https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jazzachi&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T09:42:16ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_IMO_Problems/Problem_3&diff=960852018 IMO Problems/Problem 32018-07-10T05:05:40Z<p>Jazzachi: </p>
<hr />
<div>An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from <math>1</math> to <math>10</math><br />
<br />
<cmath>4</cmath><br />
<cmath>2\quad 6</cmath><br />
<cmath>5\quad 7 \quad 1</cmath><br />
<cmath>8\quad 3 \quad 10 \quad 9</cmath><br />
<br />
Does there exist an anti-Pascal triangle with <math>2018</math> rows which contains every integer from <math>1</math> to <math>1 + 2 + 3 + \dots + 2018</math>?<br />
== Solution ==</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=2018_IMO_Problems/Problem_3&diff=960842018 IMO Problems/Problem 32018-07-10T05:05:20Z<p>Jazzachi: Started page</p>
<hr />
<div>An [i]anti-Pascal[/i] triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from <math>1</math> to <math>10</math><br />
<br />
<cmath>4</cmath><br />
<cmath>2\quad 6</cmath><br />
<cmath>5\quad 7 \quad 1</cmath><br />
<cmath>8\quad 3 \quad 10 \quad 9</cmath><br />
<br />
Does there exist an anti-Pascal triangle with <math>2018</math> rows which contains every integer from <math>1</math> to <math>1 + 2 + 3 + \dots + 2018</math>?<br />
== Solution ==</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=2018_IMO_Problems/Problem_1&diff=960832018 IMO Problems/Problem 12018-07-10T05:04:43Z<p>Jazzachi: Started page</p>
<hr />
<div>Let <math>\Gamma</math> be the circumcircle of acute triangle <math>ABC</math>. Points <math>D</math> and <math>E</math> are on segments <math>AB</math> and <math>AC</math> respectively such that <math>AD = AE</math>. The perpendicular bisectors of <math>BD</math> and <math>CE</math> intersect minor arcs <math>AB</math> and <math>AC</math> of <math>\Gamma</math> at points <math>F</math> and <math>G</math> respectively. Prove that lines <math>DE</math> and <math>FG</math> are either parallel or they are the same line.<br />
<br />
== Solution ==</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_19&diff=960541981 AHSME Problems/Problem 192018-07-09T08:00:40Z<p>Jazzachi: created solutions page (Accidentally put it in main page before) what the hell were the answer choices?</p>
<hr />
<div>==Problem 19==<br />
In <math>\triangle ABC</math>, <math>M</math> is the midpoint of side <math>BC</math>, <math>AN</math> bisects <math>\angle BAC</math>, and <math>BN\perp AN</math>. If sides <math>AB</math> and <math>AC</math> have lengths <math>14</math> and <math>19</math>, respectively, then find <math>MN</math>.<br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair B=origin, A=14*dir(42), C=intersectionpoint(B--(30,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b);<br />
draw(N--B--A--N--M--C--A^^B--M);<br />
markscalefactor=0.1;<br />
draw(rightanglemark(B,N,A));<br />
pair point=N;<br />
label("$A$", A, dir(point--A));<br />
label("$B$", B, dir(point--B));<br />
label("$C$", C, dir(point--C));<br />
label("$M$", M, dir(point--M));<br />
label("$N$", N, dir(30));<br />
label(rotate(angle(dir(A--C)))*"$19$", A--C, dir(A--C)*dir(90));<br />
label(rotate(angle(dir(A--B)))*"$14$", A--B, dir(A--B)*dir(90));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ \dfrac{5}{2}\qquad\textbf{(C)}\ \dfrac{5}{2}-\sin\theta\qquad\textbf{(D)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\theta\qquad\textbf{(E)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\left(\dfrac{1}{2}\theta\right) </math><br />
<br />
== Solution ==<br />
Extend <math>BN</math> to meet <math>AC</math> at <math>Q</math>. Then <math>\triangle BNM \sim \triangle BQC</math>, so <math>BN=NQ</math> and <math>QC=19-AQ=2MN</math>. <br />
<br />
Since <math>\angle ANB=90^\circ = \angle ANQ</math>, <math>\angle BAN=\angle NAQ</math> (since <math>AN</math> is an angle bisector) and <math>\triangle ANB</math> and <math>\triangle ANQ</math> share side <math>AN</math>, <math>\triangle ANB \cong \triangle ANQ</math>. Thus <math>AQ=14</math>, and so <math>MN=\frac{19-AQ}{2}=\frac{5}{2}</math>, hence our answer is <math>\fbox{B}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems&diff=960531981 AHSME Problems2018-07-09T07:59:47Z<p>Jazzachi: Undo revision 96052 by Jazzachi (talk)</p>
<hr />
<div>==Problem 1==<br />
If <math>\sqrt{x+2}=2</math>, then <math>(x+2)^{2}</math> equals<br />
<br />
<math> \textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16 </math><br />
<br />
[[1981 AHSME Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Point <math>E</math> is on side <math>AB</math> of square <math>ABCD</math>. If <math>EB</math> has length one and <math>EC</math> has length two, then the area of the square is<br />
<br />
<math> \textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5 </math><br />
<br />
[[1981 AHSME Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
For <math>x\neq0</math>, <math>\dfrac{1}{x}+\dfrac{1}{2x}+\dfrac{1}{3x}</math> equals<br />
<br />
<math> \textbf{(A)}\ \dfrac{1}{2x}\qquad\textbf{(B)}\ \dfrac{1}{6}\qquad\textbf{(C)}\ \dfrac{5}{6x}\qquad\textbf{(D)}\ \dfrac{11}{6x}\qquad\textbf{(E)}\ \dfrac{1}{6x^3} </math><br />
<br />
[[1981 AHSME Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
If three times the larger of two numbers is four times the smaller and the difference between the numbers is 8, the the larger of two numbers is <br />
<br />
<math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 44\qquad\textbf{(E)}\ 52 </math><br />
<br />
[[1981 AHSME Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
In trapezoid <math>ABCD</math>, sides <math>AB</math> and <math>CD</math> are parallel, and diagonal <math>BD</math> and side <math>AD</math> have equal length. If <math>m\angle DCB=110^\circ </math> and <math> m\angle CBD=30^\circ </math>, then <math> m\angle ADB=</math><br />
<br />
<math> \textbf{(A)}\ 80^\circ\qquad\textbf{(B)}\ 90^\circ\qquad\textbf{(C)}\ 100^\circ\qquad\textbf{(D)}\ 110^\circ\qquad\textbf{(E)}\ 120^\circ </math><br />
<br />
[[1981 AHSME Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
If <math> \dfrac{x}{x-1}=\dfrac{y^2+2y-1}{y^2+2y-2} </math>, then <math>x</math> equals <br />
<br />
<math> \textbf{(A)}\ y^2+2y-1\qquad\textbf{(B)}\ y^2+2y-2\qquad\textbf{(C)}\ y^2+2y+2 \qquad \\ \textbf{(D)}\ y^2+2y+1\qquad\textbf{(E)}\ -y^2-2y+1 </math><br />
<br />
[[1981 AHSME Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
How many of the first one hundred positive integers are divisible by all of the numbers <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</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 />
[[1981 AHSME Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
For all positive numbers <math>x</math>, <math>y</math>, <math>z</math>, the product <math> (x+y+z)^{-1}(x^{-1}+y^{-1}+z^{-1})(xy+yz+xz)^{-1}[(xy)^{-1}+(yz)^{-1}+(xz)^{-1}] </math> equals <math> \textbf{(A)}\ x^{-2}y^{-2}z^{-2}\qquad\textbf{(B)}\ x^{-2}+y^{-2}+z^{-2}\qquad\textbf{(C)}\ (x+y+z)^{-1}\qquad \textbf{(D)}\ \dfrac{1}{xyz}\qquad \\ \textbf{(E)}\ \dfrac{1}{xy+yz+xz} </math><br />
<br />
<br />
[[1981 AHSME Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
In the adjoining figure, <math>PQ</math> is a diagonal of the cube. If <math>PQ</math> has length <math>a</math>, then the surface area of the cube is<br />
<br />
<asy><br />
import three;<br />
unitsize(1cm);<br />
size(200);<br />
currentprojection=orthographic(1/3,-1,1/2);<br />
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle,black);<br />
draw((0,0,0)--(0,0,1),black);<br />
draw((0,1,0)--(0,1,1),black);<br />
draw((1,1,0)--(1,1,1),black);<br />
draw((1,0,0)--(1,0,1),black);<br />
draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle,black);<br />
draw((0,0,0)--(1,1,1),black);<br />
label("$P$",(0, 0, 0),NW);<br />
label("$Q$",(1, 1, 1),NE);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 2a^2\qquad\textbf{(B)}\ 2\sqrt{2}a^2\qquad\textbf{(C)}\ 2\sqrt{3}a^2\qquad\textbf{(D)}\ 3\sqrt{3}a^2\qquad\textbf{(E)}\ 6a^2 </math><br />
<br />
[[1981 AHSME Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
The lines <math>L</math> and <math>K</math> are symmetric to each other with respect to the line <math>y=x</math>. If the equation of the line <math>L</math> is <math>y=ax+b</math> with <math>a\neq 0</math> and <math>b\neq 0</math>, then the equation of <math>K</math> is <math>y=</math><br />
<br />
<math> \textbf{(A)}\ \dfrac{1}{a}x+b\qquad\textbf{(B)}\ -\dfrac{1}{a}x+b\qquad\textbf{(C)}\ \dfrac{1}{a}x-\dfrac{b}{a}\qquad\textbf{(D)}\ \dfrac{1}{a}x+\dfrac{b}{a}\qquad\textbf{(E)}\ \dfrac{1}{a}x-\dfrac{b}{a} </math><br />
<br />
[[1981 AHSME Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the sides could have length<br />
<br />
<math> \textbf{(A)}\ 22\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 81\qquad\textbf{(D)}\ 91\qquad\textbf{(E)}\ 361 </math><br />
<br />
[[1981 AHSME Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
If <math>p</math>, <math>q</math>, and <math>M</math> are positive numbers and <math>q<100</math>, then the number obtained by increasing <math>M</math> by <math>p\%</math> and decreasing the result by <math>q\%</math> exceeds <math>M</math> if and only if<br />
<br />
<math> \textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad \\ \textbf{(E)}\ p>\dfrac{100q}{100-q} </math><br />
<br />
[[1981 AHSME Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
Suppose that at the end of any year, a unit of money has lost <math>10\%</math> of the value it had at the beginning of that year. Find the smallest integer <math>n</math> such that after <math>n</math> years, the money will have lost at least <math>90\%</math> of its value (To the nearest thousandth <math>\log_{10} 3 = 0.477</math>).<br />
<br />
<math> \textbf{(A)}\ 14\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 22 </math><br />
<br />
[[1981 AHSME Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
In a geometric sequence of real numbers, the sum of the first <math>2</math> terms is <math>7</math>, and the sum of the first <math>6</math> terms is <math>91</math>. The sum of the first <math>4</math> terms is<br />
<br />
<math> \textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84 </math><br />
<br />
[[1981 AHSME Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
If <math>b>1</math>, <math>x>0</math>, and <math> (2x)^{\log_b 2}-(3x)^{\log_b 3}=0 </math>, then <math>x</math> is<br />
<br />
<math> \textbf{(A)}\ \dfrac{1}{216}\qquad\textbf{(B)}\ \dfrac{1}{6}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \text{not uniquely determined} </math><br />
<br />
[[1981 AHSME Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The base three representation of <math>x</math> is <br />
<cmath>12112211122211112222</cmath><br />
The first digit (on the left) of the base nine representation of <math>x</math> is<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math><br />
<br />
[[1981 AHSME Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
The function <math>f</math> is not defined for <math>x=0</math>, but, for all non-zero real numbers <math>x</math>, <math> f(x)+f\left(\dfrac{1}x\right)=x </math>. The equation <math> f(x)=f(-x) </math> is satisfied by<br />
<br />
<math> \textbf{(A)}\ \text{exactly one real number} \qquad \textbf{(B)}\ \text{exactly two real numbers} \qquad\textbf{(C)}\ \text{no real numbers}\qquad \\ \textbf{(D)}\ \text{infinitely many, but not all, non-zero real numbers} \qquad\textbf{(E)}\ \text{all non-zero real numbers} </math><br />
<br />
[[1981 AHSME Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
The number of real solutions to the equation<br />
<cmath>\dfrac{x}{100}=\sin x</cmath><br />
is<br />
<br />
<math> \textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65 </math><br />
<br />
[[1981 AHSME Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
In <math>\triangle ABC</math>, <math>M</math> is the midpoint of side <math>BC</math>, <math>AN</math> bisects <math>\angle BAC</math>, and <math>BN\perp AN</math>. If sides <math>AB</math> and <math>AC</math> have lengths <math>14</math> and <math>19</math>, respectively, then find <math>MN</math>.<br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair B=origin, A=14*dir(42), C=intersectionpoint(B--(30,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b);<br />
draw(N--B--A--N--M--C--A^^B--M);<br />
markscalefactor=0.1;<br />
draw(rightanglemark(B,N,A));<br />
pair point=N;<br />
label("$A$", A, dir(point--A));<br />
label("$B$", B, dir(point--B));<br />
label("$C$", C, dir(point--C));<br />
label("$M$", M, dir(point--M));<br />
label("$N$", N, dir(30));<br />
label(rotate(angle(dir(A--C)))*"$19$", A--C, dir(A--C)*dir(90));<br />
label(rotate(angle(dir(A--B)))*"$14$", A--B, dir(A--B)*dir(90));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ \dfrac{5}{2}\qquad\textbf{(C)}\ \dfrac{5}{2}-\sin\theta\qquad\textbf{(D)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\theta\qquad\textbf{(E)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\left(\dfrac{1}{2}\theta\right) </math><br />
<br />
[[1981 AHSME Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
A ray of light originates from point <math>A</math> and travels in a plane, being reflected <math>n</math> times between lines <math>AD</math> and <math>CD</math> before striking a point <math>B</math> (which may be on <math>AD</math> or <math>CD</math>) perpendicularly and retracing its path back to <math>A</math> (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for <math>n=3</math>). If <math>\measuredangle CDA=8^\circ</math>, what is the largest value <math>n</math> can have?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
pair D=origin, A=(-6,0), C=6*dir(160), E=3.2*dir(160), F=(-2.1,0), G=1.5*dir(160), B=(-1.4095,0);<br />
draw((-6.5,0)--D--C,black);<br />
draw(A--E--F--G--B,black);<br />
dotfactor=4;<br />
dot("$A$",A,S);<br />
dot("$C$",C,N);<br />
dot("$R_1$",E,N);<br />
dot("$R_2$",F,S);<br />
dot("$R_3$",G,N);<br />
dot("$B$",B,S);<br />
markscalefactor=0.015;<br />
draw(rightanglemark(G,B,D));<br />
draw(anglemark(C,E,A,12));<br />
draw(anglemark(F,E,G,12));<br />
draw(anglemark(E,F,A));<br />
draw(anglemark(E,F,A,12));<br />
