https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Therealchocolatelover&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T10:03:20ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_15&diff=1143532018 AMC 12B Problems/Problem 152020-01-06T05:33:19Z<p>Therealchocolatelover: /* Problem */</p>
<hr />
<div>== Problem ==<br />
How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?<br />
<br />
<math>\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 </math><br />
<br />
== Solution 1 (For Dummies) ==<br />
Analyze that the three-digit integers divisible by <math>3</math> start from <math>102</math>. In the <math>200</math>'s, it starts from <math>201</math>. In the <math>300</math>'s, it starts from <math>300</math>. We see that the units digits is <math>0, 1, </math> and <math>2.</math><br />
<br />
Write out the 1- and 2-digit multiples of <math>3</math> starting from <math>0, 1,</math> and <math>2.</math> Count up the ones that meet the conditions. Then, add up and multiply by <math>3</math>, since there are three sets of three from <math>1</math> to <math>9.</math> Then, subtract the amount that started from <math>0</math>, since the <math>300</math>'s ll contain the digit <math>3</math>.<br />
<br />
We get: <cmath>3(12+12+12)-12.</cmath><br />
<br />
This gives us: <cmath>\boxed{\textbf{(A) } 96}.</cmath><br />
<br />
== Solution 2==<br />
<br />
There are <math>4</math> choices for the last digit (<math>1, 5, 7, 9</math>), and <math>8</math> choices for the first digit (exclude <math>0</math>). We know what the second digit mod <math>3</math> is, so there are <math>3</math> choices for it (pick from one of the sets <math>\{0, 6, 9\},\{1, 4, 7\}, \{2, 5, 8\}</math>). The answer is <math>4\cdot 8 \cdot 3 = \boxed{96}</math> (Plasma_Vortex)<br />
<br />
== Solution 3==<br />
<br />
Consider the number of <math>2</math>-digit numbers that do not contain the digit <math>3</math>, which is <math>90-18=72</math>. For any of these <math>2</math>-digit numbers, we can append <math>1,5,7,</math> or <math>9</math> to reach a desirable <math>3</math>-digit number. However, <math>1 \equiv 7 \equiv 1</math> <math>(mod</math> <math>3)</math>, and thus we need to count any <math>2</math>-digit number <math>\equiv 2</math> <math>(mod</math> <math>3)</math> twice. There are <math>(98-11)/3+1=30</math> total such numbers that have remainder <math>2</math>, but <math>6</math> of them <math>(23,32,35,38,53,83)</math> contain <math>3</math>, so the number we want is <math>30-6=24</math>. Therefore, the final answer is <math>72+24= \boxed{96}</math>.<br />
<br />
==Solution 4==<br />
Note that this isn't a great solution, but a more practical one to achieve the answer.<br />
<br />
Notice that there are <math>300</math> numbers that have <math>3</math> digits and are divisible by <math>3</math> (from <math>102</math> to <math>999</math>). Now one by one apply the restrictions. <br />
<br />
The restriction for only odd numbers would mean that half the numbers are taken out <math>\Rightarrow 300*\frac{1}{2} = 150</math>. <br />
<br />
Next, apply the restriction of no <math>3</math>s. For the units digit, that would mean multiplying by <math>\frac{4}{5}</math> (remember that now you only have odd numbers to choose from). <br />
<br />
For the tens that would mean multiplying by <math>\frac{9}{10}</math>, and for the hundreds that would mean multiplying by <math>\frac{8}{9}</math> (because you cant have 0 here). <br />
<br />
Thus, we get <math>150*\frac{4}{5}*\frac{9}{10}*\frac{8}{9}=96</math>, which is <math>\boxed{A}</math>.<br />
<br />
Sol by IronicNinja~<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2018|ab=B|num-b=14|num-a=16}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
[[Category:Introductory Number Theory Problems]]</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_24&diff=1142402011 AMC 10B Problems/Problem 242020-01-04T03:42:55Z<p>Therealchocolatelover: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
A lattice point in an <math>xy</math>-coordinate system is any point <math>(x, y)</math> where both <math>x</math> and <math>y</math> are integers. The graph of <math>y = mx +2</math> passes through no lattice point with <math>0 < x \le 100</math> for all <math>m</math> such that <math>\frac{1}{2} < m < a</math>. What is the maximum possible value of <math>a</math>?<br />
<br />
