https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Fortytwok&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T08:49:22ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_24&diff=1403481993 AHSME Problems/Problem 242020-12-23T04:25:02Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
A box contains <math>3</math> shiny pennies and <math>4</math> dull pennies. One by one, pennies are drawn at random from the box and not replaced. If the probability is <math>a/b</math> that it will take more than four draws until the third shiny penny appears and <math>a/b</math> is in lowest terms, then <math>a+b=</math><br />
<br />
<math>\text{(A) } 11\quad<br />
\text{(B) } 20\quad<br />
\text{(C) } 35\quad<br />
\text{(D) } 58\quad<br />
\text{(E) } 66</math><br />
<br />
<br />
== Solution ==<br />
<br />
<br />
<br />
First let’s try to find the number of possible unique combinations. I’ll denote shiny coins as 1 and dull coins as 0. <br />
<br />
Now, each configuration can be represented by a string of 1s and 0s e.g. 0110100. Notice that a combination can be uniquely determined solely by the placement of their 0s OR 1s e.g. 1 - - 1 1 - - where the dashes can be replaced by 0s. This makes the number of unique combinations 7 choose 3 (if you’re counting w.r.t. shiny coins) OR 7 choose 4 (w.r.t dull coins). Both are equal to 35.<br />
<br />
Next, observe that, for the event that the third shiny coin is not within your first 4 picks, it has to be within the last three numbers. You can think of this as placing the seven coins in a vertical stack in the box and shuffling that stack randomly. Then, to pick, you extract the first coin on the top and keep repeating. It has the same effect.<br />
<br />
The sequence can have 1 shiny coin in the last 3 digits (Case 1), 2 shiny coins in the last 3 digits (Case 2) or 3 shiny coin in the last three digits (Case 3).<br />
<br />
Case 1:<br />
Let’s start with the first case of one shiny coin in its last 3 digits.<br />
<br />
Example: 0110100<br />
<br />
The first four numbers has 4 spaces and 2 shiny coins therefore the number of combinations is 4 choose 2 = 6. The last 3 digits has 3 combinations for the same reason.<br />
So, probability for Case 1 to occur is:<br />
<math>\dfrac{6*3}{35}=\dfrac{18}{35}</math><br />
<br />
Case 2:<br />
Using the fact that the combinations are uniquely determined by an order of 0s or 1s and you can just fill the rest, in you can ascertain:<br />
<math>\underbracket{0100}_{\text{4 combinations}}\, \underbracket{110}_{\text{3 combinations}}</math><br />
<br />
So, P(Case 2)=<math>12/35</math><br />
<br />
Case 3:<br />
Trivially, it is 1.<br />
P(Case 3)=<math>1/35</math><br />
<br />
Adding all these probabilites together gives you the probability that the third shiny coin will not appear in your first 4 draws:<br />
<math>\dfrac{18}{35}+\dfrac{12}{35}+\dfrac{1}{35}=\dfrac{31}{35}</math><br />
<br />
<math>\dfrac{a}{b}=\dfrac{31}{35}</math><br />
<br />
Since the fraction is irreducible:<br />
<br />
<math>a=31</math><br />
,<math>b=35</math><br />
<br />
<cmath>a+b=66</cmath><br />
<br />
The answer is E.<br />
<br />
<br />
== Solution 2==<br />
Using complementary probability, we can reduce the problem into two cases- the third shiny penny is drawn on the third draw, or on the fourth draw. For the first case, there is only one way to have the shiny pennies as the first three coins drawn, out of a possible 35 drawings (7 choose 3). <br />
For the second case, the third shiny penny has to be the fourth penny drawn, which leaves three possible orderings for the first three coins drawn (NS S S, S NS S, S S NS), out of 35 (7 choose 4). Adding these two probabilities together gives <math>\dfrac{4}{35}</math>, and subtracting this from one yields <math>\dfrac{31}{35}</math>, which makes a 31 and b 35, which sum to <math>\fbox{E}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1993|num-b=23|num-a=25}} <br />
<br />
[[Category: Intermediate Probability Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems&diff=1403451993 AHSME Problems2020-12-23T03:38:14Z<p>Fortytwok: /* Problem 22 */</p>
<hr />
<div>{{AHSME Problems<br />
|year = 1993<br />
}}<br />
== Problem 1 ==<br />
For integers <math>a, b</math> and <math>c</math>, define <math>\boxed{a,b,c}</math> to mean <math>a^b-b^c+c^a</math>. Then <math>\boxed{1,-1,2}</math> equals<br />
<br />
<math>\text{(A)} \ -4 \qquad \text{(B)} \ -2 \qquad \text{(C)} \ 0 \qquad \text{(D)} \ 2 \qquad \text{(E)} \ 4</math><br />
<br />
[[1993 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
In <math>\triangle ABC</math>, <math>\angle A=55^\circ</math>, <math>\angle C=75^\circ</math>, <math>D</math> is on side <math>\overline{AB}</math> and <math>E</math> is on side <math>\overline{BC}</math> If <math>DB=BE</math>, then <math>\angle BED=</math><br />
<br />
<math>\text{(A)}\ 50^\circ \qquad<br />
\text{(B)}\ 55^\circ \qquad<br />
\text{(C)}\ 60^\circ \qquad<br />
\text{(D)}\ 65^\circ \qquad<br />
\text{(E)}\ 70^\circ </math><br />
<br />
<br />
[[1993 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
<math>\frac{15^{30}}{45^{15}} =</math><br />
<br />
<math>\text{(A) } \left(\frac{1}{3}\right)^{15}\quad<br />
\text{(B) } \left(\frac{1}{3}\right)^{2}\quad<br />
\text{(C) } 1\quad<br />
\text{(D) } 3^{15}\quad<br />
\text{(E) } 5^{15}</math><br />
<br />
[[1993 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Define the operation "<math>\circ</math>" by <math>x\circ y=4x-3y+xy</math>, for all real numbers <math>x</math> and <math>y</math>. For how many real numbers <math>y</math> does <math>3\circ y=12</math>?<br />
<br />
<math>\text{(A) } 0\quad<br />
\text{(B) } 1\quad<br />
\text{(C) } 3\quad<br />
\text{(D) } 4\quad<br />
\text{(E) more than 4} </math><br />
<br />
<br />
[[1993 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
Last year a bicycle cost \$160 and a cycling helmet \$40. This year the cost of the bicycle increased by <math>5\%</math>, and the cost of the helmet increased by <math>10\%</math>. The percent increase in the combined cost of the bicycle and the helmet is:<br />
<br />
<math>\text{(A) } 6\%\quad<br />
\text{(B) } 7\%\quad<br />
\text{(C) } 7.5\%\quad<br />
\text{(D) } 8\%\quad<br />
\text{(E) } 15\%</math><br />
<br />
[[1993 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
<br />
<math>\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=</math><br />
<br />
<math>\text{(A) } \sqrt{2}\quad<br />
\text{(B) } 16\quad<br />
\text{(C) } 32\quad<br />
\text{(D) } (12)^{\tfrac{2}{3}}\quad<br />
\text{(E) } 512.5</math><br />
<br />
[[1993 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
The symbol <math>R_k</math> stands for an integer whose base-ten representation is a sequence of <math>k</math> ones. For example, <math>R_3=111,R_5=11111</math>, etc. When <math>R_{24}</math> is divided by <math>R_4</math>, the quotient <math>Q=R_{24}/R_4</math> is an integer whose base-ten representation is a sequence containing only ones and zeroes. The number of zeros in <math>Q</math> is:<br />
<br />
<math>\text{(A) } 10\quad<br />
\text{(B) } 11\quad<br />
\text{(C) } 12\quad<br />
\text{(D) } 13\quad<br />
\text{(E) } 15</math><br />
<br />
<br />
[[1993 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
Let <math>C_1</math> and <math>C_2</math> be circles of radius 1 that are in the same plane and tangent to each other. How many circles of radius 3 are in this plane and tangent to both <math>C_1</math> and <math>C_2</math>?<br />
<br />
<math>\text{(A) } 2\quad<br />
\text{(B) } 4\quad<br />
\text{(C) } 5\quad<br />
\text{(D) } 6\quad<br />
\text{(E) } 8</math><br />
<br />
[[1993 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
<br />
Country <math>A</math> has <math>c\%</math> of the world's population and <math>d\%</math> of the worlds wealth. Country <math>B</math> has <math>e\%</math> of the world's population and <math>f\%</math> of its wealth. Assume that the citizens of <math>A</math> share the wealth of <math>A</math> equally,and assume that those of <math>B</math> share the wealth of <math>B</math> equally. Find the ratio of the wealth of a citizen of <math>A</math> to the wealth of a citizen of <math>B</math>.<br />
<br />
<math>\text{(A) } \frac{cd}{ef}\quad<br />
\text{(B) } \frac{ce}{ef}\quad<br />
\text{(C) } \frac{cf}{de}\quad<br />
\text{(D) } \frac{de}{cf}\quad<br />
\text{(E) } \frac{df}{ce}</math><br />
<br />
[[1993 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
<br />
Let <math>r</math> be the number that results when both the base and the exponent of <math>a^b</math> are tripled, where <math>a,b>0</math>. If <math>r</math> equals the product of <math>a^b</math> and <math>x^b</math> where <math>x>0</math>, then <math>x=</math><br />
<br />
<math>\text{(A) } 3\quad<br />
\text{(B) } 3a^2\quad<br />
\text{(C) } 27a^2\quad<br />
\text{(D) } 2a^{3b}\quad<br />
\text{(E) } 3a^{2b}</math><br />
<br />
[[1993 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
<br />
If <math>\log_2(\log_2(\log_2(x)))=2</math>, then how many digits are in the base-ten representation for <math>x</math>?<br />
<br />
<math>\text{(A) } 5\quad<br />
\text{(B) } 7\quad<br />
\text{(C) } 9\quad<br />
\text{(D) } 11\quad<br />
\text{(E) } 13</math><br />
<br />
[[1993 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
If <math>f(2x)=\frac{2}{2+x}</math> for all <math>x>0</math>, then <math>2f(x)=</math><br />
<br />
<math>\text{(A) } \frac{2}{1+x}\quad<br />
\text{(B) } \frac{2}{2+x}\quad<br />
\text{(C) } \frac{4}{1+x}\quad<br />
\text{(D) } \frac{4}{2+x}\quad<br />
\text{(E) } \frac{8}{4+x}</math><br />
<br />
[[1993 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
A square of perimeter 20 is inscribed in a square of perimeter 28. What is the greatest distance between a vertex of the inner square and a vertex of the outer square?<br />
<br />
<math>\text{(A) } \sqrt{58}\quad<br />
\text{(B) } \frac{7\sqrt{5}}{2}\quad<br />
\text{(C) } 8\quad<br />
\text{(D) } \sqrt{65}\quad<br />
\text{(E) } 5\sqrt{3}</math><br />
<br />
<br />
[[1993 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<asy><br />
draw((-1,0)--(1,0)--(1+sqrt(2),sqrt(2))--(0,sqrt(2)+sqrt(13-2*sqrt(2)))--(-1-sqrt(2),sqrt(2))--cycle,black+linewidth(.75));<br />
MP("A",(-1,0),SW);MP("B",(1,0),SE);MP("C",(1+sqrt(2),sqrt(2)),E);MP("D",(0,sqrt(2)+sqrt(13-2*sqrt(2))),N);MP("E",(-1-sqrt(2),sqrt(2)),W);<br />
dot((-1,0));dot((1,0));dot((1+sqrt(2),sqrt(2)));dot((-1-sqrt(2),sqrt(2)));dot((0,sqrt(2)+sqrt(13-2*sqrt(2))));<br />
</asy><br />
<br />
The convex pentagon <math>ABCDE</math> has <math>\angle{A}=\angle{B}=120^\circ,EA=AB=BC=2</math> and <math>CD=DE=4</math>. What is the area of ABCDE?<br />
<br />
<math>\text{(A) } 10\quad<br />
\text{(B) } 7\sqrt{3}\quad<br />
\text{(C) } 15\quad<br />
\text{(D) } 9\sqrt{3}\quad<br />
\text{(E) } 12\sqrt{5}</math><br />
<br />
[[1993 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
<br />
For how many values of <math>n</math> will an <math>n</math>-sided regular polygon have interior angles with integral measures?<br />
<br />
<math>\text{(A) } 16\quad<br />
\text{(B) } 18\quad<br />
\text{(C) } 20\quad<br />
\text{(D) } 22\quad<br />
\text{(E) } 24</math><br />
<br />
[[1993 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Consider the non-decreasing sequence of positive integers<br />
<cmath>1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots</cmath><br />
in which the <math>n^{th}</math> positive integer appears <math>n</math> times. The remainder when the <math>1993^{rd} </math> term is divided by <math>5</math> is<br />
<br />
<math>\text{(A) } 0\quad<br />
\text{(B) } 1\quad<br />
\text{(C) } 2\quad<br />
\text{(D) } 3\quad<br />
\text{(E) } 4</math><br />
<br />
[[1993 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
<asy><br />
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle, black+linewidth(.75));<br />
draw((0,-1)--(0,1), black+linewidth(.75));<br />
draw((-1,0)--(1,0), black+linewidth(.75));<br />
draw((-1,-1/sqrt(3))--(1,1/sqrt(3)), black+linewidth(.75));<br />
draw((-1,1/sqrt(3))--(1,-1/sqrt(3)), black+linewidth(.75));<br />
draw((-1/sqrt(3),-1)--(1/sqrt(3),1), black+linewidth(.75));<br />
draw((1/sqrt(3),-1)--(-1/sqrt(3),1), black+linewidth(.75));<br />
</asy><br />
<br />
Amy painted a dartboard over a square clock face using the "hour positions" as boundaries.[See figure.] If <math>t</math> is the area of one of the eight triangular regions such as that between 12 o'clock and 1 o'clock, and <math>q</math> is the area of one of the four corner quadrilaterals such as that between 1 o'clock and 2 o'clock, then <math>\frac{q}{t}=</math><br />
<br />
<math>\text{(A) } 2\sqrt{3}-2\quad<br />
\text{(B) } \frac{3}{2}\quad<br />
\text{(C) } \frac{\sqrt{5}+1}{2}\quad<br />
\text{(D) } \sqrt{3}\quad<br />
\text{(E) } 2</math><br />
<br />
[[1993 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Al and Barb start their new jobs on the same day. Al's schedule is 3 work-days followed by 1 rest-day. Barb's schedule is 7 work-days followed by 3 rest-days. On how many of their first 1000 days do both have rest-days on the same day?<br />
<br />
<math>\text{(A) } 48\quad<br />
\text{(B) } 50\quad<br />
\text{(C) } 72\quad<br />
\text{(D) } 75\quad<br />
\text{(E) } 100</math><br />
<br />
[[1993 AHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
How many ordered pairs <math>(m,n)</math> of positive integers are solutions to<br />
<cmath>\frac{4}{m}+\frac{2}{n}=1?</cmath><br />
<br />
<math>\text{(A) } 1\quad<br />
\text{(B) } 2\quad<br />
\text{(C) } 3\quad<br />
\text{(D) } 4\quad<br />
\text{(E) } \text{more than }6</math><br />
<br />
[[1993 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
Consider the equation <math>10z^2-3iz-k=0</math>, where <math>z</math> is a complex variable and <math>i^2=-1</math>. Which of the following statements is true?<br />
<br />
<math>\text{(A) For all positive real numbers k, both roots are pure imaginary} \quad\\<br />
\text{(B) For all negative real numbers k, both roots are pure imaginary} \quad\\<br />
\text{(C) For all pure imaginary numbers k, both roots are real and rational} \quad\\<br />
\text{(D) For all pure imaginary numbers k, both roots are real and irrational} \quad\\<br />
\text{(E) For all complex numbers k, neither root is real} </math><br />
<br />
[[1993 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
Let <math>a_1,a_2,\cdots,a_k</math> be a finite arithmetic sequence with <math>a_4 +a_7+a_{10} = 17</math> and <math>a_4+a_5+\cdots+a_{13} +a_{14} = 77</math>. <br />
<br />
If <math>a_k = 13</math>, then <math>k =</math><br />
<br />
<math>\text{(A) } 16\quad<br />
\text{(B) } 18\quad<br />
\text{(C) } 20\quad<br />
\text{(D) } 22\quad<br />
\text{(E) } 24</math><br />
<br />
<br />
[[1993 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
<asy><br />
draw((-1,0)--(1,0)--(1,-1)--(-1,-1)--cycle,black+linewidth(.75));<br />
MP("?",(0,0),S);<br />
</asy><br />
<br />
Twenty cubical blocks are arranged as shown. First, 10 are arranged in a triangular pattern; then a layer of 6, arranged in a triangular pattern, is centered on the 10; then a layer of 3, arranged in a triangular pattern, is centered on the 6; and finally one block is centered on top of the third layer. The blocks in the bottom layer are numbered 1 through 10 in some order. Each block in layers 2,3 and 4 is assigned the number which is the sum of numbers assigned to the three blocks on which it rests. Find the smallest possible number which could be assigned to the top block.<br />
<br />
<math>\text{(A) } 55\quad<br />
\text{(B) } 83\quad<br />
\text{(C) } 114\quad<br />
\text{(D) } 137\quad<br />
\text{(E) } 144</math><br />
<br />
[[1993 AHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
<asy><br />
draw(circle((0,0),10),black+linewidth(.75));<br />
draw((-10,0)--(10,0),black+linewidth(.75));<br />
draw((-10,0)--(9,sqrt(19)),black+linewidth(.75));<br />
draw((-10,0)--(9,-sqrt(19)),black+linewidth(.75));<br />
draw((2,0)--(9,sqrt(19)),black+linewidth(.75));<br />
draw((2,0)--(9,-sqrt(19)),black+linewidth(.75));<br />
MP("X",(2,0),N);MP("A",(-10,0),W);MP("D",(10,0),E);MP("B",(9,sqrt(19)),E);MP("C",(9,-sqrt(19)),E);<br />
</asy><br />
<br />
Points <math>A,B,C</math> and <math>D</math> are on a circle of diameter <math>1</math>, and <math>X</math> is on diameter <math>\overline{AD}.</math><br />
<br />
If <math>BX=CX</math> and <math>3\angle{BAC}=\angle{BXC}=36^\circ</math>, then <math>AX=</math><br />
<br />
<br />
<math>\text{(A) } \cos(6^\circ)\cos(12^\circ)\sec(18^\circ)\quad\\<br />
\text{(B) } \cos(6^\circ)\sin(12^\circ)\csc(18^\circ)\quad\\<br />
\text{(C) } \cos(6^\circ)\sin(12^\circ)\sec(18^\circ)\quad\\<br />
\text{(D) } \sin(6^\circ)\sin(12^\circ)\csc(18^\circ)\quad\\<br />
\text{(E) } \sin(6^\circ)\sin(12^\circ)\sec(18^\circ)</math><br />
<br />
<br />
[[1993 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
A box contains <math>3</math> shiny pennies and <math>4</math> dull pennies. One by one, pennies are drawn at random from the box and not replaced. If the probability is <math>a/b</math> that it will take more than four draws until the third shiny penny appears and <math>a/b</math> is in lowest terms, then <math>a+b=</math><br />
<br />
<math>\text{(A) } 11\quad<br />
\text{(B) } 20\quad<br />
\text{(C) } 35\quad<br />
\text{(D) } 58\quad<br />
\text{(E) } 66</math><br />
<br />
[[1993 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
<asy><br />
draw((0,0)--(1,sqrt(3)),black+linewidth(.75),EndArrow);<br />
draw((0,0)--(1,-sqrt(3)),black+linewidth(.75),EndArrow);<br />
draw((0,0)--(1,0),dashed+black+linewidth(.75));<br />
dot((1,0));<br />
MP("P",(1,0),E);<br />
</asy><br />
<br />
