https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Wirecat&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T01:19:55ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=892712011 AIME I Problems/Problem 42017-12-30T05:54:12Z<p>Wirecat: Added another solution</p>
<hr />
<div>== Problem 4 ==<br />
In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>.<br />
<br />
<br />
== Solution 1 == <br />
Extend <math>{CM}</math> and <math>{CN}</math> such that they intersects lines <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. <br />
Since <math>{BM}</math> is the angle bisector of angle B,and <math>{CM}</math> is perpendicular to <math>{BM}</math> ,so , <math>BP=BC=120</math>, M is the midpoint of <math>{CP}</math> .For the same reason,<math>AQ=AC=117</math>,N is the midpoint of <math>{CQ}</math>.<br />
Hence<math>MN=\frac{PQ}{2}</math>.But <math>PQ=BP+AQ-AB=120+117-125=112</math>,so<math>MN=\boxed{56}</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>AM \perp LC</math> and <math>AN \perp KB</math>, we have <math>AMIN</math> is a cyclic quadrilateral. Consequently, <math>\frac{MN}{\sin \angle MIN} = 2R = AI</math>. Since <math>\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK</math>, we have that <math>MN = AI \cdot \cos \angle IAK</math>. Letting <math>X</math> be the point of contact of the incircle of <math>ABC</math> with side <math>AC</math>, we have <math>AX=MN</math> thus <math>MN=\frac{117+120-125}{2}=\boxed{56}</math><br />
<br />
== Solution 3 (Bash) == <br />
Project <math>I</math> onto <math>AC</math> and <math>BC</math> as <math>D</math> and <math>E</math>. <math>ID</math> and <math>IE</math> are both in-radii of <math>\triangle ABC</math> so we get right triangles with legs <math>r</math> (the in-radius length) and <math>s - c = 56</math>. Since <math>IC</math> is the hypotenuse for the 4 triangles (<math>\triangle INC, \triangle IMC, \triangle IDC,</math> and <math>\triangle IEC</math>), <math>C, D, M, I, N, E</math> are con-cyclic on a circle we shall denote as <math>\omega</math> which is also the circumcircle of <math>\triangle CMN</math> and <math>\triangle CDE</math>. To find <math>MN</math>, we can use the Law of Cosines on <math>\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})</math> where <math>O</math> is the center of <math>\omega</math>. Now, the circumradius <math>R</math> can be found with Pythagorean Theorem with <math>\triangle CDI</math> or <math>\triangle CEI</math>: <math>r^2 + 56^2 = (2R)^2</math>. To find <math>r</math>, we can use the formula <math>rs = [ABC]</math> and by Heron's, <math>[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}</math>. To find <math>\angle MCN</math>, we can find <math>\angle MIN</math> since <math>\angle MCN = 180 - \angle MIN</math>. <math>\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}</math>. Thus, <math>\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}</math> and since <math>\angle A + \angle B + \angle C = 180</math>, we have <math>\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}</math>. Plugging this into our Law of Cosines formula gives <math>MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})</math>. To find <math>\cos{\angle C}</math>, we use LoC on <math>\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}</math>. Our formula now becomes <math>MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}</math>. After simplifying, we get <math>MN^2 = 3136 \implies MN = \boxed{056}</math>. <math>\square</math><br />
<br />
--lucasxia01<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems&diff=858371950 AHSME Problems2017-05-23T23:10:28Z<p>Wirecat: /* Problem 15 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
If <math>64</math> is divided into three parts proportional to <math>2</math>, <math>4</math>, and <math>6</math>, the smallest part is:<br />
<br />
<math> \textbf{(A)}\ 5\frac{1}{3}\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\frac{2}{3}\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these answers} </math><br />
<br />
[[1950 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
Let <math> R=gS-4 </math>. When <math>S=8</math>, <math>R=16</math>. When <math>S=10</math>, <math>R</math> is equal to:<br />
<br />
<math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 21\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
The sum of the roots of the equation <math> 4x^{2}+5-8x=0 </math> is equal to:<br />
<br />
<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ -5\qquad\textbf{(C)}\ -\frac{5}{4}\qquad\textbf{(D)}\ -2\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Reduced to lowest terms, <math> \frac{a^{2}-b^{2}}{ab} - \frac{ab-b^{2}}{ab-a^{2}} </math> is equal to:<br />
<br />
<math> \textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
If five geometric means are inserted between <math>8</math> and <math>5832</math>, the fifth term in the geometric series is:<br />
<br />
<math> \textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
The values of <math>y</math> which will satisfy the equations <cmath>\begin{array}{rcl} 2x^{2}+6x+5y+1&=&0\\ 2x+y+3&=&0 \end{array}</cmath> may be found by solving:<br />
<br />
<math> \textbf{(A)}\ y^{2}+14y-7=0\qquad\textbf{(B)}\ y^{2}+8y+1=0\qquad\textbf{(C)}\ y^{2}+10y-7=0\qquad\\ \textbf{(D)}\ y^{2}+y-12=0\qquad \textbf{(E)}\ \text{None of these equations} </math><br />
<br />
[[1950 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