draw(anglemark(B,F,G));<br />
draw(anglemark(B,F,G,12));<br />
draw(anglemark(E,G,F));<br />
draw(anglemark(E,G,F,12));<br />
draw(anglemark(E,G,F,16));<br />
draw(anglemark(B,G,D));<br />
draw(anglemark(B,G,D,12));<br />
draw(anglemark(B,G,D,16));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.} </math><br />
<br />
[[1981 AHSME Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
In a triangle with sides of lengths <math>a</math>, <math>b</math>, and <math>c</math>, <math> (a+b+c)(a+b-c) = 3ab </math>. The measure of the angle opposite the side length <math>c</math> is<br />
<br />
<math> \textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ </math><br />
<br />
[[1981 AHSME Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
How many lines in a three dimensional rectangular coordiante system pass through four distinct points of the form <math>(i,j,k)</math>, where <math>i</math>, <math>j</math>, and <math>k</math> are positive integers not exceeding four?<br />
<br />
<math> \textbf{(A)}\ 60\qquad\textbf{(B)}\ 64\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 100 </math><br />
<br />
[[1981 AHSME Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
Equilateral <math>\triangle ABC</math> is inscribed in a circle. A second circle is tangent internally to the circumcircle at <math>T</math> and tangent to sides <math>AB</math> and <math>AC</math> at points <math>P</math> and <math>Q</math>. If side <math>BC</math> has length <math>12</math>, then segment <math>PQ</math> has length<br />
<br />
<asy><br />
defaultpen(linewidth(.8pt));<br />
pair B = origin;<br />
pair A = dir(60);<br />
pair C = dir(0);<br />
pair circ = circumcenter(A,B,C);<br />
pair P = intersectionpoint(circ--(circ + (-1,0)),A--B);<br />
pair Q = intersectionpoint(circ--(circ + (1,0)),A--C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$P$",P,NW);<br />
label("$Q$",Q,NE);<br />
label("$T$",(0.5,-0.3),S);<br />
draw(A--B--C--cycle);<br />
draw(circumcircle(A,B,C));<br />
draw(P--Q);<br />
draw(Circle((0.5,0.09),0.385));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 6\sqrt{3}\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 8\sqrt{3}\qquad\textbf{(E)}\ 9 </math><br />
<br />
[[1981 AHSME Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
If <math> \theta</math> is a constant such that <math> 0 < \theta < \pi</math> and <math> x + \dfrac{1}{x} = 2\cos{\theta}</math>, then for each positive integer <math> n</math>, <math> x^n + \dfrac{1}{x^n}</math> equals<br />
<br />
<math> \textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta</math><br />
<br />
[[1981 AHSME Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
In <math>\triangle ABC</math> in the adjoining figure, <math> AD</math> and <math> AE</math> trisect <math> \angle BAC</math>. The lengths of <math> BD</math>, <math> DE</math> and <math> EC</math> are <math> 2</math>, <math> 3</math>, and <math> 6</math>, respectively. The length of the shortest side of <math> \triangle ABC</math> is<br />
<br />
<asy><br />
defaultpen(linewidth(.8pt));<br />
pair A = (0,11);<br />
pair B = (2,0);<br />
pair D = (4,0);<br />
pair E = (7,0);<br />
pair C = (13,0);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,S);<br />
label("$E$",E,S);<br />
label("$2$",midpoint(B--D),N);<br />
label("$3$",midpoint(D--E),NW);<br />
label("$6$",midpoint(E--C),NW);<br />
draw(A--B--C--cycle);<br />
draw(A--D);<br />
draw(A--E);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 2\sqrt{10}\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 6\sqrt{6}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}</math><br />
<br />
[[1981 AHSME Problems/Problem 25|Solution]]<br />
<br />
==Problem 26==<br />
Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is <math> \dfrac{1}{6}</math>, independent of the outcome of any other toss.)<br />
<br />
<math> \textbf{(A)}\ \dfrac{1}{3}\qquad \textbf{(B)}\ \dfrac{2}{9}\qquad \textbf{(C)}\ \dfrac{5}{18}\qquad \textbf{(D)}\ \dfrac{25}{91}\qquad \textbf{(E)}\ \dfrac{36}{91}</math><br />
<br />
[[1981 AHSME Problems/Problem 26|Solution]]<br />
<br />
==Problem 27==<br />
In the adjoining figure triangle <math> ABC</math> is inscribed in a circle. Point <math> D</math> lies on <math> \stackrel{\frown}{AC}</math> with <math> \stackrel{\frown}{DC} = 30^\circ</math>, and point <math> G</math> lies on <math> \stackrel{\frown}{BA}</math> with <math> \stackrel{\frown}{BG}\, > \, \stackrel{\frown}{GA}</math>. Side <math> AB</math> and side <math> AC</math> each have length equal to the length of chord <math> DG</math>, and <math> \angle CAB = 30^\circ</math>. Chord <math> DG</math> intersects sides <math> AC</math> and <math> AB</math> at <math> E</math> and <math> F</math>, respectively. The ratio of the area of <math> \triangle AFE</math> to the area of <math> \triangle ABC</math> is<br />
<br />
<asy><br />
defaultpen(linewidth(.8pt));<br />
pair C = origin;<br />
pair A = 2.5*dir(75);<br />
pair B = A + 2.5*dir(-75);<br />
path circ =circumcircle(A,B,C);<br />
pair D = waypoint(circ,(7/12));<br />
pair G = waypoint(circ,(1/6));<br />
pair E = intersectionpoint(D--G,A--C);<br />
pair F = intersectionpoint(A--B,D--G);<br />
label("$A$",A,N);<br />
label("$B$",B,SE);<br />
label("$C$",C,SW);<br />
label("$D$",D,SW);<br />
label("$G$",G,NE);<br />
label("$E$",E,NW);<br />
label("$F$",F,W);<br />
label("$30^\circ$",A,12S+E,fontsize(6pt));<br />
draw(A--B--C--cycle);<br />
draw(circ);<br />
draw(Arc(A,0.25,-75,-105));<br />
draw(D--G);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \dfrac {2 - \sqrt {3}}{3}\qquad \textbf{(B)}\ \dfrac {2\sqrt {3} - 3}{3}\qquad \textbf{(C)}\ 7\sqrt {3}-12\qquad \textbf{(D)}\ 3\sqrt {3}-5\qquad\\ \textbf{(E)}\ \dfrac {9-5\sqrt {3}}{3}</math><br />
<br />
[[1981 AHSME Problems/Problem 27|Solution]]<br />
<br />
==Problem 28==<br />
Consider the set of all equations <math> x^3 + a_2x^2 + a_1x + a_0 = 0</math>, where <math> a_2</math>, <math> a_1</math>, <math> a_0</math> are real constants and <math> |a_i| < 2</math> for <math> i = 0,1,2</math>. Let <math> r</math> be the largest positive real number which satisfies at least one of these equations. Then<br />
<br />
<math> \textbf{(A)}\ 1 < r < \dfrac{3}{2}\qquad \textbf{(B)}\ \dfrac{3}{2} < r < 2\qquad \textbf{(C)}\ 2 < r < \dfrac{5}{2}\qquad \textbf{(D)}\ \dfrac{5}{2} < r < 3\qquad \\ \textbf{(E)}\ 3 < r < \dfrac{7}{2}</math><br />
<br />
[[1981 AHSME Problems/Problem 28|Solution]]<br />
<br />
==Problem 29==<br />
If <math> a > 1</math>, then the sum of the real solutions of <br />
<br />
<math> \sqrt{a - \sqrt{a + x}} = x</math><br />
<br />
is equal to <br />
<br />
<math> \textbf{(A)}\ \sqrt{a} - 1\qquad \textbf{(B)}\ \dfrac{\sqrt{a}- 1}{2}\qquad \textbf{(C)}\ \sqrt{a - 1}\qquad \textbf{(D)}\ \dfrac{\sqrt{a - 1}}{2}\qquad \textbf{(E)}\ \dfrac{\sqrt{4a- 3} - 1}{2}</math><br />
<br />
[[1981 AHSME Problems/Problem 29|Solution]]<br />
<br />
==Problem 30==<br />
If <math> a</math>, <math> b</math>, <math> c</math>, and <math> d</math> are the solutions of the equation <math> x^4 - bx - 3 = 0</math>, then an equation whose solutions are <br />
<cmath>\dfrac {a + b + c}{d^2}, \dfrac {a + b + d}{c^2}, \dfrac {a + c + d}{b^2}, \dfrac {b + c + d}{a^2}</cmath>is<br />
<br />
<math> \textbf{(A)}\ 3x^4 + bx + 1 = 0\qquad \textbf{(B)}\ 3x^4 - bx + 1 = 0\qquad \textbf{(C)}\ 3x^4 + bx^3 - 1 = 0\qquad \\\textbf{(D)}\ 3x^4 - bx^3 - 1 = 0\qquad \textbf{(E)}\ \text{none of these}</math><br />
<br />
[[1981 AHSME Problems/Problem 30|Solution]]<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME box|year=1981|before=[[1980 AHSME]]|after=[[1982 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems&diff=960521981 AHSME Problems2018-07-09T07:58:48Z<p>Jazzachi: added solution. what the hell were the answer choices?</p>
<hr />
<div>==Problem 1==<br />
If <math>\sqrt{x+2}=2</math>, then <math>(x+2)^{2}</math> equals<br />
<br />
<math> \textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16 </math><br />
<br />
[[1981 AHSME Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Point <math>E</math> is on side <math>AB</math> of square <math>ABCD</math>. If <math>EB</math> has length one and <math>EC</math> has length two, then the area of the square is<br />
<br />
<math> \textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5 </math><br />
<br />
[[1981 AHSME Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
For <math>x\neq0</math>, <math>\dfrac{1}{x}+\dfrac{1}{2x}+\dfrac{1}{3x}</math> equals<br />
<br />
<math> \textbf{(A)}\ \dfrac{1}{2x}\qquad\textbf{(B)}\ \dfrac{1}{6}\qquad\textbf{(C)}\ \dfrac{5}{6x}\qquad\textbf{(D)}\ \dfrac{11}{6x}\qquad\textbf{(E)}\ \dfrac{1}{6x^3} </math><br />
<br />
[[1981 AHSME Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
If three times the larger of two numbers is four times the smaller and the difference between the numbers is 8, the the larger of two numbers is <br />
<br />
<math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 44\qquad\textbf{(E)}\ 52 </math><br />
<br />
[[1981 AHSME Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
In trapezoid <math>ABCD</math>, sides <math>AB</math> and <math>CD</math> are parallel, and diagonal <math>BD</math> and side <math>AD</math> have equal length. If <math>m\angle DCB=110^\circ </math> and <math> m\angle CBD=30^\circ </math>, then <math> m\angle ADB=</math><br />
<br />
<math> \textbf{(A)}\ 80^\circ\qquad\textbf{(B)}\ 90^\circ\qquad\textbf{(C)}\ 100^\circ\qquad\textbf{(D)}\ 110^\circ\qquad\textbf{(E)}\ 120^\circ </math><br />
<br />
[[1981 AHSME Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
If <math> \dfrac{x}{x-1}=\dfrac{y^2+2y-1}{y^2+2y-2} </math>, then <math>x</math> equals <br />
<br />
<math> \textbf{(A)}\ y^2+2y-1\qquad\textbf{(B)}\ y^2+2y-2\qquad\textbf{(C)}\ y^2+2y+2 \qquad \\ \textbf{(D)}\ y^2+2y+1\qquad\textbf{(E)}\ -y^2-2y+1 </math><br />
<br />
[[1981 AHSME Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
How many of the first one hundred positive integers are divisible by all of the numbers <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</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 />
[[1981 AHSME Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
For all positive numbers <math>x</math>, <math>y</math>, <math>z</math>, the product <math> (x+y+z)^{-1}(x^{-1}+y^{-1}+z^{-1})(xy+yz+xz)^{-1}[(xy)^{-1}+(yz)^{-1}+(xz)^{-1}] </math> equals <math> \textbf{(A)}\ x^{-2}y^{-2}z^{-2}\qquad\textbf{(B)}\ x^{-2}+y^{-2}+z^{-2}\qquad\textbf{(C)}\ (x+y+z)^{-1}\qquad \textbf{(D)}\ \dfrac{1}{xyz}\qquad \\ \textbf{(E)}\ \dfrac{1}{xy+yz+xz} </math><br />
<br />
<br />
[[1981 AHSME Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
In the adjoining figure, <math>PQ</math> is a diagonal of the cube. If <math>PQ</math> has length <math>a</math>, then the surface area of the cube is<br />
<br />
<asy><br />
import three;<br />
unitsize(1cm);<br />
size(200);<br />
currentprojection=orthographic(1/3,-1,1/2);<br />
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle,black);<br />
draw((0,0,0)--(0,0,1),black);<br />
draw((0,1,0)--(0,1,1),black);<br />
draw((1,1,0)--(1,1,1),black);<br />
draw((1,0,0)--(1,0,1),black);<br />
draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle,black);<br />
draw((0,0,0)--(1,1,1),black);<br />
label("$P$",(0, 0, 0),NW);<br />
label("$Q$",(1, 1, 1),NE);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 2a^2\qquad\textbf{(B)}\ 2\sqrt{2}a^2\qquad\textbf{(C)}\ 2\sqrt{3}a^2\qquad\textbf{(D)}\ 3\sqrt{3}a^2\qquad\textbf{(E)}\ 6a^2 </math><br />
<br />
[[1981 AHSME Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
The lines <math>L</math> and <math>K</math> are symmetric to each other with respect to the line <math>y=x</math>. If the equation of the line <math>L</math> is <math>y=ax+b</math> with <math>a\neq 0</math> and <math>b\neq 0</math>, then the equation of <math>K</math> is <math>y=</math><br />
<br />
<math> \textbf{(A)}\ \dfrac{1}{a}x+b\qquad\textbf{(B)}\ -\dfrac{1}{a}x+b\qquad\textbf{(C)}\ \dfrac{1}{a}x-\dfrac{b}{a}\qquad\textbf{(D)}\ \dfrac{1}{a}x+\dfrac{b}{a}\qquad\textbf{(E)}\ \dfrac{1}{a}x-\dfrac{b}{a} </math><br />
<br />
[[1981 AHSME Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the sides could have length<br />
<br />
<math> \textbf{(A)}\ 22\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 81\qquad\textbf{(D)}\ 91\qquad\textbf{(E)}\ 361 </math><br />
<br />
[[1981 AHSME Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
If <math>p</math>, <math>q</math>, and <math>M</math> are positive numbers and <math>q<100</math>, then the number obtained by increasing <math>M</math> by <math>p\%</math> and decreasing the result by <math>q\%</math> exceeds <math>M</math> if and only if<br />
<br />
<math> \textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad \\ \textbf{(E)}\ p>\dfrac{100q}{100-q} </math><br />
<br />
[[1981 AHSME Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
Suppose that at the end of any year, a unit of money has lost <math>10\%</math> of the value it had at the beginning of that year. Find the smallest integer <math>n</math> such that after <math>n</math> years, the money will have lost at least <math>90\%</math> of its value (To the nearest thousandth <math>\log_{10} 3 = 0.477</math>).<br />
<br />
<math> \textbf{(A)}\ 14\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 22 </math><br />
<br />
[[1981 AHSME Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
In a geometric sequence of real numbers, the sum of the first <math>2</math> terms is <math>7</math>, and the sum of the first <math>6</math> terms is <math>91</math>. The sum of the first <math>4</math> terms is<br />
<br />
<math> \textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84 </math><br />
<br />
[[1981 AHSME Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
If <math>b>1</math>, <math>x>0</math>, and <math> (2x)^{\log_b 2}-(3x)^{\log_b 3}=0 </math>, then <math>x</math> is<br />
<br />
<math> \textbf{(A)}\ \dfrac{1}{216}\qquad\textbf{(B)}\ \dfrac{1}{6}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \text{not uniquely determined} </math><br />
<br />
[[1981 AHSME Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The base three representation of <math>x</math> is <br />
<cmath>12112211122211112222</cmath><br />
