x</math> equals: | Given that the ratio of <math>3x - 4</math> to <math>y + 15</math> is constant, and <math>y = 3</math> when <math>x = 2</math>, then, when <math>y = 12</math>, <math>x</math> equals: | ||
Line 7: | Line 10: | ||
== Problem 2 == | == Problem 2 == | ||
+ | When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is | ||
+ | <math> \text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease} </math> | ||
[[1966 AHSME Problems/Problem 2|Solution]] | [[1966 AHSME Problems/Problem 2|Solution]] | ||
Line 52: | Line 57: | ||
== Problem 8 == | == Problem 8 == | ||
− | The length of the common chord of two intersecting circles is <math>16</math> feet. If the radii are <math>10</math> feet and <math>17</math> feet, a possible value for the distance between the centers of | + | The length of the common chord of two intersecting circles is <math>16</math> feet. If the radii are <math>10</math> feet and <math>17</math> feet, a possible value for the distance between the centers of the circles, expressed in feet, is: |
<math>\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}</math> | <math>\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}</math> | ||
Line 80: | Line 85: | ||
== Problem 12 == | == Problem 12 == | ||
+ | The number of real values of <math>x</math> that satisfy the equation <cmath>(2^{6x+3})(4^{3x+6})=8^{4x+5}</cmath> is: | ||
+ | <math>\text{(A) zero} \qquad \text{(B) one} \qquad \text{(C) two} \qquad \text{(D) three} \qquad \text{(E) greater than 3}</math> | ||
[[1966 AHSME Problems/Problem 12|Solution]] | [[1966 AHSME Problems/Problem 12|Solution]] | ||
== Problem 13 == | == Problem 13 == | ||
+ | The number of points with positive rational coordinates selected from the set of points in the <math>xy</math>-plane such that <math>x+y \le 5</math>, is: | ||
+ | <math>\text{(A)} \ 9 \qquad \text{(B)} \ 10 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E) infinite}</math> | ||
[[1966 AHSME Problems/Problem 13|Solution]] | [[1966 AHSME Problems/Problem 13|Solution]] | ||
== Problem 14 == | == Problem 14 == | ||
+ | The length of rectangle <math>ABCD</math> is 5 inches and its width is 3 inches. Diagonal <math>AC</math> is divided into three equal segments by points <math>E</math> and <math>F</math>. The area of triangle <math>BEF</math>, expressed in square inches, is: | ||
+ | <math>\text{(A)} \frac{3}{2} \qquad \text{(B)} \frac {5}{3} \qquad \text{(C)} \frac{5}{2} \qquad \text{(D)} \frac{1}{3}\sqrt{34} \qquad \text{(E)} \frac{1}{3}\sqrt{68}</math> | ||
[[1966 AHSME Problems/Problem 14|Solution]] | [[1966 AHSME Problems/Problem 14|Solution]] | ||
== Problem 15 == | == Problem 15 == | ||
+ | If <math>x-y>x</math> and <math>x+y<y</math>, then | ||
+ | <math>\text{(A) } y<x \quad \text{(B) } x<y \quad \text{(C) } x<y<0 \quad \text{(D) } x<0,y<0 \quad \text{(E) } x<0,y>0</math> | ||
[[1966 AHSME Problems/Problem 15|Solution]] | [[1966 AHSME Problems/Problem 15|Solution]] | ||
== Problem 16 == | == Problem 16 == | ||
+ | If <math>\frac{4^x}{2^{x+y}}=8</math> and <math>\frac{9^{x+y}}{3^{5y}}=243</math>, <math>x</math> and <math>y</math> real numbers, then <math>xy</math> equals: | ||
+ | <math>\text{(A) } \frac{12}{5} \quad \text{(B) } 4 \quad \text{(C) } 6 \quad \text{(D)} 12 \quad \text{(E) } -4</math> | ||