If the digit <math>1</math> is placed after a two digit number whose tens' digit is <math>t</math>, and units' digit is <math>u</math>, the new number is:<br />
<br />
<math> \textbf{(A)}\ 10t+u+1\qquad\textbf{(B)}\ 100t+10u+1\qquad\textbf{(C)}\ 1000t+10u+1\qquad\textbf{(D)}\ t+u+1\qquad\\ \textbf{(E)}\ \text{None of these answers} </math><br />
<br />
[[1950 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
If the radius of a circle is increased <math>100\%</math>, the area is increased:<br />
<br />
<math> \textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
The largest area of a triangle that can be inscribed in a semi-circle whose radius is <math>r</math> is:<br />
<br />
<math> \textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2} </math><br />
<br />
[[1950 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
After rationalizing the numerator of <math> \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}} </math>, the denominator in simplest form is:<br />
<br />
<math> \textbf{(A)}\ \sqrt{3}(\sqrt{3}+\sqrt{2})\qquad\textbf{(B)}\ \sqrt{3}(\sqrt{3}-\sqrt{2})\qquad\textbf{(C)}\ 3-\sqrt{3}\sqrt{2}\qquad\\ \textbf{(D)}\ 3+\sqrt6\qquad\textbf{(E)}\ \text{None of these answers} </math><br />
<br />
[[1950 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
If in the formula <math> C =\frac{en}{R+nr} </math>, <math>n</math> is increased while <math>e</math>, <math>R</math> and <math>r</math> are kept constant, then <math>C</math>:<br />
<br />
<math> \textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Increases and then decreases}\qquad\\ \textbf{(E)}\ \text{Decreases and then increases} </math><br />
<br />
[[1950 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
As the number of sides of a polygon increases from <math>3</math> to <math>n</math>, the sum of the exterior angles formed by extending each side in succession:<br />
<br />
<math> \textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Cannot be predicted}\qquad\\ \textbf{(E)}\ \text{Becomes }(n-3)\text{ straight angles} </math><br />
<br />
[[1950 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
The roots of <math> (x^{2}-3x+2)(x)(x-4)=0 </math> are:<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 0\text{ and }4\qquad\textbf{(C)}\ 1\text{ and }2\qquad\textbf{(D)}\ 0,1,2\text{ and }4\qquad\textbf{(E)}\ 1,2\text{ and }4 </math><br />
<br />
[[1950 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
For the simultaneous equations <cmath>2x-3y=8</cmath><cmath>6y-4x=9</cmath><br />
<br />
<math> \textbf{(A)}\ x=4,y=0\qquad\textbf{(B)}\ x=0,y=\frac{3}{2}\qquad\textbf{(C)}\ x=0,y=0\qquad\\ \textbf{(D)}\ \text{There is no solution}\qquad\textbf{(E)}\ \text{There are an infinite number of solutions} </math><br />
<br />
[[1950 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
The real factors of <math>x^2+4</math> are:<br />
<br />
<math> \textbf{(A)}\ (x^{2}+2)(x^{2}+2)\qquad\textbf{(B)}\ (x^{2}+2)(x^{2}-2)\qquad\textbf{(C)}\ x^{2}(x^{2}+4)\qquad\\ \textbf{(D)}\ (x^{2}-2x+2)(x^{2}+2x+2)\qquad\textbf{(E)}\ \text{Non-existent} </math><br />
<br />
[[1950 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{2} </math> when simplified is:<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math><br />
<br />
[[1950 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
The formula which expresses the relationship between <math>x</math> and <math>y</math> as shown in the accompanying table is:<br />
<br />
<cmath> \begin{tabular}[t]{|c|c|c|c|c|c|}\hline x&0&1&2&3&4\\\hline y&100&90&70&40&0\\\hline\end{tabular} </cmath><br />
<br />
<math> \textbf{(A)}\ y=100-10x\qquad\textbf{(B)}\ y=100-5x^{2}\qquad\textbf{(C)}\ y=100-5x-5x^{2}\qquad\\ \textbf{(D)}\ y=20-x-x^{2}\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Of the following:<br />
<br />
(1) <math> a(x-y)=ax-ay </math><br />
<br />
(2) <math> a^{x-y}=a^x-a^y</math><br />
<br />
(3) <math> \log (x-y)=\log x-\log y</math><br />
<br />
(4) <math> \frac{\log x}{\log y}=\log{x}-\log{y}</math><br />
<br />
(5) <math> a(xy)=ax\times ay</math><br />
<br />
<math> \textbf{(A)}\ \text{Only 1 and 4 are true}\qquad\\ \textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\ \textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\ \textbf{(D)}\ \text{Only 1 and 2 are true}\qquad\\ \textbf{(E)}\ \text{Only 1 is true} </math><br />
<br />
[[1950 AHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
If <math>m</math> men can do a job in <math>d</math> days, then <math>m+r</math> men can do the job in:<br />
<br />
<math> \textbf{(A)}\ d+r\text{ days}\qquad\textbf{(B)}\ d-r\text{ days}\qquad\textbf{(C)}\ \frac{md}{m+r}\text{ days}\qquad\textbf{(D)}\ \frac{d}{m+r}\text{ days}\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
When <math>x^{13}+1</math> is divided by <math>x-1</math>, the remainder is:<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers} </math><br />
<br />
[[1950 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
The volume of a rectangular solid each of whose side, front, and bottom faces are <math> 12\text{ in}^{2} </math>, <math> 8\text{ in}^{2} </math>, and <math> 6\text{ in}^{2} </math> respectively is:<br />
<br />
<math> \textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
Successive discounts of <math>10\%</math> and <math>20\%</math> are equivalent to a single discount of:<br />
<br />
<math> \textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