The first digit (on the left) of the base nine representation of <math>x</math> is<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math><br />
<br />
[[1981 AHSME Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
The function <math>f</math> is not defined for <math>x=0</math>, but, for all non-zero real numbers <math>x</math>, <math> f(x)+f\left(\dfrac{1}x\right)=x </math>. The equation <math> f(x)=f(-x) </math> is satisfied by<br />
<br />
<math> \textbf{(A)}\ \text{exactly one real number} \qquad \textbf{(B)}\ \text{exactly two real numbers} \qquad\textbf{(C)}\ \text{no real numbers}\qquad \\ \textbf{(D)}\ \text{infinitely many, but not all, non-zero real numbers} \qquad\textbf{(E)}\ \text{all non-zero real numbers} </math><br />
<br />
[[1981 AHSME Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
The number of real solutions to the equation<br />
<cmath>\dfrac{x}{100}=\sin x</cmath><br />
is<br />
<br />
<math> \textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65 </math><br />
<br />
[[1981 AHSME Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
In <math>\triangle ABC</math>, <math>M</math> is the midpoint of side <math>BC</math>, <math>AN</math> bisects <math>\angle BAC</math>, and <math>BN\perp AN</math>. If sides <math>AB</math> and <math>AC</math> have lengths <math>14</math> and <math>19</math>, respectively, then find <math>MN</math>.<br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair B=origin, A=14*dir(42), C=intersectionpoint(B--(30,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b);<br />
draw(N--B--A--N--M--C--A^^B--M);<br />
markscalefactor=0.1;<br />
draw(rightanglemark(B,N,A));<br />
pair point=N;<br />
label("$A$", A, dir(point--A));<br />
label("$B$", B, dir(point--B));<br />
label("$C$", C, dir(point--C));<br />
label("$M$", M, dir(point--M));<br />
label("$N$", N, dir(30));<br />
label(rotate(angle(dir(A--C)))*"$19$", A--C, dir(A--C)*dir(90));<br />
label(rotate(angle(dir(A--B)))*"$14$", A--B, dir(A--B)*dir(90));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ \dfrac{5}{2}\qquad\textbf{(C)}\ \dfrac{5}{2}-\sin\theta\qquad\textbf{(D)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\theta\qquad\textbf{(E)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\left(\dfrac{1}{2}\theta\right) </math><br />
<br />
== Solution ==<br />
Extend <math>BN</math> to meet <math>AC</math> at <math>Q</math>. Then <math>\triangle BNM \sim \triangle BQC</math>, so <math>BN=NQ</math> and <math>QC=19-AQ=2MN</math>. <br />
<br />
Since <math>\angle ANB=90^\circ = \angle ANQ</math>, <math>\angle BAN=\angle NAQ</math> (since <math>AN</math> is an angle bisector) and <math>\triangle ANB</math> and <math>\triangle ANQ</math> share side <math>AN</math>, <math>\triangle ANB \cong \triangle ANQ</math>. Thus <math>AQ=14</math>, and so <math>MN=\frac{19-AQ}{2}=\frac{5}{2}</math>, hence our answer is <math>\fbox{B}</math>.<br />
<br />
==Problem 20==<br />
A ray of light originates from point <math>A</math> and travels in a plane, being reflected <math>n</math> times between lines <math>AD</math> and <math>CD</math> before striking a point <math>B</math> (which may be on <math>AD</math> or <math>CD</math>) perpendicularly and retracing its path back to <math>A</math> (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for <math>n=3</math>). If <math>\measuredangle CDA=8^\circ</math>, what is the largest value <math>n</math> can have?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
pair D=origin, A=(-6,0), C=6*dir(160), E=3.2*dir(160), F=(-2.1,0), G=1.5*dir(160), B=(-1.4095,0);<br />
draw((-6.5,0)--D--C,black);<br />
draw(A--E--F--G--B,black);<br />
dotfactor=4;<br />
dot("$A$",A,S);<br />
dot("$C$",C,N);<br />
dot("$R_1$",E,N);<br />
dot("$R_2$",F,S);<br />
dot("$R_3$",G,N);<br />
dot("$B$",B,S);<br />
markscalefactor=0.015;<br />
draw(rightanglemark(G,B,D));<br />
draw(anglemark(C,E,A,12));<br />
draw(anglemark(F,E,G,12));<br />
draw(anglemark(E,F,A));<br />
draw(anglemark(E,F,A,12));<br />
draw(anglemark(B,F,G));<br />
draw(anglemark(B,F,G,12));<br />
draw(anglemark(E,G,F));<br />
draw(anglemark(E,G,F,12));<br />
draw(anglemark(E,G,F,16));<br />
draw(anglemark(B,G,D));<br />
draw(anglemark(B,G,D,12));<br />
draw(anglemark(B,G,D,16));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.} </math><br />
<br />
[[1981 AHSME Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
In a triangle with sides of lengths <math>a</math>, <math>b</math>, and <math>c</math>, <math> (a+b+c)(a+b-c) = 3ab </math>. The measure of the angle opposite the side length <math>c</math> is<br />
<br />
<math> \textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ </math><br />
<br />
[[1981 AHSME Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
How many lines in a three dimensional rectangular coordiante system pass through four distinct points of the form <math>(i,j,k)</math>, where <math>i</math>, <math>j</math>, and <math>k</math> are positive integers not exceeding four?<br />
<br />
<math> \textbf{(A)}\ 60\qquad\textbf{(B)}\ 64\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 100 </math><br />
<br />
[[1981 AHSME Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
Equilateral <math>\triangle ABC</math> is inscribed in a circle. A second circle is tangent internally to the circumcircle at <math>T</math> and tangent to sides <math>AB</math> and <math>AC</math> at points <math>P</math> and <math>Q</math>. If side <math>BC</math> has length <math>12</math>, then segment <math>PQ</math> has length<br />
<br />
<asy><br />
defaultpen(linewidth(.8pt));<br />
pair B = origin;<br />
pair A = dir(60);<br />
pair C = dir(0);<br />
pair circ = circumcenter(A,B,C);<br />
pair P = intersectionpoint(circ--(circ + (-1,0)),A--B);<br />
pair Q = intersectionpoint(circ--(circ + (1,0)),A--C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$P$",P,NW);<br />
label("$Q$",Q,NE);<br />
label("$T$",(0.5,-0.3),S);<br />
draw(A--B--C--cycle);<br />
draw(circumcircle(A,B,C));<br />
draw(P--Q);<br />
draw(Circle((0.5,0.09),0.385));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 6\sqrt{3}\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 8\sqrt{3}\qquad\textbf{(E)}\ 9 </math><br />
<br />
[[1981 AHSME Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
If <math> \theta</math> is a constant such that <math> 0 < \theta < \pi</math> and <math> x + \dfrac{1}{x} = 2\cos{\theta}</math>, then for each positive integer <math> n</math>, <math> x^n + \dfrac{1}{x^n}</math> equals<br />
<br />
<math> \textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta</math><br />
<br />
[[1981 AHSME Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
In <math>\triangle ABC</math> in the adjoining figure, <math> AD</math> and <math> AE</math> trisect <math> \angle BAC</math>. The lengths of <math> BD</math>, <math> DE</math> and <math> EC</math> are <math> 2</math>, <math> 3</math>, and <math> 6</math>, respectively. The length of the shortest side of <math> \triangle ABC</math> is<br />
<br />
<asy><br />
defaultpen(linewidth(.8pt));<br />
pair A = (0,11);<br />
pair B = (2,0);<br />
pair D = (4,0);<br />
pair E = (7,0);<br />
pair C = (13,0);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,S);<br />
label("$E$",E,S);<br />
label("$2$",midpoint(B--D),N);<br />
label("$3$",midpoint(D--E),NW);<br />
label("$6$",midpoint(E--C),NW);<br />
draw(A--B--C--cycle);<br />
draw(A--D);<br />
draw(A--E);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 2\sqrt{10}\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 6\sqrt{6}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}</math><br />
<br />
[[1981 AHSME Problems/Problem 25|Solution]]<br />
<br />
==Problem 26==<br />
Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is <math> \dfrac{1}{6}</math>, independent of the outcome of any other toss.)<br />
<br />
<math> \textbf{(A)}\ \dfrac{1}{3}\qquad \textbf{(B)}\ \dfrac{2}{9}\qquad \textbf{(C)}\ \dfrac{5}{18}\qquad \textbf{(D)}\ \dfrac{25}{91}\qquad \textbf{(E)}\ \dfrac{36}{91}</math><br />
<br />
[[1981 AHSME Problems/Problem 26|Solution]]<br />
<br />
==Problem 27==<br />
In the adjoining figure triangle <math> ABC</math> is inscribed in a circle. Point <math> D</math> lies on <math> \stackrel{\frown}{AC}</math> with <math> \stackrel{\frown}{DC} = 30^\circ</math>, and point <math> G</math> lies on <math> \stackrel{\frown}{BA}</math> with <math> \stackrel{\frown}{BG}\, > \, \stackrel{\frown}{GA}</math>. Side <math> AB</math> and side <math> AC</math> each have length equal to the length of chord <math> DG</math>, and <math> \angle CAB = 30^\circ</math>. Chord <math> DG</math> intersects sides <math> AC</math> and <math> AB</math> at <math> E</math> and <math> F</math>, respectively. The ratio of the area of <math> \triangle AFE</math> to the area of <math> \triangle ABC</math> is<br />
<br />
<asy><br />
defaultpen(linewidth(.8pt));<br />
pair C = origin;<br />
pair A = 2.5*dir(75);<br />
pair B = A + 2.5*dir(-75);<br />
path circ =circumcircle(A,B,C);<br />
pair D = waypoint(circ,(7/12));<br />
pair G = waypoint(circ,(1/6));<br />
pair E = intersectionpoint(D--G,A--C);<br />
pair F = intersectionpoint(A--B,D--G);<br />
label("$A$",A,N);<br />
label("$B$",B,SE);<br />
label("$C$",C,SW);<br />
label("$D$",D,SW);<br />
label("$G$",G,NE);<br />
label("$E$",E,NW);<br />
label("$F$",F,W);<br />
label("$30^\circ$",A,12S+E,fontsize(6pt));<br />
draw(A--B--C--cycle);<br />
draw(circ);<br />
draw(Arc(A,0.25,-75,-105));<br />
draw(D--G);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \dfrac {2 - \sqrt {3}}{3}\qquad \textbf{(B)}\ \dfrac {2\sqrt {3} - 3}{3}\qquad \textbf{(C)}\ 7\sqrt {3}-12\qquad \textbf{(D)}\ 3\sqrt {3}-5\qquad\\ \textbf{(E)}\ \dfrac {9-5\sqrt {3}}{3}</math><br />
<br />
[[1981 AHSME Problems/Problem 27|Solution]]<br />
<br />
==Problem 28==<br />
Consider the set of all equations <math> x^3 + a_2x^2 + a_1x + a_0 = 0</math>, where <math> a_2</math>, <math> a_1</math>, <math> a_0</math> are real constants and <math> |a_i| < 2</math> for <math> i = 0,1,2</math>. Let <math> r</math> be the largest positive real number which satisfies at least one of these equations. Then<br />
<br />
<math> \textbf{(A)}\ 1 < r < \dfrac{3}{2}\qquad \textbf{(B)}\ \dfrac{3}{2} < r < 2\qquad \textbf{(C)}\ 2 < r < \dfrac{5}{2}\qquad \textbf{(D)}\ \dfrac{5}{2} < r < 3\qquad \\ \textbf{(E)}\ 3 < r < \dfrac{7}{2}</math><br />
<br />
[[1981 AHSME Problems/Problem 28|Solution]]<br />
<br />
==Problem 29==<br />
If <math> a > 1</math>, then the sum of the real solutions of <br />
<br />
<math> \sqrt{a - \sqrt{a + x}} = x</math><br />
<br />
is equal to <br />
<br />
<math> \textbf{(A)}\ \sqrt{a} - 1\qquad \textbf{(B)}\ \dfrac{\sqrt{a}- 1}{2}\qquad \textbf{(C)}\ \sqrt{a - 1}\qquad \textbf{(D)}\ \dfrac{\sqrt{a - 1}}{2}\qquad \textbf{(E)}\ \dfrac{\sqrt{4a- 3} - 1}{2}</math><br />
<br />
[[1981 AHSME Problems/Problem 29|Solution]]<br />
<br />
==Problem 30==<br />
If <math> a</math>, <math> b</math>, <math> c</math>, and <math> d</math> are the solutions of the equation <math> x^4 - bx - 3 = 0</math>, then an equation whose solutions are <br />
<cmath>\dfrac {a + b + c}{d^2}, \dfrac {a + b + d}{c^2}, \dfrac {a + c + d}{b^2}, \dfrac {b + c + d}{a^2}</cmath>is<br />
<br />
<math> \textbf{(A)}\ 3x^4 + bx + 1 = 0\qquad \textbf{(B)}\ 3x^4 - bx + 1 = 0\qquad \textbf{(C)}\ 3x^4 + bx^3 - 1 = 0\qquad \\\textbf{(D)}\ 3x^4 - bx^3 - 1 = 0\qquad \textbf{(E)}\ \text{none of these}</math><br />
<br />
[[1981 AHSME Problems/Problem 30|Solution]]<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME box|year=1981|before=[[1980 AHSME]]|after=[[1982 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_28&diff=960511983 AHSME Problems/Problem 282018-07-09T07:07:47Z<p>Jazzachi: Added question, made the solution a bit clearer.</p>
<hr />
<div>== Problem 28 ==<br />
<br />
Triangle <math>\triangle ABC</math> in the figure has area <math>10</math>. Points <math>D, E</math> and <math>F</math>, all distinct from <math>A, B</math> and <math>C</math>, <br />
are on sides <math>AB, BC</math> and <math>CA</math> respectively, and <math>AD = 2, DB = 3</math>. If triangle <math>\triangle ABE</math> and quadrilateral <math>DBEF</math> <br />
have equal areas, then that area is<br />
<br />
<asy><br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B);<br />
draw(A--B--C--A--E--F--D);<br />
pair point=incenter(A,B,C);<br />
label("$A$", A, dir(point--A));<br />
label("$B$", B, dir(point--B));<br />
label("$C$", C, dir(point--C));<br />
label("$D$", D, dir(point--D));<br />
label("$E$", E, dir(point--E));<br />
label("$F$", F, dir(point--F));<br />
label("$2$", (2,0), S);<br />
label("$3$", (7,0), S);</asy><br />
<br />
<math>\textbf{(A)}\ 4\qquad<br />
\textbf{(B)}\ 5\qquad<br />
\textbf{(C)}\ 6\qquad<br />
\textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad<br />
\textbf{(E)}\ \text{not uniquely determined} </math> <br />
<br />
== Solution == <br />
<br />
Clearly since <math>[DBEF] = [ABE]</math> it follows that <math>[ADF] = [AFE]</math>. This implies that <math>AC \parallel DE</math> and so <math>\frac{BE}{BC} = \frac{BD}{DA} = \frac{3}{5}</math>. Since <math>\triangle ABE</math> and <math>\triangle ABC</math> have the same height, <math>[ABE] = \frac{3}{5} \cdot [ABC]=\frac{3}{5}\cdot 10 = 6</math>, hence our answer is <math>\fbox{C}</math></div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_27&diff=960431983 AHSME Problems/Problem 272018-07-09T01:01:41Z<p>Jazzachi: Added solution - nice problem</p>
<hr />
<div>== Problem 27 ==<br />
<br />
A large sphere is on a horizontal field on a sunny day. At a certain time the shadow of the sphere reaches out a distance <br />
of <math>10</math> m from the point where the sphere touches the ground. At the same instant a meter stick <br />
(held vertically with one end on the ground) casts a shadow of length <math>2</math> m. What is the radius of the sphere in meters? <br />
(Assume the sun's rays are parallel and the meter stick is a line segment.)<br />
<br />
<math>\textbf{(A)}\ \frac{5}{2}\qquad<br />
\textbf{(B)}\ 9 - 4\sqrt{5}\qquad<br />
\textbf{(C)}\ 8\sqrt{10} - 23\qquad<br />
\textbf{(D)}\ 6-\sqrt{15}\qquad<br />
\textbf{(E)}\ 10\sqrt{5}-20 </math><br />
<br />
== Solution == <br />