Let <math>S</math> be the set of points on the rays forming the sides of a <math>120^{\circ}</math> angle, and let <math>P</math> be a fixed point inside the angle <br />
on the angle bisector. Consider all distinct equilateral triangles <math>PQR</math> with <math>Q</math> and <math>R</math> in <math>S</math>. <br />
(Points <math>Q</math> and <math>R</math> may be on the same ray, and switching the names of <math>Q</math> and <math>R</math> does not create a distinct triangle.) <br />
There are<br />
<br />
<math>\text{(A) exactly 2 such triangles} \quad\\<br />
\text{(B) exactly 3 such triangles} \quad\\<br />
\text{(C) exactly 7 such triangles} \quad\\<br />
\text{(D) exactly 15 such triangles} \quad\\<br />
\text{(E) more than 15 such triangles} </math><br />
<br />
[[1993 AHSME Problems/Problem 25|Solution]]<br />
<br />
== Problem 26 ==<br />
<br />
Find the largest positive value attained by the function<br />
<cmath>f(x)=\sqrt{8x-x^2}-\sqrt{14x-x^2-48} ,\quad x \text{ a real number}</cmath><br />
<br />
<math>\text{(A) } \sqrt{7}-1\quad<br />
\text{(B) } 3\quad<br />
\text{(C) } 2\sqrt{3}\quad<br />
\text{(D) } 4\quad<br />
\text{(E) } \sqrt{55}-\sqrt{5}</math><br />
<br />
[[1993 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27 ==<br />
<br />
<asy><br />
draw(circle((4,1),1),black+linewidth(.75));<br />
draw((0,0)--(8,0)--(8,6)--cycle,black+linewidth(.75));<br />
MP("A",(0,0),SW);MP("B",(8,0),SE);MP("C",(8,6),NE);MP("P",(4,1),NW);<br />
MP("8",(4,0),S);MP("6",(8,3),E);MP("10",(4,3),NW);<br />
MP("->",(5,1),E);<br />
dot((4,1));<br />
</asy><br />
The sides of <math>\triangle ABC</math> have lengths <math>6,8,</math> and <math>10</math>. A circle with center <math>P</math> and radius <math>1</math> rolls around the inside of <math>\triangle ABC</math>, always remaining tangent to at least one side of the triangle. When <math>P</math> first returns to its original position, through what distance has <math>P</math> traveled?<br />
<br />
<math>\text{(A) } 10\quad<br />
\text{(B) } 12\quad<br />
\text{(C) } 14\quad<br />
\text{(D) } 15\quad<br />
\text{(E) } 17</math><br />
<br />
[[1993 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28 ==<br />
How many triangles with positive area are there whose vertices are points in the <math>xy</math>-plane whose coordinates are integers <math>(x,y)</math> satisfying <math>1\le x\le 4</math> and <math>1\le y\le 4</math>?<br />
<br />
<math>\text{(A) } 496\quad<br />
\text{(B) } 500\quad<br />
\text{(C) } 512\quad<br />
\text{(D) } 516\quad<br />
\text{(E) } 560</math><br />
<br />
<br />
[[1993 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29 ==<br />
<br />
Which of the following could NOT be the lengths of the external diagonals of a right regular prism [a "box"]? (An <math>\textit{external diagonal}</math> is a diagonal of one of the rectangular faces of the box.)<br />
<br />
<math>\text{(A) }\{4,5,6\} \quad<br />
\text{(B) } \{4,5,7\} \quad<br />
\text{(C) } \{4,6,7\} \quad<br />
\text{(D) } \{5,6,7\} \quad<br />
\text{(E) } \{5,7,8\} </math><br />
<br />
[[1993 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30 ==<br />
<br />
Given <math>0\le x_0<1</math>, let <br />
<cmath><br />
x_n=\left\{ \begin{array}{ll}<br />
2x_{n-1} &\text{ if }2x_{n-1}<1 \\<br />
2x_{n-1}-1 &\text{ if }2x_{n-1}\ge 1<br />
\end{array}\right.<br />
</cmath><br />
for all integers <math>n>0</math>. For how many <math>x_0</math> is it true that <math>x_0=x_5</math>?<br />
<br />
<math>\text{(A) 0} \quad<br />
\text{(B) 1} \quad<br />
\text{(C) 5} \quad<br />
\text{(D) 31} \quad<br />
\text{(E) }\infty </math><br />
<br />
[[1993 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=1993|before=[[1992 AHSME]]|after=[[1994 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_19&diff=1403441993 AHSME Problems/Problem 192020-12-23T03:30:46Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
How many ordered pairs <math>(m,n)</math> of positive integers are solutions to<br />
<cmath>\frac{4}{m}+\frac{2}{n}=1?</cmath><br />
<br />
<math>\text{(A) } 1\quad<br />
\text{(B) } 2\quad<br />
\text{(C) } 3\quad<br />
\text{(D) } 4\quad<br />
\text{(E) } \text{more than }6</math><br />
<br />
== Solution ==<br />
Multiply both sides by <math>mn</math> to clear the denominator. Moving all the terms to the right hand side, the equation becomes <math>0 = mn-4n-2m</math>. Adding 8 to both sides allows us to factor the equation as follows: <math>(m-4)(n-2) = 8</math>. Since the problem only wants integer pairs <math>(m,n)</math>, the pairs are given by the factors of 8, which are 1,2,4, and 8. This results in four pairs, <math>(5,10), (6,6), (8,4),</math> and <math>(12,3)</math>, which is answer <br />
<math>\fbox{D}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1993|num-b=18|num-a=20}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_19&diff=1403431993 AHSME Problems/Problem 192020-12-23T03:30:29Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
How many ordered pairs <math>(m,n)</math> of positive integers are solutions to<br />
<cmath>\frac{4}{m}+\frac{2}{n}=1?</cmath><br />
<br />
<math>\text{(A) } 1\quad<br />
\text{(B) } 2\quad<br />
\text{(C) } 3\quad<br />
\text{(D) } 4\quad<br />
\text{(E) } \text{more than }6</math><br />
<br />
== Solution ==<br />
Multiply both sides by <math>mn</math> to clear the denominator. Moving all the terms to the right hand side, the equation becomes <math>0 = mn-4n-2m</math>. Adding 8 to both sides allows us to factor the equation as follows: <math>(m-4)(n-2) = 8</math>. Since the problem only wants integer pairs <math>(m,n)</math>, the pairs are given by the factors of 8, which are 1,2,4, and 8. This results in four pairs, <math>(5,10), (6,6), (8,4),</math> and <math>(12,3)</math>, which is answer <br />
<math>\fbox{D}</math>.<br />
<br />
~42k<br />
<br />
== See also ==<br />
{{AHSME box|year=1993|num-b=18|num-a=20}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_19&diff=1403421993 AHSME Problems/Problem 192020-12-23T03:28:57Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
How many ordered pairs <math>(m,n)</math> of positive integers are solutions to<br />
<cmath>\frac{4}{m}+\frac{2}{n}=1?</cmath><br />
<br />
<math>\text{(A) } 1\quad<br />
\text{(B) } 2\quad<br />
\text{(C) } 3\quad<br />
\text{(D) } 4\quad<br />
\text{(E) } \text{more than }6</math><br />
<br />
== Solution ==<br />
Multiply both sides by <math>mn</math> to clear the denominator. Moving all the terms to the right hand side, the equation becomes <math>0 = mn-4n-2m</math>. Adding 8 to both sides allows us to factor the equation as follows: <math>(m-4)(n-2) = 8</math>. Since the problem only wants integer pairs <math>(m,n)</math>, the pairs are given by the factors of 8, which are 1,2,4, and 8. This results in four pairs, <math>(5,10), (6,6), (8,4),</math> and <math>(12,3)</math>, which is answer <br />
<math>\fbox{D}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1993|num-b=18|num-a=20}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_5&diff=1220161955 AHSME Problems/Problem 52020-05-04T17:36:53Z<p>Fortytwok: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<math>5y</math> varies inversely as the square of <math>x</math>. When <math>y=16, x=1</math>. When <math>x=8, y</math> equals: <br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024 </math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|num-b=4|num-a=6}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_5&diff=1220151955 AHSME Problems/Problem 52020-05-04T17:34:59Z<p>Fortytwok: /* See Also */</p>
<hr />
<div>== Problem ==<br />
<br />
==Solution==<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|num-b=4|num-a=6}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_5&diff=1220141955 AHSME Problems/Problem 52020-05-04T17:34:43Z<p>Fortytwok: Created page with "== Problem == ==Solution== ==See Also== {{AHSME 50p box|year=1955|num-b=|num-a=}} {{MAA Notice}}"</p>
<hr />
<div>== Problem ==<br />
<br />
==Solution==<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|num-b=|num-a=}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_3&diff=1220131955 AHSME Problems/Problem 32020-05-04T17:18:51Z<p>Fortytwok: /* See Also */</p>
<hr />
<div>== Problem==<br />
If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the ten numbers: <br />
<br />
<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math><br />
==Solution 1==<br />
Let the sum of the 10 numbers be x. The mean is then <math>\frac{x}{10}</math>. Then, since you're adding 20 to each number, the new sum of the numbers is x+200, since there are 10 numbers. Then, the new mean is <math>\frac{x+200}{10}</math>, which simplifies to <math>\frac{x}{10}+20</math>, which is <math>\fbox{B}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|num-b=2|num-a=4}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_3&diff=1220121955 AHSME Problems/Problem 32020-05-04T17:18:24Z<p>Fortytwok: /* See Also */</p>
<hr />
<div>== Problem==<br />
If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the ten numbers: <br />
<br />
<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math><br />
==Solution 1==<br />
Let the sum of the 10 numbers be x. The mean is then <math>\frac{x}{10}</math>. Then, since you're adding 20 to each number, the new sum of the numbers is x+200, since there are 10 numbers. Then, the new mean is <math>\frac{x+200}{10}</math>, which simplifies to <math>\frac{x}{10}+20</math>, which is <math>\fbox{B}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|num-b=1|num-a=3}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_3&diff=1220111955 AHSME Problems/Problem 32020-05-04T17:17:42Z<p>Fortytwok: /* See Also */</p>
<hr />
<div>== Problem==<br />
If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the ten numbers: <br />
<br />
<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math><br />
==Solution 1==<br />
Let the sum of the 10 numbers be x. The mean is then <math>\frac{x}{10}</math>. Then, since you're adding 20 to each number, the new sum of the numbers is x+200, since there are 10 numbers. Then, the new mean is <math>\frac{x+200}{10}</math>, which simplifies to <math>\frac{x}{10}+20</math>, which is <math>\fbox{B}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|num+a=1|num-a=3}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_3&diff=1220101955 AHSME Problems/Problem 32020-05-04T17:17:17Z<p>Fortytwok: /* Solution 1 */</p>
<hr />
<div>== Problem==<br />
If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the ten numbers: <br />
<br />
<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math><br />
==Solution 1==<br />
Let the sum of the 10 numbers be x. The mean is then <math>\frac{x}{10}</math>. Then, since you're adding 20 to each number, the new sum of the numbers is x+200, since there are 10 numbers. Then, the new mean is <math>\frac{x+200}{10}</math>, which simplifies to <math>\frac{x}{10}+20</math>, which is <math>\fbox{B}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|before=num-a=1|num-a=3}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_3&diff=1220091955 AHSME Problems/Problem 32020-05-04T16:47:01Z<p>Fortytwok: /* Solution 1 */</p>
<hr />
<div>== Problem==<br />
If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the ten numbers: <br />
<br />
<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math><br />
==Solution 1==<br />
Let the sum of the 10 numbers be x. The mean is then <math>\frac{x}{10}</math>. Then, since you're adding 20 to each number, the new sum of the numbers is x+200, since there are 10 numbers. Then, the new mean is <math>\frac{x+200}{10}</math>, which simplifies to <math>\frac{x}{10}+20</math>, which is <math>\fbox{B}</math>.</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_3&diff=1220081955 AHSME Problems/Problem 32020-05-04T16:44:32Z<p>Fortytwok: /* Solution 1 */</p>
<hr />
<div>== Problem==<br />
If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the ten numbers: <br />
<br />
<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math><br />
==Solution 1==<br />
Let the sum of the 10 numbers be x. The mean is then <math>\frac{x}{10}</math>. Then, since you're adding 20 to each number, the new sum of the numbers is x+200, since there are 10 numbers. Then, the new mean is <math>\frac{x+200}{10}, which simplifies to \frac{x}{10}+20</math>, which is <math>\fbox{B}</math>.</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_3&diff=1220071955 AHSME Problems/Problem 32020-05-04T16:40:23Z<p>Fortytwok: </p>
<hr />
<div>== Problem==<br />
If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the ten numbers: <br />
<br />
<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math><br />
==Solution 1==</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_2&diff=1220061955 AHSME Problems/Problem 22020-05-04T16:37:23Z<p>Fortytwok: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
The smaller angle between the hands of a clock at <math>12:25</math> p.m. is: <br />
<br />
<math> \textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ </math> <br />
<br />
==Solution==<br />
At <math>12:25</math>, the minute hand is at <math>\frac{25}{60}\cdot 360^\circ</math>, or <math>150^\circ</math>.<br />
The hour hand moves <math>5</math> 'minutes' every hour, or <math>\frac{5}{60}\cdot 360 = 30^\circ</math> every hour. At <math>30^\circ</math> every hour, the hour hand moves <math>\frac{1}{2}</math> minutes on the clock every minute. At <math>12:25</math>, the hour hand is at <math>\frac{25}{2}^\circ</math>. Therefore, the angle between the hands is <math>150^\circ - \frac{25}{2}^\circ</math>, <math>137.5^\circ</math>, or <math>137^{\circ} 30^{'} \implies</math> <math>\boxed{\mathrm{(B) 137^\circ 30'}}</math>.<br />
<br />
==Solution 2==<br />
<br />
Using the formula <math>\frac{|(60h-11m)|}{2}</math>, and plugging in 12 for h and 25 for m, we get the angle between to be 222.5<math>^{\circ}</math>. Since the problem wants the smaller angle, we do <math>360-222.5 = 137.5 \Rightarrow 137^{\circ}30'\Rightarrow \fbox{B}</math><br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|num-b=1|num-a=3}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_2&diff=1220041955 AHSME Problems/Problem 22020-05-04T16:33:43Z<p>Fortytwok: </p>
<hr />
<div>== Problem ==<br />
The smaller angle between the hands of a clock at <math>12:25</math> p.m. is: <br />
<br />
<math> \textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ </math> <br />
<br />
==Solution==<br />
At <math>12:25</math>, the minute hand is at <math>\frac{25}{60}\cdot 360^\circ</math>, or <math>150^\circ</math>.<br />
The hour hand moves <math>5</math> 'minutes' every hour, or <math>\frac{5}{60}\cdot 360 = 30^\circ</math> every hour. At <math>30^\circ</math> every hour, the hour hand moves <math>\frac{1}{2}</math> minutes on the clock every minute. At <math>12:25</math>, the hour hand is at <math>\frac{25}{2}^\circ</math>. Therefore, the angle between the hands is <math>150^\circ - \frac{25}{2}^\circ</math>, <math>137.5^\circ</math>, or <math>137^{\circ} 30^{'} \implies</math> <math>\boxed{\mathrm{(B) 137^\circ 30'}}</math>.<br />
<br />
==Solution 2==<br />
<br />
Using the formula <math>\frac{|(60h-11m)|}{2}</math>, and plugging in 12 for h and 25 for m, we get the angle between to be 222.5°. Since the problem wants the smaller angle, we do <math>360-222.5 = 137.5 \Rightarrow 137°30'</math>.<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|num-b=1|num-a=3}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_1&diff=1220011955 AHSME Problems/Problem 12020-05-04T16:21:21Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Which one of the following is not equivalent to <math>0.000000375</math>? <br />
<br />
<math> \textbf{(A)}\ 3.75\times 10^{-7}\qquad\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}\qquad\textbf{(C)}\ 375\times 10^{-9}\qquad \textbf{(D)}\ \frac{3}{8}\times 10^{-7}\qquad\textbf{(E)}\ \frac{3}{80000000} </math><br />
<br />
==Solution==<br />
First of all, <math>0.000000375 = 3.75 \times 10^{-7}</math> in scientific notation. This eliminates <math>\textbf{(A)}\ 3.75\times 10^{-7}</math>, <math>\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}</math>, and <math>\textbf{(C)}\ 375\times 10^{-9}</math> immediately. <math>\textbf{(E)}\ \frac{3}{80000000}</math> is a bit harder, but it can be rewritten as <math>\frac{3}{8} \cdot 10^{-7}</math>. Thus, as entered, the answer is <math>\boxed{\mathrm{(D) \frac{3}{8} \times 10^{-7}}}</math><br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1955|before=First Question|num-a=2}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems&diff=1219991955 AHSME Problems2020-05-04T15:30:45Z<p>Fortytwok: /* Problem 26 */</p>
<hr />
<div>{{AHSC 50 Problems<br />
|year=1955<br />
}}<br />
== Problem 1==<br />
Which one of the following is not equivalent to <math>0.000000375</math>? <br />
<br />
<math> \textbf{(A)}\ 3.75\times 10^{-7}\qquad\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}\qquad\textbf{(C)}\ 375\times 10^{-9}\\ \textbf{(D)}\ \frac{3}{8}\times 10^{-7}\qquad\textbf{(E)}\ \frac{3}{80000000} </math> <br />
<br />
[[1955 AHSME Problems/Problem 1|Solution]]<br />
== Problem 2==<br />
<br />
The smaller angle between the hands of a clock at <math>12:25</math> p.m. is: <br />
<br />
<math> \textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ </math> <br />
<br />
[[1955 AHSME Problems/Problem 2|Solution]]<br />
== Problem 3==<br />
<br />
If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the ten numbers: <br />
<br />
<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math><br />
<br />
[[1955 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4==<br />
The equality <math>\frac{1}{x-1}=\frac{2}{x-2}</math> is satisfied by: <br />
<br />