[[1966 AHSME Problems/Problem 16|Solution]] | [[1966 AHSME Problems/Problem 16|Solution]] | ||
== Problem 17 == | == Problem 17 == | ||
+ | The number of distinct points common to the curves <math>x^2+4y^2=1</math> and <math>4x^2+y^2=4</math> is: | ||
+ | <math>\text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3 \quad \text{(E) } 4</math> | ||
[[1966 AHSME Problems/Problem 17|Solution]] | [[1966 AHSME Problems/Problem 17|Solution]] | ||
== Problem 18 == | == Problem 18 == | ||
+ | In a given arithmetic sequence the first term is <math>2</math>, the last term is <math>29</math>, and the sum of all the terms is <math>155</math>. The common difference is: | ||
+ | <math>\text{(A) } 3 \qquad \text{(B) } 2 \qquad \text{(C) } \frac{27}{19} \qquad \text{(D) } \frac{13}{9} \qquad \text{(E) } \frac{23}{38}</math> | ||
[[1966 AHSME Problems/Problem 18|Solution]] | [[1966 AHSME Problems/Problem 18|Solution]] | ||
== Problem 19 == | == Problem 19 == | ||
+ | Let <math>s_1</math> be the sum of the first <math>n</math> terms of the arithmetic sequence <math>8,12,\cdots</math> and let <math>s_2</math> be the sum of the first <math>n</math> terms of the arithmetic sequence <math>17,19,\cdots</math>. Assume <math>n \ne 0</math>. Then <math>s_1=s_2</math> for: | ||
+ | <math>\text{(A) no value of } n \quad \text{(B) one value of } n \quad \text{(C) two values of } n \quad \text{(D) four values of } n \quad \text{(E) more than four values of } n</math> | ||
[[1966 AHSME Problems/Problem 19|Solution]] | [[1966 AHSME Problems/Problem 19|Solution]] | ||
== Problem 20 == | == Problem 20 == | ||
+ | The negation of the proposition "For all pairs of real numbers <math>a,b</math>, if <math>a=0</math>, then <math>ab=0</math>" is: There are real numbers <math>a,b</math> such that | ||
+ | <math>\text{(A) } a\ne 0 \text{ and } ab\ne 0 \qquad \text{(B) } a\ne 0 \text{ and } ab=0 \qquad \text{(C) } a=0 \text{ and } ab\ne 0</math> | ||
+ | |||
+ | <math>\text{(D) } ab\ne 0 \text{ and } a\ne 0 \qquad \text{(E) } ab=0 \text{ and } a\ne 0</math> | ||
[[1966 AHSME Problems/Problem 20|Solution]] | [[1966 AHSME Problems/Problem 20|Solution]] | ||
Line 126: | Line 151: | ||
== Problem 21 == | == Problem 21 == | ||
+ | <asy> | ||
+ | draw((0,-5)--(-6,10),black+dashed+linewidth(1)); | ||
+ | draw((-6,10)--(10,0),black+dashed+linewidth(1)); | ||
+ | draw((10,0)--(-10,0),black+dashed+linewidth(1)); | ||
+ | draw((-10,-0)--(10,10),black+dashed+linewidth(1)); | ||
+ | draw((10,10)--(0,-5),black+dashed+linewidth(1)); | ||
+ | draw((-2,0)--(-10/3,10/3),black+linewidth(2)); | ||
+ | draw((-10/3,10/3)--(10/9,50/9),black+linewidth(2)); | ||
+ | draw((10/9,50/9)--(90/17,50/17),black+linewidth(2)); | ||
+ | draw((90/17,50/17)--(10/3,0),black+linewidth(2)); | ||
+ | draw((10/3,0)--(-2,0),black+linewidth(2)); | ||
+ | MP("1", (1,0), N); MP("2", (-2.5,2), E); MP("3", (-.5,4.6), S); MP("4",(3.5,3.6), W);MP("5",(3.5,1), N); | ||
+ | </asy> | ||
+ | |||