A man buys a house for <math>10000</math> dollars and rents it. He puts <math> 12\frac{1}{2}\% </math> of each month's rent aside for repairs and upkeep; pays <math>325</math> dollars a year taxes and realizes <math> 5\frac{1}{2}\% </math> on his investment. The monthly rent is:<br />
<br />
<math> \textbf{(A)}\ 64.82\text{ dollars}\qquad\textbf{(B)}\ 83.33\text{ dollars}\qquad\textbf{(C)}\ 72.08\text{ dollars}\qquad\textbf{(D)}\ 45.83\text{ dollars}\qquad\textbf{(E)}\ 177.08\text{ dollars} </math><br />
<br />
[[1950 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
The equation <math> x+\sqrt{x-2}=4 </math> has:<br />
<br />
<math> \textbf{(A)}\ \text{2 real roots}\qquad\textbf{(B)}\ \text{1 real and 1 imaginary root}\qquad\textbf{(C)}\ \text{2 imaginary roots}\qquad\textbf{(D)}\ \text{No roots}\qquad\textbf{(E)}\ \text{1 real root} </math><br />
<br />
[[1950 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
The value of <math> \log_{5}\frac{(125)(625)}{25} </math> is equal to:<br />
<br />
<math> \textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 25|Solution]]<br />
<br />
== Problem 26 ==<br />
<br />
if <math> \log_{10}{m}= b-\log_{10}{n} </math>, then <math>m=</math><br />
<br />
<math> \textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n} </math><br />
<br />
[[1950 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27 ==<br />
<br />
A car travels <math>120</math> miles from <math>A</math> to <math>B</math> at <math>30</math> miles per hour but returns the same distance at <math>40</math> miles per hour. The average speed for the round trip is closest to:<br />
<br />
<math> \textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37\text{ mph} </math><br />
<br />
[[1950 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28 ==<br />
<br />
Two boys <math>A</math> and <math>B</math> start at the same time to ride from Port Jervis to Poughkeepsie, <math>60</math> miles away. <math>A</math> travels <math>4</math> miles an hour slower than <math>B</math>. <math>B</math> reaches Poughkeepsie and at once turns back meeting <math>A</math> <math>12</math> miles from Poughkeepsie. The rate of <math>A</math> was:<br />
<br />
<math> \textbf{(A)}\ 4\text{ mph}\qquad\textbf{(B)}\ 8\text{ mph}\qquad\textbf{(C)}\ 12\text{ mph}\qquad\textbf{(D)}\ 16\text{ mph}\qquad\textbf{(E)}\ 20\text{ mph} </math><br />
<br />
[[1950 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29 ==<br />
<br />
A manufacturer built a machine which will address <math>500</math> envelopes in <math>8</math> minutes. He wishes to build another machine so that when both are operating together they will address <math>500</math> envelopes in <math>2</math> minutes. The equation used to find how many minutes <math>x</math> it would require the second machine to address <math>500</math> envelopes alone is:<br />
<br />
<math> \textbf{(A)}\ 8-x=2\qquad\textbf{(B)}\ \frac{1}{8}+\frac{1}{x}=\frac{1}{2}\qquad\textbf{(C)}\ \frac{500}{8}+\frac{500}{x}=500\qquad\textbf{(D)}\ \frac{x}{2}+\frac{x}{8}=1\qquad\\ \textbf{(E)}\ \text{None of these answers} </math><br />
<br />
[[1950 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30 ==<br />
<br />
From a group of boys and girls, <math>15</math> girls leave. There are then left two boys for each girl. After this <math>45</math> boys leave. There are then <math>5</math> girls for each boy. The number of girls in the beginning was:<br />
<br />
<math> \textbf{(A)}\ 40\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 30|Solution]]<br />
<br />
== Problem 31 ==<br />
<br />
John ordered <math>4</math> pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by <math>50\%</math>. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:<br />
<br />
<math> \textbf{(A)}\ 4:1\qquad\textbf{(B)}\ 2:1\qquad\textbf{(C)}\ 1:4\qquad\textbf{(D)}\ 1:2\qquad\textbf{(E)}\ 1:8 </math><br />
<br />
[[1950 AHSME Problems/Problem 31|Solution]]<br />
<br />
== Problem 32 ==<br />
<br />
A <math>25</math> foot ladder is placed against a vertical wall of a building. The foot of the ladder is <math>7</math> feet from the base of the building. If the top of the ladder slips <math>4</math> feet, then the foot of the ladder will slide:<br />
<br />
<math> \textbf{(A)}\ 9\text{ ft}\qquad\textbf{(B)}\ 15\text{ ft}\qquad\textbf{(C)}\ 5\text{ ft}\qquad\textbf{(D)}\ 8\text{ ft}\qquad\textbf{(E)}\ 4\text{ ft} </math><br />
<br />
[[1950 AHSME Problems/Problem 32|Solution]]<br />
<br />
== Problem 33 ==<br />
<br />
The number of circular pipes with an inside diameter of <math>1</math> inch which will carry the same amount of water as a pipe with an inside diameter of <math>6</math> inches is:<br />
<br />
<math> \textbf{(A)}\ 6\pi\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 36\pi </math><br />
<br />
[[1950 AHSME Problems/Problem 33|Solution]]<br />
<br />
== Problem 34 ==<br />
<br />
When the circumference of a toy balloon is increased from <math>20</math> inches to <math>25</math> inches, the radius is increased by:<br />
<br />
<math> \textbf{(A)}\ 5\text{ in}\qquad\textbf{(B)}\ 2\frac{1}{2}\text{ in}\qquad\textbf{(C)}\ \frac{5}{\pi}\text{ in}\qquad\textbf{(D)}\ \frac{5}{2\pi}\text{ in}\qquad\textbf{(E)}\ \frac{\pi}{5}\text{ in} </math><br />
<br />
[[1950 AHSME Problems/Problem 34|Solution]]<br />
<br />
== Problem 35 ==<br />
<br />
In triangle <math>ABC</math>, <math>AC=24</math> inches, <math>BC=10</math> inches, <math>AB=26</math> inches. The radius of the inscribed circle is:<br />
<br />
<math> \textbf{(A)}\ 26\text{ in}\qquad\textbf{(B)}\ 4\text{ in}\qquad\textbf{(C)}\ 13\text{ in}\qquad\textbf{(D)}\ 8\text{ in}\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 35|Solution]]<br />