<math> \textbf{(A)}\ \text{no real values of }x\qquad\textbf{(B)}\ \text{either }x=1\text{ or }x=2\qquad\textbf{(C)}\ \text{only }x=1\\ \textbf{(D)}\ \text{only }x=2\qquad\textbf{(E)}\ \text{only }x=0 </math><br />
<br />
[[1955 AHSME Problems/Problem 4|Solution]]<br />
== Problem 5==<br />
<br />
<math>5y</math> varies inversely as the square of <math>x</math>. When <math>y=16, x=1</math>. When <math>x=8, y</math> equals: <br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024 </math><br />
<br />
[[1955 AHSME Problems/Problem 5|Solution]]<br />
== Problem 6==<br />
<br />
A merchant buys a number of oranges at <math>3</math> for <math>10</math> cents and an equal number at <math>5</math> for <math>20</math> cents. To "break even" he must sell all at: <br />
<br />
<math> \textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents} </math><br />
<br />
[[1955 AHSME Problems/Problem 6|Solution]]<br />
==Problem 7==<br />
If a worker receives a <math>20</math>% cut in wages, he may regain his original pay exactly by obtaining a raise of:<br />
<br />
<math>\textbf{(A)}\ \text{20\%}\qquad\textbf{(B)}\ \text{25\%}\qquad\textbf{(C)}\ 22\frac{1}{2}\text{\%}\qquad\textbf{(D)}\ \textdollar{20}\qquad\textbf{(E)}\ \textdollar{25}</math><br />
<br />
[[1955 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8==<br />
<br />
The graph of <math>x^2-4y^2=0</math>: <br />
<br />
<math> \textbf{(A)}\ \text{is a hyperbola intersecting only the }x\text{-axis}\\ \textbf{(B)}\ \text{is a hyperbola intersecting only the }y\text{-axis}\\ \textbf{(C)}\ \text{is a hyperbola intersecting neither axis}\\ \textbf{(D)}\ \text{is a pair of straight lines}\\ \textbf{(E)}\ \text{does not exist} </math><br />
<br />
[[1955 AHSME Problems/Problem 8|Solution]]<br />
== Problem 9==<br />
<br />
A circle is inscribed in a triangle with sides <math>8, 15</math>, and <math>17</math>. The radius of the circle is: <br />
<br />
<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 7 </math> <br />
<br />
[[1955 AHSME Problems/Problem 9|Solution]]<br />
== Problem 10==<br />
<br />
How many hours does it take a train traveling at an average rate of 40 mph between stops to travel a miles it makes n stops of m minutes each? <br />
<br />
<math> \textbf{(A)}\ \frac{3a+2mn}{120}\qquad\textbf{(B)}\ 3a+2mn\qquad\textbf{(C)}\ \frac{3a+2mn}{12}\qquad\textbf{(D)}\ \frac{a+mn}{40}\qquad\textbf{(E)}\ \frac{a+40mn}{40} </math><br />
<br />
[[1955 AHSME Problems/Problem 10|Solution]]<br />
== Problem 11==<br />
<br />
The negation of the statement "No slow learners attend this school" is: <br />
<br />
<math> \textbf{(A)}\ \text{All slow learners attend this school}\\ \textbf{(B)}\ \text{All slow learners do not attend this school}\\ \textbf{(C)}\ \text{Some slow learners attend this school}\\ \textbf{(D)}\ \text{Some slow learners do not attend this school}\\ \textbf{(E)}\ \text{No slow learners do not attend this school} </math><br />
<br />
[[1955 AHSME Problems/Problem 11|Solution]]<br />
== Problem 12==<br />
<br />
The solution of <math>\sqrt{5x-1}+\sqrt{x-1}=2</math> is: <br />
<br />
<math> \textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0 </math><br />
<br />
[[1955 AHSME Problems/Problem 12|Solution]]<br />
== Problem 13==<br />
<br />
The fraction <math>\frac{a^{-4}-b^{-4}}{a^{-2}-b^{-2}}</math> is equal to: <br />
<br />
<math> \textbf{(A)}\ a^{-6}-b^{-6}\qquad\textbf{(B)}\ a^{-2}-b^{-2}\qquad\textbf{(C)}\ a^{-2}+b^{-2}\\ \textbf{(D)}\ a^2+b^2\qquad\textbf{(E)}\ a^2-b^2 </math><br />
<br />
[[1955 AHSME Problems/Problem 13|Solution]]<br />
== Problem 14==<br />
<br />
The length of rectangle <math>R</math> is <math>10</math>% more than the side of square <math>S</math>. The width of the rectangle is <math>10</math>% less than the side of the square. <br />
The ratio of the areas, <math>R:S</math>, is: <br />
<br />
<math> \textbf{(A)}\ 99: 100\qquad\textbf{(B)}\ 101: 100\qquad\textbf{(C)}\ 1: 1\qquad\textbf{(D)}\ 199: 200\qquad\textbf{(E)}\ 201: 200 </math> <br />
<br />
[[1955 AHSME Problems/Problem 14|Solution]]<br />
== Problem 15==<br />
<br />
The ratio of the areas of two concentric circles is <math>1: 3</math>. If the radius of the smaller is <math>r</math>, then the difference between the <br />
radii is best approximated by: <br />
<br />
<math>\textbf{(A)}\ 0.41r \qquad \textbf{(B)}\ 0.73 \qquad \textbf{(C)}\ 0.75 \qquad \textbf{(D)}\ 0.73r \qquad \textbf{(E)}\ 0.75r </math> <br />
<br />
[[1955 AHSME Problems/Problem 15|Solution]]<br />
== Problem 16==<br />
<br />
The value of <math>\frac{3}{a+b}</math> when <math>a=4</math> and <math>b=-4</math> is: <br />
<br />
<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \text{any finite number}\qquad\textbf{(E)}\ \text{meaningless} </math><br />
<br />
[[1955 AHSME Problems/Problem 16|Solution]]<br />
== Problem 17==<br />
<br />
If <math>\log x-5 \log 3=-2</math>, then <math>x</math> equals: <br />
<br />
<math> \textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25} </math><br />
<br />
[[1955 AHSME Problems/Problem 17|Solution]]<br />
== Problem 18==<br />
<br />
The discriminant of the equation <math>x^2+2x\sqrt{3}+3=0</math> is zero. Hence, its roots are: <br />
<br />
<math> \textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}\ \text{rational and equal}\qquad\textbf{(C)}\ \text{rational and unequal}\\ \textbf{(D)}\ \text{irrational and unequal}\qquad\textbf{(E)}\ \text{imaginary} </math><br />
<br />
[[1955 AHSME Problems/Problem 18|Solution]]<br />
== Problem 19==<br />
<br />
Two numbers whose sum is <math>6</math> and the absolute value of whose difference is <math>8</math> are roots of the equation: <br />
<br />
<math> \textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0 </math><br />
<br />
[[1955 AHSME Problems/Problem 19|Solution]]<br />
== Problem 20==<br />
<br />
The expression <math>\sqrt{25-t^2}+5</math> equals zero for: <br />
<br />
<math> \textbf{(A)}\ \text{no real or imaginary values of }t\qquad\textbf{(B)}\ \text{no real values of }t\text{ only}\\ \textbf{(C)}\ \text{no imaginary values of }t\text{ only}\qquad\textbf{(D)}\ t=0\qquad\textbf{(E)}\ t=\pm 5 </math><br />
<br />
[[1955 AHSME Problems/Problem 20|Solution]]<br />
== Problem 21==<br />
<br />
Represent the hypotenuse of a right triangle by <math>c</math> and the area by <math>A</math>. The altitude on the hypotenuse is: <br />
<br />
<math> \textbf{(A)}\ \frac{A}{c}\qquad\textbf{(B)}\ \frac{2A}{c}\qquad\textbf{(C)}\ \frac{A}{2c}\qquad\textbf{(D)}\ \frac{A^2}{c}\qquad\textbf{(E)}\ \frac{A}{c^2} </math><br />
<br />
[[1955 AHSME Problems/Problem 21|Solution]]<br />
== Problem 22==<br />
<br />
On a \textdollar{10000} order a merchant has a choice between three successive discounts of <math>20</math>%, <math>20</math>%, and <math>10</math>% and <br />
three successive discounts of <math>40</math>%, <math>5</math>%, and <math>5</math>%. By choosing the better offer, he can save: <br />
<br />
<math> \textbf{(A)}\ \text{nothing at all}\qquad\textbf{(B)}\ \$440\qquad\textbf{(C)}\ \$330\qquad\textbf{(D)}\ \$345\qquad\textbf{(E)}\ \$360 </math><br />
<br />
[[1955 AHSME Problems/Problem 22|Solution]]<br />
== Problem 23==<br />
<br />
In checking the petty cash a clerk counts <math>q</math> quarters, <math>d</math> dimes, <math>n</math> nickels, and <math>c</math> cents. <br />
Later he discovers that <math>x</math> of the nickels were counted as quarters and <math>x</math> of the dimes were counted as cents. <br />
To correct the total obtained the clerk must: <br />
<br />
<math> \textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text{subtract 11 cents}\qquad\textbf{(C)}\ \text{subtract 11}x\text{ cents}\\ \textbf{(D)}\ \text{add 11}x\text{ cents}\qquad\textbf{(E)}\ \text{add }x\text{ cents} </math><br />
<br />
[[1955 AHSME Problems/Problem 23|Solution]]<br />
== Problem 24==<br />
<br />
The function <math>4x^2-12x-1</math>: <br />
<br />
<math> \textbf{(A)}\ \text{always increases as }x\text{ increases}\\ \textbf{(B)}\ \text{always decreases as }x\text{ decreases to 1}\\ \textbf{(C)}\ \text{cannot equal 0}\\ \textbf{(D)}\ \text{has a maximum value when }x\text{ is negative}\\ \textbf{(E)}\ \text{has a minimum value of-10} </math><br />
<br />
[[1955 AHSME Problems/Problem 24|Solution]]<br />
== Problem 25==<br />
<br />
One of the factors of <math>x^4+2x^2+9</math> is: <br />
<br />
<math> \textbf{(A)}\ x^2+3\qquad\textbf{(B)}\ x+1\qquad\textbf{(C)}\ x^2-3\qquad\textbf{(D)}\ x^2-2x-3\qquad\textbf{(E)}\ \text{none of these} </math><br />
<br />
[[1955 AHSME Problems/Problem 25|Solution]]<br />
== Problem 26==<br />
<br />
Mr. A owns a house worth <math>\textdollar{10000}</math>. He sells it to Mr. <math>B</math> at <math>10</math>% profit. Mr. <math>B</math> sells the house back to Mr. <math>A</math> at a <math>10</math>% loss. Then: <br />
<br />
<math> \textbf{(A)}\ \text{Mr. A comes out even}\qquad\textbf{(B)}\ \text{Mr. A makes } \textdollar{ 100}\qquad\textbf{(C)}\ \text{Mr. A makes } \textdollar{ 1000}\\ \textbf{(D)}\ \text{Mr. B loses } \textdollar{ 100}\qquad\textbf{(E)}\ \text{none of the above is correct} </math><br />
<br />
[[1955 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27==<br />
<br />
If <math>r</math> and <math>s</math> are the roots of <math>x^2-px+q=0</math>, then <math>r^2+s^2</math> equals: <br />
<br />
<math> \textbf{(A)}\ p^2+2q\qquad\textbf{(B)}\ p^2-2q\qquad\textbf{(C)}\ p^2+q^2\qquad\textbf{(D)}\ p^2-q^2\qquad\textbf{(E)}\ p^2 </math><br />
<br />
[[1955 AHSME Problems/Problem 27|Solution]]<br />
== Problem 28==<br />
<br />
On the same set of axes are drawn the graph of <math>y=ax^2+bx+c</math> and the graph of the equation obtained by replacing <math>x</math> by <math>-x</math> in the given equation. <br />
If <math>b \neq 0</math> and <math>c \neq 0</math> these two graphs intersect: <br />
<br />
<math> \textbf{(A)}\ \text{in two points, one on the x-axis and one on the y-axis}\\ \textbf{(B)}\ \text{in one point located on neither axis}\\ \textbf{(C)}\ \text{only at the origin}\\ \textbf{(D)}\ \text{in one point on the x-axis}\\ \textbf{(E)}\ \text{in one point on the y-axis} </math><br />
<br />
[[1955 AHSME Problems/Problem 28|Solution]]<br />
== Problem 29==<br />
<br />
In the figure, <math>PA</math> is tangent to semicircle <math>SAR</math>; <math>PB</math> is tangent to semicircle <math>RBT</math>; <math>SRT</math> is a straight line; <br />
the arcs are indicated in the figure. <math>\angle APB</math> is measured by: <br />
<br />
<asy><br />
unitsize(1.2cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=3;<br />
pair O1=(0,0), O2=(3,0), Sp=(-2,0), R=(2,0), T=(4,0);<br />
pair A=O1+2*dir(60), B=O2+dir(85);<br />
pair Pa=rotate(90,A)*O1, Pb=rotate(-90,B)*O2;<br />
pair P=extension(A,Pa,B,Pb);<br />
pair[] dots={Sp,R,T,A,B,P};<br />
draw(P--P+5*(A-P));<br />
draw(P--P+5*(B-P));<br />
clip((-2,0)--(-2,2.5)--(4,2.5)--(4,0)--cycle);<br />
draw(Arc(O1,2,0,180)--cycle);<br />
draw(Arc(O2,1,0,180)--cycle);<br />
dot(dots);<br />
label("$S$",Sp,S);<br />
label("$R$",R,S);<br />
label("$T$",T,S);<br />
label("$A$",A,NE);<br />
label("$B$",B,N);<br />
label("$P$",P,NNE);<br />
label("$a$",midpoint(Arc(O1,2,0,60)),SW);<br />
label("$b$",midpoint(Arc(O2,1,85,180)),SE);<br />
label("$c$",midpoint(Arc(O1,2,60,180)),SE);<br />
label("$d$",midpoint(Arc(O2,1,0,85)),SW);</asy><br />
<br />
<math> \textbf{(A)}\ \frac{1}{2}(a-b)\qquad\textbf{(B)}\ \frac{1}{2}(a+b)\qquad\textbf{(C)}\ (c-a)-(d-b)\qquad\textbf{(D)}\ a-b\qquad\textbf{(E)}\ a+b </math><br />
<br />
[[1955 AHSME Problems/Problem 29|Solution]]<br />
== Problem 30==<br />
<br />
Each of the equations <math>3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}</math> has: <br />
<br />
<math> \textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root} </math><br />
<br />
[[1955 AHSME Problems/Problem 30|Solution]]<br />
== Problem 31==<br />
<br />
An equilateral triangle whose side is <math>2</math> is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. <br />
If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is: <br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ 2+\sqrt{2}\qquad\textbf{(D)}\ \frac{2+\sqrt{2}}{2}\qquad\textbf{(E)}\ \frac{2\sqrt{3}-\sqrt{6}}{2} </math><br />
<br />
[[1955 AHSME Problems/Problem 31|Solution]]<br />
== Problem 32==<br />
<br />
If the discriminant of <math>ax^2+2bx+c=0</math> is zero, then another true statement about <math>a, b</math>, and <math>c</math> is that: <br />
<br />
<math> \textbf{(A)}\ \text{they form an arithmetic progression}\\ \textbf{(B)}\ \text{they form a geometric progression}\\ \textbf{(C)}\ \text{they are unequal}\\ \textbf{(D)}\ \text{they are all negative numbers}\\ \textbf{(E)}\ \text{only b is negative and a and c are positive} </math><br />
<br />
[[1955 AHSME Problems/Problem 32|Solution]]<br />
== Problem 33==<br />
<br />
Henry starts a trip when the hands of the clock are together between <math>8</math> a.m. and <math>9</math> a.m. <br />
He arrives at his destination between <math>2</math> p.m. and <math>3</math> p.m. when the hands of the clock are exactly <math>180^\circ</math> apart. The trip takes: <br />
<br />
<math> \textbf{(A)}\ \text{6 hr.}\qquad\textbf{(B)}\ \text{6 hr. 43-7/11 min.}\qquad\textbf{(C)}\ \text{5 hr. 16-4/11 min.}\qquad\textbf{(D)}\ \text{6 hr. 30 min.}\qquad\textbf{(E)}\ \text{none of these} </math><br />
<br />
[[1955 AHSME Problems/Problem 33|Solution]]<br />
== Problem 34==<br />
<br />
A <math>6</math>-inch and <math>18</math>-inch diameter pole are placed together and bound together with wire. <br />
The length of the shortest wire that will go around them is: <br />
<br />
<math> \textbf{(A)}\ 12\sqrt{3}+16\pi\qquad\textbf{(B)}\ 12\sqrt{3}+7\pi\qquad\textbf{(C)}\ 12\sqrt{3}+14\pi\\ \textbf{(D)}\ 12+15\pi\qquad\textbf{(E)}\ 24\pi </math><br />
<br />
[[1955 AHSME Problems/Problem 34|Solution]]<br />
== Problem 35==<br />
<br />
Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles. <br />
The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy. <br />
The original number of marbles: <br />
<br />
<math> \textbf{(A)}\ \text{is none of the following}\qquad\textbf{(B)}\ \text{cannot be determined from the given data}\\ \textbf{(C)}\ \text{is 20 or 26}\qquad\textbf{(D)}\ \text{is 14 or 32}\qquad\textbf{(E)}\ \text{is 8 or 38} </math><br />
<br />
[[1955 AHSME Problems/Problem 35|Solution]]<br />
== Problem 36==<br />
<br />
A cylindrical oil tank, lying horizontally, has an interior length of <math>10</math> feet and an interior diameter of <math>6</math> feet. <br />
If the rectangular surface of the oil has an area of <math>40</math> square feet, the depth of the oil is: <br />
<br />
<math> \textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 3-\sqrt{5}\qquad\textbf{(D)}\ 3+\sqrt{5}\\ \textbf{(E)}\ \text{either }3-\sqrt{5}\text{ or }3+\sqrt{5} </math><br />
<br />
[[1955 AHSME Problems/Problem 36|Solution]]<br />
== Problem 37==<br />
<br />
A three-digit number has, from left to right, the digits <math>h, t</math>, and <math>u</math>, with <math>h>u</math>. <br />
When the number with the digits reversed is subtracted from the original number, the units' digit in the difference of r. <br />
The next two digits, from right to left, are: <br />
<br />
<math> \textbf{(A)}\ \text{5 and 9}\qquad\textbf{(B)}\ \text{9 and 5}\qquad\textbf{(C)}\ \text{impossible to tell}\qquad\textbf{(D)}\ \text{5 and 4}\qquad\textbf{(E)}\ \text{4 and 5} </math><br />
<br />
[[1955 AHSME Problems/Problem 37|Solution]]<br />
== Problem 38==<br />
<br />
Four positive integers are given. Select any three of these integers, find their arithmetic average, <br />
and add this result to the fourth integer. Thus the numbers <math>29, 23, 21</math>, and <math>17</math> are obtained. One of the original integers is: <br />
<br />
<math>\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 29 \qquad \textbf{(E)}\ 17 </math> <br />
<br />
[[1955 AHSME Problems/Problem 38|Solution]]<br />
== Problem 39==<br />
<br />
If <math>y=x^2+px+q</math>, then if the least possible value of <math>y</math> is zero <math>q</math> is equal to: <br />
<br />
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{p^2}{4}\qquad\textbf{(C)}\ \frac{p}{2}\qquad\textbf{(D)}\ -\frac{p}{2}\qquad\textbf{(E)}\ \frac{p^2}{4}-q </math><br />
<br />
[[1955 AHSME Problems/Problem 39|Solution]]<br />
== Problem 40==<br />
<br />
The fractions <math>\frac{ax+b}{cx+d}</math> and <math>\frac{b}{d}</math> are unequal if: <br />
<br />
<math> \textbf{(A)}\ a=c=1, x\neq 0\qquad\textbf{(B)}\ a=b=0\qquad\textbf{(C)}\ a=c=0\\ \textbf{(D)}\ x=0\qquad\textbf{(E)}\ ad=bc </math><br />
<br />
[[1955 AHSME Problems/Problem 40|Solution]]<br />
== Problem 41==<br />
<br />
A train traveling from Aytown to Beetown meets with an accident after <math>1</math> hr. It is stopped for <math>\frac{1}{2}</math> hr., <br />
after which it proceeds at four-fifths of its usual rate, arriving at Beetown <math>2</math> hr. late. <br />
If the train had covered <math>80</math> miles more before the accident, it would have been just <math>1</math> hr. late. <br />
The usual rate of the train is: <br />
<br />
<math> \textbf{(A)}\ \text{20 mph}\qquad\textbf{(B)}\ \text{30 mph}\qquad\textbf{(C)}\ \text{40 mph}\qquad\textbf{(D)}\ \text{50 mph}\qquad\textbf{(E)}\ \text{60 mph} </math><br />
<br />
[[1955 AHSME Problems/Problem 41|Solution]]<br />
== Problem 42==<br />
<br />
If <math>a, b</math>, and <math>c</math> are positive integers, the radicals <math>\sqrt{a+\frac{b}{c}}</math> and <math>a\sqrt{\frac{b}{c}}</math> are equal when and only when: <br />
<br />
<math> \textbf{(A)}\ a=b=c=1\qquad\textbf{(B)}\ a=b\text{ and }c=a=1\qquad\textbf{(C)}\ c=\frac{b(a^2-1)}{a}\\ \textbf{(D)}\ a=b\text{ and }c\text{ is any value}\qquad\textbf{(E)}\ a=b\text{ and }c=a-1 </math><br />
<br />