+ | An "<math>n</math>-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively <math>1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5</math>; for all <math>n</math> values of <math>k</math>, sides <math>k</math> and <math>k+2</math> are non-parallel, sides <math>n+1</math> and <math>n+2</math> being respectively identical with sides <math>1</math> and <math>2</math>; prolong the <math>n</math> pairs of sides numbered <math>k</math> and <math>k+2</math> until they meet. (A figure is shown for the case <math>n=5</math>). | ||
+ | |||
+ | Let <math>S</math> be the degree-sum of the interior angles at the <math>n</math> points of the star; then <math>S</math> equals: | ||
+ | |||
+ | <math>\text{(A) } 180 \quad \text{(B) } 360 \quad \text{(C) } 180(n+2) \quad \text{(D) } 180(n-2) \quad \text{(E) } 180(n-4)</math> | ||
[[1966 AHSME Problems/Problem 21|Solution]] | [[1966 AHSME Problems/Problem 21|Solution]] | ||
== Problem 22 == | == Problem 22 == | ||
+ | Consider the statements: (I)<math>\sqrt{a^2+b^2}=0</math>, (II) <math>\sqrt{a^2+b^2}=ab</math>, (III) <math>\sqrt{a^2+b^2}=a+b</math>, (IV) <math>\sqrt{a^2+b^2}=a - b</math>, where we allow <math>a</math> and <math>b</math> to be real or complex numbers. Those statements for which there exist solutions other than <math>a=0</math> and <math>b=0</math>, are: | ||
+ | <math>\text{(A) (I),(II),(III),(IV)}\quad \text{(B) (II),(III),(IV) only} \quad \text{(C) (I),(III),(IV) only}\quad \text{(D) (III),(IV) only} \quad \text{(E) (I) only}</math> | ||
[[1966 AHSME Problems/Problem 22|Solution]] | [[1966 AHSME Problems/Problem 22|Solution]] | ||
== Problem 23 == | == Problem 23 == | ||
+ | If <math>x</math> is real and <math>4y^2+4xy+x+6=0</math>, then the complete set of values of <math>x</math> for which <math>y</math> is real, is: | ||
+ | <math>\text{(A) } x\le-2 \text{ or } x\ge3 \quad \text{(B) } x\le2 \text{ or } x\ge3 \quad \text{(C) } x\le-3 \text{ or } x\ge2 \quad \\ \text{(D) } -3\le x\le2 \quad \text{(E) } -2\le x\le3</math> | ||
[[1966 AHSME Problems/Problem 23|Solution]] | [[1966 AHSME Problems/Problem 23|Solution]] | ||
== Problem 24 == | == Problem 24 == | ||
+ | If <math>Log_M{N}=Log_N{M},M \ne N,MN>0,M \ne 1, N \ne 1</math>, then <math>MN</math> equals: | ||
+ | <math>\text{(A) } \frac{1}{2} \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 10 \\ \text{(E) a number greater than 2 and less than 10}</math> | ||
[[1966 AHSME Problems/Problem 24|Solution]] | [[1966 AHSME Problems/Problem 24|Solution]] | ||
== Problem 25 == | == Problem 25 == | ||
+ | If <math>F(n+1)=\frac{2F(n)+1}{2}</math> for <math>n=1,2,\cdots</math> and <math>F(1)=2</math>, then <math>F(101)</math> equals: | ||
+ | <math>\text{(A) } 49 \quad \text{(B) } 50 \quad \text{(C) } 51 \quad \text{(D) } 52 \quad \text{(E) } 53</math> | ||
[[1966 AHSME Problems/Problem 25|Solution]] | [[1966 AHSME Problems/Problem 25|Solution]] | ||
== Problem 26 == | == Problem 26 == | ||
+ | Let <math>m</math> be a positive integer and let the lines <math>13x+11y=700</math> and <math>y=mx-1</math> intersect in a point whose coordinates are integers. Then m can be: | ||
+ | <math>\text{(A) 4 only} \quad \text{(B) 5 only} \quad \text{(C) 6 only} \quad \text{(D) 7 only} \\ \text{(E) one of the integers 4,5,6,7 and one other positive integer}</math> | ||
[[1966 AHSME Problems/Problem 26|Solution]] | [[1966 AHSME Problems/Problem 26|Solution]] | ||
== Problem 27 == | == Problem 27 == | ||