<math> \textbf{(A)}\ \frac{51}{101} \qquad\textbf{(B)}\ \frac{50}{99} \qquad\textbf{(C)}\ \frac{51}{100} \qquad\textbf{(D)}\ \frac{52}{101} \qquad\textbf{(E)}\ \frac{13}{25}</math><br />
<br />
== Solution 1==<br />
For <math>y=mx+2</math> to not pass through any lattice points with <math>0<x\leq 100</math> is the same as saying that <math>mx\notin\mathbb Z</math> for <math>x\in\{1,2,\dots,100\}</math>, or in other words, <math>m</math> is not expressible as a ratio of positive integers <math>s/t</math> with <math>t\leq 100</math>. Hence the maximum possible value of <math>a</math> is the first real number after <math>1/2</math> that is so expressible.<br />
<br />
For each <math>d=2,\dots,100</math>, the smallest multiple of <math>1/d</math> which exceeds <math>1/2</math> is <math>1,\frac23,\frac34,\frac35,\dots,\frac{50}{98},\frac{50}{99},\frac{51}{100}</math> respectively, and the smallest of these is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.<br />
<br />
==Solution 2==<br />
We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points, the denominator of <math>m</math> must be greater than <math>100</math>, or else it would be canceled by some <math>0<x\le100</math> which would make <math>y</math> an integer. By using common denominators, we find that the order of the fractions from smallest to largest is <math>\text{(A), (B), (C), (D), (E)}</math>. We can see that when <math>m=\frac{50}{99}</math>, <math>y</math> would be an integer, so therefore any fraction greater than <math>\frac{50}{99}</math> would not work, as substituting our fraction <math>\frac{50}{99}</math> for <math>m</math> would produce an integer for <math>y</math>. So now we are left with only <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>. But since <math>\frac{51}{101}=\frac{5049}{9999}</math> and <math>\frac{50}{99}=\frac{5050}{9999}</math>, we can be absolutely certain that there isn't a number between <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math> that can reduce to a fraction whose denominator is less than or equal to <math>100</math>. Since we are looking for the maximum value of <math>a</math>, we take the larger of <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>, which is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.<br />
<br />
==Solution 3==<br />
We want to find the smallest <math>m</math> such that there will be an integral solution to <math>y=mx+2</math> with <math>0<x\le100</math>. We first test A, but since the denominator has a <math>101</math>, <math>x</math> must be a nonzero multiple of <math>101</math>, but it then will be greater than <math>100</math>. We then test B. <math>y=\frac{50}{99}x+2</math> yields the solution <math>(99,52)</math> which satisfies <math>0<x\le100</math>. Checking the answer choices, we know that the smallest possible <math>a</math> must be <math>\frac{50}{99}\implies\boxed{\textbf{(B)}}</math><br />
<br />
<br />
== Solution 4 ==<br />
<br />
Notice that for <math>y=\frac{1}{2}x+2=\frac{50}{100}x+2</math>, <math>x=99</math> is one of the integral values of <math>x</math> such that the value of <math>\frac{50}{100}x</math> is the closest to its next integral value.<br />
<br />
<br />
Thus the maximum value for <math>a</math> is the value of <math>m</math> when the equation <math>y=99m+2</math> goes through its next lattice point, which occurs when <math>m=\frac{b}{99}</math> for some positive integer <math>b</math>.<br />
<br />
<br />
Finding the common denominator, we have <cmath>\frac{50}{100}=\frac{4950}{9900}, \frac{b}{99}=\frac{100b}{9900}</cmath> Since <math>a>\frac{1}{2}</math>, the smallest value for <math>b</math> such that <math>100b>4950</math> is <math>b=50</math>. <br />
<br />
Thus the maximum value of <math>a</math> is <math>\frac{50}{99}.\boxed{\mathrm{(B)}}</math><br />
<br />
~ Nafer<br />
<br />
==See Also==<br />
{{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}}<br />
{{MAA Notice}}</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_22&diff=1104912012 AMC 10B Problems/Problem 222019-10-21T04:40:32Z<p>Therealchocolatelover: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Let (<math>a_1</math>, <math>a_2</math>, ... <math>a_{10}</math>) be a list of the first 10 positive integers such that for each <math>2\le</math> <math>i</math> <math>\le10</math> either <math>a_i + 1</math> or <math>a_i-1</math> or both appear somewhere before <math>a_i</math> in the list. How many such lists are there?<br />
<br />
<br />
<math>\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880</math><br />