Consider the angle that the shadow makes with the ground. Since the sun's rays are parallel, it's the same as the angle made with the shadow of the stick. We know that the angle of the stick is <math>\arctan \left(\frac{1}{2}\right )</math> since the stick is <math>1</math>m high and its shadow is <math>2</math>m long, so <math>\frac{r}{10}=\tan \left(\arctan \left(\frac{1}{2}\right )\right )=\tan \left (\frac{1}{2} \right )</math>. Since <math>\tan \left(\frac{\theta}{2} \right ) = \frac{1-\cos \left(\theta \right )}{\sin \left(\theta \right )}</math>, we find that <math>\frac{r}{10}=\sqrt{5}-2</math>. Hence <math>r=10\sqrt{5}-20</math>, and our answer is <math>\fbox{E}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_19&diff=960131954 AHSME Problems/Problem 192018-07-08T02:44:32Z<p>Jazzachi: /* Made solution clearer */</p>
<hr />
<div>== Problem 19==<br />
<br />
If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle: <br />
<br />
<math> \textbf{(A)}\ \text{are always equal to }60^\circ\\ \textbf{(B)}\ \text{are always one obtuse angle and two unequal acute angles}\\ \textbf{(C)}\ \text{are always one obtuse angle and two equal acute angles}\\ \textbf{(D)}\ \text{are always acute angles}\\ \textbf{(E)}\ \text{are always unequal to each other} </math><br />
<br />
== Solution ==<br />
For the sake of clarity, let the outermost triangle be <math>ABC</math> with incircle tangency points <math>D</math>, <math>E</math>, and <math>F</math> on <math>BC</math>, <math>AC</math> and <math>AB</math> respectively. Let <math>\angle A=\alpha</math>, and <math>\angle B=\beta</math>. Because <math>\triangle AFE</math> and <math>\triangle BDF</math> are isosceles, <math>\angle AFE=\frac{180-\alpha}{2}</math> and <math>\angle BFD=\frac{180-\beta}{2}</math>. So <math>\angle DFE=180-\frac{180-\alpha}{2}-\frac{180-\beta}{2}=\frac{\alpha+\beta}{2}</math>, and since <math>\alpha + \beta <180^\circ</math>, <math>\angle DFE</math> is acute. <br />
<br />
The same method applies to <math>\angle FED</math> and <math>\angle FDE</math>, which means <math>\triangle DEF</math> is acute - hence our answer is <math>\fbox{D}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1956_AHSME_Problems/Problem_44&diff=960121956 AHSME Problems/Problem 442018-07-08T02:17:19Z<p>Jazzachi: Added solution</p>
<hr />
<div>== Problem 44==<br />
<br />
If <math>x < a < 0</math> means that <math>x</math> and <math>a</math> are numbers such that <math>x</math> is less than <math>a</math> and <math>a</math> is less than zero, then: <br />
<br />
<math>\textbf{(A)}\ x^2 < ax < 0 \qquad<br />
\textbf{(B)}\ x^2 > ax > a^2 \qquad<br />
\textbf{(C)}\ x^2 < a^2 < 0 \\<br />
\textbf{(D)}\ x^2 > ax\text{ but }ax < 0 \qquad<br />
\textbf{(E)}\ x^2 > a^2\text{ but }a^2 < 0 </math><br />
<br />
== Solution == <br />
Note that since <math>x</math> and <math>a</math> are negative, <math>x^2, a^2,</math> and <math>ax</math> are all positive. Hence the only correct answer is <math>\fbox{B}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1972_AHSME_Problems/Problem_13&diff=960111972 AHSME Problems/Problem 132018-07-08T02:13:48Z<p>Jazzachi: Added solution.</p>
<hr />
<div>== Problem 13 ==<br />
<br />
<asy><br />
draw(unitsquare);draw((0,0)--(.4,1)^^(0,.6)--(1,.2));<br />
label("D",(0,1),NW);label("E",(.4,1),N);label("C",(1,1),NE);<br />
label("P",(0,.6),W);label("M",(.25,.55),E);label("Q",(1,.2),E);<br />
label("A",(0,0),SW);label("B",(1,0),SE);<br />
//Credit to Zimbalono for the diagram<br />
</asy><br />
<br />
Inside square <math>ABCD</math> (See figure) with sides of length <math>12</math> inches, segment <math>AE</math> is drawn where <math>E</math> is the point on <math>DC</math> which is <math>5</math> inches from <math>D</math>. <br />
The perpendicular bisector of <math>AE</math> is drawn and intersects <math>AE, AD</math>, and <math>BC</math> at points <math>M, P</math>, and <math>Q</math> respectively. The ratio of segment <math>PM</math> to <math>MQ</math> is<br />
<br />
<math>\textbf{(A) }5:12\qquad<br />
\textbf{(B) }5:13\qquad<br />
\textbf{(C) }5:19\qquad<br />
\textbf{(D) }1:4\qquad <br />
\textbf{(E) }5:21 </math><br />
<br />
== Solution == <br />
Let the line passing through <math>M</math> parallel to <math>AB</math> intersect <math>AD</math> and <math>BC</math> and <math>S</math> and <math>T</math> respectively. Since <math>M</math> is the midpoint of <math>AE</math>, <math>SM=\frac{5}{2}</math> and <math>TM=12-\frac{5}{2}=\frac{19}{2}</math>. Since <math>\triangle PSM\sim \triangle QTM</math>, <math>PM:MQ=SM:MT=5:9</math>, hence our answer is <math>\fbox{C}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems/Problem_31&diff=960101966 AHSME Problems/Problem 312018-07-08T01:55:58Z<p>Jazzachi: /* Added solution - very useful lemma! */</p>
<hr />
<div>== Problem ==<br />
<asy><br />
draw(circle((0,0),10),black+linewidth(1));<br />
draw(circle((-1.25,2.5),4.5),black+linewidth(1));<br />
dot((0,0));<br />
dot((-1.25,2.5));<br />
draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5));<br />
draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5));<br />
draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5));<br />
draw((-sqrt(96),-2)--(sqrt(96),-2),black+linewidth(.5));<br />
MP("O'", (0,0), W);<br />
MP("O", (-2,2), W);<br />
MP("A", (-10,-2), W);<br />
MP("B", (10,-2), E);<br />
MP("C", (-2,sqrt(96)), N);<br />
MP("D", (sqrt(96)-2.5,7), NE);<br />
</asy><br />
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:<br />
<br />
<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><br />
<br />
== Solution ==<br />
We will prove that <math>\triangle DOB</math> and <math>\triangle COD</math> is isosceles, meaning that <math>CD=OD=BD</math> and hence <math>\fbox{D}</math>. <br />
<br />
Let <math>\angle A=2\alpha</math> and <math>\angle B=2\beta</math>. Since the incentre of a triangle is the intersection of its angle bisectors, <math>\angle OAB=\alpha</math> and <math>\angle ABO=\beta</math>. Hence <math>\angle DOB=\alpha +\beta</math>. Since quadrilateral <math>ABCD</math> is cyclic, <math>\angle CAD=\alpha=\angle CBD</math>. So <math>\angle OBD=\angle OBC+\angle CBD=\alpha + \beta = \angle DOB</math>. This means that <math>\triangle DOB</math> is isosceles, and hence <math>BD=OD</math>. <br />
<br />
Now let <math>\angle C=2\gamma</math> which means <math>angle ACO=COD=\gamma</math>. Since <math>ABCD</math> is cyclic, <math>\angle DAB=\alpha=\angle DCB.</math> Also, <math>\angle DAC\alpha</math> so <math>\angle DOC=\alpha + \gamma</math>. Thus, <math>\angle OCD=\angle OCB+\angle BCD=\gamma + \alpha=\angle DOC</math> which means <math>\triangle COD</math> is isosceles, and hence <math>CD=OD=BD</math>.<br />
<br />
Thus our answer is <math>\fbox{D}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1966|num-b=30|num-a=32}} <br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1967_AHSME_Problems/Problem_39&diff=959401967 AHSME Problems/Problem 392018-07-07T03:36:36Z<p>Jazzachi: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let <math>S_n</math> be the sum of the elements in the nth set. Then <math>S_n</math> equals:<br />
<br />
<math>\textbf{(A)}\ 1113\qquad<br />
\textbf{(B)}\ 4641 \qquad<br />
\textbf{(C)}\ 5082\qquad<br />
\textbf{(D)}\ 53361\qquad<br />
\textbf{(E)}\ \text{none of these}</math><br />
<br />
== Solution ==<br />
The last element of the 21st set is <math>\frac{21\times 22}{2}=231</math>, and hence the first element of the 21st set is <math>211</math>.<br />
So <math>S_{21}=\frac{231\times 232}{2}-\frac{210\times 211}{2}=4641</math>, hence our answer is <math>\fbox{B}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1967|num-b=38|num-a=40}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1967_AHSME_Problems/Problem_39&diff=959391967 AHSME Problems/Problem 392018-07-07T03:36:24Z<p>Jazzachi: /* minor clarification */</p>
<hr />
<div>== Problem ==<br />
Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let <math>S_n</math> be the sum of the elements in the nth set. Then <math>S_n</math> equals:<br />
<br />
<math>\textbf{(A)}\ 1113\qquad<br />
\textbf{(B)}\ 4641 \qquad<br />
\textbf{(C)}\ 5082\qquad<br />
\textbf{(D)}\ 53361\qquad<br />
\textbf{(E)}\ \text{none of these}</math><br />
<br />
== Solution ==<br />
The last element of the 21st set is <math>\frac{21\times 22}{2}=231</math>, and hence the first element of the 21st set is <math>211</math>.<br />
So <math>S_{21}=\frac{231\times 232}{2}-\frac{210\times 211}{2}=4616</math>, hence our answer is <math>\fbox{B}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1967|num-b=38|num-a=40}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1967_AHSME_Problems/Problem_39&diff=959381967 AHSME Problems/Problem 392018-07-07T03:35:40Z<p>Jazzachi: Added solution</p>
<hr />
<div>== Problem ==<br />
Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let <math>S_n</math> be the sum of the elements in the nth set. Then <math>S_n</math> equals:<br />
<br />
<math>\textbf{(A)}\ 1113\qquad<br />
\textbf{(B)}\ 4641 \qquad<br />
\textbf{(C)}\ 5082\qquad<br />
\textbf{(D)}\ 53361\qquad<br />
\textbf{(E)}\ \text{none of these}</math><br />
<br />
== Solution ==<br />
The last element of the 21st set is <math>\frac{21\times 22}{2}=231</math>, and hence the first element of the 21st set is <math>211</math>. So <math>S_{21}=\frac{231\times 232}{2}-\frac{210\times 211}{2}</math>, hence our answer is <math>\fbox{B}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1967|num-b=38|num-a=40}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1967_AHSME_Problems&diff=959371967 AHSME Problems2018-07-07T03:30:49Z<p>Jazzachi: /* Fixed problem */</p>
<hr />
<div>==Problem 1==<br />
<br />
The three-digit number <math>2a3</math> is added to the number <math>326</math> to give the three-digit number <math>5b9</math>. If <math>5b9</math> is divisible by 9, then <math>a+b</math> equals<br />
<br />
<math> \text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9</math><br />
<br />
[[1967 AHSME Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<br />
An equivalent of the expression<br />
<br />
<math>\left(\frac{x^2+1}{x}\right)\left(\frac{y^2+1}{y}\right)+\left(\frac{x^2-1}{y}\right)\left(\frac{y^2-1}{x}\right)</math>, <math>xy \not= 0</math>,<br />
<br />
is:<br />
<br />
<math> \text{(A)}\ 1\qquad\text{(B)}\ 2xy\qquad\text{(C)}\ 2x^2y^2+2\qquad\text{(D)}\ 2xy+\frac{2}{xy}\qquad\text{(E)}\ \frac{2x}{y}+\frac{2y}{x} </math><br />
<br />
[[1967 AHSME Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
The side of an equilateral triangle is <math>s</math>. A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is:<br />
<br />
<math> \text{(A)}\ \frac{s^2}{24}\qquad\text{(B)}\ \frac{s^2}{6}\qquad\text{(C)}\ \frac{s^2\sqrt{2}}{6}\qquad\text{(D)}\ \frac{s^2\sqrt{3}}{6}\qquad\text{(E)}\ \frac{s^2}{3} </math><br />
<br />
[[1967 AHSME Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
Given <math>\frac{\log{a}}{p}=\frac{\log{b}}{q}=\frac{\log{c}}{r}=\log{x}</math>, all logarithms to the same base and <math>x \not= 1</math>. If <math>\frac{b^2}{ac}=x^y</math>, then <math>y</math> is:<br />
<br />
<math> \text{(A)}\ \frac{q^2}{p+r}\qquad\text{(B)}\ \frac{p+r}{2q}\qquad\text{(C)}\ 2q-p-r\qquad\text{(D)}\ 2q-pr\qquad\text{(E)}\ q^2-pr</math><br />
<br />
<br />
[[1967 AHSME Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
A triangle is circumscribed about a circle of radius <math>r</math> inches. If the perimeter of the triangle is <math>P</math> inches and the area is <math>K</math> square inches, then <math>\frac{P}{K}</math> is:<br />
<br />
<math> \text{(A)}\text{independent of the value of} \; r\qquad\text{(B)}\ \frac{\sqrt{2}}{r}\qquad\text{(C)}\ \frac{2}{\sqrt{r}}\qquad\text{(D)}\ \frac{2}{r}\qquad\text{(E)}\ \frac{r}{2} </math><br />
<br />
[[1967 AHSME Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
If <math>f(x)=4^x</math> then <math>f(x+1)-f(x)</math> equals:<br />
<br />
<math> \text{(A)}\ 4\qquad\text{(B)}\ f(x)\qquad\text{(C)}\ 2f(x)\qquad\text{(D)}\ 3f(x)\qquad\text{(E)}\ 4f(x) </math><br />
<br />
[[1967 AHSME Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
If <math>\frac{a}{b}<-\frac{c}{d}</math> where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are real numbers and <math>bd \not= 0</math>, then:<br />
<br />
<math> \text{(A)}\ a \; \text{must be negative} \qquad<br />
\text{(B)}\ a \; \text{must be positive} \qquad</math><br />
<math>\text{(C)}\ a \; \text{must not be zero} \qquad<br />
\text{(D)}\ a \; \text{can be negative or zero, but not positive } \\<br />
\text{(E)}\ a \; \text{can be positive, negative, or zero}</math><br />
<br />
[[1967 AHSME Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
To <math>m</math> ounces of a <math>m\%</math> solution of acid, <math>x</math> ounces of water are added to yield a <math>(m-10)\%</math> solution. If <math>m>25</math>, then <math>x</math> is <br />
<br />
<math>\textbf{(A)}\ \frac{10m}{m-10} \qquad<br />
\textbf{(B)}\ \frac{5m}{m-10} \qquad<br />
\textbf{(C)}\ \frac{m}{m-10} \qquad<br />
\textbf{(D)}\ \frac{5m}{m-20} \\<br />
\textbf{(E)}\ \text{not determined by the given information}</math><br />
<br />
[[1967 AHSME Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
Let <math>K</math>, in square units, be the area of a trapezoid such that the shorter base, the altitude, and the longer base, in that order, are in arithmetic progression. Then:<br />
<br />
<math>\textbf{(A)}\ K \; \text{must be an integer} \qquad<br />
\textbf{(B)}\ K \; \text{must be a rational fraction} \\<br />
\textbf{(C)}\ K \; \text{must be an irrational number} \qquad<br />
\textbf{(D)}\ K\; \text{must be an integer or a rational fraction} \qquad</math><br />
<math>\textbf{(E)}\ \text{taken alone neither} \; \textbf{(A)} \; \text{nor} \; \textbf{(B)} \; \text{nor} \; \textbf{(C)} \; \text{nor} \; \textbf{(D)} \; \text{is true}</math><br />
<br />
[[1967 AHSME Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
If <math>\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)}</math> is an identity for positive rational values of <math>x</math>, then the value of <math>a-b</math> is:<br />
<br />
<math>\textbf{(A)}\ \frac{4}{3} \qquad<br />
\textbf{(B)}\ \frac{5}{3} \qquad<br />
\textbf{(C)}\ 2 \qquad<br />
\textbf{(D)}\ \frac{11}{4} \qquad<br />