+ | At his usual rate a man rows 15 miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream's current is: | ||
+ | <math>\text{(A) } 2 \quad \text{(B) } \frac{5}{2} \quad \text{(C) } 3 \quad \text{(D) } \frac{7}{2} \quad \text{(E) } 4</math> | ||
[[1966 AHSME Problems/Problem 27|Solution]] | [[1966 AHSME Problems/Problem 27|Solution]] | ||
Line 168: | Line 224: | ||
== Problem 29 == | == Problem 29 == | ||
+ | The number of positive integers less than <math>1000</math> divisible by neither <math>5</math> nor <math>7</math> is: | ||
+ | <math>\text{(A) } 688 \quad \text{(B) } 686 \quad \text{(C) } 684 \quad \text{(D) } 658 \quad \text{(E) } 630</math> | ||
[[1966 AHSME Problems/Problem 29|Solution]] | [[1966 AHSME Problems/Problem 29|Solution]] | ||
== Problem 30 == | == Problem 30 == | ||
+ | If three of the roots of <math>x^4+ax^2+bx+c=0</math> are <math>1</math>, <math>2</math>, and <math>3</math> then the value of <math>a+c</math> is: | ||
+ | <math>\text{(A) } 35 \quad \text{(B) } 24 \quad \text{(C) } -12 \quad \text{(D) } -61 \quad \text{(E) } -63</math> | ||
[[1966 AHSME Problems/Problem 30|Solution]] | [[1966 AHSME Problems/Problem 30|Solution]] | ||
+ | |||
+ | == Problem 31 == | ||
+ | <asy> | ||
+ | draw(circle((0,0),10),black+linewidth(1)); | ||
+ | draw(circle((-1.25,2.5),4.5),black+linewidth(1)); | ||
+ | dot((0,0)); | ||
+ | dot((-1.25,2.5)); | ||
+ | draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5)); | ||
+ | draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5)); | ||
+ | draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5)); | ||
+ | draw((-sqrt(96),-2)--(sqrt(96),-2),black+linewidth(.5)); | ||
+ | MP("O'", (0,0), W); | ||
+ | MP("O", (-2,2), W); | ||
+ | MP("A", (-10,-2), W); | ||
+ | MP("B", (10,-2), E); | ||
+ | MP("C", (-2,sqrt(96)), N); | ||
+ | MP("D", (sqrt(96)-2.5,7), NE); | ||
+ | </asy> | ||
+ | Triangle <math>ABC</math> is inscribed in a circle with center <math>O'</math>. A circle with center <math>O</math> is inscribed in triangle <math>ABC</math>. <math>AO</math> is drawn, and extended to intersect the larger circle in <math>D</math>. Then we must have: | ||
+ | |||
+ | <math>\text{(A) } CD=BD=O'D \quad \text{(B) } AO=CO=OD \quad \text{(C) } CD=CO=BD \\ \text{(D) } CD=OD=BD \quad \text{(E) } O'B=O'C=OD</math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 31|Solution]] | ||
+ | |||
+ | == Problem 32 == | ||
+ | Let <math>M</math> be the midpoint of side <math>AB</math> of triangle <math>ABC</math>. Let <math>P</math> be a point on <math>AB</math> between <math>A</math> and <math>M</math>, and let <math>MD</math> be drawn parallel to <math>PC</math> and intersecting <math>BC</math> at <math>D</math>. If the ratio of the area of triangle <math>BPD</math> to that of triangle <math>ABC</math> is denoted by <math>r</math>, then | ||
+ | |||
+ | <math>\text{(A) } \frac{1}{2}<r<1 \text{, depending upon the position of P} \\ \text{(B) } r=\frac{1}{2} \text{, independent of the position of P} \\ \text{(C) } \frac{1}{2} \le r <1 \text{, depending upon the position of P} \\ \text{(D) } \frac{1}{3}<r<\frac{2}{3} \text{, depending upon the position of P}\\ \text{(E) } r=\frac{1}{3} \text{, independent of the position of P}</math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 32|Solution]] | ||