[[1955 AHSME Problems/Problem 42|Solution]]<br />
<br />
== Problem 43==<br />
<br />
The pairs of values of <math>x</math> and <math>y</math> that are the common solutions of the equations <math>y=(x+1)^2</math> and <math>xy+y=1</math> are: <br />
<br />
<math> \textbf{(A)}\ \text{3 real pairs}\qquad\textbf{(B)}\ \text{4 real pairs}\qquad\textbf{(C)}\ \text{4 imaginary pairs}\\ \textbf{(D)}\ \text{2 real and 2 imaginary pairs}\qquad\textbf{(E)}\ \text{1 real and 2 imaginary pairs} </math><br />
<br />
[[1955 AHSME Problems/Problem 43|Solution]]<br />
== Problem 44==<br />
<br />
In circle <math>O</math> chord <math>AB</math> is produced so that <math>BC</math> equals a radius of the circle. <math>CO</math> is drawn and extended to <math>D</math>. <br />
<math>AO</math> is drawn. Which of the following expresses the relationship between <math>x</math> and <math>y</math>?<br />
<br />
<asy><br />
size(200);defaultpen(linewidth(0.7)+fontsize(10));<br />
pair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0);<br />
draw(O--A--C--D);<br />
dot(A^^B^^C^^D^^O);<br />
pair point=O;<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("$O$", O, dir(285));<br />
label("$x$", O+0.1*dir(172.5), dir(172.5));<br />
label("$y$", C+0.4*dir(187.5), dir(187.5));<br />
draw(Circle(O,1));</asy><br />
<br />
<br />
<math> \textbf{(A)}\ x=3y\\ \textbf{(B)}\ x=2y\\ \textbf{(C)}\ x=60^\circ\\ \textbf{(D)}\ \text{there is no special relationship between }x\text{ and }y\\ \textbf{(E)}\ x=2y\text{ or }x=3y\text{, depending upon the length of }AB </math><br />
<br />
[[1955 AHSME Problems/Problem 44|Solution]]<br />
== 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 />
[[1955 AHSME Problems/Problem 45|Solution]]<br />
== Problem 46==<br />
<br />
The graphs of <math>2x+3y-6=0, 4x-3y-6=0, x=2</math>, and <math>y=\frac{2}{3}</math> intersect in: <br />
<br />
<math> \textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points} </math><br />
<br />
[[1955 AHSME Problems/Problem 46|Solution]]<br />
== Problem 47==<br />
<br />
The expressions <math>a+bc</math> and <math>(a+b)(a+c)</math> are: <br />
<br />
<math> \textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0 </math><br />
<br />
[[1955 AHSME Problems/Problem 47|Solution]]<br />
== Problem 48==<br />
<br />
Given <math>\triangle ABC</math> with medians <math>AE, BF, CD</math>; <math>FH</math> parallel and equal to <math>AE</math>; <math>BH and HE</math> are drawn; <math>FE</math> extended meets <math>BH</math> in <math>G</math>. <br />
Which one of the following statements is not necessarily correct? <br />
<br />
<math> \textbf{(A)}\ AEHF\text{ is a parallelogram}\qquad\textbf{(B)}\ HE=HG\\ \textbf{(C)}\ BH=DC\qquad\textbf{(D)}\ FG=\frac{3}{4}AB\qquad\textbf{(E)}\ FG\text{ is a median of triangle }BFH </math><br />
<br />
[[1955 AHSME Problems/Problem 48|Solution]]<br />
== Problem 49==<br />
<br />
The graphs of <math>y=\frac{x^2-4}{x-2}</math> and <math>y=2x</math> intersect in: <br />
<br />
<math> \textbf{(A)}\ \text{1 point whose abscissa is 2}\qquad\textbf{(B)}\ \text{1 point whose abscissa is 0}\\ \textbf{(C)}\ \text{no points}\qquad\textbf{(D)}\ \text{two distinct points}\qquad\textbf{(E)}\ \text{two identical points} </math><br />
<br />
[[1955 AHSME Problems/Problem 49|Solution]]<br />
== Problem 50==<br />
<br />
In order to pass <math>B</math> going <math>40</math> mph on a two-lane highway, <math>A</math>, going <math>50</math> mph, must gain <math>30</math> feet. <br />
Meantime, <math>C, 210</math> feet from <math>A</math>, is headed toward him at <math>50</math> mph. If <math>B</math> and <math>C</math> maintain their speeds, <br />
then, in order to pass safely, <math>A</math> must increase his speed by: <br />
<br />
<math> \textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph} </math><br />
<br />
[[1955 AHSME Problems/Problem 50|Solution]]<br />
<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME 50p box|year=1955|before=[[1954 AHSME|1954 AHSC]]|after=[[1956 AHSME|1956 AHSC]]}} <br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_37&diff=1219391954 AHSME Problems/Problem 372020-05-03T14:11:54Z<p>Fortytwok: </p>
<hr />
<div>== Problem 37==<br />
<br />
Given <math>\triangle PQR</math> with <math>\overline{RS}</math> bisecting <math>\angle R</math>, <math>PQ</math> extended to <math>D</math> and <math>\angle n</math> a right angle, then: <br />
<br />
<asy><br />
path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false)<br />
{<br />
pair M,N;<br />
path mark;<br />
M=t*0.03*unit(A-B)+B;<br />
N=t*0.03*unit(C-B)+B;<br />
if(flip)<br />
mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B));<br />
else<br />
mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B));<br />
return mark;<br />
}<br />
unitsize(1.5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
pair P=(0,0), R=(3,2), Q=(4,0);<br />
pair S0=bisectorpoint(P,R,Q);<br />
pair Sp=extension(P,Q,S0,R);<br />
pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp);<br />
pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R);<br />
draw(P--R--Q);<br />
draw(R--Sp);<br />
draw(P--D--M);<br />
draw(anglemark2(Sp,P,R,17));<br />
label("$p$",P+(0.35,0.1));<br />
draw(anglemark2(R,Q,P,11));<br />
label("$q$",Q+(-0.17,0.1));<br />
draw(anglemark2(R,Np,D,8,true));<br />
label("$n$",Np+(+0.12,0.07));<br />
draw(anglemark2(R,M,D,13,true));<br />
label("$m$",M+(+0.25,0.03));<br />
draw(anglemark2(M,D,P,29));<br />
label("$d$",D+(-0.75,0.095));<br />
pen f=fontsize(10pt);<br />
label("$R$",R,N,f);<br />
label("$P$",P,S,f);<br />
label("$S$",Sp,S,f);<br />
label("$Q$",Q,S,f);<br />
label("$D$",D,S,f);</asy><br />
<br />
<math>\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad <br />
\textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) \qquad<br />
\textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad<br />
\textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad<br />
\textbf{(E)}\ \text{none of these is correct} </math><br />
<br />
== Solution ==<br />
Let <math>\angle PRS</math> be <math>\theta</math>.<br />
<br />
<math>p+ q + 2\theta = 180</math><br />
<br />
<math>m+\theta+90=180 \implies m+\theta=90 \implies 2m+2\theta=180</math><br />
<br />
<math>p+q+2\theta=2m+2\theta \implies \frac{p+q}{2}=m \implies \boxed{\textbf{(B) \ } \angle m = \frac{1}{2}(\angle p + \angle q)}</math><br />
<br />
== Partial Solution ==<br />
<asy><br />
import math;<br />
path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false)<br />
{<br />
pair M,N;<br />
path mark;<br />
M=t*0.03*unit(A-B)+B;<br />
N=t*0.03*unit(C-B)+B;<br />
if(flip)<br />
mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B));<br />
else<br />
mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B));<br />
return mark;<br />
}<br />
unitsize(1.5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
pair P=(0,0), R=(3,2), Q=(4,0);<br />
pair S0=bisectorpoint(P,R,Q);<br />
pair Sp=extension(P,Q,S0,R);<br />
pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp);<br />
pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R);<br />
draw(P--R--Q);<br />
draw(R--Sp);<br />
draw(P--D--M);<br />
pen f=fontsize(10pt);<br />
pair pI=extension(D,M,R,Q);<br />
label("$O$",pI+(-0.2,0.166),f);<br />
draw(anglemark2(Sp,P,R,17));<br />
label("$p$",P+(0.35,0.1));<br />
draw(anglemark2(R,Q,P,11));<br />
label("$q$",Q+(-0.17,0.1));<br />
draw(anglemark2(R,Np,D,8,true));<br />
label("$n$",Np+(+0.12,0.07));<br />
draw(anglemark2(R,M,D,13,true));<br />
label("$m$",M+(+0.25,0.03));<br />
draw(anglemark2(M,D,P,29));<br />
label("$d$",D+(-0.75,0.095));<br />
label("$R$",R,N,f);<br />
label("$M$",M+(-.07,.07),f);<br />
label("$N$",Np+(-.08,.15),f);<br />
label("$P$",P,S,f);<br />
label("$S$",Sp,S,f);<br />
label("$Q$",Q,S,f);<br />
label("$D$",D,S,f);</asy><br />
Looking at triangle PRQ, we have <math>\angle RPD+\angle RQS+\angle MRN=180</math> and from the given statement <math>\angle NMR=\frac{1}{2}\angle MRN</math>, so looking at triangle MOR <math>\angle NMR=90-\frac{\angle RPD+\angle RQS}{2}</math>, which rules out<br />
<br />
<br />
{{AHSME 50p box|year=1954|num-b=36|num-a=38}}<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_46&diff=1217541954 AHSME Problems/Problem 462020-04-27T21:01:26Z<p>Fortytwok: </p>
<hr />
<div>== Problem 46==<br />
<br />
In the diagram, if points <math>A, B</math> and <math>C</math> are points of tangency, then <math>x</math> equals: <br />
<br />
<asy><br />
unitsize(5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=3;<br />
pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16);<br />
pair O=(0,3/8);<br />
draw((-2/3,9/16)--(2/3,9/16));<br />
draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2));<br />
draw(Circle(O,3/16));<br />
draw((-2/3,0)--(2/3,0));<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,N);<br />
label("$\frac{3}{8}$",O);<br />
draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3));<br />
draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3));<br />
label("$\frac{1}{2}$",(.5,.25));<br />
draw((.5,.33)--(.5,.5),EndArrow(3));<br />
draw((.5,.17)--(.5,0),EndArrow(3));<br />
label("$x$",midpoint((.5,.5)--(.5,9/16)));<br />
draw((.5,5/8)--(.5,9/16),EndArrow(3));<br />
label("$60^{\circ}$",(0.01,0.12));<br />
dot(A);<br />
dot(B);<br />
dot(C);</asy><br />
<br />
<math> \textbf{(A)}\ \frac{3}{16}"\qquad\textbf{(B)}\ \frac{1}{8}"\qquad\textbf{(C)}\ \frac{1}{32}"\qquad\textbf{(D)}\ \frac{3}{32}"\qquad\textbf{(E)}\ \frac{1}{16}" </math><br />
<br />
==Solution 1==</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_46&diff=1217531954 AHSME Problems/Problem 462020-04-27T21:00:39Z<p>Fortytwok: </p>
<hr />
<div><br />
==Solution 1==</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_42&diff=1217301954 AHSME Problems/Problem 422020-04-27T03:28:50Z<p>Fortytwok: /* Solution 1 */</p>
<hr />
<div>== Problem 42==<br />
<br />
Consider the graphs of <br />
<cmath>(1)\qquad y=x^2-\frac{1}{2}x+2</cmath> <br />
and <br />
<cmath>(2)\qquad y=x^2+\frac{1}{2}x+2</cmath> <br />
on the same set of axis. <br />
These parabolas are exactly the same shape. Then: <br />
<br />
<math> \textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).} </math><br />
<br />
<br />
==Solution 1==<br />
Let us consider the vertices of the two parabolas. The x-coordinate of parabola 1 is given by <math>\frac{\frac{1}{2}}{2} = \frac{1}{4}</math>.<br />
<br />
Similarly, the x-coordinate of parabola 2 is given by <math>\frac{\frac{-1}{2}}{2} = -\frac{1}{4}</math>.<br />
<br />
From this information, we can deduce that <math>\textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}</math>, since the parabolas are the same shape.</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_42&diff=1217291954 AHSME Problems/Problem 422020-04-27T03:28:13Z<p>Fortytwok: </p>
<hr />
<div>== Problem 42==<br />
<br />
Consider the graphs of <br />
<cmath>(1)\qquad y=x^2-\frac{1}{2}x+2</cmath> <br />
and <br />
<cmath>(2)\qquad y=x^2+\frac{1}{2}x+2</cmath> <br />
on the same set of axis. <br />
These parabolas are exactly the same shape. Then: <br />
<br />
<math> \textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).} </math><br />
<br />
<br />
==Solution 1==<br />
Let us consider the vertices of the two parabolas. The x-coordinate of parabola 1 is given by <math>\frac{\frac{1}{2}}{2} = \frac{1}{4}</math>.<br />
<br />
Similarly, the x-coordinate of parabola 2 is given by <math>\frac{\frac{-1}{2}}{2} = -\frac{1}{4}</math>.<br />
<br />
From this information, we can deduce that <math>\textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}</math></div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_42&diff=1217281954 AHSME Problems/Problem 422020-04-27T03:26:50Z<p>Fortytwok: /* Solution 1 */</p>
<hr />
<div>==Solution 1==<br />
Let us consider the vertices of the two parabolas. The x-coordinate of parabola 1 is given by <math>\frac{\frac{1}{2}}{2} = \frac{1}{4}</math>.<br />
<br />
Similarly, the x-coordinate of parabola 2 is given by <math>\frac{\frac{-1}{2}}{2} = -\frac{1}{4}</math>.<br />
<br />
From this information, we can deduce that <math>\textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}</math></div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_42&diff=1217271954 AHSME Problems/Problem 422020-04-27T03:24:01Z<p>Fortytwok: </p>
<hr />
<div>==Solution 1==</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_42&diff=1217261954 AHSME Problems/Problem 422020-04-27T03:23:38Z<p>Fortytwok: Created page with "a"</p>
<hr />
<div>a</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_40&diff=1217251954 AHSME Problems/Problem 402020-04-27T03:21:22Z<p>Fortytwok: /* Solution 3 */</p>
<hr />
<div>== Problem 40==<br />
<br />
If <math>\left (a+\frac{1}{a} \right )^2=3</math>, then <math>a^3+\frac{1}{a^3}</math> equals: <br />
<br />
<math> \textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3} </math><br />
<br />
== Solution 1 ==<br />
<math>a+\frac{1}{a}=\sqrt{3}\implies (a+\frac{1}{a})^3=3\sqrt{3}\implies a^3+3\frac{a^2}{a}+\frac{3a}{a^2}+\frac{1}{a^3}\implies a^3+3(a+\frac{1}{a})+\frac{1}{a^3}=3\sqrt{3}\implies a^3+\frac{1}{a^3}+3\sqrt{3}=3\sqrt{3}\implies a^3+\frac{1}{a^3}=0</math>, <math>\fbox{C}</math><br />
<br />
== Solution 2 ==<br />
<math>\left (a+\frac{1}{a} \right )^2=3 \implies a+\frac{1}{a} =\sqrt{3}</math><br />
<br />
<math>a+\frac{1}{a} =\sqrt{3} \implies (a+\frac{1}{a})^3=\sqrt{3^3}</math><br />
<br />
<math>(a+\frac{1}{a})^3=\sqrt{3^3}\implies a^3+3a^2\frac{1}{a}+3a\frac{1}{a^2}+\frac{1}{a^3}=\sqrt{27}</math><br />
<br />
<math>a^3+3a^2\frac{1}{a}+3a\frac{1}{a^2}+\frac{1}{a^3}=\sqrt{27} \implies a^3+\frac{1}{a^3}+3a+\frac{3}{a}=\sqrt{27}</math><br />
<br />
<math>a^3+\frac{1}{a^3}+3a+\frac{3}{a}=\sqrt{27} \implies a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=\sqrt{27}</math><br />
<br />
<math>a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=\sqrt{27} \implies a^3+\frac{1}{a^3}+3(\sqrt{3})=\sqrt{27}</math><br />
<br />
<math>3(\sqrt{3})=\sqrt{27}</math><br />
<br />
<math>a^3+\frac{1}{a^3}=0</math><br />
<br />
==Solution 3==<br />
<math>(a+\frac{1}{a})=\sqrt{3}</math><br />
<br />
<math>a^{2}+2+\frac{1}{a^{2}}=3 \Rightarrow a^{2}+\frac{1}{a^{2}}=1</math><br />
<br />
<math>(a^{3}+\frac{1}{a^{3}})=(a+\frac{1}{a})(a^{2}-1+\frac{1}{a^{2}}) \Rightarrow (a^{3}+\frac{1}{a^{3}})=\sqrt{3}(1-1) \Rightarrow \fbox{C}</math></div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_40&diff=1217241954 AHSME Problems/Problem 402020-04-27T03:15:19Z<p>Fortytwok: </p>
<hr />
<div>== Problem 40==<br />
<br />
If <math>\left (a+\frac{1}{a} \right )^2=3</math>, then <math>a^3+\frac{1}{a^3}</math> equals: <br />
<br />
<math> \textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3} </math><br />
<br />
== Solution 1 ==<br />
<math>a+\frac{1}{a}=\sqrt{3}\implies (a+\frac{1}{a})^3=3\sqrt{3}\implies a^3+3\frac{a^2}{a}+\frac{3a}{a^2}+\frac{1}{a^3}\implies a^3+3(a+\frac{1}{a})+\frac{1}{a^3}=3\sqrt{3}\implies a^3+\frac{1}{a^3}+3\sqrt{3}=3\sqrt{3}\implies a^3+\frac{1}{a^3}=0</math>, <math>\fbox{C}</math><br />
<br />
== Solution 2 ==<br />
<math>\left (a+\frac{1}{a} \right )^2=3 \implies a+\frac{1}{a} =\sqrt{3}</math><br />
<br />
<math>a+\frac{1}{a} =\sqrt{3} \implies (a+\frac{1}{a})^3=\sqrt{3^3}</math><br />
<br />
<math>(a+\frac{1}{a})^3=\sqrt{3^3}\implies a^3+3a^2\frac{1}{a}+3a\frac{1}{a^2}+\frac{1}{a^3}=\sqrt{27}</math><br />
<br />
<math>a^3+3a^2\frac{1}{a}+3a\frac{1}{a^2}+\frac{1}{a^3}=\sqrt{27} \implies a^3+\frac{1}{a^3}+3a+\frac{3}{a}=\sqrt{27}</math><br />
<br />
<math>a^3+\frac{1}{a^3}+3a+\frac{3}{a}=\sqrt{27} \implies a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=\sqrt{27}</math><br />
<br />
<math>a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=\sqrt{27} \implies a^3+\frac{1}{a^3}+3(\sqrt{3})=\sqrt{27}</math><br />
<br />
<math>3(\sqrt{3})=\sqrt{27}</math><br />
<br />
<math>a^3+\frac{1}{a^3}=0</math><br />
<br />
==Solution 3==</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems&diff=1216271954 AHSME Problems2020-04-24T20:29:07Z<p>Fortytwok: /* Problem 25 */</p>
<hr />
<div>{{AHSC 50 Problems<br />
|year=1954<br />
}}<br />
== Problem 1==<br />
<br />
The square of <math>5-\sqrt{y^2-25}</math> is: <br />
<br />
<math>\textbf{(A)}\ y^2-5\sqrt{y^2-25} \qquad \textbf{(B)}\ -y^2 \qquad \textbf{(C)}\ y^2 \\ \textbf{(D)}\ (5-y)^2\qquad\textbf{(E)}\ y^2-10\sqrt{y^2-25} </math><br />
<br />
[[1954 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2==<br />
<br />
The equation <math>\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0</math> can be transformed by eliminating fractions to the equation <math>x^2-5x+4=0</math>. <br />
The roots of the latter equation are <math>4</math> and <math>1</math>. Then the roots of the first equation are: <br />
<br />
<math>\textbf{(A)}\ 4 \text{ and }1 \qquad \textbf{(B)}\ \text{only }1 \qquad \textbf{(C)}\ \text{only }4 \qquad \textbf{(D)}\ \text{neither 4 nor 1}\qquad\textbf{(E)}\ \text{4 and some other root} </math><br />
<br />
[[1954 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3==<br />
<br />
If <math>x</math> varies as the cube of <math>y</math>, and <math>y</math> varies as the fifth root of <math>z</math>, then <math>x</math> varies as the nth power of <math>z</math>, where n is:<br />
<br />
<math>\textbf{(A)}\ \frac{1}{15} \qquad\textbf{(B)}\ \frac{5}{3} \qquad\textbf{(C)}\ \frac{3}{5} \qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 8 </math> <br />
<br />
[[1954 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4==<br />
<br />
If the Highest Common Divisor of <math>6432</math> and <math>132</math> is diminished by <math>8</math>, it will equal: <br />
<br />
<math>\textbf{(A)}\ -6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4 </math><br />
<br />