\textbf{(E)}\ 3</math><br />
<br />
[[1967 AHSME Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
If the perimeter of rectangle <math>ABCD</math> is <math>20</math> inches, the least value of diagonal <math>\overline{AC}</math>, in inches, is:<br />
<br />
<math>\textbf{(A)}\ 0\qquad<br />
\textbf{(B)}\ \sqrt{50}\qquad<br />
\textbf{(C)}\ 10\qquad<br />
\textbf{(D)}\ \sqrt{200}\qquad<br />
\textbf{(E)}\ \text{none of these}</math><br />
<br />
[[1967 AHSME Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
If the (convex) area bounded by the x-axis and the lines <math>y=mx+4</math>, <math>x=1</math>, and <math>x=4</math> is <math>7</math>, then <math>m</math> equals:<br />
<br />
<math>\textbf{(A)}\ -\frac{1}{2}\qquad<br />
\textbf{(B)}\ -\frac{2}{3}\qquad<br />
\textbf{(C)}\ -\frac{3}{2} \qquad<br />
\textbf{(D)}\ -2 \qquad<br />
\textbf{(E)}\ \text{none of these}</math><br />
<br />
[[1967 AHSME Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
A triangle <math>ABC</math> is to be constructed given a side <math>a</math> (oppisite angle <math>A</math>). angle <math>B</math>, and <math>h_c</math>, the altitude from <math>C</math>. If <math>N</math> is the number of noncongruent solutions, then <math>N</math><br />
<br />
<math>\textbf{(A)}\ \text{is} \; 1\qquad<br />
\textbf{(B)}\ \text{is} \; 2\qquad<br />
\textbf{(C)}\ \text{must be zero}\qquad<br />
\textbf{(D)}\ \text{must be infinite}\qquad<br />
\textbf{(E)}\ \text{must be zero or infinite}</math><br />
<br />
[[1967 AHSME Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
Let <math>f(t)=\frac{t}{1-t}</math>, <math>t \not= 1</math>. If <math>y=f(x)</math>, then <math>x</math> can be expressed as <br />
<br />
<math>\textbf{(A)}\ f\left(\frac{1}{y}\right)\qquad<br />
\textbf{(B)}\ -f(y)\qquad<br />
\textbf{(C)}\ -f(-y)\qquad<br />
\textbf{(D)}\ f(-y)\qquad<br />
\textbf{(E)}\ f(y)</math><br />
<br />
[[1967 AHSME Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
The difference in the areas of two similar triangles is <math>18</math> square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triange, in square feet, is an integer, and one of its sides is <math>3</math> feet. The corresponding side of the larger triangle, in feet, is:<br />
<br />
<math>\textbf{(A)}\ 12\quad<br />
\textbf{(B)}\ 9\qquad<br />
\textbf{(C)}\ 6\sqrt{2}\qquad<br />
\textbf{(D)}\ 6\qquad<br />
\textbf{(E)}\ 3\sqrt{2}</math><br />
<br />
[[1967 AHSME Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
Let the product <math>(12)(15)(16)</math>, each factor written in base <math>b</math>, equal <math>3146</math> in base <math>b</math>. Let <math>s=12+15+16</math>, each term expressed in base <math>b</math>. Then <math>s</math>, in base <math>b</math>, is <br />
<br />
<math>\textbf{(A)}\ 43\qquad<br />
\textbf{(B)}\ 44\qquad<br />
\textbf{(C)}\ 45\qquad<br />
\textbf{(D)}\ 46\qquad<br />
\textbf{(E)}\ 47</math><br />
<br />
[[1967 AHSME Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
If <math>r_1</math> and <math>r_2</math> are the distinct real roots of <math>x^2+px+8=0</math>, then it must follow that:<br />
<br />
<math>\textbf{(A)}\ |r_1+r_2|>4\sqrt{2}\qquad<br />
\textbf{(B)}\ |r_1|>3 \; \text{or} \; |r_2| >3 \\<br />
\textbf{(C)}\ |r_1|>2 \; \text{and} \; |r_2|>2\qquad<br />
\textbf{(D)}\ r_1<0 \; \text{and} \; r_2<0\qquad<br />
\textbf{(E)}\ |r_1+r_2|<4\sqrt{2}</math><br />
<br />
[[1967 AHSME Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
If <math>x^2-5x+6<0</math> and <math>P=x^2+5x+6</math> then<br />
<br />
<math>\textbf{(A)}\ P \; \text{can take any real value} \qquad<br />
\textbf{(B)}\ 20<P<30\\<br />
\textbf{(C)}\ 0<P<20 \qquad<br />
\textbf{(D)}\ P<0 \qquad<br />
\textbf{(E)}\ P>30</math><br />
<br />
[[1967 AHSME Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
The area of a rectangle remains unchanged when it is made <math>2 \frac{1}{2}</math> inches longer and <math>\frac{2}{3}</math> inch narrower, or when it is made <math>2 \frac{1}{2}</math> inches shorter and <math>\frac{4}{3}</math> inch wider. Its area, in square inches, is:<br />
<br />
<math>\textbf{(A)}\ 30\qquad<br />
\textbf{(B)}\ \frac{80}{3}\qquad<br />
\textbf{(C)}\ 24\qquad<br />
\textbf{(D)}\ \frac{45}{2}\qquad<br />
\textbf{(E)}\ 20</math><br />
<br />
[[1967 AHSME Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
A circle is inscribed in a square of side <math>m</math>, then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If <math>S_n</math> is the sum of the areas of the first <math>n</math> circles so inscribed, then, as <math>n</math> grows beyond all bounds, <math>S_n</math> approaches:<br />
<br />
<math>\textbf{(A)}\ \frac{\pi m^2}{2}\qquad<br />
\textbf{(B)}\ \frac{3\pi m^2}{8}\qquad<br />
\textbf{(C)}\ \frac{\pi m^2}{3}\qquad<br />
\textbf{(D)}\ \frac{\pi m^2}{4}\qquad<br />
\textbf{(E)}\ \frac{\pi m^2}{8}</math><br />
<br />
[[1967 AHSME Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
In right triangle <math>ABC</math> the hypotenuse <math>\overline{AB}=5</math> and leg <math>\overline{AC}=3</math>. The bisector of angle <math>A</math> meets the opposite side in <math>A_1</math>. A second right triangle <math>PQR</math> is then constructed with hypotenuse <math>\overline{PQ}=A_1B</math> and leg <math>\overline{PR}=A_1C</math>. If the bisector of angle <math>P</math> meets the opposite side in <math>P_1</math>, the length of <math>PP_1</math> is:<br />
<br />
<math>\textbf{(A)}\ \frac{3\sqrt{6}}{4}\qquad<br />
\textbf{(B)}\ \frac{3\sqrt{5}}{4}\qquad<br />
\textbf{(C)}\ \frac{3\sqrt{3}}{4}\qquad<br />
\textbf{(D)}\ \frac{3\sqrt{2}}{2}\qquad<br />
\textbf{(E)}\ \frac{15\sqrt{2}}{16}</math><br />
<br />
[[1967 AHSME Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
For natural numbers, when <math>P</math> is divided by <math>D</math>, the quotient is <math>Q</math> and the remainder is <math>R</math>. When <math>Q</math> is divided by <math>D'</math>, the quotient is <math>Q'</math> and the remainder is <math>R'</math>. Then, when <math>P</math> is divided by <math>DD'</math>, the remainder is:<br />
<br />
<math>\textbf{(A)}\ R+R'D\qquad<br />
\textbf{(B)}\ R'+RD\qquad<br />
\textbf{(C)}\ RR'\qquad<br />
\textbf{(D)}\ R\qquad<br />
\textbf{(E)}\ R'</math><br />
<br />
[[1967 AHSME Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
If <math>x</math> is real and positive and grows beyond all bounds, then <math>\log_3{(6x-5)}-\log_3{(2x+1)}</math> approaches:<br />
<br />
<math>\textbf{(A)}\ 0\qquad<br />
\textbf{(B)}\ 1\qquad<br />
\textbf{(C)}\ 3\qquad<br />
\textbf{(D)}\ 4\qquad<br />
\textbf{(E)}\ \text{no finite number}</math><br />
<br />
[[1967 AHSME Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
The number of solution-pairs in the positive integers of the equation <math>3x+5y=501</math> is:<br />
<br />
<math>\textbf{(A)}\ 33\qquad<br />
\textbf{(B)}\ 34\qquad<br />
\textbf{(C)}\ 35\qquad<br />
\textbf{(D)}\ 100\qquad<br />
\textbf{(E)}\ \text{none of these}</math><br />
<br />
[[1967 AHSME Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
For every odd number <math>p>1</math> we have:<br />
<br />
<math>\textbf{(A)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-2\qquad<br />
\textbf{(B)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p\\<br />
\textbf{(C)}\ (p-1)^{\frac{1}{2}(p-1)} \; \text{is divisible by} \; p\qquad<br />
\textbf{(D)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p+1\\<br />
\textbf{(E)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-1</math><br />
<br />
[[1967 AHSME Problems/Problem 25|Solution]]<br />
<br />
==Problem 26==<br />
<br />
If one uses only the tabular information <math>10^3=1000</math>, <math>10^4=10,000</math>, <math>2^{10}=1024</math>, <math>2^{11}=2048</math>, <math>2^{12}=4096</math>, <math>2^{13}=8192</math>, then the strongest statement one can make for <math>\log_{10}{2}</math> is that it lies between:<br />
<br />
<math>\textbf{(A)}\ \frac{3}{10} \; \text{and} \; \frac{4}{11}\qquad<br />
\textbf{(B)}\ \frac{3}{10} \; \text{and} \; \frac{4}{12}\qquad<br />
\textbf{(C)}\ \frac{3}{10} \; \text{and} \; \frac{4}{13}\qquad<br />
\textbf{(D)}\ \frac{3}{10} \; \text{and} \; \frac{40}{132}\qquad<br />
\textbf{(E)}\ \frac{3}{11} \; \text{and} \; \frac{40}{132}</math><br />
<br />
[[1967 AHSME Problems/Problem 26|Solution]]<br />
<br />
==Problem 27==<br />
<br />
Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in <math>3</math> hours and the other in <math>4</math> hours. At what time P.M. should the candles be lighted so that, at 4 P.M., one stub is twice the length of the other?<br />
<br />
<math>\textbf{(A)}\ 1:24\qquad<br />
\textbf{(B)}\ 1:28\qquad<br />
\textbf{(C)}\ 1:36\qquad<br />
\textbf{(D)}\ 1:40\qquad<br />
\textbf{(E)}\ 1:48</math><br />
<br />
[[1967 AHSME Problems/Problem 27|Solution]]<br />
<br />
==Problem 28==<br />
<br />
Given the two hypotheses: <math>\text{I}</math> Some Mems are not Ens and <math>\text{II}</math> No Ens are Veens. If "some" means "at least one," we can conclude that:<br />
<br />
<math>\textbf{(A)}\ \text{Some Mems are not Veens}\qquad<br />
\textbf{(B)}\ \text{Some Vees are not Mems}\\<br />
\textbf{(C)}\ \text{No Mem is a Vee}\qquad<br />
\textbf{(D)}\ \text{Some Mems are Vees}\\<br />
\textbf{(E)}\ \text{Neither} \; \textbf{(A)} \; \text{nor} \; \textbf{(B)} \; \text{nor} \; \textbf{(C)} \; \text{nor} \; \textbf{(D)} \; \text{is deducible from the given statements}</math><br />
<br />
[[1967 AHSME Problems/Problem 28|Solution]]<br />
<br />
==Problem 29==<br />
<br />
<math>\overline{AB}</math> is a diameter of a circle. Tangents <math>\overline{AD}</math> and <math>\overline{BC}</math> are drawn so that <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect in a point on the circle. If <math>\overline{AD}=a</math> and <math>\overline{BD}=b</math>, <math>a \not= b</math>, the diameter of the circle is:<br />
<br />
<math>\textbf{(A)}\ |a-b|\qquad<br />
\textbf{(B)}\ \frac{1}{2}(a+b)\qquad<br />
\textbf{(C)}\ \sqrt{ab} \qquad<br />
\textbf{(D)}\ \frac{ab}{a+b}\qquad<br />
\textbf{(E)}\ \frac{1}{2}\frac{ab}{a+b}</math><br />
<br />
[[1967 AHSME Problems/Problem 29|Solution]]<br />
<br />
==Problem 30==<br />
<br />
A dealer bought <math>n</math> radios for <math>d</math> dollars, <math>d</math> a positive integer. He contributed two radios to a community bazaar at half their cost. The rest he sold at a profit of &#036;8 on each radio sold. If the overall profit was &#036;72, then the least possible value of <math>n</math> for the given information is:<br />
<br />
<math>\textbf{(A)}\ 18\qquad<br />
\textbf{(B)}\ 16\qquad<br />
\textbf{(C)}\ 15\qquad<br />
\textbf{(D)}\ 12\qquad<br />
\textbf{(E)}\ 11</math><br />
<br />
[[1967 AHSME Problems/Problem 30|Solution]]<br />
<br />
==Problem 31==<br />
<br />
Let <math>D=a^2+b^2+c^2</math>, where <math>a</math>, <math>b</math>, are consecutive integers and <math>c=ab</math>. Then <math>\sqrt{D}</math> is:<br />
<br />
<math>\textbf{(A)}\ \text{always an even integer}\qquad<br />
\textbf{(B)}\ \text{sometimes an odd integer, sometimes not}\\<br />
\textbf{(C)}\ \text{always an odd integer}\qquad<br />
\textbf{(D)}\ \text{sometimes rational, sometimes not}\\<br />
\textbf{(E)}\ \text{always irrational}</math><br />
<br />
[[1967 AHSME Problems/Problem 31|Solution]]<br />
<br />
==Problem 32==<br />
<br />
In quadrilateral <math>ABCD</math> with diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersecting at <math>O</math>, <math>\overline{BO}=4</math>, <math>\overline{AO}=8</math>, <math>\overline{OC}=3</math>, and <math>\overline{AB}=6</math>. The length of <math>\overline{AD}</math> is:<br />
<br />
<math>\textbf{(A)}\ 9\qquad<br />
\textbf{(B)}\ 10\qquad<br />
\textbf{(C)}\ 6\sqrt{3}\qquad<br />
\textbf{(D)}\ 8\sqrt{2}\qquad<br />
\textbf{(E)}\ \sqrt{166}</math><br />
<br />
[[1967 AHSME Problems/Problem 32|Solution]]<br />
<br />
==Problem 33==<br />
<br />
<asy><br />
fill(circle((4,0),4),grey);<br />
fill((0,0)--(8,0)--(8,-4)--(0,-4)--cycle,white);<br />
fill(circle((7,0),1),white);<br />
fill(circle((3,0),3),white);<br />
draw((0,0)--(8,0),black+linewidth(1));<br />
draw((6,0)--(6,sqrt(12)),black+linewidth(1));<br />
MP("A", (0,0), W); MP("B", (8,0), E); MP("C", (6,0), S); MP("D",(6,sqrt(12)), N);<br />
</asy><br />
<br />
In this diagram semi-circles are constructed on diameters <math>\overline{AB}</math>, <math>\overline{AC}</math>, and <math>\overline{CB}</math>, so that they are mutually tangent. If <math>\overline{CD} \bot \overline{AB}</math>, then the ratio of the shaded area to the area of a circle with <math>\overline{CD}</math> as radius is:<br />
<br />
<math>\textbf{(A)}\ 1:2\qquad<br />
\textbf{(B)}\ 1:3\qquad<br />
\textbf{(C)}\ \sqrt{3}:7\qquad<br />
\textbf{(D)}\ 1:4\qquad<br />
\textbf{(E)}\ \sqrt{2}:6</math><br />
<br />
[[1967 AHSME Problems/Problem 33|Solution]]<br />
<br />
==Problem 34==<br />
<br />
Points <math>D</math>, <math>E</math>, <math>F</math> are taken respectively on sides <math>AB</math>, <math>BC</math>, and <math>CA</math> of triangle <math>ABC</math> so that <math>AD:DB=BE:CE=CF:FA=1:n</math>. The ratio of the area of triangle <math>DEF</math> to that of triangle <math>ABC</math> is:<br />
<br />
<math>\textbf{(A)}\ \frac{n^2-n+1}{(n+1)^2}\qquad<br />
\textbf{(B)}\ \frac{1}{(n+1)^2}\qquad<br />
\textbf{(C)}\ \frac{2n^2}{(n+1)^2}\qquad<br />
\textbf{(D)}\ \frac{n^2}{(n+1)^2}\qquad<br />
\textbf{(E)}\ \frac{n(n-1)}{n+1}</math><br />
<br />
[[1967 AHSME Problems/Problem 34|Solution]]<br />
<br />
==Problem 35==<br />
<br />
The roots of <math>64x^3-144x^2+92x-15=0</math> are in arithmetic progression. The difference between the largest and smallest roots is:<br />
<br />
<math>\textbf{(A)}\ 2\qquad<br />
\textbf{(B)}\ 1\qquad<br />
\textbf{(C)}\ \frac{1}{2}\qquad<br />
\textbf{(D)}\ \frac{3}{8}\qquad<br />
\textbf{(E)}\ \frac{1}{4}</math><br />
<br />
[[1967 AHSME Problems/Problem 35|Solution]]<br />
<br />
==Problem 36==<br />
<br />
Given a geometric progression of five terms, each a positive integer less than <math>100</math>. The sum of the five terms is <math>211</math>. If <math>S</math> is the sum of those terms in the progression which are squares of integers, then <math>S</math> is:<br />
<br />
<math>\textbf{(A)}\ 0\qquad<br />