<br />
==Solution 1==<br />
If we have 1 as the first number, then the only possible list is <math>(1,2,3,4,5,6,7,8,9,10)</math>. <br />
<br />
If we have 2 as the first number, then we have 9 ways to choose where the <math>1</math> goes, and the numbers ascend from the first number, <math>2</math>, with the exception of the <math>1</math>.<br />
For example, <math>(2,3,1,4,5,6,7,8,9,10)</math>, or <math>(2,3,4,1,5,6,7,8,9,10)</math>. There are <math>\dbinom{9}{1}</math> ways to do so.<br />
<br />
If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are <math>\dbinom{9}{2}</math> ways to do this.<br />
<br />
In the same way, the total number of lists is:<br />
<math>\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}</math><br />
<br />
By the binomial theorem, this is <math>2^{9}</math> = <math>512</math>, or <math>\boxed{\textbf{(B)}}</math><br />
<br />
==Solution 2==<br />
Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is <math>(1,2,3,4,5...10)</math>. There are 9 arrows, so the answer is <math>2^{9}</math> = <math>512</math> <math>\boxed{\textbf{(B)}}</math><br />
<br />
NOTE:<br />
Solution cited from: http://www.artofproblemsolving.com/Videos/external.php?video_id=269.<br />
<br />
==Solution 3==<br />
Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, <math>(1, 2)</math> and <math>(2, 1)</math>. If we replace it with 3, there are four lists, <math>(1, 2, 3), (2, 1, 3), (2, 3, 1),</math> and <math>(3, 2, 1)</math>. Since 2 and 4 are both powers of 2, it is likely that the number of lists is <math>2^{n-1}</math>, where <math>n</math> is the length of the lists. <math>2^{10-1}=512=\boxed{\textbf{(B)}}</math><br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=User:Therealchocolatelover&diff=109628User:Therealchocolatelover2019-09-02T22:09:04Z<p>Therealchocolatelover: Blanked the page</p>
<hr />
<div></div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=User:Therealchocolatelover&diff=109627User:Therealchocolatelover2019-09-02T22:08:50Z<p>Therealchocolatelover: </p>
<hr />
<div>[[therealchocolatelover]]</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=User:Therealchocolatelover&diff=109626User:Therealchocolatelover2019-09-02T22:08:32Z<p>Therealchocolatelover: </p>
<hr />
<div>[[hi]]</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_25&diff=1096252016 AMC 10B Problems/Problem 252019-09-02T22:06:38Z<p>Therealchocolatelover: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)</math>, where <math>\lfloor r \rfloor</math> denotes the greatest integer less than or equal to <math>r</math>. How many distinct values does <math>f(x)</math> assume for <math>x \ge 0</math>?<br />
<br />
<math>\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}</math><br />
<br />
==Solution==<br />
<br />
Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have <br />
<br />
<cmath>f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath><br />
<br />
The function can then be simplified into <br />
<br />
<cmath>f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath><br />
<br />
which becomes<br />
<br />
<cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor</cmath><br />
<br />
We can see that for each value of <math>k</math>, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from <math>0</math> to <math>k-1</math>. <br />
<br />
Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when <math>x</math> is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>.<br />
<br />
So we want to count how many distinct fractions less than <math>1</math> have the form <math>\frac{m}{n}</math> where <math>n \le 10</math>. We can find this easily by computing<br />
<br />
<cmath>\sum_{k=2}^{10} \phi(k)</cmath><br />
<br />
where <math>\phi(k)</math> is the [[Euler Totient Function]]. Basically <math>\phi(k)</math> counts the number of fractions with <math>k</math> as its denominator (after simplification). This comes out to be <math>31</math>.<br />
<br />
Because the value of <math>f(x)</math> is at least 0 and can increase 31 times, there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=User:Therealchocolatelover&diff=109624User:Therealchocolatelover2019-09-02T22:05:32Z<p>Therealchocolatelover: Blanked the page</p>
<hr />
<div></div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=User:Therealchocolatelover&diff=109623User:Therealchocolatelover2019-09-02T22:05:23Z<p>Therealchocolatelover: </p>
<hr />
<div>ok<br />
<br />