[[1954 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5==<br />
<br />
A regular hexagon is inscribed in a circle of radius <math>10</math> inches. Its area is: <br />
<br />
<math>\textbf{(A)}\ 150\sqrt{3} \text{ sq. in.} \qquad \textbf{(B)}\ \text{150 sq. in.} \qquad \textbf{(C)}\ 25\sqrt{3}\text{ sq. in.}\qquad\textbf{(D)}\ \text{600 sq. in.}\qquad\textbf{(E)}\ 300\sqrt{3}\text{ sq. in.} </math> <br />
<br />
[[1954 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6==<br />
<br />
The value of <math>\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}</math> is: <br />
<br />
<math>\textbf{(A)}\ 1 \frac{13}{16} \qquad \textbf{(B)}\ 1 \frac{3}{16} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{1}{16} </math> <br />
<br />
[[1954 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7==<br />
<br />
A housewife saved <math>\textdollar{2.50}</math> in buying a dress on sale. If she spent <math>\textdollar{25}</math> for the dress, she saved about: <br />
<br />
<math>\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\% </math> <br />
<br />
[[1954 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8==<br />
<br />
The base of a triangle is twice as long as a side of a square and their areas are the same. <br />
Then the ratio of the altitude of the triangle to the side of the square is: <br />
<br />
<math>\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4 </math><br />
<br />
[[1954 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9==<br />
<br />
A point <math>P</math> is outside a circle and is <math>13</math> inches from the center. A secant from <math>P</math> cuts the circle at <math>Q</math> and <math>R</math> <br />
so that the external segment of the secant <math>PQ</math> is <math>9</math> inches and <math>QR</math> is <math>7</math> inches. The radius of the circle is: <br />
<br />
<math>\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)}\ 6"\qquad\textbf{(E)}\ 7" </math> <br />
<br />
[[1954 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10==<br />
<br />
The sum of the numerical coefficients in the expansion of the binomial <math>(a+b)^6</math> is: <br />
<br />
<math>\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 7 </math> <br />
<br />
[[1954 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11==<br />
<br />
A merchant placed on display some dresses, each with a marked price. He then posted a sign “<math>\frac{1}{3}</math> off on these dresses.” <br />
The cost of the dresses was <math>\frac{3}{4}</math> of the price at which he actually sold them. Then the ratio of the cost to the marked price was: <br />
<br />
<math>\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
[[1954 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12==<br />
<br />
The solution of the equations <br />
<br />
<cmath>\begin{align*}2x-3y &=7 \\ 4x-6y &=20\end{align*}</cmath> <br />
<br />
is: <br />
<br />
<math>\textbf{(A)}\ x=18, y=12 \qquad <br />
\textbf{(B)}\ x=0, y=0 \qquad <br />
\textbf{(C)}\ \text{There is no solution} \\ <br />
\textbf{(D)}\ \text{There are an unlimited number of solutions}\qquad<br />
\textbf{(E)}\ x=8, y=5 </math> <br />
<br />
[[1954 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13==<br />
<br />
A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off <br />
by the sides of the quadrilateral, their sum will be: <br />
<br />
<math>\textbf{(A)}\ 180^\circ \qquad \textbf{(B)}\ 540^\circ \qquad \textbf{(C)}\ 360^\circ \qquad \textbf{(D)}\ 450^\circ\qquad\textbf{(E)}\ 1080^\circ </math> <br />
<br />
[[1954 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14==<br />
<br />
When simplified <math>\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}</math> equals: <br />
<br />
<math>\textbf{(A)}\ \frac{x^4+2x^2-1}{2x^2} \qquad \textbf{(B)}\ \frac{x^4-1}{2x^2} \qquad \textbf{(C)}\ \frac{\sqrt{x^2+1}}{2}\\ \textbf{(D)}\ \frac{x^2}{\sqrt{2}}\qquad\textbf{(E)}\ \frac{x^2}{2}+\frac{1}{2x^2} </math> <br />
<br />
[[1954 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15==<br />
<br />
<math>\log 125</math> equals: <br />
<br />
<math>\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25 \\<br />
\textbf{(D)}\ 3 - 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5) </math> <br />
<br />
[[1954 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16==<br />
<br />
If <math>f(x) = 5x^2 - 2x - 1</math>, then <math>f(x + h) - f(x)</math> equals: <br />
<br />
<math>\textbf{(A)}\ 5h^2 - 2h \qquad \textbf{(B)}\ 10xh - 4x + 2 \qquad \textbf{(C)}\ 10xh - 2x - 2 \\ \textbf{(D)}\ h(10x+5h-2)\qquad\textbf{(E)}\ 3h </math> <br />
<br />
[[1954 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17==<br />
<br />
The graph of the function <math>f(x) = 2x^3 - 7</math> goes: <br />
<br />
<math>\textbf{(A)}\ \text{up to the right and down to the left} \\ \textbf{(B)}\ \text{down to the right and up to the left}\\ \textbf{(C)}\ \text{up to the right and up to the left}\\ \textbf{(D)}\ \text{down to the right and down to the left}\\ \textbf{(E)}\ \text{none of these ways.} </math><br />
<br />
[[1954 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18==<br />
<br />
Of the following sets, the one that includes all values of <math>x</math> which will satisfy <math>2x - 3 > 7 - x</math> is: <br />
<br />
<math>\textbf{(A)}\ x > 4 \qquad \textbf{(B)}\ x < \frac {10}{3} \qquad \textbf{(C)}\ x = \frac {10}{3} \qquad \textbf{(D)}\ x >\frac{10}{3}\qquad\textbf{(E)}\ x < 0 </math> <br />
<br />
[[1954 AHSME Problems/Problem 18|Solution]]<br />
<br />
== 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 />
[[1954 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20==<br />
<br />
The equation <math>x^3+6x^2+11x+6=0</math> has: <br />
<br />
<math> \textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root} </math><br />
<br />
[[1954 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21==<br />
<br />
The roots of the equation <math>2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5</math> can be found by solving: <br />
<br />
<math> \textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}\ 4x^2-25x+4 = 0\qquad\textbf{(C)}\ 4x^2-17x+4 = 0\\ \textbf{(D)}\ 2x^2-21x+2 = 0\qquad\textbf{(E)}\ 4x^2-25x-4 = 0 </math><br />
<br />
[[1954 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22==<br />
<br />
The expression <math>\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}</math> cannot be evaluated for <math>x=-1</math> or <math>x=2</math>, <br />
since division by zero is not allowed. For other values of <math>x</math>: <br />
<br />
<math> \textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1. </math><br />
<br />
[[1954 AHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23==<br />
<br />
If the margin made on an article costing <math>C</math> dollars and selling for <math>S</math> dollars is <math>M=\frac{1}{n}C</math>, then the margin is given by: <br />
<br />
<math> \textbf{(A)}\ M=\frac{1}{n-1}S\qquad\textbf{(B)}\ M=\frac{1}{n}S\qquad\textbf{(C)}\ M=\frac{n}{n+1}S\\ \textbf{(D)}\ M=\frac{1}{n+1}S\qquad\textbf{(E)}\ M=\frac{n}{n-1}S </math><br />
<br />
[[1954 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24==<br />
<br />
The values of <math>k</math> for which the equation <math>2x^2-kx+x+8=0</math> will have real and equal roots are: <br />
<br />
<math> \textbf{(A)}\ 9\text{ and }-7\qquad\textbf{(B)}\ \text{only }-7\qquad\textbf{(C)}\ \text{9 and 7}\\ \textbf{(D)}\ -9\text{ and }-7\qquad\textbf{(E)}\ \text{only 9} </math><br />
<br />
[[1954 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25==<br />
<br />
The two roots of the equation <math>a(b-c)x^2+b(c-a)x+c(a-b)=0</math> are <math>1</math> and: <br />
<br />
<math> \textbf{(A)}\ \frac{b(c-a)}{a(b-c)}\qquad\textbf{(B)}\ \frac{a(b-c)}{c(a-b)}\qquad\textbf{(C)}\ \frac{a(b-c)}{b(c-a)}\qquad\textbf{(D)}\ \frac{c(a-b)}{a(b-c)}\qquad\textbf{(E)}\ \frac{c(a-b)}{b(c-a)} </math><br />
<br />
[[1954 AHSME Problems/Problem 25|Solution]]<br />
<br />
== Problem 26==<br />
<br />
The straight line <math>\overline{AB}</math> is divided at <math>C</math> so that <math>AC=3CB</math>. Circles are described on <math>\overline{AC}</math> <br />
and <math>\overline{CB}</math> as diameters and a common tangent meets <math>AB</math> produced at <math>D</math>. Then <math>BD</math> equals: <br />
<br />
<math> \textbf{(A)}\ \text{diameter of the smaller circle}\\ \textbf{(B)}\ \text{radius of the smaller circle}\\ \textbf{(C)}\ \text{radius of the larger circle}\\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii} </math><br />
<br />
[[1954 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27==<br />
<br />
A right circular cone has for its base a circle having the same radius as a given sphere. <br />
The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:<br />
<br />
<math> \textbf{(A)}\ \frac{1}{1}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{2}{1}\qquad\textbf{(E)}\ \sqrt{\frac{5}{4}} </math><br />
<br />
[[1954 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28==<br />
<br />
If <math>\frac{m}{n}=\frac{4}{3}</math> and <math>\frac{r}{t}=\frac{9}{14}</math>, the value of <math>\frac{3mr-nt}{4nt-7mr}</math> is: <br />
<br />
<math> \textbf{(A)}\ -5\frac{1}{2}\qquad\textbf{(B)}\ -\frac{11}{14}\qquad\textbf{(C)}\ -1\frac{1}{4}\qquad\textbf{(D)}\ \frac{11}{14}\qquad\textbf{(E)}\ -\frac{2}{3} </math><br />
<br />
[[1954 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29==<br />
<br />
If the ratio of the legs of a right triangle is <math>1: 2</math>, then the ratio of the corresponding segments of <br />
the hypotenuse made by a perpendicular upon it from the vertex is: <br />
<br />
<math> \textbf{(A)}\ 1: 4\qquad\textbf{(B)}\ 1:\sqrt{2}\qquad\textbf{(C)}\ 1: 2\qquad\textbf{(D)}\ 1:\sqrt{5}\qquad\textbf{(E)}\ 1: 5 </math><br />
<br />
[[1954 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30==<br />
<br />
<math>A</math> and <math>B</math> together can do a job in <math>2</math> days; <math>B</math> and <math>C</math> can do it in four days; and <math>A</math> and <math>C</math> in <math>2\frac{2}{5}</math> days. <br />
The number of days required for A to do the job alone is: <br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8 </math> <br />
<br />
[[1954 AHSME Problems/Problem 30|Solution]]<br />
<br />
== Problem 31==<br />
<br />
In <math>\triangle ABC</math>, <math>AB=AC</math>, <math>\angle A=40^\circ</math>. Point <math>O</math> is within the triangle with <math>\angle OBC \cong \angle OCA</math>. <br />
The number of degrees in <math>\angle BOC</math> is: <br />
<br />
<math>\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}</math> <br />
<br />
[[1954 AHSME Problems/Problem 31|Solution]]<br />
<br />
== Problem 32==<br />
<br />
The factors of <math>x^4+64</math> are: <br />
<br />
<math> \textbf{(A)}\ (x^2+8)^2\qquad\textbf{(B)}\ (x^2+8)(x^2-8)\qquad\textbf{(C)}\ (x^2+2x+4)(x^2-8x+16)\\ \textbf{(D)}\ (x^2-4x+8)(x^2-4x-8)\qquad\textbf{(E)}\ (x^2-4x+8)(x^2+4x+8) </math><br />
<br />
[[1954 AHSME Problems/Problem 32|Solution]]<br />
<br />
== Problem 33==<br />
<br />
A bank charges <math>\textdollar{6}</math> for a loan of <math>\textdollar{120}</math>. The borrower receives <math>\textdollar{114}</math> and <br />
repays the loan in <math>12</math> easy installments of <math>\textdollar{10}</math> a month. The interest rate is approximately: <br />
<br />
<math>\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \% </math><br />
<br />
[[1954 AHSME Problems/Problem 33|Solution]]<br />
<br />
== Problem 34==<br />
<br />
The fraction <math>\frac{1}{3}</math>: <br />
<br />
<math> \textbf{(A)}\ \text{equals 0.33333333}\qquad\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(C)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^9}\\ \textbf{(D)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(E)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^9} </math><br />
<br />
[[1954 AHSME Problems/Problem 34|Solution]]<br />
<br />
== Problem 35==<br />
<br />
In the right triangle shown the sum of the distances <math>BM</math> and <math>MA</math> is equal to the sum of the distances <math>BC</math> and <math>CA</math>. <br />
If <math>MB = x, CB = h</math>, and <math>CA = d</math>, then <math>x</math> equals: <br />
<br />
<asy><br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
draw((0,0)--(8,0)--(0,5)--cycle);<br />
label("C",(0,0),SW);<br />
label("A",(8,0),SE);<br />
label("M",(0,5),N);<br />
dot((0,3.5));<br />
label("B",(0,3.5),W);<br />
label("$x$",(0,4.25),W);<br />
label("$h$",(0,1),W);<br />
label("$d$",(4,0),S);</asy><br />
<br />
<math> \textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h </math><br />
<br />
[[1954 AHSME Problems/Problem 35|Solution]]<br />
<br />
== Problem 36==<br />
<br />
A boat has a speed of <math>15</math> mph in still water. In a stream that has a current of <math>5</math> mph it travels a certain <br />
distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is: <br />
<br />
<math> \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{1}{1}\qquad\textbf{(C)}\ \frac{8}{9}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{9}{8} </math><br />
<br />
[[1954 AHSME Problems/Problem 36|Solution]]<br />
<br />
== Problem 37==<br />
<br />
Given <math>\triangle PQR</math> with <math>\overline{RS}</math> bisecting <math>\angle R</math>, <math>PQ</math> extended to <math>D</math> and <math>\angle n</math> a right angle, then: <br />
<br />
<asy><br />
path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false)<br />
{<br />
pair M,N;<br />
path mark;<br />
M=t*0.03*unit(A-B)+B;<br />
N=t*0.03*unit(C-B)+B;<br />
if(flip)<br />
mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B));<br />
else<br />
mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B));<br />
return mark;<br />
}<br />
unitsize(1.5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
pair P=(0,0), R=(3,2), Q=(4,0);<br />
pair S0=bisectorpoint(P,R,Q);<br />
pair Sp=extension(P,Q,S0,R);<br />
pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp);<br />
pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R);<br />
draw(P--R--Q);<br />
draw(R--Sp);<br />
draw(P--D--M);<br />
draw(anglemark2(Sp,P,R,17));<br />
label("$p$",P+(0.35,0.1));<br />
draw(anglemark2(R,Q,P,11));<br />
label("$q$",Q+(-0.17,0.1));<br />
draw(anglemark2(R,Np,D,8,true));<br />
label("$n$",Np+(+0.12,0.07));<br />
draw(anglemark2(R,M,D,13,true));<br />
label("$m$",M+(+0.25,0.03));<br />
draw(anglemark2(M,D,P,29));<br />
label("$d$",D+(-0.75,0.095));<br />
pen f=fontsize(10pt);<br />
label("$R$",R,N,f);<br />
label("$P$",P,S,f);<br />
label("$S$",Sp,S,f);<br />
label("$Q$",Q,S,f);<br />
label("$D$",D,S,f);</asy><br />
<br />
<math>\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad <br />
\textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) </math><br />
<math> \textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad<br />
\textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad<br />
\textbf{(E)}\ \text{none of these is correct} </math><br />
<br />
[[1954 AHSME Problems/Problem 37|Solution]]<br />
<br />
== Problem 38==<br />
<br />
If <math>\log 2=.3010</math> and <math>\log 3=.4771</math>, the value of <math>x</math> when <math>3^{x+3}=135</math> is approximately: <br />
<br />
<math>\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 1.47 \qquad \textbf{(C)}\ 1.67 \qquad \textbf{(D)}\ 1.78 \qquad \textbf{(E)}\ 1.63 </math> <br />
<br />
[[1954 AHSME Problems/Problem 38|Solution]]<br />
<br />
== Problem 39==<br />
<br />
The locus of the midpoint of a line segment that is drawn from a given external point <math>P</math> to a given circle with center <math>O</math> and radius <math>r</math>, is: <br />
<br />
<math> \textbf{(A)}\ \text{a straight line perpendicular to }\overline{PO}\\ \textbf{(B)}\ \text{a straight line parallel to }\overline{PO}\\ \textbf{(C)}\ \text{a circle with center }P\text{ and radius }r\\ \textbf{(D)}\ \text{a circle with center at the midpoint of }\overline{PO}\text{ and radius }2r\\ \textbf{(E)}\ \text{a circle with center at the midpoint }\overline{PO}\text{ and radius }\frac{1}{2}r </math><br />
<br />
[[1954 AHSME Problems/Problem 39|Solution]]<br />
<br />
== Problem 40==<br />
<br />
If <math>\left (a+\frac{1}{a} \right )^2=3</math>, then <math>a^3+\frac{1}{a^3}</math> equals: <br />
<br />
<math> \textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3} </math><br />
<br />
[[1954 AHSME Problems/Problem 40|Solution]]<br />
<br />
== Problem 41==<br />
<br />
The sum of all the roots of <math>4x^3-8x^2-63x-9=0</math> is: <br />
<br />
<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0 </math> <br />
<br />
[[1954 AHSME Problems/Problem 41|Solution]]<br />
<br />
== Problem 42==<br />
<br />
Consider the graphs of <br />
<cmath>(1)\qquad y=x^2-\frac{1}{2}x+2</cmath> <br />
and <br />
<cmath>(2)\qquad y=x^2+\frac{1}{2}x+2</cmath> <br />
on the same set of axis. <br />
These parabolas are exactly the same shape. Then: <br />
<br />
<math> \textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).} </math><br />
<br />
[[1954 AHSME Problems/Problem 42|Solution]]<br />
<br />
== Problem 43==<br />
<br />
The hypotenuse of a right triangle is <math>10</math> inches and the radius of the inscribed circle is <math>1</math> inch. The perimeter of the triangle in inches is: <br />
<br />
<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30 </math> <br />
<br />
[[1954 AHSME Problems/Problem 43|Solution]]<br />
<br />
== Problem 44==<br />
<br />
A man born in the first half of the nineteenth century was <math>x</math> years old in the year <math>x^2</math>. He was born in: <br />
<br />
<math>\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806 </math> <br />
<br />
[[1954 AHSME Problems/Problem 44|Solution]]<br />
<br />
== Problem 45==<br />