<br />
== Problem 36 ==<br />
<br />
A merchant buys goods at <math>25\%</math> off the list price. He desires to mark the goods so that he can give a discount of <math>20\%</math> on the marked price and still clear a profit of <math>25\%</math> on the selling price. What percent of the list price must he mark the goods?<br />
<br />
<math> \textbf{(A)}\ 125\%\qquad\textbf{(B)}\ 100\%\qquad\textbf{(C)}\ 120\%\qquad\textbf{(D)}\ 80\%\qquad\textbf{(E)}\ 75\% </math><br />
<br />
[[1950 AHSME Problems/Problem 36|Solution]]<br />
<br />
== Problem 37 ==<br />
<br />
If <math> y =\log_{a}{x} </math>, <math>a>1</math>, which of the following statements is incorrect?<br />
<br />
<math> \textbf{(A)}\ \text{If }x=1,y=0\qquad\\ \textbf{(B)}\ \text{If }x=a,y=1\qquad\\ \textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)}\qquad\\ \textbf{(D)}\ \text{If }0<x<a,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero}\qquad\\ \textbf{(E)}\ \text{Only some of the above statements are correct} </math><br />
<br />
[[1950 AHSME Problems/Problem 37|Solution]]<br />
<br />
== Problem 38 ==<br />
<br />
If the expression <math> \begin{pmatrix}a & c\\ d & b\end{pmatrix} </math> has the value <math>ab-cd</math> for all values of <math>a, b, c</math> and <math>d</math>, then the equation <math> \begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3 </math>:<br />
<br />
<math> \textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \textbf{(C)}\ \text{Is satisified for no values of }x\qquad\\ \textbf{(D)}\ \text{Is satisfied for an infinite number of values of }x\qquad\\ \textbf{(E)}\ \text{None of these.} </math><br />
<br />
[[1950 AHSME Problems/Problem 38|Solution]]<br />
<br />
== Problem 39 ==<br />
<br />
Given the series <math> 2+1+\frac{1}{2}+\frac{1}{4}+... </math> and the following five statements:<br />
(1) the sum increases without limit<br />
(2) the sum decreases without limit<br />
(3) the difference between any term of the sequence and zero can be made less than any positive quantity no matter how small<br />
(4) the difference between the sum and 4 can be made less than any positive quantity no matter how small<br />
(5) the sum approaches a limit<br />
Of these statements, the correct ones are:<br />
<br />
<math> \textbf{(A)}\ \text{Only }3\text{ and }4\qquad\textbf{(B)}\ \text{Only }5\qquad\textbf{(C)}\ \text{Only }2\text{ and }4\qquad\textbf{(D)}\ \text{Only }2,3\text{ and }4\qquad\textbf{(E)}\ \text{Only }4\text{ and }5 </math><br />
<br />
[[1950 AHSME Problems/Problem 39|Solution]]<br />
<br />
== Problem 40 ==<br />
<br />
The limit of <math> \frac{x^{2}-1}{x-1} </math> as <math>x</math> approaches <math>1</math> as a limit is:<br />
<br />
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ \text{Indeterminate}\qquad\textbf{(C)}\ x-1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 1 </math><br />
<br />
[[1950 AHSME Problems/Problem 40|Solution]]<br />
<br />
== Problem 41 ==<br />
<br />
The least value of the function <math>ax^2+bx+c</math> with <math>a>0</math> is:<br />
<br />
<math> \textbf{(A)}\ -\frac{b}{a}\qquad\textbf{(B)}\ -\frac{b}{2a}\qquad\textbf{(C)}\ b^{2}-4ac\qquad\textbf{(D)}\ \frac{4ac-b^{2}}{4a}\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 41|Solution]]<br />
<br />
== Problem 42 ==<br />
<br />
The equation <math> x^{x^{x^{.^{.^.}}}}=2 </math> is satisfied when <math>x</math> is equal to:<br />
<br />
<math> \textbf{(A)}\ \infty\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt[4]{2}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 42|Solution]]<br />
<br />
== Problem 43 ==<br />
<br />
The sum to infinity of <math> \frac{1}{7}+\frac{2}{7^{2}}+\frac{1}{7^{3}}+\frac{2}{7^{4}}+... </math> is:<br />
<br />
<math> \textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{1}{24}\qquad\textbf{(C)}\ \frac{5}{48}\qquad\textbf{(D)}\ \frac{1}{16}\qquad\textbf{(E)}\ \text{None of these} </math><br />
<br />
[[1950 AHSME Problems/Problem 43|Solution]]<br />
<br />
== Problem 44 ==<br />
<br />
The graph of <math>y=\log x</math><br />
<br />
<math> \textbf{(A)}\ \text{Cuts the }y\text{-axis}\qquad\\ \textbf{(B)}\ \text{Cuts all lines perpendicular to the }x\text{-axis}\qquad\\ \textbf{(C)}\ \text{Cuts the }x\text{-axis}\qquad\\ \textbf{(D)}\ \text{Cuts neither axis}\qquad\\ \textbf{(E)}\ \text{Cuts all circles whose center is at the origin} </math><br />
<br />
[[1950 AHSME Problems/Problem 44|Solution]]<br />
<br />
== Problem 45 ==<br />
<br />
The number of diagonals that can be drawn in a polygon of <math>100</math> sides is:<br />
<br />
<math> \textbf{(A)}\ 4850\qquad\textbf{(B)}\ 4950\qquad\textbf{(C)}\ 9900\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 8800 </math><br />
<br />
[[1950 AHSME Problems/Problem 45|Solution]]<br />
<br />
== Problem 46 ==<br />
<br />
In triangle <math>ABC</math>, <math>AB=12</math>, <math>AC=7</math>, and <math>BC=10</math>. If sides <math>AB</math> and <math>AC</math> are doubled while <math>BC</math> remains the same, then:<br />
<br />
<math> \textbf{(A)}\ \text{The area is doubled}\qquad\\ \textbf{(B)}\ \text{The altitude is doubled}\qquad\\ \textbf{(C)}\ \text{The area is four times the original area}\qquad\\ \textbf{(D)}\ \text{The median is unchanged}\qquad\\ \textbf{(E)}\ \text{The area of the triangle is 0} </math><br />
<br />
[[1950 AHSME Problems/Problem 46|Solution]]<br />
<br />
== Problem 47 ==<br />
<br />