\textbf{(B)}\ 91\qquad<br />
\textbf{(C)}\ 133\qquad<br />
\textbf{(D)}\ 195\qquad<br />
\textbf{(E)}\ 211</math><br />
<br />
[[1967 AHSME Problems/Problem 36|Solution]]<br />
<br />
==Problem 37==<br />
<br />
Segments <math>AD=10</math>, <math>BE=6</math>, <math>CF=24</math> are drawn from the vertices of triangle <math>ABC</math>, each perpendicular to a straight line <math>RS</math>, not intersecting the triangle. Points <math>D</math>, <math>E</math>, <math>F</math> are the intersection points of <math>RS</math> with the perpendiculars. If <math>x</math> is the length of the perpendicular segment <math>GH</math> drawn to <math>RS</math> from the intersection point <math>G</math> of the medians of the triangle, then <math>x</math> is:<br />
<br />
<math>\textbf{(A)}\ \frac{40}{3}\qquad<br />
\textbf{(B)}\ 16\qquad<br />
\textbf{(C)}\ \frac{56}{3}\qquad<br />
\textbf{(D)}\ \frac{80}{3}\qquad<br />
\textbf{(E)}\ \text{undetermined}</math><br />
<br />
[[1967 AHSME Problems/Problem 37|Solution]]<br />
<br />
==Problem 38==<br />
<br />
Given a set <math>S</math> consisting of two undefined elements "pib" and "maa", and the four postulates: <math>P_1</math>: Every pib is a collection of maas, <math>P_2</math>: Any two distinct pibs have one and only one maa in common, <math>P_3</math>: Every maa belongs to two and only two pibs, <math>P_4</math>: There are exactly four pibs. Consider the three theorems: <math>T_1</math>: There are exactly six maas, <math>T_2</math>: There are exactly three maas in each pib, <math>T_3</math>: For each maa there is exactly one other maa not in the same pid with it. The theorems which are deducible from the postulates are:<br />
<br />
<math>\textbf{(A)}\ T_3 \; \text{only}\qquad<br />
\textbf{(B)}\ T_2 \; \text{and} \; T_3 \; \text{only} \qquad<br />
\textbf{(C)}\ T_1 \; \text{and} \; T_2 \; \text{only}\\<br />
\textbf{(D)}\ T_1 \; \text{and} \; T_3 \; \text{only}\qquad<br />
\textbf{(E)}\ \text{all}</math><br />
<br />
[[1967 AHSME Problems/Problem 38|Solution]]<br />
<br />
==Problem 39==<br />
<br />
Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let <math>S_n</math> be the sum of the elements in the nth set. Then <math>S_{21}</math> equals:<br />
<br />
<math>\textbf{(A)}\ 1113\qquad<br />
\textbf{(B)}\ 4641 \qquad<br />
\textbf{(C)}\ 5082\qquad<br />
\textbf{(D)}\ 53361\qquad<br />
\textbf{(E)}\ \text{none of these}</math><br />
<br />
[[1967 AHSME Problems/Problem 39|Solution]]<br />
<br />
==Problem 40==<br />
<br />
Located inside equilateral triangle <math>ABC</math> is a point <math>P</math> such that <math>PA=8</math>, <math>PB=6</math>, and <math>PC=10</math>. To the nearest integer the area of triangle <math>ABC</math> is:<br />
<br />
<math>\textbf{(A)}\ 159\qquad<br />
\textbf{(B)}\ 131\qquad<br />
\textbf{(C)}\ 95\qquad<br />
\textbf{(D)}\ 79\qquad<br />
\textbf{(E)}\ 50</math><br />
<br />
[[1967 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=1967|before=[[1966 AHSME]]|after=[[1968 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1967_AHSME_Problems/Problem_39&diff=959361967 AHSME Problems/Problem 392018-07-07T03:30:00Z<p>Jazzachi: /* fixed problem */</p>
<hr />
<div>== Problem ==<br />
Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let <math>S_n</math> be the sum of the elements in the nth set. Then <math>S_{21}</math> equals:<br />
<br />
<math>\textbf{(A)}\ 1113\qquad<br />
\textbf{(B)}\ 4641 \qquad<br />
\textbf{(C)}\ 5082\qquad<br />
\textbf{(D)}\ 53361\qquad<br />
\textbf{(E)}\ \text{none of these}</math><br />
<br />
== Solution ==<br />
<math>\fbox{B}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1967|num-b=38|num-a=40}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_19&diff=959352012 AMC 10B Problems/Problem 192018-07-07T02:59:29Z<p>Jazzachi: </p>
<hr />
<div>==Problem==<br />
In rectangle <math>ABCD</math>, <math>AB=6</math>, <math>AD=30</math>, and <math>G</math> is the midpoint of <math>\overline{AD}</math>. Segment <math>AB</math> is extended 2 units beyond <math>B</math> to point <math>E</math>, and <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>. What is the area of <math>BFDG</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
<br />
Note that the area of <math>BFDG</math> equals the area of <math>ABCD-\triangle AGB-\triangle DCF</math>. <br />
Since <math>AG=\frac{AD}{2}=15,</math> <math>\triangle AGB=\frac{15\times 6}{2}=45</math>. Now, <math>\triangle AED\sim \triangle BEF</math>, so <math>\frac{AE}{BE}=4=\frac{AD}{BF}=\frac{30}{BF}\implies BF=7.5</math> and <math>FC=22.5,</math> so <math>\triangle DCF=\frac{22.5\times6}{2}=\frac{135}{2}. </math><br />
<br />
Therefore, <cmath>\begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*}</cmath><br />
hence our answer is <math>\fbox{C}</math><br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1961_AHSME_Problems/Problem_39&diff=959341961 AHSME Problems/Problem 392018-07-07T02:56:09Z<p>Jazzachi: Added solution</p>
<hr />
<div>== 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 />
== Solution == <br />
Partition the unit square into four smaller squares of sidelength <math>\frac{1}{2}</math>. Each of the five points lies in one of these squares, and so by the Pigeonhole Principle, there exists two points in the same <math>\frac{1}{2}\times \frac{1}{2}</math> square - the maximum possible distance between them being <math>\frac{\sqrt{2}}{2}</math> by Pythagoras. <br />
<br />
Hence our answer is <math>\fbox{B}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1959_AHSME_Problems/Problem_49&diff=959331959 AHSME Problems/Problem 492018-07-07T02:46:48Z<p>Jazzachi: Added solution</p>
<hr />
<div>== Problem 49==<br />
For the infinite series <math>1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots</math> let <math>S</math> be the (limiting) sum. Then <math>S</math> equals:<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac27\qquad\textbf{(C)}\ \frac67\qquad\textbf{(D)}\ \frac{9}{32}\qquad\textbf{(E)}\ \frac{27}{32} </math> <br />
<br />
== Solution == <br />
Notice that we can arrange the sequence like so: <cmath>\begin{align*}1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots &= 1-\frac{5}{8}-\frac{5}{64}-\frac{5}{512}-\dots \\ &= 1-\left(\sum_{i=1}^{\infty}\frac{5}{8^n}\right) \\ &= 1-\frac{5}{7} \text{ (by the convergence of a geometric series)} \\ &=\frac{2}{7}\end{align*}</cmath><br />
hence our answer is <math>\fbox{B}</math></div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_19&diff=959322002 AMC 8 Problems/Problem 192018-07-07T02:35:21Z<p>Jazzachi: Formalised solution.</p>
<hr />
<div>==Problem==<br />
<br />
How many whole numbers between 99 and 999 contain exactly one 0?<br />
<br />
<math> \text{(A)}\ 72\qquad\text{(B)}\ 90\qquad\text{(C)}\ 144\qquad\text{(D)}\ 162\qquad\text{(E)}\ 180 </math><br />
<br />
==Solution==<br />
Numbers with exactly one zero have the form <math>\overline{a0b}</math> or <math>\overline{ab0}</math>, where the <math>a,b \neq 0</math>. There are <math>(9\cdot1\cdot9)+(9\cdot9\cdot1) = 81+81 = \boxed{162}</math> such numbers, hence our answer is <math>\fbox{D}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_45&diff=959311955 AHSME Problems/Problem 452018-07-07T02:30:50Z<p>Jazzachi: Added solution</p>
<hr />
<div>== Problem 45==<br />
<br />
Given a geometric sequence with the first term <math>\neq 0</math> and <math>r \neq 0</math> and an arithmetic sequence with the first term <math>=0</math>. <br />
A third sequence <math>1,1,2\ldots</math> is formed by adding corresponding terms of the two given sequences. <br />
The sum of the first ten terms of the third sequence is: <br />
<br />
<math> \textbf{(A)}\ 978\qquad\textbf{(B)}\ 557\qquad\textbf{(C)}\ 467\qquad\textbf{(D)}\ 1068\\ \textbf{(E)}\ \text{not possible to determine from the information given} </math><br />
<br />
== Solution == <br />
Let our geometric sequence be <math>a,ar,ar^2</math> and let our arithmetic sequence be <math>0,d,2d</math>. We know that <cmath>\begin{cases} a+0=1 \\ ar+d=1\\ ar^2+2d=1\end{cases}</cmath><br />
This implies that <math>a=1</math>, hence <math>r+d=1</math> and <math>r^2+2d=1</math>. Solving this system yields <math>r(r-2)=0</math>, so <math>r=0</math> or <math>2</math>. But since <math>r\neq 0</math>, <math>r=2</math> and <math>d=-1</math>. <br />
So our two sequences are <math>\{1-n\}</math> and <math>\{2^{n-1}\}</math>, which means the third sequence will be <cmath>\{2^{n-1}-n+1\}=\{1,1,2,5,12,27,\dots\}</cmath>Calculating the sum of the first 10 terms and adding them up yields 978, hence our answer is <math>\fbox{A}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1965_AHSME_Problems/Problem_30&diff=959001965 AHSME Problems/Problem 302018-07-06T02:01:17Z<p>Jazzachi: Added two solutions</p>
<hr />
<div>== Problem 30==<br />
<br />
Let <math>BC</math> of right triangle <math>ABC</math> be the diameter of a circle intersecting hypotenuse <math>AB</math> in <math>D</math>. <br />
At <math>D</math> a tangent is drawn cutting leg <math>CA</math> in <math>F</math>. This information is not sufficient to prove that <br />
<br />
<math>\textbf{(A)}\ DF \text{ bisects }CA \qquad <br />
\textbf{(B) }\ DF \text{ bisects }\angle CDA \\<br />
\textbf{(C) }\ DF = FA \qquad <br />
\textbf{(D) }\ \angle A = \angle BCD \qquad <br />
\textbf{(E) }\ \angle CFD = 2\angle A </math> <br />
<br />
<br />
<br />
== Solution 1 == <br />
We will prove every result except for <math>\fbox{B}</math>. <br />
<br />
By Thales' Theorem, <math>\angle CDB=90^\circ </math> and so <math>\angle CDA= 90^\circ </math>. <math>FC</math> and <math>FD</math> are both tangents to the same circle, and hence equal. Let <math>\angle CFD=\alpha</math>. Then <math>\angle FDC = \frac{180^\circ - \alpha}{2}</math>, and so <math>\angle FDA = \frac{\alpha}{2}</math>. We also have <math>\angle AFD = 180^\circ - \alpha</math>, which implies <math>\angle FAD=\frac{\alpha}{2}</math>. This means that <math>CF=DF=FA</math>, so <math>DF</math> indeed bisects <math>CA</math>. We also know that <math>\angle BCD=90-\frac{180^\circ - \alpha}{2}=\frac{\alpha}{2}</math>, hence <math>\angle A = \angle BCD</math>. And <math>\angle CFD=2\angle A</math> as <math>\alpha = \frac{\alpha}{2}\times 2</math>. <br />
<br />
Since all of the results except for <math>B</math> are true, our answer is <math>\fbox{B}</math>.<br />
<br />
== Solution 2 == <br />
It's easy to verify that <math>\angle CDA</math> always equals <math>90^\circ</math>. Since <math>\angle CDF</math> changes depending on the sidelengths of the triangle, we cannot be certain that <math>\angle CDF=45^\circ</math>. Hence our answer is <math>\fbox{B}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1962_AHSME_Problems/Problem_23&diff=958481962 AHSME Problems/Problem 232018-07-05T11:27:52Z<p>Jazzachi: Added solution</p>
<hr />
<div>==Problem==<br />
In triangle <math>ABC</math>, <math>CD</math> is the altitude to <math>AB</math> and <math>AE</math> is the altitude to <math>BC</math>. If the lengths of <math>AB</math>, <math>CD</math>, and <math>AE</math> are known, the length of <math>DB</math> is: <br />
<br />
<br />
<math>\textbf{(A)}\ \text{not determined by the information given} \qquad</math><br />
<br />
<math>\textbf{(B)}\ \text{determined only if A is an acute angle} \qquad</math> <br />
<br />
<math>\textbf{(C)}\ \text{determined only if B is an acute angle} \qquad</math> <br />
<br />
<math>\textbf{(D)}\ \text{determined only in ABC is an acute triangle} \qquad</math> <br />
<br />
<math>\textbf{(E)}\ \text{none of these is correct} </math><br />
<br />
==Solution==<br />
<br />
We can actually determine the length of <math>DB</math> no matter what type of angles <math>A</math> and <math>B</math> are. This can be easily proved through considering all possible cases, though for the purposes of this solution, we'll show that we can determine <math>DB</math> if <math>A</math> is an obtuse angle. <br />
<br />
Let's see what happens when <math>A</math> is an obtuse angle. <math>\triangle AEB\sim \triangle CDB</math> by SSS, so <math>\frac{AE}{CD}=\frac{AB}{DB}</math>. Hence <math>DB=\frac{AE}{CD\times AB}</math>. Since we've determined the length of <math>DB</math> even though we have an obtuse angle, <math>DB</math> is not <math>\bf{only}</math> determined by what type of angle <math>A</math> may be. Hence our answer is <math>\fbox{E}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1960_AHSME_Problems/Problem_32&diff=957691960 AHSME Problems/Problem 322018-07-03T14:22:24Z<p>Jazzachi: Added a rigorous solution</p>
<hr />
<div>==Problem==<br />
<br />
In this figure the center of the circle is <math>O</math>. <math>AB \perp BC</math>, <math>ADOE</math> is a straight line, <math>AP = AD</math>, and <math>AB</math> has a length twice the radius. Then:<br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.8)+fontsize(10));<br />
real e=350,c=55;<br />
pair O=origin,E=dir(e),C=dir(c),B=dir(180+c),D=dir(180+e), rot=rotate(90,B)*O,A=extension(E,D,B,rot);<br />
path tangent=A--B;<br />
pair P=waypoint(tangent,abs(A-D)/abs(A-B));<br />
draw(unitcircle^^C--B--A--E);<br />
dot(A^^B^^C^^D^^E^^P,linewidth(2));<br />
label("$O$",O,dir(290));<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,NE);<br />
label("$D$",D,dir(120));<br />
label("$E$",E,SE);<br />
label("$P$",P,SW);</asy><br />
<br />
<math>\textbf{(A)} AP^2 = PB \times AB\qquad \\<br />
\textbf{(B)}\ AP \times DO = PB \times AD\qquad \\<br />
\textbf{(C)}\ AB^2 = AD \times DE\qquad \\<br />
\textbf{(D)}\ AB \times AD = OB \times AO\qquad \\<br />
\textbf{(E)}\ \text{none of these} </math><br />
<br />
== Solution 1== <br />
<br />
We claim that <math>\fbox{A}</math> is the right answer.<br />
<br />
Let the radius of circle <math>O</math> be <math>r</math>, and let the length of <math>AD=x</math>. Since <math>AB \perp BC</math>, <math>AB</math> is a tangent to circle <math>O</math>. Thus, by the tangent-secant theorem, we have <math>AB^2=AD\times AE</math>, or, <math>(2r^2)=x(x+2r)</math>. Through some algebraic manipulation, we find <cmath>\begin{align}AD^2=x^2=2r(2r-x)\end{align}</cmath>. <br />
<br />
Since <math>AD^2=AP^2=x^2</math>, <math>PB=AB-AP=2r-x</math>, and <math>AB=2r</math>, we see that <math>(1)</math> is identical to <math>AP^2=PB\times AB</math>, hence our answer is <math>\fbox{A}</math><br />