<br />
In a rhombus, <math>ABCD</math>, line segments are drawn within the rhombus, parallel to diagonal <math>BD</math>, <br />
and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment <br />
as a function of its distance from vertex <math>A</math>. The graph is: <br />
<br />
<math> \textbf{(A)}\ \text{A straight line passing through the origin.}\\ \textbf{(B)}\ \text{A straight line cutting across the upper right quadrant.}\\ \textbf{(C)}\ \text{Two line segments forming an upright V.}\\ \textbf{(D)}\ \text{Two line segments forming an inverted V.}\\ \textbf{(E)}\ \text{None of these.} </math><br />
<br />
[[1954 AHSME Problems/Problem 45|Solution]]<br />
<br />
== Problem 46==<br />
<br />
In the diagram, if points <math>A, B</math> and <math>C</math> are points of tangency, then <math>x</math> equals: <br />
<br />
<asy><br />
unitsize(5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=3;<br />
pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16);<br />
pair O=(0,3/8);<br />
draw((-2/3,9/16)--(2/3,9/16));<br />
draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2));<br />
draw(Circle(O,3/16));<br />
draw((-2/3,0)--(2/3,0));<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,N);<br />
label("$\frac{3}{8}$",O);<br />
draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3));<br />
draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3));<br />
label("$\frac{1}{2}$",(.5,.25));<br />
draw((.5,.33)--(.5,.5),EndArrow(3));<br />
draw((.5,.17)--(.5,0),EndArrow(3));<br />
label("$x$",midpoint((.5,.5)--(.5,9/16)));<br />
draw((.5,5/8)--(.5,9/16),EndArrow(3));<br />
label("$60^{\circ}$",(0.01,0.12));<br />
dot(A);<br />
dot(B);<br />
dot(C);</asy><br />
<br />
<math> \textbf{(A)}\ \frac{3}{16}"\qquad\textbf{(B)}\ \frac{1}{8}"\qquad\textbf{(C)}\ \frac{1}{32}"\qquad\textbf{(D)}\ \frac{3}{32}"\qquad\textbf{(E)}\ \frac{1}{16}" </math><br />
<br />
[[1954 AHSME Problems/Problem 46|Solution]]<br />
<br />
== Problem 47==<br />
<br />
At the midpoint of line segment <math>AB</math> which is <math>p</math> units long, a perpendicular <math>MR</math> is erected with length <math>q</math> units. <br />
An arc is described from <math>R</math> with a radius equal to <math>\frac{1}{2}AB</math>, meeting <math>AB</math> at <math>T</math>. Then <math>AT</math> and <math>TB</math> are the roots of: <br />
<br />
<math> \textbf{(A)}\ x^2+px+q^2=0\\ \textbf{(B)}\ x^2-px+q^2=0\\ \textbf{(C)}\ x^2+px-q^2=0\\ \textbf{(D)}\ x^2-px-q^2=0\\ \textbf{(E)}\ x^2-px+q=0 </math><br />
<br />
[[1954 AHSME Problems/Problem 47|Solution]]<br />
<br />
== Problem 48==<br />
<br />
A train, an hour after starting, meets with an accident which detains it a half hour, after which it <br />
proceeds at <math>\frac{3}{4}</math> of its former rate and arrives <math>3\tfrac{1}{2}</math> hours late. <br />
Had the accident happened <math>90</math> miles farther along the line, it would have arrived only <math>3</math> hours late. The length of the trip in miles was: <br />
<br />
<math>\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550 </math> <br />
<br />
[[1954 AHSME Problems/Problem 48|Solution]]<br />
<br />
== Problem 49==<br />
<br />
The difference of the squares of two odd numbers is always divisible by <math>8</math>. If <math>a>b</math>, and <math>2a+1</math> and <math>2b+1</math> are the odd numbers, <br />
to prove the given statement we put the difference of the squares in the form: <br />
<br />
<math> \textbf{(A)}\ (2a+1)^2-(2b+1)^2\\ \textbf{(B)}\ 4a^2-4b^2+4a-4b\\ \textbf{(C)}\ 4[a(a+1)-b(b+1)]\\ \textbf{(D)}\ 4(a-b)(a+b+1)\\ \textbf{(E)}\ 4(a^2+a-b^2-b) </math><br />
<br />
[[1954 AHSME Problems/Problem 49|Solution]]<br />
<br />
== Problem 50==<br />
<br />
The times between <math>7</math> and <math>8</math> o'clock, correct to the nearest minute, when the hands of a clock will form an angle of <math>84^{\circ}</math> are: <br />
<br />
<math> \textbf{(A)}\ \text{7: 23 and 7: 53}\qquad<br />
\textbf{(B)}\ \text{7: 20 and 7: 50}\qquad<br />
\textbf{(C)}\ \text{7: 22 and 7: 53}\\ <br />
\textbf{(D)}\ \text{7: 23 and 7: 52}\qquad<br />
\textbf{(E)}\ \text{7: 21 and 7: 49} </math><br />
<br />
[[1954 AHSME Problems/Problem 50|Solution]]<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME 50p box|year=1954|before=[[1953 AHSME|1953 AHSC]]|after=[[1955 AHSME|1955 AHSC]]}} <br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_20&diff=1215861954 AHSME Problems/Problem 202020-04-23T20:55:10Z<p>Fortytwok: /* Solution 2 */</p>
<hr />
<div>== Problem 20==<br />
<br />
The equation <math>x^3+6x^2+11x+6=0</math> has: <br />
<br />
<math> \textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root} </math><br />
<br />
== Solution ==<br />
By the rational root theorem, <math>1, -1, 2, -2, 3, -3, 6, -6</math> are possible rational roots. Because <math>x^3+6x^2+11x+6>0</math> for <math>x>0</math>, so there are no positive roots. We try <math>-1, -2, -3, -6</math>, so <math>x=-1, x=-3, x=-2</math>, so there are no positive real roots; <math>\fbox{B}</math><br />
== Solution 2==<br />
Note that there are no sign changes (all coefficients of terms are positive), so by Descartes' rule of signs, there are no positive real roots <math>\Rightarrow \fbox{B}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1954|num-b=19|num-a=21}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_20&diff=1215851954 AHSME Problems/Problem 202020-04-23T20:54:48Z<p>Fortytwok: /* Solution 2 */</p>
<hr />
<div>== Problem 20==<br />
<br />
The equation <math>x^3+6x^2+11x+6=0</math> has: <br />
<br />
<math> \textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root} </math><br />
<br />
== Solution ==<br />
By the rational root theorem, <math>1, -1, 2, -2, 3, -3, 6, -6</math> are possible rational roots. Because <math>x^3+6x^2+11x+6>0</math> for <math>x>0</math>, so there are no positive roots. We try <math>-1, -2, -3, -6</math>, so <math>x=-1, x=-3, x=-2</math>, so there are no positive real roots; <math>\fbox{B}</math><br />
== Solution 2==<br />
Note that there are no sign changes (all coefficients of terms are positive), so by Descartes' rule of signs, there are no positive real roots, or <math>\fbox{B}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1954|num-b=19|num-a=21}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_20&diff=1215841954 AHSME Problems/Problem 202020-04-23T20:54:10Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem 20==<br />
<br />
The equation <math>x^3+6x^2+11x+6=0</math> has: <br />
<br />
<math> \textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root} </math><br />
<br />
== Solution ==<br />
By the rational root theorem, <math>1, -1, 2, -2, 3, -3, 6, -6</math> are possible rational roots. Because <math>x^3+6x^2+11x+6>0</math> for <math>x>0</math>, so there are no positive roots. We try <math>-1, -2, -3, -6</math>, so <math>x=-1, x=-3, x=-2</math>, so there are no positive real roots; <math>\fbox{B}</math><br />
== Solution 2==<br />
By Descartes' Rule of Signs, there are no sign changes (all coefficients of terms are positive), so there are no positive real roots, or <math>\fbox{B}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1954|num-b=19|num-a=21}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_47&diff=1215411953 AHSME Problems/Problem 472020-04-22T22:51:45Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
If <math>x>0</math>, then the correct relationship is: <br />
<br />
<math>\textbf{(A)}\ \log (1+x) = \frac{x}{1+x} \qquad<br />
\textbf{(B)}\ \log (1+x) < \frac{x}{1+x} \\ <br />
\textbf{(C)}\ \log(1+x) > x\qquad<br />
\textbf{(D)}\ \log (1+x) < x\qquad<br />
\textbf{(E)}\ \text{none of these} </math><br />
<br />
==Solution 1(Cheap)==<br />
Plug in <math>x=9</math>. Then, you can see that the answer is <math>\fbox{D}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1953|num-b=46|num-a=48}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_39&diff=1215341953 AHSME Problems/Problem 392020-04-22T21:57:28Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The product, <math>\log_a b \cdot \log_b a</math> is equal to: <br />
<br />
<math>\textbf{(A)}\ 1 \qquad<br />
\textbf{(B)}\ a \qquad<br />
\textbf{(C)}\ b \qquad<br />
\textbf{(D)}\ ab \qquad<br />
\textbf{(E)}\ \text{none of these} </math><br />
<br />
==Solution==<br />
<cmath>a^x=b</cmath><br />
<cmath>b^y=a</cmath><br />
<cmath>{a^x}^y=a</cmath><br />
<cmath>xy=1</cmath><br />
<cmath>\log_a b\log_b a=1</cmath><br />
As a result, the answer should be <math>\boxed{\textbf{(A) }1}</math>.<br />
<br />
==Solution 2==<br />
Apply the change of base formula to <math>\log_a b</math> and <math>\log_b a</math>. For simplicity, let us convert to base-10 log.<br />
By change of base, the expression becomes <math>\frac{\log b}{\log a} * \frac{\log a}{\log b} = \boxed{\textbf{(A) }1}</math>.<br />
<br />
==See Also==<br />
{{AHSME 50p box|year=1953|num-b=38|num-a=40}}<br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_30&diff=1215331953 AHSME Problems/Problem 302020-04-22T21:33:41Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>==Problem 30==<br />
<br />
A house worth \$ <math>9000</math> is sold by Mr. A to Mr. B at a <math>10</math> % loss. Mr. B sells the house back to Mr. A at a <math>10</math> % gain. <br />
The result of the two transactions is: <br />
<br />
<math>\textbf{(A)}\ \text{Mr. A breaks even} \qquad<br />
\textbf{(B)}\ \text{Mr. B gains }\$900 \qquad<br />
\textbf{(C)}\ \text{Mr. A loses }\$900\\ \textbf{(D)}\ \text{Mr. A loses }\$810\qquad<br />
\textbf{(E)}\ \text{Mr. B gains }\$1710 </math><br />
<br />
==Solution==<br />
When Mr.A sells the house at a <math>10</math>% loss, he sells it for <math>9000(1 - .1) = 8100</math>. When Mr.B sells the house back to Mr. A at a <math>10</math> % gain he sells it for <math>8100(1 + .1) = 8910</math>. Therefore Mr. A has lost <math>8100-8910 = 810</math> dollars, so the answer is <math>\boxed{D}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1953|num-b=29|num-a=31}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_16&diff=1214831953 AHSME Problems/Problem 162020-04-22T16:30:52Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>==Problem 16==<br />
<br />
Adams plans a profit of <math>10</math> % on the selling price of an article and his expenses are <math>15</math> % of sales. The rate of markup on an article that sells for \$ <math>5.00</math> is: <br />
<br />
<math>\textbf{(A)}\ 20\% \qquad<br />
\textbf{(B)}\ 25\% \qquad<br />
\textbf{(C)}\ 30\% \qquad<br />
\textbf{(D)}\ 33\frac {1}{3}\% \qquad<br />
\textbf{(E)}\ 35\% </math><br />
<br />
==Solution==<br />
Let the cost of the product be C, and the selling price be S. Since Adams wants a 10% profit on the sale, and his expenses are 15% of the sale, we add <math>0.10S</math> and <math>0.15S</math> to the cost to get the selling price.<br />
Then, <math>S=C+0.15S+0.10S \Rightarrow 0.75S=C \Rightarrow S=\frac{4}{3}C \Rightarrow \fbox{D}</math>.<br />
<br />
==See Also==<br />
{{AHSME 50p box|year=1953|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_11&diff=1214781953 AHSME Problems/Problem 112020-04-22T16:23:53Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>A running track is the ring formed by two concentric circles. It is <math>10</math> feet wide. The circumference of the two circles differ by about: <br />
<math>\textbf{(A)}\ 10\text{ feet} \qquad<br />
\textbf{(B)}\ 30\text{ feet} \qquad<br />
\textbf{(C)}\ 60\text{ feet} \qquad<br />
\textbf{(D)}\ 100\text{ feet}\\ \textbf{(E)}\ \text{none of these} </math><br />
<br />
== Solution ==<br />
<br />
Since the track is 10 feet wide, the diameter of the outer circle will be 20 feet more than the inner circle. Since the circumference of a circle is directly proportional to its diameter, the difference in the circles' diameters is simply <math>20\pi </math> feet. Using <math>\pi \approx 3</math>, the answer is <math>\fbox{C}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1953|num-b=10|num-a=12}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_11&diff=1214771953 AHSME Problems/Problem 112020-04-22T16:23:20Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>A running track is the ring formed by two concentric circles. It is <math>10</math> feet wide. The circumference of the two circles differ by about: <br />
<math>\textbf{(A)}\ 10\text{ feet} \qquad<br />
\textbf{(B)}\ 30\text{ feet} \qquad<br />
\textbf{(C)}\ 60\text{ feet} \qquad<br />
\textbf{(D)}\ 100\text{ feet}\\ \textbf{(E)}\ \text{none of these} </math><br />
<br />
== Solution ==<br />
<br />
Since the track is 10 feet wide, the diameter of the outer circle will be 20 feet more than the inner circle. Since the circumference of a circle is directly proportional to its diameter, the difference in the circles' diameters is simply <math>20\pi</math>. Using <math>\pi \approx 3</math>, the answer is <math>\fbox{C}</math>.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1953|num-b=10|num-a=12}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_11&diff=1214761953 AHSME Problems/Problem 112020-04-22T16:19:20Z<p>Fortytwok: </p>
<hr />
<div>A running track is the ring formed by two concentric circles. It is <math>10</math> feet wide. The circumference of the two circles differ by about: <br />
<math>\textbf{(A)}\ 10\text{ feet} \qquad<br />
\textbf{(B)}\ 30\text{ feet} \qquad<br />
\textbf{(C)}\ 60\text{ feet} \qquad<br />
\textbf{(D)}\ 100\text{ feet}\\ \textbf{(E)}\ \text{none of these} </math><br />
<br />
== Solution ==<br />
<br />
Call the radius of the outer circle <math>r_1</math> and that of the inner circle <math>r_2</math>. The width of the track is <math>r_1-r_2</math>. The circumference of a circle is <math>2\pi</math> times the radius, so the difference in circumferences is <math>2\pi r_1-2\pi r_2=10\pi</math> feet. If we divide each side by <math>2\pi</math>, we get <math>r_1-r_2=\boxed{5}</math> feet.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1953|num-b=10|num-a=12}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_11&diff=1214751953 AHSME Problems/Problem 112020-04-22T16:18:27Z<p>Fortytwok: </p>
<hr />
<div>A running track is the ring formed by two concentric circles. It is <math> 10</math> feet wide. The circumference of the two circles differ by about<br />
<br />
== Solution ==<br />
<br />
Call the radius of the outer circle <math>r_1</math> and that of the inner circle <math>r_2</math>. The width of the track is <math>r_1-r_2</math>. The circumference of a circle is <math>2\pi</math> times the radius, so the difference in circumferences is <math>2\pi r_1-2\pi r_2=10\pi</math> feet. If we divide each side by <math>2\pi</math>, we get <math>r_1-r_2=\boxed{5}</math> feet.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1953|num-b=10|num-a=12}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_4&diff=1212891953 AHSME Problems/Problem 42020-04-20T23:58:58Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>The roots of <math>x(x^2+8x+16)(4-x)=0</math> are:<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 0,4 \qquad \textbf{(C)}\ 0,4,-4 \qquad \textbf{(D)}\ 0,4,-4,-4 \qquad \textbf{(E)}\ \text{none of these}</math><br />
<br />
==Solution==<br />
<br />
We factor the middle part into <math>(x+4)^2</math>.<br />
<br />
The equation now becomes <math>x(x+4)^2(4-x)=0</math><br />
<br />
The solutions are then <math>0, -4, 4</math>. So the answer is <math>\boxed{\text{C}}</math><br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1953|num-b=3|num-a=5}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_48&diff=1212881952 AHSME Problems/Problem 482020-04-20T23:46:24Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Two cyclists, <math>k</math> miles apart, and starting at the same time, would be together in <math>r</math> hours if they traveled in the same direction, but would pass each other in <math>t</math> hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is: <br />
<br />
<math>\text{(A) } \frac {r + t}{r - t} \qquad<br />
\text{(B) } \frac {r}{r - t} \qquad<br />
\text{(C) } \frac {r + t}{r} \qquad<br />
\text{(D) } \frac{r}{t}\qquad<br />
\text{(E) } \frac{r+k}{t-k}</math><br />
<br />
== Solution ==<br />
<asy><br />
pair A,B,C;<br />
A=(0,0); B=(8,0); C=(4,1);<br />
draw((A)--(B));<br />
label("$A$",A,S); label("$B$",B,SE); label("$k$",C,SW);<br />
</asy><br />
<br />
== See also ==<br />
{{AHSME 50p box|year=1952|num-b=47|num-a=49}} <br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_46&diff=1212871952 AHSME Problems/Problem 462020-04-20T22:59:36Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
The base of a new rectangle equals the sum of the diagonal and the greater side of a given rectangle, while the altitude of the new rectangle equals the difference of the diagonal and the greater side of the given rectangle. The area of the new rectangle is: <br />
<br />
<math>\text{(A) greater than the area of the given rectangle} \quad\\<br />
\text{(B) equal to the area of the given rectangle} \quad\\<br />
\text{(C) equal to the area of a square with its side equal to the smaller side of the given rectangle} \quad\\<br />
\text{(D) equal to the area of a square with its side equal to the greater side of the given rectangle} \quad\\<br />
\text{(E) equal to the area of a rectangle whose dimensions are the diagonal and the shorter side of the given rectangle}</math><br />
<br />
== Solution ==<br />
Let the larger side of the rectangle be y and the smaller side of the rectangle be x. Then, by pythagorean theorem, the diagonal of the rectangle has length <math>\sqrt{x^{2}+y^{2}}</math>. <br />
<br />
By the definition of the problem, the area of the new rectangle is <math>(\sqrt{x^{2}+y^{2}}+y)(\sqrt{x^{2}+y^{2}}-y)</math><br />
<br />