A rectangle inscribed in a triangle has its base coinciding with the base <math>b</math> of the triangle. If the altitude of the triangle is <math>h</math>, and the altitude <math>x</math> of the rectangle is half the base of the rectangle, then:<br />
<br />
<math> \textbf{(A)}\ x=\frac{1}{2}h\qquad\textbf{(B)}\ x=\frac{bh}{b+h}\qquad\textbf{(C)}\ x=\frac{bh}{2h+b}\qquad\textbf{(D)}\ x=\sqrt{\frac{hb}{2}}\qquad\\ \textbf{(E)}\ x=\frac{1}{2}b </math><br />
<br />
[[1950 AHSME Problems/Problem 47|Solution]]<br />
<br />
== Problem 48 ==<br />
<br />
A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:<br />
<br />
<math> \textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle}\qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity} </math><br />
<br />
[[1950 AHSME Problems/Problem 48|Solution]]<br />
<br />
== Problem 49 ==<br />
<br />
A triangle has a fixed base <math>AB</math> that is <math>2</math> inches long. The median from <math>A</math> to side <math>BC</math> is <math>1 \frac{1}{2}</math> inches long and can have any position emanating from <math>A</math>. The locus of the vertex <math>C</math> of the triangle is:<br />
<br />
<math> \textbf{(A)}\ \text{A straight line }AB,1\frac{1}{2}\text{ inches from }A\qquad\\ \textbf{(B)}\ \text{A circle with }A\text{ as center and radius }2\text{ inches}\qquad\\ \textbf{(C)}\ \text{A circle with }A\text{ as center and radius }3\text{ inches}\qquad\\ \textbf{(D)}\ \text{A circle with radius }3\text{ inches and center }4\text{ inches from }B\text{ along }BA\qquad\\ \textbf{(E)}\ \text{An ellipse with }A\text{ as focus} </math><br />
<br />
[[1950 AHSME Problems/Problem 49|Solution]]<br />
<br />
== Problem 50 ==<br />
<br />
A privateer discovers a merchantman <math>10</math> miles to leeward at <math>11\text{:}45 \text{ a.m.}</math> and with a good breeze bears down upon her at <math>11</math> mph, while the merchantman can only make <math>8</math> mph in his attempt to escape. After a two hour chase, the top sail of the privateer is carried away; she can now make only <math>17</math> miles while the merchantman makes <math>15</math>. The privateer will overtake the merchantman at:<br />
<br />
<math> \textbf{(A)}\ 3\text{:}45\text{ p.m.}\qquad\textbf{(B)}\ 3\text{:}30\text{ p.m.}\qquad\textbf{(C)}\ 5\text{:}00\text{ p.m.}\qquad\textbf{(D)}\ 2\text{:}45\text{ p.m.}\qquad\textbf{(E)}\ 5\text{:}30\text{ p.m.} </math><br />
<br />
[[1950 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=1950|before=First AHSME|after=[[1951 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_13&diff=821381980 AHSME Problems/Problem 132017-01-02T22:15:14Z<p>Wirecat: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
A bug (of negligible size) starts at the origin on the coordinate plane. First, it moves one unit right to <math>(1,0)</math>. Then it makes a <math>90^\circ</math> counterclockwise and travels <math>\frac 12</math> a unit to <math>\left(1, \frac 12 \right)</math>. If it continues in this fashion, each time making a <math>90^\circ</math> degree turn counterclockwise and traveling half as far as the previous move, to which of the following points will it come closest? <br />
<br />
<math>\text{(A)} \ \left(\frac 23, \frac 23 \right) \qquad \text{(B)} \ \left( \frac 45, \frac 25 \right) \qquad \text{(C)} \ \left( \frac 23, \frac 45 \right) \qquad \text{(D)} \ \left(\frac 23, \frac 13 \right) \qquad \text{(E)} \ \left(\frac 25, \frac 45 \right)</math><br />
<br />
<br />
== Solution ==<br />
Writing out the change in <math>x</math> coordinates and then in <math>y</math> coordinates gives the infinite sum <math>1-\frac{1}{4}+\frac{1}{16}-\dots</math> and <math>\frac{1}{2}-\frac{1}{8}+\dots</math> respectively. Using the infinite geometric sum formula, we have <math>\frac{1}{1+\frac{1}{4}}=\frac{4}{5}</math> and <math>\frac{\frac{1}{2}}{1+\frac{1}{4}}=\frac{2}{5}</math>, thus the answer is <math>\left( \frac{4}{5}, \frac{2}{5} \right)</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1980|num-b=12|num-a=14}} <br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_9&diff=821351980 AHSME Problems/Problem 92017-01-02T21:52:44Z<p>Wirecat: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A man walks <math>x</math> miles due west, turns <math>150^\circ</math> to his left and walks 3 miles in the new direction. If he finishes a a point <math>\sqrt{3}</math> from his starting point, then <math>x</math> is<br />
<br />
<math>\text{(A)} \ \sqrt 3 \qquad \text{(B)} \ 2\sqrt{5} \qquad \text{(C)} \ \frac 32 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}</math><br />
<br />
== Solution ==<br />
Let us think about this. We only know that he ends up <math>\sqrt{3}</math> away from the origin. However, think about the locus of points <math>\sqrt{3}</math> away from the origin, a circle. However, his path could end on any part of the circle below the <math>x-</math>axis, so therefore, the answer is<br />
<math>\fbox{E: not uniquely determined}.</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1980|num-b=8|num-a=10}} <br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_8&diff=820782009 AMC 10A Problems/Problem 82016-12-30T20:23:58Z<p>Wirecat: /* Solution */</p>
<hr />
<div>== Problem 8 ==<br />
Three Generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a <math>50</math>% discount as children. The two members of the oldest generation receive a <math>25\%</math> discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs <dollar/><math>6.00</math>, is paying for everyone. How many dollars must he pay?<br />