<br />
==Solution 2 (guess and check) ==<br />
Let <math>r</math> be the radius of the circle, so <math>AB = 2r</math>. By the [[Pythagorean Theorem]], <math>AO = \sqrt{(2r)^2 + r^2} = r \sqrt{5}</math>. That means, <math>AD = AP = r \sqrt{5} - r</math>, so <math>PB = 2r - r\sqrt{5} + r = 3r - r\sqrt{5}</math>.<br />
<br />
Substitute values for each answer choice to determine which one is correct for all <math>r</math>.<br />
<br />
For option A, substitution results in<br />
<cmath>(r \sqrt{5} - r)^2 = (3r - r\sqrt{5})2r</cmath><br />
<cmath>5r^2 - 2r^2 \sqrt{5} + r^2 = 6r^2 - 2r^2 \sqrt{5}</cmath><br />
<cmath>6r^2 - 2r^2 \sqrt{5} = 6r^2 - 2r^2 \sqrt{5}</cmath><br />
<br />
For option B, substitution results in<br />
<cmath>(r \sqrt{5} - r)r = (3r - r\sqrt{5})(r \sqrt{5} - r)</cmath><br />
<cmath>r^2 \sqrt{5} - r^2 = 4r^2 \sqrt{5} - 8r^2</cmath><br />
<br />
For option C, substitution results in<br />
<cmath>(2r)^2 = (r\sqrt{5} - r)2r</cmath><br />
<cmath>4r^2 = 2r^2 \sqrt{5} - 2r^2</cmath><br />
<br />
For option D, substitution results in<br />
<cmath>(r \sqrt{5} - r) \cdot 2r = r \cdot r\sqrt{5}</cmath><br />
<cmath>2r^2 \sqrt{5} - 2r^2 = r^2 \sqrt{5}</cmath><br />
<br />
From each option, only option A has both sides equaling each other, so the answer is <math>\boxed{\textbf{(A)}}</math>.<br />
<br />
==See Also==<br />
{{AHSME 40p box|year=1960|num-b=31|num-a=33}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1960_AHSME_Problems/Problem_32&diff=957681960 AHSME Problems/Problem 322018-07-03T14:12:40Z<p>Jazzachi: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
In this figure the center of the circle is <math>O</math>. <math>AB \perp BC</math>, <math>ADOE</math> is a straight line, <math>AP = AD</math>, and <math>AB</math> has a length twice the radius. Then:<br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.8)+fontsize(10));<br />
real e=350,c=55;<br />
pair O=origin,E=dir(e),C=dir(c),B=dir(180+c),D=dir(180+e), rot=rotate(90,B)*O,A=extension(E,D,B,rot);<br />
path tangent=A--B;<br />
pair P=waypoint(tangent,abs(A-D)/abs(A-B));<br />
draw(unitcircle^^C--B--A--E);<br />
dot(A^^B^^C^^D^^E^^P,linewidth(2));<br />
label("$O$",O,dir(290));<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,NE);<br />
label("$D$",D,dir(120));<br />
label("$E$",E,SE);<br />
label("$P$",P,SW);</asy><br />
<br />
<math>\textbf{(A)} AP^2 = PB \times AB\qquad \\<br />
\textbf{(B)}\ AP \times DO = PB \times AD\qquad \\<br />
\textbf{(C)}\ AB^2 = AD \times DE\qquad \\<br />
\textbf{(D)}\ AB \times AD = OB \times AO\qquad \\<br />
\textbf{(E)}\ \text{none of these} </math><br />
<br />
==Solution 2 (guess and check) ==<br />
Let <math>r</math> be the radius of the circle, so <math>AB = 2r</math>. By the [[Pythagorean Theorem]], <math>AO = \sqrt{(2r)^2 + r^2} = r \sqrt{5}</math>. That means, <math>AD = AP = r \sqrt{5} - r</math>, so <math>PB = 2r - r\sqrt{5} + r = 3r - r\sqrt{5}</math>.<br />
<br />
Substitute values for each answer choice to determine which one is correct for all <math>r</math>.<br />
<br />
For option A, substitution results in<br />
<cmath>(r \sqrt{5} - r)^2 = (3r - r\sqrt{5})2r</cmath><br />
<cmath>5r^2 - 2r^2 \sqrt{5} + r^2 = 6r^2 - 2r^2 \sqrt{5}</cmath><br />
<cmath>6r^2 - 2r^2 \sqrt{5} = 6r^2 - 2r^2 \sqrt{5}</cmath><br />
<br />
For option B, substitution results in<br />
<cmath>(r \sqrt{5} - r)r = (3r - r\sqrt{5})(r \sqrt{5} - r)</cmath><br />
<cmath>r^2 \sqrt{5} - r^2 = 4r^2 \sqrt{5} - 8r^2</cmath><br />
<br />
For option C, substitution results in<br />
<cmath>(2r)^2 = (r\sqrt{5} - r)2r</cmath><br />
<cmath>4r^2 = 2r^2 \sqrt{5} - 2r^2</cmath><br />
<br />
For option D, substitution results in<br />
<cmath>(r \sqrt{5} - r) \cdot 2r = r \cdot r\sqrt{5}</cmath><br />
<cmath>2r^2 \sqrt{5} - 2r^2 = r^2 \sqrt{5}</cmath><br />
<br />
From each option, only option A has both sides equaling each other, so the answer is <math>\boxed{\textbf{(A)}}</math>.<br />
<br />
==See Also==<br />
{{AHSME 40p box|year=1960|num-b=31|num-a=33}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_13&diff=957671964 AHSME Problems/Problem 132018-07-03T13:59:12Z<p>Jazzachi: Added solution</p>
<hr />
<div>== Problem 13==<br />
<br />
A circle is inscribed in a triangle with side lengths <math>8, 13</math>, and <math>17</math>. Let the segments of the side of length <math>8</math>, <br />
made by a point of tangency, be <math>r</math> and <math>s</math>, with <math>r<s</math>. What is the ratio <math>r:s</math>? <br />
<br />
<math>\textbf{(A)}\ 1:3 \qquad<br />
\textbf{(B)}\ 2:5 \qquad<br />
\textbf{(C)}\ 1:2 \qquad<br />
\textbf{(D)}\ 2:3 \qquad<br />
\textbf{(E)}\ 3:4 </math> <br />
<br />
== Solution ==<br />
<br />
Label our triangle <math>ABC</math> where that <math>AB=17</math>, <math>AC=13</math>, and <math>BC=8</math>. Let <math>L</math>, <math>J</math>, and <math>K</math> be the tangency points of <math>BC</math>, <math>AC</math>, and <math>AB</math> respectively. Let <math>AK=x</math>, which implies <math>AJ=x</math>. Thus <math>KB=BL=17-x</math> and <math>JC=LC=13-x</math>. <br />
<br />
Since <math>BL+LC=(17-x)+(13-x)=8</math>, <math>x=11</math>. Thus <math>r:s=CL:BL=13-x:17-x=2:6=1:3</math>, hence our answer is <math>\fbox{A}</math>.</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1970_AHSME_Problems/Problem_24&diff=956721970 AHSME Problems/Problem 242018-06-29T06:32:36Z<p>Jazzachi: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
<br />
An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is <math>2</math>, then the area of the hexagon is<br />
<br />
<math>\text{(A) } 2\quad<br />
\text{(B) } 3\quad<br />
\text{(C) } 4\quad<br />
\text{(D) } 6\quad<br />
\text{(E) } 12</math><br />
<br />
== Solution 1 ==<br />
Let <math>ABCDEF</math> be our regular hexagon, with centre <math>O</math> - and join <math>AO, BO, CO, DO, EO, </math> and <math>FO</math>. Note that we form six equilateral triangles with sidelength <math>\frac{s}{2}</math>, where <math>s</math> is the sidelength of the triangle (since the perimeter of the two polygons are equal). If the area of the original equilateral triangle is <math>2</math>, then the area of each of the six smaller triangles is <math>1/4\times 2 = \frac{1}{2}</math> (by similarity area ratios). <br />
<br />
Thus, the area of the hexagon is <math>6\times \frac{1}{2}=3</math>, hence our answer is <math>\fbox{B}</math><br />
<br />
== See also ==<br />
{{AHSME 35p box|year=1970|num-b=23|num-a=25}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1970_AHSME_Problems/Problem_24&diff=956711970 AHSME Problems/Problem 242018-06-29T06:32:17Z<p>Jazzachi: Added solution</p>
<hr />
<div>== Problem ==<br />
<br />
An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is <math>2</math>, then the area of the hexagon is<br />
<br />
<math>\text{(A) } 2\quad<br />
\text{(B) } 3\quad<br />
\text{(C) } 4\quad<br />
\text{(D) } 6\quad<br />
\text{(E) } 12</math><br />
<br />
== Solution 1 ==<br />
Let <math>ABCDEF</math> be our regular hexagon, with centre <math>O</math> - and join <math>AO, BO, CO, DO, EO, </math> and <math>FO</math>. Note that we form six equilateral triangles with sidelength <math>\frac{s}{2}</math>, where <math>s</math> is the sidelength of the triangle (since the perimeter of the two polygons are equal.) If the area of the original equilateral triangle is <math>2</math>, then the area of each of the six smaller triangles is <math>1/4\times 2 = \frac{1}{2}</math> (by similarity area ratios). <br />
<br />
Thus, the area of the hexagon is <math>6\times \frac{1}{2}=3</math>, hence our answer is <br />
<math>\fbox{B}</math><br />
<br />
== See also ==<br />
{{AHSME 35p box|year=1970|num-b=23|num-a=25}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_45&diff=956701950 AHSME Problems/Problem 452018-06-29T05:08:28Z<p>Jazzachi: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
The number of diagonals that can be drawn in a polygon of 100 sides is:<br />
<br />
<math>\textbf{(A)}\ 4850 \qquad<br />
\textbf{(B)}\ 4950\qquad<br />
\textbf{(C)}\ 9900 \qquad<br />
\textbf{(D)}\ 98 \qquad<br />
\textbf{(E)}\ 8800</math><br />
<br />
== Solution ==<br />
Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be <math>\binom{100}{2}=4950</math>. However this also counts the 100 sides of the polygon, so the actual answer is <math>4950-100=\boxed{\textbf{(A)}\ 4850 }</math>.<br />
<br />
== Solution 2 ==<br />
For each vertex we can choose <math>100 - 3 = 97</math> vertices to draw the diagonal, as we cannot connect a vertex to itself or either adjacent vertices. Thus, the answer is <math>(100)(97)/2=4850</math>, as we are overcounting by a factor of <math>2</math> (we are counting each diagonal twice - one for each endpoint). Thus our answer is <math>\fbox{A}</math>.<br />
<br />
== See Also ==<br />
{{AHSME 50p box|year=1950|num-b=44|num-a=46}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_45&diff=956691950 AHSME Problems/Problem 452018-06-29T05:08:03Z<p>Jazzachi: Clarified point, removed label</p>
<hr />
<div>==Problem==<br />
<br />
The number of diagonals that can be drawn in a polygon of 100 sides is:<br />
<br />
<math>\textbf{(A)}\ 4850 \qquad<br />
\textbf{(B)}\ 4950\qquad<br />
\textbf{(C)}\ 9900 \qquad<br />
\textbf{(D)}\ 98 \qquad<br />
\textbf{(E)}\ 8800</math><br />
<br />
== Solution ==<br />
Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be <math>\binom{100}{2}=4950</math>. However this also counts the 100 sides of the polygon, so the actual answer is <math>4950-100=\boxed{\textbf{(A)}\ 4850 }</math>.<br />
<br />
== Solution 2 ==<br />
For each vertex we can choose <math>100 - 3 = 97</math> vertices to draw the diagonal, as we cannot connect a vertex to itself or either adjacent vertices. Thus, the answer is <math>(100)(97)/2=4850</math>, as we are overcounting by a factor of 2 (we are counting each diagonal twice - one for each endpoint).<br />
<br />
== See Also ==<br />
{{AHSME 50p box|year=1950|num-b=44|num-a=46}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_24&diff=956581992 AHSME Problems/Problem 242018-06-28T13:08:19Z<p>Jazzachi: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>ABCD</math> be a parallelogram of area <math>10</math> with <math>AB=3</math> and <math>BC=5</math>. Locate <math>E,F</math> and <math>G</math> on segments <math>\overline{AB},\overline{BC}</math> and <math>\overline{AD}</math>, respectively, with <math>AE=BF=AG=2</math>. Let the line through <math>G</math> parallel to <math>\overline{EF}</math> intersect <math>\overline{CD}</math> at <math>H</math>. The area of quadrilateral <math>EFHG</math> is<br />
<br />
<math>\text{(A) } 4\quad<br />
\text{(B) } 4.5\quad<br />
\text{(C) } 5\quad<br />
\text{(D) } 5.5\quad<br />
\text{(E) } 6</math><br />
<br />
== Solution 1 ==<br />
<math>\fbox{C}</math> Use vectors. Place an origin at <math>A</math>, with <math>B = p, D = q, C = p + q</math>. We know that <math>\|p \times q\|=10</math>, and also <math>E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q</math>, and now we can find the area of <math>EFHG</math> by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero).<br />
<br />
== Solution 2 ==<br />
By noting the following: <cmath>\begin{align*}|EFHG| &= |\triangle EGH| + |\triangle EFH| \\ &= \frac{1}{2}|AEHD|+\frac{1}{2}|EHCB| \\ &= \frac{1}{2}[|AEHD|+|EHCB|] \\ &= \frac{1}{2}|ABCD| \\ &= 5 \end{align*}</cmath> we see that the answer is <math>\fbox{C}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1992|num-b=23|num-a=25}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1970_AHSME_Problems/Problem_30&diff=956571970 AHSME Problems/Problem 302018-06-28T12:57:00Z<p>Jazzachi: Added solution</p>
<hr />
<div>== Problem ==<br />
<br />
<asy><br />
draw((0,0)--(2,2)--(5/2,1/2)--(2,0)--cycle,dot);<br />
MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(5/2,1/2),SE);MP("D",(2,0),S);<br />
MP("a",(1,0),N);MP("b",(17/8,1/8),N);<br />
</asy><br />
<br />
In the accompanying figure, segments <math>AB</math> and <math>CD</math> are parallel, the measure of angle <math>D</math> is twice that of angle <math>B</math>, and the measures of segments <math>AD</math> and <math>CD</math> are <math>a</math> and <math>b</math> respectively. Then the measure of <math>AB</math> is equal to<br />
<br />
<math>\text{(A) } \tfrac{1}{2}a+2b\quad<br />
\text{(B) } \tfrac{3}{2}b+\tfrac{3}{4}a\quad<br />
\text{(C) } 2a-b\quad<br />
\text{(D) } 4b-\tfrac{1}{2}a\quad<br />
\text{(E) } a+b</math><br />
<br />
== Solution ==<br />
With reference to the diagram above, let <math>E</math> be the point on <math>AB</math> such that <math>DE||BC</math>. Let <math>\angle ABC=\alpha</math>. We then have <math>\alpha =\angle AED = \angle EDC</math> since <math>AB||CD</math>, so <math>\angle ADE=\angle ADC-\angle BDC=2\alpha-\alpha = \alpha</math>, which means <math>\triangle AED</math> is isosceles. <br />
<br />
Therefore, <math>AB=AE+EB=a+b</math>, hence our answer is <math>\fbox{E}</math>.<br />
== See also ==<br />
{{AHSME 35p box|year=1970|num-b=29|num-a=31}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1977_AHSME_Problems/Problem_19&diff=956561977 AHSME Problems/Problem 192018-06-28T12:38:02Z<p>Jazzachi: Added solution</p>
<hr />
<div>== Problem 19 ==<br />
<br />
Let <math>E</math> be the point of intersection of the diagonals of convex quadrilateral <math>ABCD</math>, and let <math>P,Q,R</math>, and <math>S</math> be the centers of the circles <br />
circumscribing triangles <math>ABE, BCE, CDE</math>, and <math>ADE</math>, respectively. Then<br />
<br />
<math>\textbf{(A) }PQRS\text{ is a parallelogram}\\<br />
\textbf{(B) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rhombus}\\<br />
\textbf{(C) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rectangle}\\<br />
\textbf{(D) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a parallelogram}\\<br />
\textbf{(E) }\text{none of the above are true} </math><br />
<br />
<br />
<br />
== Solution ==<br />
<br />