Expanding this gives the area of the new rectangle to be <math>x^{2}</math>, or <br />
<math>\fbox{C}</math>.<br />
<br />
== See also ==<br />
{{AHSME 50p box|year=1952|num-b=45|num-a=47}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_43&diff=1212861952 AHSME Problems/Problem 432020-04-20T22:45:59Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
The diameter of a circle is divided into <math>n</math> equal parts. On each part a semicircle is constructed. As <math>n</math> becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length: <br />
<math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle<br />
<math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle<br />
<math>\textbf{(C) } \qquad</math> greater than the diameter, but less than the semi-circumference of the original circle<br />
<math>\textbf{(D) } \qquad</math> that is infinite<br />
<math>\textbf{(E) } </math> greater than the semi-circumference<br />
<br />
== Solution ==<br />
Note that the circumference of a circle is <math>\pi*d</math>. <br />
<br />
Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter <math>\frac{D}{n}.</math> Since there are n circles, each with diameter <math>\frac{D}{n}</math>, the sum of the circumferences of the small circles is D. Since we are only drawing semicircles and not full circles, the requested sum is <br />
<math>\boxed{A}</math>.<br />
<br />
== See also ==<br />
{{AHSME 50p box|year=1952|num-b=42|num-a=44}} <br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems&diff=1210521952 AHSME Problems2020-04-16T16:44:34Z<p>Fortytwok: /* Problem 43 */</p>
<hr />
<div>{{AHSC 50 Problems<br />
|year=1952<br />
}}<br />
== Problem 1 ==<br />
<br />
If the radius of a circle is a rational number, its area is given by a number which is:<br />
<br />
<math> \textbf{(A)\ } \text{rational} \qquad \textbf{(B)\ } \text{irrational} \qquad \textbf{(C)\ } \text{integral} \qquad \textbf{(D)\ } \text{a perfect square }\qquad \textbf{(E)\ } \text{none of these} </math><br />
<br />
[[1952 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
Two high school classes took the same test. One class of <math> 20 </math> students made an average grade of <math> 80\% </math>; the other class of <math> 30 </math> students made an average grade of <math> 70\% </math>. The average grade for all students in both classes is:<br />
<br />
<math> \textbf{(A)}\ 75\%\qquad \textbf{(B)}\ 74\%\qquad \textbf{(C)}\ 72\%\qquad \textbf{(D)}\ 77\%\qquad \textbf{(E)\ }\text{none of these} </math><br />
<br />
[[1952 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
The expression <math> a^3-a^{-3} </math> equals: <br />
<br />
<math> \textbf{(A) \ }\left(a-\frac{1}{a}\right)\left(a^2+1+\frac{1}{a^2}\right) \qquad \textbf{(B) \ }\left(\frac{1}{a}-a\right)\left(a^2-1+\frac{1}{a^2}\right) \qquad \textbf{(C) \ }\left(a-\frac{1}{a}\right)\left(a^2-2+\frac{1}{a^2}\right) \qquad </math><br />
<math> \textbf{(D) \ }\left(\frac{1}{a}-a\right)\left(\frac{1}{a^2}+1+a^2\right) \qquad \textbf{(E) \ }\text{none of these} </math><br />
<br />
[[1952 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
The cost <math> C </math> of sending a parcel post package weighing <math> P </math> pounds, <math> P </math> an integer, is <math> 10 </math> cents for the first pound and <math> 3 </math> cents for each additional pound. The formula for the cost is:<br />
<br />
<math> \textbf{(A) \ }C=10+3P \qquad \textbf{(B) \ }C=10P+3 \qquad \textbf{(C) \ }C=10+3(P-1) \qquad </math><br />
<br />
<math> \textbf{(D) \ }C=9+3P \qquad \textbf{(E) \ }C=10P-7 </math><br />
<br />
[[1952 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
The points <math> (6,12) </math> and <math> (0,-6) </math> are connected by a straight line. Another point on this line is:<br />
<br />
<math> \textbf{(A) \ }(3,3) \qquad \textbf{(B) \ }(2,1) \qquad \textbf{(C) \ }(7,16) \qquad \textbf{(D) \ }(-1,-4) \qquad \textbf{(E) \ }(-3,-8) </math><br />
<br />
[[1952 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
The difference of the roots of <math> x^2-7x-9=0 </math> is:<br />
<br />
<math> \textbf{(A) \ }+7 \qquad \textbf{(B) \ }+\frac{7}{2} \qquad \textbf{(C) \ }+9 \qquad \textbf{(D) \ }2\sqrt{85} \qquad \textbf{(E) \ }\sqrt{85} </math><br />
<br />
[[1952 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
When simplified, <math> (x^{-1}+y^{-1})^{-1} </math> is equal to:<br />
<br />
<math> \textbf{(A) \ }x+y \qquad \textbf{(B) \ }\frac{xy}{x+y} \qquad \textbf{(C) \ }xy \qquad \textbf{(D) \ }\frac{1}{xy} \qquad \textbf{(E) \ }\frac{x+y}{xy} </math><br />
<br />
[[1952 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
Two equal circles in the same plane cannot have the following number of common tangents.<br />
<br />
<math> \textbf{(A) \ }1 \qquad \textbf{(B) \ }2 \qquad \textbf{(C) \ }3 \qquad \textbf{(D) \ }4 \qquad \textbf{(E) \ }\text{none of these} </math><br />
<br />
[[1952 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
If <math> m=\frac{cab}{a-b} </math>, then <math> b </math> equals:<br />
<br />
<math> \textbf{(A) \ }\frac{m(a-b)}{ca} \qquad \textbf{(B) \ }\frac{cab-ma}{-m} \qquad \textbf{(C) \ }\frac{1}{1+c} \qquad \textbf{(D) \ }\frac{ma}{m+ca} \qquad \textbf{(E) \ }\frac{m+ca}{ma} </math><br />
<br />
[[1952 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
An automobile went up a hill at a speed of <math> 10 </math> miles an hour and down the same distance at a speed of <math> 20 </math> miles an hour. The average speed for the round trip was:<br />
<br />
<math> \textbf{(A) \ }12\frac{1}{2}\text{mph} \qquad \textbf{(B) \ }13\frac{1}{3}\text{mph} \qquad \textbf{(C) \ }14\frac{1}{2}\text{mph} \qquad \textbf{(D) \ }15\text{mph} \qquad \textbf{(E) \ }\text{none of these} </math><br />
<br />
[[1952 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
If <math> y=f(x)=\frac{x+2}{x-1} </math>, then it is incorrect to say:<br />
<br />
<math> \textbf{(A)\ }x=\frac{y+2}{y-1}\qquad\textbf{(B)\ }f(0)=-2\qquad\textbf{(C)\ }f(1)=0\qquad </math><br />
<br />
<math> \textbf{(D)\ }f(-2)=0\qquad\textbf{(E)\ }f(y)=x </math><br />
<br />
[[1952 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The sum to infinity of the terms of an infinite geometric progression is <math> 6 </math>. The sum of the first two terms is <math> 4\frac{1}{2} </math>. The first term of the progression is:<br />
<br />
<math> \textbf{(A) \ }3 \text{ or } 1\frac{1}{2} \qquad \textbf{(B) \ }1 \qquad \textbf{(C) \ }2\frac{1}{2} \qquad \textbf{(D) \ }6 \qquad \textbf{(E) \ }9\text{ or }3 </math><br />
<br />
[[1952 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
The function <math> x^2+px+q </math> with <math> p </math> and <math> q </math> greater than zero has its minimum value when:<br />
<br />
<math> \textbf{(A) \ }x=-p \qquad \textbf{(B) \ }x=\frac{p}{2} \qquad \textbf{(C) \ }x=-2p \qquad \textbf{(D) \ }x=\frac{p^2}{4q} \qquad </math><br />
<br />
<math> \textbf{(E) \ }x=\frac{-p}{2} </math><br />
<br />
[[1952 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
A house and store were sold for <math> \textdollar 12,000 </math> each. The house was sold at a loss of <math> 20\% </math> of the cost, and the store at a gain of <math> 20\% </math> of the cost. The entire transaction resulted in:<br />
<br />
<math> \textbf{(A) \ }\text{no loss or gain} \qquad \textbf{(B) \ }\text{loss of }\textdollar 1000 \qquad \textbf{(C) \ }\text{gain of }\textdollar 1000 \qquad \textbf{(D) \ }\text{gain of }\textdollar 2000 \qquad \textbf{(E) \ }\text{none of these} </math><br />
<br />
[[1952 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
The sides of a triangle are in the ratio <math> 6:8:9 </math>. Then:<br />
<br />
<math> \textbf{(A) \ }\text{the triangle is obtuse} </math><br />
<br />
<math> \textbf{(B) \ }\text{the angles are in the ratio }6:8:9 </math> <br />
<br />
<math> \textbf{(C) \ }\text{the triangle is acute} </math><br />
<br />
<math> \textbf{(D) \ }\text{the angle opposite the largest side is double the angle opposite the smallest side} </math><br />
<br />
<math> \textbf{(E) \ }\text{none of these} </math><br />
<br />
[[1952 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
If the base of a rectangle is increased by <math> 10\% </math> and the area is unchanged, then the altitude is decreased by:<br />
<br />
<math> \textbf{(A) \ }9\% \qquad \textbf{(B) \ }10\% \qquad \textbf{(C) \ }11\% \qquad \textbf{(D) \ }11\frac{1}{9}\% \qquad \textbf{(E) \ }9\frac{1}{11}\% </math><br />
<br />
[[1952 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
A merchant bought some goods at a discount of <math> 20\% </math> of the list price. He wants to mark them at such a price that he can give a discount of <math> 20\% </math> of the marked price and still make a profit of <math> 20\% </math> of the selling price.. The per cent of the list price at which he should mark them is:<br />
<br />
<math> \textbf{(A) \ }20 \qquad \textbf{(B) \ }100 \qquad \textbf{(C) \ }125 \qquad \textbf{(D) \ }80 \qquad \textbf{(E) \ }120 </math><br />
<br />
[[1952 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
<math> \log p+\log q=\log(p+q) </math> only if:<br />
<br />
<math> \textbf{(A) \ }p=q=\text{zero} \qquad \textbf{(B) \ }p=\frac{q^2}{1-q} \qquad \textbf{(C) \ }p=q=1 \qquad </math><br />
<br />
<math> \textbf{(D) \ }p=\frac{q}{q-1} \qquad \textbf{(E) \ }p=\frac{q}{q+1} </math><br />
<br />
[[1952 AHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
Angle <math> B </math> of triangle <math> ABC </math> is trisected by <math> BD </math> and <math> BE </math> which meet <math> AC </math> at <math> D </math> and <math> E </math> respectively. Then:<br />
<br />
<math> \textbf{(A) \ }\frac{AD}{EC}=\frac{AE}{DC} \qquad \textbf{(B) \ }\frac{AD}{EC}=\frac{AB}{BC} \qquad \textbf{(C) \ }\frac{AD}{EC}=\frac{BD}{BE} \qquad </math><br />
<br />
<math> \textbf{(D) \ }\frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)} \qquad \textbf{(E) \ }\frac{AD}{EC}=\frac{(AE)(BD)}{(DC)(BE)} </math><br />
<br />
[[1952 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
If <math> \frac{x}{y}=\frac{3}{4} </math>, then the incorrect expression in the following is:<br />
<br />
<math> \textbf{(A) \ }\frac{x+y}{y}=\frac{7}{4} \qquad \textbf{(B) \ }\frac{y}{y-x}=\frac{4}{1} \qquad \textbf{(C) \ }\frac{x+2y}{x}=\frac{11}{3} \qquad </math><br />
<br />
<math> \textbf{(D) \ }\frac{x}{2y}=\frac{3}{8} \qquad \textbf{(E) \ }\frac{x-y}{y}=\frac{1}{4} </math><br />
<br />
[[1952 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
The sides of a regular polygon of <math> n </math> sides, <math> n>4 </math>, are extended to form a star. The number of degrees at each point of the star is:<br />
<br />
<math> \textbf{(A) \ }\frac{360}{n} \qquad \textbf{(B) \ }\frac{(n-4)180}{n} \qquad \textbf{(C) \ }\frac{(n-2)180}{n} \qquad </math><br />
<br />
<math> \textbf{(D) \ }180-\frac{90}{n} \qquad \textbf{(E) \ }\frac{180}{n} </math><br />
<br />
[[1952 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
On hypotenuse <math> AB </math> of a right triangle <math> ABC </math> a second right triangle <math> ABD </math> is constructed with hypotenuse <math> AB </math>. If <math> \overline{BC}=1 </math>, <math> \overline{AC}=b </math>, and <math> \overline{AD}=2 </math>, then <math> \overline{BD} </math> equals:<br />
<br />
<math> \textbf{(A) \ }\sqrt{b^2+1} \qquad \textbf{(B) \ }\sqrt{b^2-3} \qquad \textbf{(C) \ }\sqrt{b^2+1}+2 \qquad </math><br />
<br />
<math> \textbf{(D) \ }b^2+5 \qquad \textbf{(E) \ }\sqrt{b^2+3} </math><br />
<br />
[[1952 AHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
If <math> \frac{x^2-bx}{ax-c}=\frac{m-1}{m+1} </math> has roots which are numerically equal but of opposite signs, the value of <math> m </math> must be:<br />
<br />
<math> \textbf{(A) \ }\frac{a-b}{a+b} \qquad \textbf{(B) \ }\frac{a+b}{a-b} \qquad \textbf{(C) \ }c \qquad \textbf{(D) \ }\frac{1}{c} \qquad \textbf{(E) \ }1 </math><br />
<br />
[[1952 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
In the figure, it is given that angle <math> C = 90^{\circ} </math>,<math> \overline{AD} = \overline{DB} </math>,<math> DE \perp AB </math>, <math> \overline{AB} = 20 </math>, and <math> \overline{AC} = 12 </math>. The area of quadrilateral <math> ADEC </math> is: <br />
<asy><br />
unitsize(7);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
pair A,B,C,D,E;<br />
A=(0,0); B=(20,0); C=(36/5,48/5); D=(10,0); E=(10,75/10);<br />
draw(A--B--C--cycle); draw(D--E);<br />
label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE);<br />
draw(rightanglemark(B,D,E,30));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 75\qquad\textbf{(B)}\ 58\frac{1}{2}\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 37\frac{1}{2}\qquad\textbf{(E)}\ \text{none of these} </math><br />
<br />
[[1952 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
A powderman set a fuse for a blast to take place in <math>30</math> seconds. He ran away at a rate of <math>8</math> yards per second. Sound travels at the rate of <math>1080</math> feet per second. When the powderman heard the blast, he had run approximately:<br />
<br />
<math> \textbf{(A)}\ \text{200 yd.}\qquad\textbf{(B)}\ \text{352 yd.}\qquad\textbf{(C)}\ \text{300 yd.}\qquad\textbf{(D)}\ \text{245 yd.}\qquad\textbf{(E)}\ \text{512 yd.} </math><br />
<br />
[[1952 AHSME Problems/Problem 25|Solution]]<br />
<br />
== Problem 26 ==<br />
If <math>\left(r+\frac1r\right)^2=3</math>, then <math>r^3+\frac1{r^3}</math> equals<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6</math><br />
<br />
[[1952 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27 ==<br />
The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:<br />
<br />
<math> \textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3</math><br />
<br />
[[1952 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28 ==<br />
In the table shown, the formula relating <math>x</math> and <math>y</math> is:<br />
<br />
<math>\begin{tabular}{|l|l|l|l|l|l|}<br />
\hline<br />
x & 1 & 2 & 3 & 4 & 5 \\ \hline<br />
y & 3 & 7 & 13 & 21 & 31 \\<br />
\hline<br />
\end{tabular}</math><br />
<br />
<br />
<math> \textbf{(A)}\ y=4x-1 \qquad \textbf{(B)}\ y=x^{3}-x^{2}+x+2 \qquad \textbf{(C)}\ y=x^2+x+1 \\ \qquad \textbf{(D)}\ y=(x^2+x+1)(x-1) \qquad \textbf{(E)}\ \text{None of these}</math><br />
<br />
[[1952 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29 ==<br />
In a circle of radius <math>5</math> units, <math>CD</math> and <math>AB</math> are perpendicular diameters. A chord <math>CH</math> cutting <math>AB</math> at <math>K</math> is <math>8</math> units long. The diameter <math>AB</math> is divided into two segments whose dimensions are:<br />
<br />
<math> \textbf{(A)}\ 1.25, 8.75 \qquad \textbf{(B)}\ 2.75, 7.25\qquad \textbf{(C)}\ 2,8 \qquad \textbf{(D)} \ 4,6 \qquad \textbf{(E)} \text{None of these}</math><br />
<br />
[[1952 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30 ==<br />
<br />
When the sum of the first ten terms of an arithmetic progression is four times the sum of the first five terms, the ratio of the first term to the common difference is: <br />
<br />
<math>\textbf{(A)}\ 1: 2 \qquad<br />
\textbf{(B)}\ 2: 1 \qquad<br />
\textbf{(C)}\ 1: 4 \qquad<br />
\textbf{(D)}\ 4: 1 \qquad<br />
\textbf{(E)}\ 1: 1 </math><br />
<br />
[[1952 AHSME Problems/Problem 30|Solution]]<br />
<br />
== Problem 31 ==<br />
<br />
Given <math>12</math> points in a plane no three of which are collinear, the number of lines they determine is: <br />
<br />
<math>\textbf{(A)}\ 24 \qquad<br />
\textbf{(B)}\ 54 \qquad<br />
\textbf{(C)}\ 120 \qquad<br />
\textbf{(D)}\ 66 \qquad<br />
\textbf{(E)}\ \text{none of these}</math><br />
[[1952 AHSME Problems/Problem 31|Solution]]<br />
<br />
== Problem 32 ==<br />
<br />
<math>K</math> takes <math>30</math> minutes less time than <math>M</math> to travel a distance of <math>30</math> miles. <math>K</math> travels <math>\frac {1}{3}</math> mile per hour faster than <math>M</math>. If <math>x</math> is <math>K</math>'s rate of speed in miles per hours, then <math>K</math>'s time for the distance is: <br />
<br />
<math>\textbf{(A)}\ \dfrac{x + \frac {1}{3}}{30} \qquad<br />
\textbf{(B)}\ \dfrac{x - \frac {1}{3}}{30} \qquad<br />
\textbf{(C)}\ \frac{30}{x+\frac{1}{3}}\qquad<br />
\textbf{(D)}\ \frac{30}{x}\qquad<br />
\textbf{(E)}\ \frac{x}{30} </math><br />
<br />
[[1952 AHSME Problems/Problem 32|Solution]]<br />
<br />
== Problem 33 ==<br />
<br />
A circle and a square have the same perimeter. Then: <br />
<br />
<math>\text{(A) their areas are equal}\qquad\\<br />
\text{(B) the area of the circle is the greater} \qquad\\<br />
\text{(C) the area of the square is the greater} \qquad\\<br />
\text{(D) the area of the circle is } \pi \text{ times the area of the square}\qquad\\<br />
\text{(E) none of these}</math><br />
<br />
[[1952 AHSME Problems/Problem 33|Solution]]<br />
<br />
== Problem 34 ==<br />
<br />
The price of an article was increased <math>p\%</math>. Later the new price was decreased <math>p\%</math>. If the last price was one dollar, the original price was: <br />
<br />
<math>\textbf{(A)}\ \frac{1-p^2}{200}\qquad<br />
\textbf{(B)}\ \frac{\sqrt{1-p^2}}{100}\qquad<br />
\textbf{(C)}\ \text{one dollar}\qquad\\<br />
\textbf{(D)}\ 1-\frac{p^2}{10000-p^2}\qquad<br />
\textbf{(E)}\ \frac{10000}{10000-p^2} </math><br />
<br />
<br />
[[1952 AHSME Problems/Problem 34|Solution]]<br />
<br />
== Problem 35 ==<br />
<br />
With a rational denominator, the expression <math>\frac {\sqrt {2}}{\sqrt {2} + \sqrt {3} - \sqrt {5}}</math> is equivalent to: <br />
<br />
<math>\textbf{(A)}\ \frac {3 + \sqrt {6} + \sqrt {15}}{6} \qquad<br />
\textbf{(B)}\ \frac {\sqrt {6} - 2 + \sqrt {10}}{6} \qquad<br />
\textbf{(C)}\ \frac{2+\sqrt{6}+\sqrt{10}}{10} \qquad\\<br />