<br />
<math><br />
\mathrm{(A)}\ 34<br />
\qquad<br />
\mathrm{(B)}\ 36<br />
\qquad<br />
\mathrm{(C)}\ 42<br />
\qquad<br />
\mathrm{(D)}\ 46<br />
\qquad<br />
\mathrm{(E)}\ 48<br />
</math><br />
<br />
== Solution ==<br />
<br />
A senior ticket costs <math>\$6.00</math>, so a regular ticket costs <math>6 \cdot \frac{1}{\frac{3}{4}}\:=\:6\cdot\frac{4}{3}\:=\:8</math> dollars. Therefore children's tickets cost half that, or <math>\$4.00</math>, so we have:<br />
<br />
<math>2(6+8+4)\:=\:36</math><br />
<br />
So Grandfather Wen pays <math>\$36</math>, or <math>B</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2009|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_8&diff=820772009 AMC 10A Problems/Problem 82016-12-30T20:23:07Z<p>Wirecat: /* Solution */</p>
<hr />
<div>== Problem 8 ==<br />
Three Generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a <math>50</math>% discount as children. The two members of the oldest generation receive a <math>25\%</math> discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs <dollar/><math>6.00</math>, is paying for everyone. How many dollars must he pay?<br />
<br />
<math><br />
\mathrm{(A)}\ 34<br />
\qquad<br />
\mathrm{(B)}\ 36<br />
\qquad<br />
\mathrm{(C)}\ 42<br />
\qquad<br />
\mathrm{(D)}\ 46<br />
\qquad<br />
\mathrm{(E)}\ 48<br />
</math><br />
<br />
== Solution ==<br />
<br />
A senior ticket costs <dollar/><math>6.00</math>, so a regular ticket costs <math>6 \cdot \frac{1}{\frac{3}{4}}\:=\:6\cdot\frac{4}{3}\:=\:8</math> dollars. Therefore children's tickets cost half that, or <dollar/><math>4.00</math>, so we have:<br />
<br />
<math>2(6+8+4)\:=\:36</math><br />
<br />
So Grandfather Wen pays <math>\$36</math>, or <math>B</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2009|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_2&diff=817191989 AIME Problems/Problem 22016-11-29T04:23:40Z<p>Wirecat: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Ten [[point]]s are marked on a [[circle]]. How many distinct [[convex polygon]]s of three or more sides can be drawn using some (or all) of the ten points as [[vertex | vertices]]?<br />
<br />
== Solution ==<br />
Any [[subset]] of the ten points with three or more members can be made into exactly one such polygon. Thus, we need to count the number of such subsets. There are <math>2^{10} = 1024</math> total subsets of a ten-member [[set]], but of these <math>{10 \choose 0} = 1</math> have 0 members, <math>{10 \choose 1} = 10</math> have 1 member and <math>{10 \choose 2} = 45</math> have 2 members. Thus the answer is <math>1024 - 1 - 10 - 45 = 968</math>.<br />
<br />
Note <math>{N \choose 0}+{N \choose 1} + {N \choose 2} + \dots + {N \choose N}</math> is equivalent to <math>2^N</math><br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_18&diff=815362016 AMC 8 Problems/Problem 182016-11-23T22:29:05Z<p>Wirecat: /* Solution */</p>
<hr />
<div>In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?<br />
<br />
<math>\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72</math><br />
==Solution==<br />
From any <math>n-</math>th race, only <math>\frac{1}{6}</math> will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.<br />
Starting with the first race:<br />
<cmath>\frac{216}{6}=36</cmath><br />
<cmath>\frac{36}{6}=6</cmath><br />
<cmath>\frac{6}{6}=1</cmath><br />
Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math><br />
==Solution 2==<br />
Every race eliminates <math>5</math> players. The winner is decided when there is only <math>1</math> runner left. Thus, <math>215</math> players have to be eliminated. Therefore, we need <math>\frac{215}{5}</math> games to decide the winner, or <math>43 \rightarrow \boxed{C}</math><br />
{{AMC8 box|year=2016|num-b=17|num-a=19}}<br />
{{MAA Notice}}\end{align}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_13&diff=814592016 AMC 8 Problems/Problem 132016-11-23T16:40:12Z<p>Wirecat: /* Solution */</p>
<hr />
<div>Two different numbers are randomly selected from the set <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>?<br />
<br />
<math>\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}</math><br />
<br />
==Solution==<br />
The product can only be <math>0</math> if one of the numbers is 0. Once we chose <math>0</math>, there are <math>5</math> ways we can chose the second number, or <math>6-1</math>. There are <math>\dbinom{6}{2}</math> ways we can chose <math>2</math> numbers randomly, and that is <math>15</math>. So, <math>\frac{5}{15}=\frac{1}{3}=\textbf{(D)}</math><br />
{{AMC8 box|year=2016|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_17&diff=814512016 AMC 8 Problems/Problem 172016-11-23T16:36:45Z<p>Wirecat: /* Solution */</p>
<hr />
<div>17. An ATM password at Fred's Bank is composed of four digits from <math>0</math> to <math>9</math>, with repeated digits allowable. If no password may begin with the sequence <math>9,1,1,</math> then how many passwords are possible?<br />
<math>(A)\mbox{ }30\mbox{ }(B)\mbox{ }7290\mbox{ }(C)\mbox{ }9000\mbox{ }(D)\mbox{ }9990\mbox{ }(E)\mbox{ }9999\mbox{ }</math><br />
==Solution==<br />
For the first three digits, there are <math>10^3-1=999</math> combinations since <math>911</math> is not allowed. For the final digit, any of the <math>10</math> numbers are allowed. <math>999 \cdot 10 = 9990 \rightarrow \boxed{D}</math><br />
{{AMC8 box|year=2016|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_17&diff=814492016 AMC 8 Problems/Problem 172016-11-23T16:36:23Z<p>Wirecat: </p>