Let <math>L</math>, <math>M</math>, <math>N</math>, and <math>O</math> be the intersections of <math>AC</math> and <math>PS</math>, <math>BD</math> and <math>PQ</math>, <math>CA</math> and <math>QR</math>, and <math>DB</math> and <math>RS</math> respectively. Since the [[circumcenter]] of a triangle is determined by the intersection of its perpendicular bisectors, each of the pairs mentioned earlier are perpendicular to each other. Let <math>\angle SPQ=\alpha</math>. Then <math>\angle LEM=180^\circ-\alpha</math> since <math>\angle PLC=PMD=90^\circ</math>. This means <math>\angle DEC=180^\circ-\alpha</math>, which similarly implies <math>\angle QRS=\alpha</math>. Through a similar method, we find that <math>\angle PSR=\angle PQR=180^\circ-\alpha</math>.<br />
<br />
Since the opposite angles of <math>PQRS</math> are congruent, <math>PQRS</math> is indeed a parallelogram - and we showed this without any assumption about the type of quadrilateral <math>ABCD</math> was. Therefore, <math>PQRS</math> is always a parallelogram, and so our answer is <math>\fbox{A}</math>. <br />
== See Also ==<br />
{{AMC12 box|year=2018|ab=A|before = First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_24&diff=956551992 AHSME Problems/Problem 242018-06-28T11:29:06Z<p>Jazzachi: Added 2nd solution</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>ABCD</math> be a parallelogram of area <math>10</math> with <math>AB=3</math> and <math>BC=5</math>. Locate <math>E,F</math> and <math>G</math> on segments <math>\overline{AB},\overline{BC}</math> and <math>\overline{AD}</math>, respectively, with <math>AE=BF=AG=2</math>. Let the line through <math>G</math> parallel to <math>\overline{EF}</math> intersect <math>\overline{CD}</math> at <math>H</math>. The area of quadrilateral <math>EFHG</math> is<br />
<br />
<math>\text{(A) } 4\quad<br />
\text{(B) } 4.5\quad<br />
\text{(C) } 5\quad<br />
\text{(D) } 5.5\quad<br />
\text{(E) } 6</math><br />
<br />
== Solution 1 ==<br />
<math>\fbox{C}</math> Use vectors. Place an origin at <math>A</math>, with <math>B = p, D = q, C = p + q</math>. We know that <math>\|p \times q\|=10</math>, and also <math>E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q</math>, and now we can find the area of <math>EFHG</math> by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero).<br />
<br />
== Solution 2 ==<br />
By noting the following: <cmath>\begin{align*}|EFHG| &= |\triangle EGH| + |\triangle EFH| \\ &= \frac{1}{2}|AEHD|+\frac{1}{2}|EHCB| \\ &= \frac{1}{2}[|AEHD|+EHCB|] \\ &= \frac{1}{2}|ABCD| \\ &= 5 \end{align*}</cmath> we see that the answer is <math>\fbox{C}</math>.<br />
== See also ==<br />
{{AHSME box|year=1992|num-b=23|num-a=25}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_29&diff=956311986 AHSME Problems/Problem 292018-06-26T10:28:44Z<p>Jazzachi: </p>
<hr />
<div>==Problem==<br />
<br />
Two of the altitudes of the scalene triangle <math>ABC</math> have length <math>4</math> and <math>12</math>. <br />
If the length of the third altitude is also an integer, what is the biggest it can be?<br />
<br />
<math>\textbf{(A)}\ 4\qquad<br />
\textbf{(B)}\ 5\qquad<br />
\textbf{(C)}\ 6\qquad<br />
\textbf{(D)}\ 7\qquad<br />
\textbf{(E)}\ \text{none of these} </math> <br />
<br />
==Solution 1==<br />
Assume we have a scalene triangle <math>ABC</math>. Arbitrarily, let <math>12</math> be the height to base <math>AB</math> and <math>4</math> be the height to base <math>AC</math>. Due to area equivalences, the base <math>AC</math> must be three times the length of <math>AB</math>. <br />
<br />
Let the base <math>AB</math> be <math>x</math>, thus making <math>AC = 3x</math>. Thus, setting the final height to base <math>BC</math> to <math>h</math>, we note that (by area equivalence) <math>\frac{BC \cdot h}{2} = \frac{3x \cdot 4}{2} = 6x</math>. Thus, <math>h = \frac{12x}{BC}</math>. We note that to maximize <math>h</math> we must minimize <math>BC</math>. Using the triangle inequality, <math>BC + AB > AC</math>, thus <math>BC + x > 3x</math> or <math>BC > 2x</math>. The minimum value of <math>BC</math> is <math>2x</math>, which would output <math>h = 6</math>. However, because <math>BC</math> must be larger than <math>2x</math>, the minimum integer height must be <math>5</math>.<br />
<br />
<math>\fbox{B}</math><br />
<br />
<br />
==Solution 2== <br />
<br />
The reciprocals of the altitudes of a triangle themselves form a triangle - this can be easily proven. Let our desired altitude be <math>a</math>. <br />
<br />
We have <math>\frac{1}{a}<\frac{1}{4}+\frac{1}{12}=\frac{1}{3}</math>, which implies <math>a>3</math>. We also have <math>\frac{1}{a}<\frac{1}{4}-\frac{1}{12}=\frac{1}{6}</math>, which implies <math>a<6</math>. Therefore the maximum integral value of <math>a</math> is 5.<br />
<br />
<math>\fbox{B}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1986|num-b=28|num-a=30}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_26&diff=956281953 AHSME Problems/Problem 262018-06-26T07:10:32Z<p>Jazzachi: Created page with "==Problem 26== The base of a triangle is <math>15</math> inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into..."</p>
<hr />
<div>==Problem 26==<br />
<br />
The base of a triangle is <math>15</math> inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is: <br />
<br />
<math>\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad<br />
\textbf{(B)}\ 10\text{ inches} \qquad<br />
\textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad<br />
\textbf{(D)}\ 7.5\text{ inches}\\ <br />
\textbf{(E)}\ \text{none of these} </math><br />
<br />
==Solution==<br />
Let the triangle be <math>\triangle ABC</math> where <math>BC</math> is the base. Then let the parallels be <math>MN</math> and <math>PQ</math>, where <math>PQ</math> is closer to our base <math>BC</math>. <br />
<br />
It's obvious that <math>\triangle APQ \sim \triangle ABC</math>, where <math>|\triangle APQ|:|\triangle ABC|=2:3</math>, so <math>\frac{PQ}{BC}=\sqrt{\frac{2}{3}}</math>. Since we know <math>BC=15</math>, <math>PQ=\frac{15\sqrt{2}}{\sqrt{3}}=5\sqrt{6}</math><br />
<br />
Hence our answer is <math>\textbf{(A)}</math></div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems/Problem_11&diff=956251966 AHSME Problems/Problem 112018-06-26T04:50:42Z<p>Jazzachi: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The sides of triangle <math>BAC</math> are in the ratio <math>2:3:4</math>. <math>BD</math> is the angle-bisector<br />
drawn to the shortest side <math>AC</math>, dividing it into segments <math>AD</math> and <math>CD</math>.<br />
If the length of <math>AC</math> is <math>10</math>, then the length of the longer segment of <math>AC</math> is:<br />
<br />
<math>\text{(A)} \ 3\frac{1}{2} \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac{5}{7} \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac{1}{2}</math><br />
<br />
== Solution ==<br />
By the [[Angle Bisector Theorem]], we have <math>\frac{BA}{AD}=\frac{BC}{CD}</math> which implies <math>\frac{AD}{DC}=\frac{BA}{BC}=\frac{3}{4}</math>. So <math>AC=10=AD+DC=\frac{7}{4}DC</math>, and thus <math>DC=\frac{40}{7}=5\frac{5}{7}</math>.<br />
Hence<br />
<math>\fbox{C}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1966|num-b=10|num-a=12}} <br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=2010_IMO_Problems/Problem_5&diff=955972010 IMO Problems/Problem 52018-06-25T10:25:15Z<p>Jazzachi: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<br />
Each of the six boxes <math>B_1</math>, <math>B_2</math>, <math>B_3</math>, <math>B_4</math>, <math>B_5</math>, <math>B_6</math> initially contains one coin. The following operations are allowed<br />
<br />
Type 1) Choose a non-empty box <math>B_j</math>, <math>1\leq j \leq 5</math>, remove one coin from <math>B_j</math> and add two coins to <math>B_{j+1}</math>; <br />
<br />
Type 2) Choose a non-empty box <math>B_k</math>, <math>1\leq k \leq 4</math>, remove one coin from <math>B_k</math> and swap the contents (maybe empty) of the boxes <math>B_{k+1}</math> and <math>B_{k+2}</math>.<br />
<br />
Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes <math>B_1</math>, <math>B_2</math>, <math>B_3</math>, <math>B_4</math>, <math>B_5</math> become empty, while box <math>B_6</math> contains exactly <math>2010^{2010^{2010}}</math> coins.<br />
<br />
''Author: Hans Zantema''<br />
<br />
== Solution ==<br />
<br />
Let the notation <math>[a_1,a_2,a_3,a_4,a_5,a_6]</math> be the configuration in which the <math>x</math>-th box has <math>a_x</math> coin,<br />
<br />
Let <math>T=2010^{2010^{2010}}</math>.<br />
<br />
<br/><br />
<br />
Our starting configuration is <math>[1,1,1,1,1,1]</math><br />
<br />
<br /><br />
<br />
We need these compound moves:<br />
<br />
Compound move 1: <math>[a,0]\rightarrow[0,2a]</math>, this is just repeated type 1 move on all <math>a</math> coins.<br />
<br />
Compound move 2: <math>[a,0,0]\rightarrow[0,2^a,0]</math>, apply type 1 move on 1 of the coin to get <math>[a-1,2,0]</math>, then apply compound move 1 to the 2 coins to get <math>[a-1,0,2^2]</math>, apply type 2 move and get <math>[a-2,2^2,0]</math>, and repeat compound move 1 and type 2 move until <math>[0,2^a,0]</math> is achieve.<br />
<br />
Compound move 3: <math>[a,b,0,0]\rightarrow[a-1,2^b,0,0]</math>, apply compound move 2 to obtain <math>[a,0,2^b,0]</math> and use type 2 move to get <math>[a-1,2^b,0,0]</math><br />
<br />
Compound move 4: <math>[a,0,0,0]\rightarrow[0,2^{2^{.^{.^{.^2}}}},0,0]</math> with <math>a</math> <math>2</math>'s. Apply compound move 3 <math>a</math> times.<br />
<br />
Compound move 5: <math>[a,0,0]\rightarrow[b,0,0]</math> with <math>a>b</math>, use type 2 move <math>(a-b)</math> times.<br />
<br/><br/><br />
<br />
Let's follow this move:<br />
<br />
<math>[1,1,1,1,1,1] \rightarrow[0,3,1,1,1,1] \rightarrow[0,2,3,1,1,1]\rightarrow[0,2,1,5,1,1]\rightarrow[0,2,1,1,9,1]\rightarrow[0,2,1,1,0,19]\rightarrow[0,1,19,0,0,0]</math><br />
<br />
<br />
Using Compound move 1, 4 and 5, We can obtain:<br />
<br />
<math>[0,1,19,0,0,0]\rightarrow[0,1,0,X,0,0]\rightarrow[0,0,X,0,0,0]\rightarrow[0,0,0,Y,0,0]\rightarrow[0,0,0,T/4,0,0]\rightarrow[0,0,0,0,T/2,0]\rightarrow[0,0,0,0,0,T]</math><br />
<br />
, where <math>X, Y = 2^{2^{.^{.^{.^2}}}}</math> where X has <math>19</math> 2's and Y has <math>X</math> 2's, and <math>Y</math> is clearly bigger then <math>T/4</math><br />
<br />
<br />
== See Also ==<br />
{{IMO box|year=2010|num-b=4|num-a=6}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_19&diff=930622012 AMC 10B Problems/Problem 192018-03-09T11:26:59Z<p>Jazzachi: /* Solution */</p>
<hr />
<div>==Problem==<br />
In rectangle <math>ABCD</math>, <math>AB=6</math>, <math>AD=30</math>, and <math>G</math> is the midpoint of <math>\overline{AD}</math>. Segment <math>AB</math> is extended 2 units beyond <math>B</math> to point <math>E</math>, and <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>. What is the area of <math>BFDG</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
<br />
Note that the area of <math>BFDG</math> equals the area of <math>ABCD-\triangle AGB-\triangle DCF</math>. <br />
Since <math>AG=\frac{AD}{2}=15,</math> <math>\triangle AGB=\frac{15\times 6}{2}=45</math>. Now, <math>\triangle AED\sim \triangle BEF</math>, so <math>\frac{AE}{BE}=4=\frac{AD}{BF}=\frac{30}{BF}\implies BF=7.5</math> and <math>FC=22.5,</math> so <math>\triangle DCF=\frac{22.5\times6}{2}=\frac{135}{2}. </math><br />
<br />
Therefore, <cmath>\begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*}</cmath><br />
which is answer choice (C).<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_19&diff=930612012 AMC 10B Problems/Problem 192018-03-09T11:26:42Z<p>Jazzachi: /* Solution */</p>
<hr />
<div>==Problem==<br />
In rectangle <math>ABCD</math>, <math>AB=6</math>, <math>AD=30</math>, and <math>G</math> is the midpoint of <math>\overline{AD}</math>. Segment <math>AB</math> is extended 2 units beyond <math>B</math> to point <math>E</math>, and <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>. What is the area of <math>BFDG</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
<br />
Note that the area of <math>BFDG</math> equals the area of <math>ABCD-\triangle AGB-\triangle DCF</math>. <br />
Since <math>AG=\frac{AD}{2}=15,</math> <math>\triangle AGB=\frac{15\times 6}{2}=45</math>. Now, <math>\triangle AED\sim \triangle BEF</math>, so <math>\frac{AE}{BE}=4=\frac{AD}{BF}=\frac{30}{BF}\implies BF=7.5</math> and <math>FC=22.5,</math> so <math>\triangle DCF=\frac{22.5\times6}{2}=\frac{135}{2}. </math><br />
<br />
Therefore, <cmath>\begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*}</cmath><br />
which is answer choice (C).<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_19&diff=930602012 AMC 10B Problems/Problem 192018-03-09T11:25:40Z<p>Jazzachi: /* Solution */</p>
<hr />
<div>==Problem==<br />
In rectangle <math>ABCD</math>, <math>AB=6</math>, <math>AD=30</math>, and <math>G</math> is the midpoint of <math>\overline{AD}</math>. Segment <math>AB</math> is extended 2 units beyond <math>B</math> to point <math>E</math>, and <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>. What is the area of <math>BFDG</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
<br />
Note that the area of <math>BFDG</math> equals the area of <math>ABCD-\triangle AGB-\triangle DCF</math>. <br />
Since <math>AG=\frac{AD}{2}=15,</math> <math>\triangle AGB=\frac{15\times 6}{2}=45</math>. Now, <math>\triangle AED\sim \triangle BEF</math>, so <math>\frac{AE}{BE}=4=\frac{AD}{BF}=\frac{30}{BF}\implies BF=7.5</math> and <math>FC=22.5</math> So <math>\triangle DCF=\frac{22.5\times6}{2}=\frac{135}{2}. </math> Therefore, <cmath>\begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*}</cmath><br />
which is answer choice (C).<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Jazzachihttps://artofproblemsolving.com/wiki/index.php?title=1956_AHSME_Problems/Problem_44&diff=901631956 AHSME Problems/Problem 442018-02-03T02:30:14Z<p>Jazzachi: made question</p>
<hr />
<div>== Problem 44==<br />
<br />
If <math>x < a < 0</math> means that <math>x</math> and <math>a</math> are numbers such that <math>x</math> is less than <math>a</math> and <math>a</math> is less than zero, then: <br />
<br />
<math>\textbf{(A)}\ x^2 < ax < 0 \qquad<br />
\textbf{(B)}\ x^2 > ax > a^2 \qquad<br />
\textbf{(C)}\ x^2 < a^2 < 0 \\<br />
\textbf{(D)}\ x^2 > ax\text{ but }ax < 0 \qquad<br />
\textbf{(E)}\ x^2 > a^2\text{ but }a^2 < 0 </math></div>Jazzachi