+ | == Problem 33 == | ||
+ | If <math>ab \ne 0</math> and <math>|a| \ne |b|</math>, the number of distinct values of <math>x</math> satisfying the equation | ||
+ | |||
+ | <cmath>\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},</cmath> | ||
+ | |||
+ | is: | ||
+ | |||
+ | <math>\text{(A) zero} \quad \text{(B) one} \quad \text{(C) two} \quad \text{(D) three} \quad \text{(E) four} </math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 33|Solution]] | ||
+ | == Problem 34 == | ||
+ | Let <math>r</math> be the speed in miles per hour at which a wheel, <math>11</math> feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by <math>\frac{1}{4}</math> of a second, the speed <math>r</math> is increased by <math>5</math> miles per hour. Then <math>r</math> is: | ||
+ | |||
+ | <math>\text{(A) } 9 \quad \text{(B) } 10 \quad \text{(C) } 10\frac{1}{2} \quad \text{(D) } 11 \quad \text{(E) } 12</math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 34|Solution]] | ||
+ | == Problem 35 == | ||
+ | Let <math>O</math> be an interior point of triangle <math>ABC</math>, and let <math>s_1=OA+OB+OC</math>. If <math>s_2=AB+BC+CA</math>, then | ||
+ | |||
+ | <math>\text{(A) for every triangle } s_2>2s_1,s_1 \le s_2 \\ | ||
+ | \text{(B) for every triangle } s_2>2s_1,s_1 < s_2 \\ | ||
+ | \text{(C) for every triangle } s_1> \tfrac{1}{2}s_2,s_1 < s_2 \\ | ||
+ | \text{(D) for every triangle } s_2\ge 2s_1,s_1 \le s_2 \\ | ||
+ | \text{(E) neither (A) nor (B) nor (C) nor (D) applies to every triangle}</math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 35|Solution]] | ||
+ | == Problem 36 == | ||
+ | Let <math>(1+x+x^2)^n=a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}</math> be an identity in <math>x</math>. If we let <math>s=a_0+a_2+a_4+\cdots +a_{2n}</math>, then <math>s</math> equals: | ||
+ | |||
+ | <math>\text{(A) } 2^n \quad \text{(B) } 2^n+1 \quad \text{(C) } \frac{3^n-1}{2} \quad \text{(D) } \frac{3^n}{2} \quad \text{(E) } \frac{3^n+1}{2}</math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 36|Solution]] | ||
+ | == Problem 37 == | ||
+ | Three men, Alpha, Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let <math>h</math> be the number of hours needed by Alpha and Beta, working together, to do the job. Then <math>h</math> equals: | ||
+ | |||
+ | <math>\text{(A) } \frac{5}{2} \quad \text{(B) } \frac{3}{2} \quad \text{(C) } \frac{4}{3} \quad \text{(D) } \frac{5}{4} \quad \text{(E) } \frac{3}{4}</math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 37|Solution]] | ||
+ | == Problem 38 == | ||
+ | In triangle <math>ABC</math> the medians <math>AM</math> and <math>CN</math> to sides <math>BC</math> and <math>AB</math>, respectively, intersect in point <math>O</math>. <math>P</math> is the midpoint of side <math>AC</math>, and <math>MP</math> intersects <math>CN</math> in <math>Q</math>. If the area of triangle <math>OMQ</math> is <math>n</math>, then the area of triangle <math>ABC</math> is: | ||
+ | |||
+ | <math>\text{(A) } 16n \quad \text{(B) } 18n \quad \text{(C) } 21n \quad \text{(D) } 24n \quad \text{(E) } 27n</math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 38|Solution]] | ||
+ | == Problem 39 == | ||