<hr />
<div>17. An ATM password at Fred's Bank is composed of four digits from <math>0</math> to <math>9</math>, with repeated digits allowable. If no password may begin with the sequence <math>9,1,1,</math> then how many passwords are possible?<br />
<math>(A)\mbox{ }30\mbox{ }(B)\mbox{ }7290\mbox{ }(C)\mbox{ }9000\mbox{ }(D)\mbox{ }9990\mbox{ }(E)\mbox{ }9999\mbox{ }</math><br />
==Solution==<br />
For the first three digits, there are <math>10^3-1=999</math> combinations since <math>911</math> is not allowed. For the finaly digit, any of the <math>10</math> numbers are allowed. <math>999 \cdot 10 = 9990 \rightarrow \boxed{D}</math><br />
{{AMC8 box|year=2016|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_18&diff=814422016 AMC 8 Problems/Problem 182016-11-23T16:33:23Z<p>Wirecat: /* Solution */</p>
<hr />
<div>18. In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?<br />
<br />
<br />
<math>(A)\mbox{ }36\mbox{ }(B)\mbox{ }42\mbox{ }(C)\mbox{ }43\mbox{ }(D)\mbox{ }60\mbox{ }(E)\mbox{ }72\mbox{ }</math><br />
==Solution==<br />
From any nth race, only <math>\frac{1}{6}</math> will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.<br />
Starting with the first race:<br />
<cmath>\frac{216}{6}=36</cmath><br />
<cmath>\frac{36}{6}=6</cmath><br />
<cmath>\frac{6}{6}=1</cmath><br />
Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math><br />
{{AMC8 box|year=2016|num-b=17|num-a=19}}<br />
{{MAA Notice}}\end{align}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_18&diff=814392016 AMC 8 Problems/Problem 182016-11-23T16:21:30Z<p>Wirecat: </p>
<hr />
<div>18. In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?<br />
<br />
<br />
<math>(A)\mbox{ }36\mbox{ }(B)\mbox{ }42\mbox{ }(C)\mbox{ }43\mbox{ }(D)\mbox{ }60\mbox{ }(E)\mbox{ }72\mbox{ }</math><br />
==Solution==<br />
From any nth race, only <math>\frac{1}{6}</math> will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.<br />
Starting with the first race:<br />
<math><br />
\frac{216}{6}=36 \\<br />
\frac{36}{6}=6 \\<br />
\frac{6}{6}=1</math><br />
Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math><br />
{{AMC8 box|year=2016|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_19&diff=814362016 AMC 8 Problems/Problem 192016-11-23T16:15:56Z<p>Wirecat: </p>
<hr />
<div>19. The sum of <math>25</math> consecutive even integers is <math>10,000</math>. What is the largest of these <math>25</math> consecutive integers?<br />
<br />
<math>(A)\mbox{ }360\mbox{ }(B)\mbox{ }388\mbox{ }(C)\mbox{ }412\mbox{ }(D)\mbox{ }416\mbox{ }(E)\mbox{ }424\mbox{ }</math><br />
<br />
==Solution==<br />
Let <math>n</math> be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simply by <math>25n</math> since <math>(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n</math>. Now, <math>25n=10000 \rightarrow n=400</math> Remembering that this is the 13th integer, we wish to find the 25th, which is <math>400+2 \cdot (15-13)=424</math>.<br />
{{AMC8 box|year=2016|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_19&diff=814352016 AMC 8 Problems/Problem 192016-11-23T16:15:27Z<p>Wirecat: </p>
<hr />
<div>19. The sum of <math>25</math> consecutive even integers is <math>10,000</math>. What is the largest of these <math>25</math> consecutive integers?<br />
<br />
<math>(A)\mbox{ }360\mbox{ }(B)\mbox{ }388\mbox{ }(C)\mbox{ }412\mbox{ }(D)\mbox{ }416\mbox{ }(E)\mbox{ }424\mbox{ }</math><br />
<br />
==Solution==<br />
Let <math>n</math> be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simply by <math>25n</math> since <math>(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n</math>. Now, <math>25n=10000 \rightarrow n=400</math> Remembering that this is the 13th integer, we wish to find the 25th, which is <math>400+2 \cdot (15-13)=424</math>.</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_20&diff=814312016 AMC 8 Problems/Problem 202016-11-23T16:07:39Z<p>Wirecat: Added solution</p>
<hr />
<div>The least common multiple of <math>a</math> and <math>b</math> is <math>12</math>, and the least common multiple of <math>b</math> and <math>c</math> is <math>15</math>. What is the least possible value of the least common multiple of <math>a</math> and <math>c</math>?<br />
<br />
<math>\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180</math><br />
<br />
==Solution==<br />
We wish to find possible values of <math>a</math>,<math>b</math>, and <math>c</math>. By finding the greatest common factor of <math>12</math> and <math>15</math>, algebraically, it's some multiple of <math>b</math> and from looking at the numbers, we are sure that it is 3, thus b is 3. Moving on to <math>a</math> and <math>c</math>, in order to minimize them, we wish to find the least such that the least common multiple of <math>a</math> and <math>3</math> is <math>12</math>, <math>\rightarrow 4</math>. Similarly with <math>3</math> and <math>c</math>, we obtain <math>5</math>. The least common multiple of <math>4</math> and <math>5</math> is <math>15 \rightarrow \boxed{A}</math><br />
{{AMC8 box|year=2016|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Wirecathttps://artofproblemsolving.com/wiki/index.php?title=Integral&diff=79355Integral2016-07-15T20:12:45Z<p>Wirecat: /* Definite Integral */</p>
<hr />
<div>The '''integral''' is one of the two base concepts of [[calculus]], along with the [[derivative]].<br />
<br />
==Beginner Level==<br />