\textbf{(D)}\ \frac {2 + \sqrt {6} - \sqrt {10}}{6} \qquad<br />
\textbf{(E)}\ \text{none of these} </math><br />
<br />
[[1952 AHSME Problems/Problem 35|Solution]]<br />
<br />
== Problem 36 ==<br />
<br />
To be continuous at <math>x = - 1</math>, the value of <math>\frac {x^3 + 1}{x^2 - 1}</math> is taken to be: <br />
<br />
<math>\textbf{(A)}\ - 2 \qquad<br />
\textbf{(B)}\ 0 \qquad<br />
\textbf{(C)}\ \frac {3}{2} \qquad<br />
\textbf{(D)}\ \infty \qquad<br />
\textbf{(E)}\ -\frac{3}{2} </math><br />
<br />
[[1952 AHSME Problems/Problem 36|Solution]]<br />
<br />
== Problem 37 ==<br />
<br />
Two equal parallel chords are drawn <math>8</math> inches apart in a circle of radius <math>8</math> inches. The area of that part of the circle that lies between the chords is:<br />
<br />
<math>\textbf{(A)}\ 21\frac{1}{3}\pi-32\sqrt{3}\qquad<br />
\textbf{(B)}\ 32\sqrt{3}+21\frac{1}{3}\pi\qquad<br />
\textbf{(C)}\ 32\sqrt{3}+42\frac{2}{3}\pi \qquad\\<br />
\textbf{(D)}\ 16\sqrt {3} + 42\frac {2}{3}\pi \qquad<br />
\textbf{(E)}\ 42\frac {2}{3}\pi </math><br />
<br />
[[1952 AHSME Problems/Problem 37|Solution]]<br />
<br />
== Problem 38 ==<br />
<br />
The area of a trapezoidal field is <math>1400</math> square yards. Its altitude is <math>50</math> yards. Find the two bases, if the number of yards in each base is an integer divisible by <math>8</math>. The number of solutions to this problem is: <br />
<br />
<math>\textbf{(A)}\ \text{none} \qquad<br />
\textbf{(B)}\ \text{one} \qquad<br />
\textbf{(C)}\ \text{two} \qquad<br />
\textbf{(D)}\ \text{three} \qquad<br />
\textbf{(E)}\ \text{more than three}</math><br />
<br />
<br />
[[1952 AHSME Problems/Problem 38|Solution]]<br />
<br />
== Problem 39 ==<br />
<br />
If the perimeter of a rectangle is <math>p</math> and its diagonal is <math>d</math>, the difference between the length and width of the rectangle is: <br />
<br />
<math>\textbf{(A)}\ \frac {\sqrt {8d^2 - p^2}}{2} \qquad<br />
\textbf{(B)}\ \frac {\sqrt {8d^2 + p^2}}{2} \qquad<br />
\textbf{(C)}\ \frac{\sqrt{6d^2-p^2}}{2}\qquad\\<br />
\textbf{(D)}\ \frac {\sqrt {6d^2 + p^2}}{2} \qquad<br />
\textbf{(E)}\ \frac {8d^2 - p^2}{4} </math><br />
<br />
[[1952 AHSME Problems/Problem 39|Solution]]<br />
<br />
== Problem 40 ==<br />
<br />
In order to draw a graph of <math>ax^2+bx+c</math>, a table of values was constructed. These values of the function for a set of equally spaced increasing values of <math>x</math> were <math>3844, 3969, 4096, 4227, 4356, 4489, 4624</math>, and <math>4761</math>. The one which is incorrect is: <br />
<br />
<math>\text{(A) } 4096 \qquad<br />
\text{(B) } 4356 \qquad<br />
\text{(C) } 4489 \qquad<br />
\text{(D) } 4761 \qquad<br />
\text{(E) } \text{none of these}</math><br />
<br />
[[1952 AHSME Problems/Problem 40|Solution]]<br />
<br />
== Problem 41 ==<br />
<br />
Increasing the radius of a cylinder by <math>6</math> units increased the volume by <math>y</math> cubic units. Increasing the altitude of the cylinder by <math>6</math> units also increases the volume by <math>y</math> cubic units. If the original altitude is <math>2</math>, then the original radius is: <br />
<br />
<math>\text{(A) } 2 \qquad<br />
\text{(B) } 4 \qquad<br />
\text{(C) } 6 \qquad<br />
\text{(D) } 6\pi \qquad<br />
\text{(E) } 8 </math><br />
<br />
[[1952 AHSME Problems/Problem 41|Solution]]<br />
<br />
== Problem 42 ==<br />
Let <math>D</math> represent a repeating decimal. If <math>P</math> denotes the <math>r</math> figures of <math>D</math> which do not repeat themselves, and <math>Q</math> denotes the <math>s</math> figures of <math>D</math> which do repeat themselves, then the incorrect expression is:<br />
<br />
<math>\text{(A) } D = .PQQQ\ldots \qquad\\ \text{(B) } 10^rD = P.QQQ\ldots \\ \text{(C) } 10^{r + s}D = PQ.QQQ\ldots \qquad\\ \text{(D) } 10^r(10^s - 1)D = Q(P - 1) \\ \text{(E) } 10^r\cdot10^{2s}D = PQQ.QQQ\ldots</math><br />
<br />
[[1952 AHSME Problems/Problem 42|Solution]]<br />
<br />
== Problem 43 ==<br />
<br />
The diameter of a circle is divided into <math>n</math> equal parts. On each part a semicircle is constructed. As <math>n</math> becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length: <br />
<br />
<math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle<br />
<br />
<math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle<br />
<br />
<math>\textbf{(C) } \qquad</math> greater than the diameter, but less than the semi-circumference of the original circle<br />
<br />
<math>\textbf{(D) } \qquad</math> that is infinite<br />
<br />
<math>\textbf{(E) } </math> greater than the semi-circumference<br />
<br />
[[1952 AHSME Problems/Problem 43|Solution]]<br />
<br />
== Problem 44 ==<br />
<br />
If an integer of two digits is <math>k</math> times the sum of its digits, the number formed by interchanging the the digits is the sum of the digits multiplied by<br />
<br />
<math> \textbf{(A) \ } 9-k \qquad \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1 </math><br />
<br />
[[1952 AHSME Problems/Problem 44|Solution]]<br />
<br />
== Problem 45 ==<br />
<br />
If <math>a</math> and <math>b</math> are two unequal positive numbers, then: <br />
<br />
<math>\text{(A) } \frac{2ab}{a+b}>\sqrt{ab}>\frac{a+b}{2}\qquad<br />
\text{(B) } \sqrt{ab}>\frac{2ab}{a+b}>\frac{a+b}{2} \\<br />
\text{(C) } \frac{2ab}{a+b}>\frac{a+b}{2}>\sqrt{ab}\qquad<br />
\text{(D) } \frac{a+b}{2}>\frac{2ab}{a+b}>\sqrt{ab} \\<br />
\text{(E) } \frac {a + b}{2} > \sqrt {ab} > \frac {2ab}{a + b} </math><br />
<br />
[[1952 AHSME Problems/Problem 45|Solution]]<br />
<br />
== Problem 46 ==<br />
<br />
<br />
The base of a new rectangle equals the sum of the diagonal and the greater side of a given rectangle, while the altitude of the new rectangle equals the difference of the diagonal and the greater side of the given rectangle. The area of the new rectangle is: <br />
<br />
<math>\text{(A) greater than the area of the given rectangle} \quad\\<br />
\text{(B) equal to the area of the given rectangle} \quad\\<br />
\text{(C) equal to the area of a square with its side equal to the smaller side of the given rectangle} \quad\\<br />
\text{(D) equal to the area of a square with its side equal to the greater side of the given rectangle} \quad\\<br />
\text{(E) equal to the area of a rectangle whose dimensions are the diagonal and the shorter side of the given rectangle}</math><br />
<br />
<br />
[[1952 AHSME Problems/Problem 46|Solution]]<br />
<br />
== Problem 47 ==<br />
<br />
In the set of equations <math>z^x = y^{2x},\quad 2^z = 2\cdot4^x, \quad x + y + z = 16</math>, the integral roots in the order <math>x,y,z</math> are: <br />
<br />
<math>\textbf{(A) } 3,4,9 \qquad<br />
\textbf{(B) } 9,-5,-12 \qquad<br />
\textbf{(C) } 12,-5,9 \qquad<br />
\textbf{(D) } 4,3,9 \qquad<br />
\textbf{(E) } 4,9,3 </math><br />
<br />
[[1952 AHSME Problems/Problem 47|Solution]]<br />
<br />
== Problem 48 ==<br />
<br />
Two cyclists, <math>k</math> miles apart, and starting at the same time, would be together in <math>r</math> hours if they traveled in the same direction, but would pass each other in <math>t</math> hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is: <br />
<br />
<math>\text{(A) } \frac {r + t}{r - t} \qquad<br />
\text{(B) } \frac {r}{r - t} \qquad<br />
\text{(C) } \frac {r + t}{r} \qquad<br />
\text{(D) } \frac{r}{t}\qquad<br />
\text{(E) } \frac{r+k}{t-k}</math><br />
<br />
<br />
[[1952 AHSME Problems/Problem 48|Solution]]<br />
<br />
== Problem 49 ==<br />
<br />
<br />
<asy><br />
unitsize(27);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
pair A,B,C,D,E,F,X,Y,Z;<br />
A=(3,3); B=(0,0); C=(6,0); D=(4,0); E=(4,2); F=(1,1);<br />
draw(A--B--C--cycle);<br />
draw(A--D); draw(B--E); draw(C--F);<br />
X=intersectionpoint(A--D,C--F); Y=intersectionpoint(B--E,A--D); Z=intersectionpoint(B--E,C--F);<br />
label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE);<br />
label("$D$",D,S); label("$E$",E,NE); label("$F$",F,NW);<br />
label("$N_1$",X,NE); label("$N_2$",Y,WNW); label("$N_3$",Z,S);<br />
</asy><br />
<br />
In the figure, <math>\overline{CD}</math>, <math>\overline{AE}</math> and <math>\overline{BF}</math> are one-third of their respective sides. It follows that <math>\overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} = 3: 3: 1</math>, and similarly for lines BE and CF. Then the area of triangle <math>N_1N_2N_3</math> is: <br />
<br />
<math>\text{(A) } \frac {1}{10} \triangle ABC \qquad<br />
\text{(B) } \frac {1}{9} \triangle ABC \qquad<br />
\text{(C) } \frac{1}{7}\triangle ABC\qquad<br />
\text{(D) } \frac{1}{6}\triangle ABC\qquad<br />
\text{(E) } \text{none of these}</math><br />
<br />
[[1952 AHSME Problems/Problem 49|Solution]]<br />
<br />
== Problem 50 ==<br />
<br />
A line initially 1 inch long grows according to the following law, where the first term is the initial length. <br />
<cmath> 1+\frac{1}{4}\sqrt{2}+\frac{1}{4}+\frac{1}{16}\sqrt{2}+\frac{1}{16}+\frac{1}{64}\sqrt{2}+\frac{1}{64}+\cdots </cmath><br />
<br />
If the growth process continues forever, the limit of the length of the line is: <br />
<br />
<math>\textbf{(A) } \infty\qquad<br />
\textbf{(B) } \frac{4}{3}\qquad<br />
\textbf{(C) } \frac{8}{3}\qquad<br />
\textbf{(D) } \frac{1}{3}(4+\sqrt{2})\qquad<br />
\textbf{(E) } \frac{2}{3}(4+\sqrt{2})</math><br />
<br />
[[1952 AHSME Problems/Problem 50|Solution]]<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME 50p box|year=1952|before=[[1951 AHSME|1951 AHSC]]|after=[[1953 AHSME|1953 AHSC]]}} <br />
<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_39&diff=1210271952 AHSME Problems/Problem 392020-04-16T02:36:10Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
If the perimeter of a rectangle is <math>p</math> and its diagonal is <math>d</math>, the difference between the length and width of the rectangle is: <br />
<br />
<math>\textbf{(A)}\ \frac {\sqrt {8d^2 - p^2}}{2} \qquad<br />
\textbf{(B)}\ \frac {\sqrt {8d^2 + p^2}}{2} \qquad<br />
\textbf{(C)}\ \frac{\sqrt{6d^2-p^2}}{2}\qquad\\<br />
\textbf{(D)}\ \frac {\sqrt {6d^2 + p^2}}{2} \qquad<br />
\textbf{(E)}\ \frac {8d^2 - p^2}{4} </math><br />
<br />
== Solution ==<br />
<asy><br />
pair A,B,C,D,E,F,G,H;<br />
A=(0,0); B=(10,0); C=(10,5); D=(0,5); E=(5,5.5); F=(5,-0.5); G=(-0.5,2.75); H=(10.5,2.75);<br />
draw(A--B--C--D--cycle); draw (B--D);<br />
label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",(-0.5,5),N); <br />
label("$x$",E); label("$x$",F); label("$y$",G); label("$y$",H);<br />
</asy><br />
Let the sides of the rectangle be x and y. WLOG, assume x>y. Then, <math>2x+2y=p \Rightarrow x+y=\frac{p}{2}</math>.<br />
<br />
By pythagorean theorem, <math>x^2 + y^2 =d^2</math>.<br />
<br />
Since <math>x+y=\frac{p}{2}</math>, <math>(x+y)^2=\frac{p^2}{4} \Rightarrow x^2+2xy+y^2=\frac{p^2}{4}</math>.<br />
<br />
Rearranging to solve for <math>2xy</math> gives <math>2xy = \frac{p^2}{4}-d^2</math>.<br />
<br />
Rearranging <math>(x-y)^2</math> in terms of the defined variables becomes <math>(x-y)^2 = d^2 - (\frac{p^2}{4}-d^2) </math>. <br />
<br />
In order to get (x-y), we have to take the square root of the expression and simplify. <br />
<br />
<math>(x-y)=\sqrt{2d^2-\frac{p^2}{4}} \Rightarrow (x-y)=\sqrt{\frac{8d^2-p^2}{4}}</math> <math>\Rightarrow</math> <math>(x-y)=\frac{\sqrt{8d^2+p^2}}{2}</math> <math>\Rightarrow</math><br />
<math>\fbox{A}</math>.<br />
<br />
== See also ==<br />
{{AHSME 50p box|year=1952|num-b=38|num-a=40}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_37&diff=1210141952 AHSME Problems/Problem 372020-04-16T00:23:55Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Two equal parallel chords are drawn <math>8</math> inches apart in a circle of radius <math>8</math> inches. The area of that part of the circle that lies between the chords is:<br />
<br />
<math>\textbf{(A)}\ 21\frac{1}{3}\pi-32\sqrt{3}\qquad<br />
\textbf{(B)}\ 32\sqrt{3}+21\frac{1}{3}\pi\qquad<br />
\textbf{(C)}\ 32\sqrt{3}+42\frac{2}{3}\pi \qquad\\<br />
\textbf{(D)}\ 16\sqrt {3} + 42\frac {2}{3}\pi \qquad<br />
\textbf{(E)}\ 42\frac {2}{3}\pi </math><br />
<br />
== Solution ==<br />
<asy><br />
pair A,B,C,D,E,F,G;<br />
A=(0,0); B=(-5,4sqrt(3)+2); C=(5,4sqrt(3)+2); D=(-5,-4sqrt(3)-0.5); E=(5,-4sqrt(3)-0.5); F=(-4,0); G=(4,-sqrt(2)/2);<br />
label("$A$",(0,-0.5),S); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,SE); label("$E$",E,SW); label("$F$",F,W); label("$G$",G,NE);<br />
draw(circle(A,8)); <br />
draw((-4,-4sqrt(3))--(-4,4sqrt(3)));<br />
draw((4,-4sqrt(3))--(4,4sqrt(3)));<br />
draw((-4,-4sqrt(3))--(4,4sqrt(3)));<br />
draw((4,-4sqrt(3))--(-4,4sqrt(3)));<br />
draw((4,0)--(-4,0));<br />
label("$4$",(-2.5,0.5)); label("$4$",(2.5,0.5)); label("$8$",(-2.5,3)); label("$8$",(2.5,3)); label("$8$",(-2.5,-3)); label("$8$",(2.5,-3));<br />
</asy><br />
Draw in the diameter through A perpendicular to chords BD and CE. Label the intersection points of the diameter with the chords F and G. Then, AF=4, AG=4, and AFB ,AFD, AEG. and AGC are right triangles. We're going to find the area inside sectors BAD and CAE but outside triangles BAD and CAE.<br />
<br />
By pythagorean theorem, BF=<math>4\sqrt{3}</math>, as are DF, EG, and GC.<br />
<br />
It then follows that the area of triangles BAD and CAE are <math>16\sqrt{3}</math>. <br />
<br />
Since AFB and AFD are 30-60-90 triangles, the area of sector BAD is <math>\frac{64\pi}{3}</math>, as is sector CAE.<br />
<br />
Thus, the area outside of the two chords is <math>\frac{128\pi}{3}-32\sqrt{3}</math>.<br />
Since we want the area inside the two chords, you can subtract the outside from the whole circle, which is <math>64\pi-(\frac{128\pi}{3}-32\sqrt{3})</math> <math>\Rightarrow \frac{64\pi}{3}+32\sqrt{3}</math>, or <br />
<math>\fbox{B}</math>.<br />
<br />
== See also ==<br />
{{AHSME 50p box|year=1952|num-b=36|num-a=38}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_37&diff=1209991952 AHSME Problems/Problem 372020-04-15T02:59:11Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Two equal parallel chords are drawn <math>8</math> inches apart in a circle of radius <math>8</math> inches. The area of that part of the circle that lies between the chords is:<br />
<br />
<math>\textbf{(A)}\ 21\frac{1}{3}\pi-32\sqrt{3}\qquad<br />
\textbf{(B)}\ 32\sqrt{3}+21\frac{1}{3}\pi\qquad<br />
\textbf{(C)}\ 32\sqrt{3}+42\frac{2}{3}\pi \qquad\\<br />
\textbf{(D)}\ 16\sqrt {3} + 42\frac {2}{3}\pi \qquad<br />
\textbf{(E)}\ 42\frac {2}{3}\pi </math><br />
<br />
== Solution ==<br />
<asy><br />
pair A,B,C,D,E,F,G;<br />
A=(0,0); B=(-5,4sqrt(3)+2); C=(5,4sqrt(3)+2); D=(-5,-4sqrt(3)-0.5); E=(5,-4sqrt(3)-0.5); F=(-4,0); G=(4,-sqrt(2)/2);<br />
label("$A$",(0,-0.5),S); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,SE); label("$E$",E,SW); label("$F$",F,W); label("$G$",G,NE);<br />
draw(circle(A,8)); <br />
draw((-4,-4sqrt(3))--(-4,4sqrt(3)));<br />
draw((4,-4sqrt(3))--(4,4sqrt(3)));<br />
draw((-4,-4sqrt(3))--(4,4sqrt(3)));<br />
draw((4,-4sqrt(3))--(-4,4sqrt(3)));<br />
draw((4,0)--(-4,0));<br />
label("$4$",(-2.5,0.5)); label("$4$",(2.5,0.5)); label("$8$",(-2.5,3)); label("$8$",(2.5,3)); label("$8$",(-2.5,-3)); label("$8$",(2.5,-3));<br />
</asy><br />
Draw in the diameter through A perpendicular to chords BD and CE. Then, AF=4, AG=4, and AFB ,AFD, AEG. and AGC are right triangles. We're going to find the area inside sectors BAD and CAE but outside triangles BAD and CAE.<br />
<br />
By pythagorean theorem, BF=<math>4\sqrt{3}</math>, as are DF, EG, and GC.<br />
<br />
It then follows that the area of triangles BAD and CAE are <math>16\sqrt{3}</math>. <br />
<br />
Since AFB and AFD are 30-60-90 triangles, the area of sector BAD is <math>\frac{64\pi}{3}</math>, as is sector CAE.<br />
<br />
Thus, the area outside of the two chords is <math>\frac{128\pi}{3}-32\sqrt{3}</math>.<br />
Since we want the area inside the two chords, you can subtract the outside from the whole circle, which is <math>64\pi-(\frac{128\pi}{3}-32\sqrt{3})</math> <math>\Rightarrow \frac{64\pi}{3}+32\sqrt{3}</math>, or <br />
<math>\fbox{B}</math>.<br />
<br />
== See also ==<br />
{{AHSME 50p box|year=1952|num-b=36|num-a=38}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Fortytwokhttps://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_37&diff=1209471952 AHSME Problems/Problem 372020-04-14T14:48:41Z<p>Fortytwok: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Two equal parallel chords are drawn <math>8</math> inches apart in a circle of radius <math>8</math> inches. The area of that part of the circle that lies between the chords is:<br />
<br />
<math>\textbf{(A)}\ 21\frac{1}{3}\pi-32\sqrt{3}\qquad<br />
\textbf{(B)}\ 32\sqrt{3}+21\frac{1}{3}\pi\qquad<br />
\textbf{(C)}\ 32\sqrt{3}+42\frac{2}{3}\pi \qquad\\<br />
\textbf{(D)}\ 16\sqrt {3} + 42\frac {2}{3}\pi \qquad<br />
\textbf{(E)}\ 42\frac {2}{3}\pi </math><br />
<br />
== Solution ==<br />
<asy><br />
pair A,B,C,D,E,F,G;<br />
A=(0,0); B=(-5,4sqrt(3)+2); C=(5,4sqrt(3)+2); D=(-5,-4sqrt(3)-0.5); E=(5,-4sqrt(3)-0.5); F=(-4,0); G=(4,-sqrt(2)/2);<br />
label("$A$",(0,-0.5),S); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,SE); label("$E$",E,SW); label("$F$",F,W); label("$G$",G,NE);<br />
draw(circle(A,8)); <br />
draw((-4,-4sqrt(3))--(-4,4sqrt(3)));<br />
draw((4,-4sqrt(3))--(4,4sqrt(3)));<br />
draw((-4,-4sqrt(3))--(4,4sqrt(3)));<br />
draw((4,-4sqrt(3))--(-4,4sqrt(3)));<br />
draw((4,0)--(-4,0));<br />
label("$4$",(-2.5,0.5)); label("$4$",(2.5,0.5)); label("$8$",(-2.5,3)); label("$8$",(2.5,3)); label("$8$",(-2.5,-3)); label("$8$",(2.5,-3));<br />
</asy><br />
<br />
<math>\fbox{B}</math><br />
<br />
== See also ==<br />
{{AHSME 50p box|year=1952|num-b=36|num-a=38}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Fortytwok