+ | In base <math>R_1</math> the expanded fraction <math>F_1</math> becomes <math>.373737\cdots</math>, and the expanded fraction <math>F_2</math> becomes <math>.737373\cdots</math>. In base <math>R_2</math> fraction <math>F_1</math>, when expanded, becomes <math>.252525\cdots</math>, while the fraction <math>F_2</math> becomes <math>.525252\cdots</math>. The sum of <math>R_1</math> and <math>R_2</math>, each written in the base ten, is: | ||
+ | |||
+ | <math>\text{(A) } 24 \quad \text{(B) } 22 \quad \text{(C) } 21 \quad \text{(D) } 20 \quad \text{(E) } 19</math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 39|Solution]] | ||
+ | == Problem 40 == | ||
+ | <asy> | ||
+ | draw(circle((0,0),10),black+linewidth(1)); | ||
+ | MP("O", (0,0), S);MP("A", (-10,0), W);MP("B", (10,0), E);MP("C", (10,10), E);MP("D", (6,8), N); | ||
+ | MP("a", (-5,0), S);MP("E", (-6,3), N); | ||
+ | dot((0,0));dot((-6,2)); | ||
+ | draw((-10,0)--(10,0),black+linewidth(1)); | ||
+ | draw((-10,0)--(10,10),black+linewidth(1)); | ||
+ | draw((-10,-12)--(-10,12),black+linewidth(1)); | ||
+ | draw((10,-12)--(10,12),black+linewidth(1)); | ||
+ | </asy> | ||
+ | In this figure <math>AB</math> is a diameter of a circle, centered at <math>O</math>, with radius <math>a</math>. A chord <math>AD</math> is drawn and extended to meet the tangent to the circle at <math>B</math> in point <math>C</math>. Point <math>E</math> is taken on <math>AC</math> so the <math>AE=DC</math>. Denoting the distances of <math>E</math> from the tangent through <math>A</math> and from the diameter <math>AB</math> by <math>x</math> and <math>y</math>, respectively, we can deduce the relation: | ||
+ | |||
+ | <math>\text{(A) } y^2=\frac{x^3}{2a-x} \quad \text{(B) } y^2=\frac{x^3}{2a+x} \quad \text{(C) } y^4=\frac{x^2}{2a-x} \\ \text{(D) } x^2=\frac{y^2}{2a-x} \quad \text{(E) } x^2=\frac{y^2}{2a+x}</math> | ||
+ | |||
+ | [[1966 AHSME Problems/Problem 40|Solution]] | ||
== See also == | == See also == | ||
− | + | ||
− | * [[ | + | * [[AMC 12 Problems and Solutions]] |
− | |||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 40p box|year=1966|before=[[1965 AHSME|1965 AHSC]]|after=[[1967 AHSME|1967 AHSC]]}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 22:10, 11 March 2024
1966 AHSC (Answer Key) Printable versions: • AoPS Resources • PDF | ||
Instructions
| ||
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 |
Contents
- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 Problem 26
- 27 Problem 27
- 28 Problem 28
- 29 Problem 29
- 30 Problem 30
- 31 Problem 31
- 32 Problem 32
- 33 Problem 33
- 34 Problem 34
- 35 Problem 35
- 36 Problem 36
- 37 Problem 37
- 38 Problem 38
- 39 Problem 39
- 40 Problem 40
- 41 See also
Problem 1
Given that the ratio of to
is constant, and
when
, then, when
,
equals:
Problem 2
When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is
Problem 3
If the arithmetic mean of two numbers is and their geometric mean is
, then an equation with the given two numbers as roots is:
Problem 4
Circle I is circumscribed about a given square and circle II is inscribed in the given square. If is the ratio of the area of circle
to that of circle
, then
equals:
Problem 5
The number of values of satisfying the equation
is:
Problem 6
is the diameter of a circle centered at
.