In introductory, high-school level texts, the integral is often presented in two parts, the '''indefinite integral''' and '''definite integral'''. Although this approach lacks mathematical formality, it has the advantage of being easy to grasp and convenient to use in most of its applications, especially in [[Physics]].<br />
<br />
===Indefinite Integral===<br />
The indefinite integral, or antiderivative, is a partial [[inverse]] of the [[derivative]]. That is, if the derivative of a [[function ]]<math>f(x)</math> is written as <math>f'(x)</math>, then the indefinite integral of <math>f'(x)</math> is <math>f(x)+c</math>, where <math>c</math> is a [[real]] [[constant]]. This is because the derivative of a constant is <math>0</math>.<br />
====Notation====<br />
*The integral of a function <math>f(x)</math> is written as <math>\int f(x)\,dx</math>, where the <math>dx</math> means that the function is being integrated in relation to <math>x</math>.<br />
*Often, to save space, the integral of <math>f(x)</math> is written as <math>F(x)</math>, the integral of <math>h(x)</math> as <math>H(x)</math>, etc.<br />
====Rules of Indefinite Integrals====<br />
*<math>\int c\,dx=cx+C</math> for a constant <math>c</math> and another constant <math>C</math>.<br />
*<math>\int f(x)+g(x)...+z(x)\,dx=\int f(x)\,dx+\int g(x)\,dx...+\int z(x)\,dx</math><br />
*<math>\int x^n\,dx=\frac{1}{n+1}x^{n+1}+C</math>, <math>n \ne -1</math><br />
*<math>\int x^{-1}\,dx=\ln |x|+C</math><br />
*<math>\int \sin x\,dx = -\cos x + C</math><br />
*<math>\int \cos x\,dx = \sin x + C</math><br />
*<math>\int\tan x\,dx = \ln |\cos x| + C</math><br />
*<math>\int \sec x\,dx = \ln |\sec x + \tan x| + C</math><br />
*<math>\int \csc \, dx =\ln |\csc x + \cot x| + C</math><br />
*<math>\int \cot x\,dx = \ln |\sin x| + C</math><br />
*<math>\int cf(x)\, dx=c\int f(x)\,dx</math><br />
<br />
===Definite Integral===<br />
The definite integral is also the [[area]] under a [[curve]] between two [[points]] <math>a</math> and <math>b</math>. For example, the area under the curve <math>f(x)=\sin x</math> between <math>-\frac{\pi}{2}</math> and <math>\frac{\pi}{2}</math> is <math>0</math>, as area below the x-axis is taken as negative area. <br />
====Definition and Notation====<br />
*The definite integral of a function between <math>a</math> and <math>b</math> is written as <math>\int^{b}_{a}f(x)\,dx</math>. <br />
*<math>\int^{b}_{a}f(x)\,dx=F(b)-F(a)</math>, where <math>F(x)</math> is the antiderivative of <math>f(x)</math>. This is also notated <math>\int f(x)\,dx \eval^{b}_{a}</math>, read as "The integral of <math>f(x)</math> evaluated at <math>a</math> and <math>b</math>." Note that this means in definite integration, one need not add a constant, as the constants from the functions cancel out.<br />
====Rules of Definite Integrals====<br />
*<math>\int^{b}_{a}f(x)\,dx=\int^{b}_{c}f(x)\,dx+\int^{c}_{a}f(x)\,dx</math> for any <math>c</math>.<br />
<br />
==Formal Use==<br />
The notion of an '''integral''' is one of the key ideas in severel areas of higher mathematics including [[analysis]] and [[topology]]. The integral can be defined in several ways which can be applied to several different settings. However, the most common definition, and the one which most closely resembles the the 'definite integral' is the '''Riemann Integral'''<br />
<br />
===Riemann Integral===<br />
Let <math>f:[a,b]\rightarrow\mathbb{R}</math><br />
<br />
Let <math>L\in\mathbb{R}</math><br />
<br />
We say that <math>f</math> is '''Riemann Integrable''' on <math>[a,b]</math> if and only if<br />
<br />
<math>\forall \epsilon>0\:\exists\delta>0</math> such that if <math>\mathcal{\dot{P}}</math> is a [[Partition|tagged partition]] on <math>[a,b]</math> with <math>\|\mathcal{\dot{P}}\|<\delta</math> <math>\implies</math> <math>|S(f,\mathcal{\dot{P}})-L|<\epsilon</math>, where <math>S(f,\mathcal{\dot{P}})</math> is the [[Riemann sum]] of <math>f</math> with respect to <math>\mathcal{\dot{P}}</math><br />
<br />
<math>L</math> is said to be the '''integral''' of <math>f</math> on <math>[a,b]</math> and is written as <math>L=\int_a^b f(x)dx=\int_a^b f</math><br />
<br />
<geogebra>2f3876024e3b8d9e4506f2173c591cbfaca665de</geogebra><br />
<br />
Another integral commonly used in introductory texts is the '''Darboux Integral''' (which is often called the Riemann Integral)<br />
<br />
===Darboux Integral===<br />
Let <math>f:[a,b]\rightarrow\mathbb{R}</math><br />
<br />
We say that <math>f</math> is '''Darboux Integrable''' on <math>[a,b]</math> if and only if <math>\inf_{\mathcal{P}}U(f,\mathcal{P})=\sup_{\mathcal{P}}L(f,\mathcal{P})</math>, where <math>L(f,\mathcal{P})</math> and <math>U(f,\mathcal{P})</math> are respectively the [[Riemann sum|lower sum]] and [[Riemann sum|upper sum]] of <math>f</math> with respect to [[partition]] <math>\mathcal{P}</math><br />
<br />
The notation used for the Darboux integral is the same as that for the Riemann integral. <br />
{{image}}<br />
<br />
===Other Definitions===<br />
Other important definitions of integration include the [[Riemann-Stieltjes integral]], [[Lebesgue integral]], [[Henstock-Kurzweil integral]] etc.<br />
<br />
==Disambiguation==<br />
*The word ''integral'' is the adjectival form of the noun "[[integer]]." Thus, <math>3</math> is integral while <math>\pi</math> is not.<br />
<br />
==See also==<br />
*[[Calculus]]<br />
*[[Derivative]]<br />
*[[Function]]<br />
*[[Measure theory]]<br />
<br />
[[Category:Calculus]]<br />
[[Category:Definition]]</div>Wirecat