is a point on the circle such that angle
is
. If the diameter of the circle is
inches, the length of chord
, expressed in inches, is:
Problem 7
Let be an identity in
. The numerical value of
is:
Problem 8
The length of the common chord of two intersecting circles is feet. If the radii are
feet and
feet, a possible value for the distance between the centers of the circles, expressed in feet, is:
Problem 9
If , then
equals:
Problem 10
If the sum of two numbers is 1 and their product is 1, then the sum of their cubes is:
Problem 11
The sides of triangle are in the ratio
.
is the angle-bisector drawn to the shortest side
, dividing it into segments
and
. If the length of
is
, then the length of the longer segment of
is:
Problem 12
The number of real values of that satisfy the equation
is:
Problem 13
The number of points with positive rational coordinates selected from the set of points in the -plane such that
, is:
Problem 14
The length of rectangle is 5 inches and its width is 3 inches. Diagonal
is divided into three equal segments by points
and
. The area of triangle
, expressed in square inches, is:
Problem 15
If and
, then
Problem 16
If and
,
and
real numbers, then
equals:
Problem 17
The number of distinct points common to the curves and
is:
Problem 18
In a given arithmetic sequence the first term is , the last term is
, and the sum of all the terms is
. The common difference is:
Problem 19
Let be the sum of the first
terms of the arithmetic sequence
and let
be the sum of the first
terms of the arithmetic sequence
. Assume
. Then
for:
Problem 20
The negation of the proposition "For all pairs of real numbers , if
, then
" is: There are real numbers
such that
Problem 21
An "-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively
; for all
values of
, sides
and
are non-parallel, sides
and
being respectively identical with sides
and
; prolong the
pairs of sides numbered
and
until they meet. (A figure is shown for the case
).
Let be the degree-sum of the interior angles at the
points of the star; then
equals:
Problem 22
Consider the statements: (I), (II)
, (III)
, (IV)
, where we allow
and
to be real or complex numbers. Those statements for which there exist solutions other than
and
, are:
Problem 23
If is real and
, then the complete set of values of
for which
is real, is:
Problem 24
If , then
equals:
Problem 25
If for
and
, then
equals:
Problem 26
Let be a positive integer and let the lines
and
intersect in a point whose coordinates are integers. Then m can be:
Problem 27
At his usual rate a man rows 15 miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream's current is:
Problem 28
Five points are taken in order on a straight line with distances
,
,
, and
.
is a point on the line between
and
and such that
. Then
equals:
Problem 29
The number of positive integers less than divisible by neither
nor
is:
Problem 30
If three of the roots of are
,
, and
then the value of
is:
Problem 31
Triangle
is inscribed in a circle with center
. A circle with center
is inscribed in triangle
.
is drawn, and extended to intersect the larger circle in
. Then we must have:
Problem 32
Let be the midpoint of side
of triangle
. Let
be a point on
between
and
, and let
be drawn parallel to
and intersecting
at
. If the ratio of the area of triangle
to that of triangle
is denoted by
, then
Problem 33
If and
, the number of distinct values of
satisfying the equation
is:
Problem 34
Let be the speed in miles per hour at which a wheel,
feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by
of a second, the speed
is increased by
miles per hour. Then
is:
Problem 35
Let be an interior point of triangle
, and let
. If
, then
Problem 36
Let be an identity in
. If we let
, then
equals:
Problem 37
Three men, Alpha, Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let be the number of hours needed by Alpha and Beta, working together, to do the job. Then
equals:
Problem 38
In triangle the medians
and
to sides
and
, respectively, intersect in point
.
is the midpoint of side
, and
intersects
in
. If the area of triangle
is
, then the area of triangle
is:
Problem 39
In base the expanded fraction
becomes
, and the expanded fraction
becomes
. In base
fraction
, when expanded, becomes
, while the fraction
becomes
. The sum of
and
, each written in the base ten, is:
Problem 40
In this figure
is a diameter of a circle, centered at
, with radius
. A chord
is drawn and extended to meet the tangent to the circle at
in point
. Point
is taken on
so the
. Denoting the distances of
from the tangent through
and from the diameter
by
and
, respectively, we can deduce the relation:
See also
1966 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1965 AHSC |
Followed by 1967 AHSC | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.