[url=http://www.google.com]hi[/url]</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=User:Therealchocolatelover&diff=109622User:Therealchocolatelover2019-09-02T22:05:11Z<p>Therealchocolatelover: Created page with "ok [url=google.com]hi[/url]"</p>
<hr />
<div>ok<br />
<br />
[url=google.com]hi[/url]</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_25&diff=1096212016 AMC 10B Problems/Problem 252019-09-02T21:58:26Z<p>Therealchocolatelover: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)</math>, where <math>\lfloor r \rfloor</math> denotes the greatest integer less than or equal to <math>r</math>. How many distinct values does <math>f(x)</math> assume for <math>x \ge 0</math>?<br />
<br />
<math>\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}</math><br />
<br />
==Solution==<br />
<br />
Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have <br />
<br />
<cmath>f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath><br />
<br />
The function can then be simplified into <br />
<br />
<cmath>f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath><br />
<br />
which becomes<br />
<br />
<cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor</cmath><br />
<br />
We can see that for each value of <math>k</math>, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from <math>0</math> to <math>k-1</math>. <br />
<br />
Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when <math>x</math> is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>.<br />
<br />
So we want to count how many distinct fractions less than <math>1</math> have the form <math>\frac{m}{n}</math> where <math>n \le 10</math>. We can find this easily by computing<br />
<br />
<cmath>\sum_{k=2}^{10} \phi(k)</cmath><br />
<br />
where <math>\phi(k)</math> is the Euler Totient Function. Basically <math>\phi(k)</math> counts the number of fractions with <math>k</math> as its denominator (after simplification). This comes out to be <math>31</math>.<br />
<br />
Because the value of <math>f(x)</math> is at least 0 and can increase 31 times, there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_23&diff=1011822005 AMC 10A Problems/Problem 232019-02-03T17:53:37Z<p>Therealchocolatelover: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?<br />
<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br />
draw(A--B--D--cycle);<br />
draw(D--E--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",E,SSE);<br />
label("$B$",B,E);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math><br />
<br />
==Solution 1==<br />
[[File:Circlenc1.png]]<br />
<br />
Let us assume that the diameter is of length <math>1</math>. <br />
<br />
<math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math>. <br />
<br />
<math>OD</math> is the radius of the circle, so using the Pythagorean theorem height <math>CD</math> of <math>\triangle DCO</math> is <math>\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{6}\right)^2} = \frac{\sqrt{2}}{3}</math>. This is also the height of the <math>\triangle ABD</math>.<br />
<br />
Area of the <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>.<br />
<br />
The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. <br />
<br />
Hence the height of <math>\triangle DCE</math> is <math>\dfrac{\dfrac{\sqrt{2}}{36}}{\dfrac{1}{2}\cdot\dfrac{1}{2}}</math> = <math>\dfrac{\sqrt{2}}{9}</math>. <br />
<br />
The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math>. <br />
<br />
Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is<br />
<math>\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}</math> = <math>\dfrac{1}{3} \Rightarrow C</math><br />
<br />
==Solution 2==<br />
<br />
Since <math>\triangle DCE</math> and <math>\triangle ABD</math> share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from <math>C</math> to <math>DE</math>. <br />
<br />
<asy><br />
import graph;<br />
import olympiad;<br />
pair O,A,B,C,D,E,F;<br />
O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774);<br />
draw(Circle((0,0),15)); <br />
draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B);<br />
label("A",A,NE);label("B",B,W);label("C",C,SE);label("D",D,NE);label("E",E,SW);label("O",O,SW);label("F",F,NW);<br />
markscalefactor=0.2;<br />
draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue);<br />
</asy><br />
<math>OD=r, OC=\frac{1}{3}r</math>.<br />
<br />
Since <math>m\angle DCO=m\angle DFC=90^\circ</math>, then <math>\triangle DCO\cong \triangle DFC</math>. So the ratio of the two altitudes is <math>\frac{CF}{DC}=\frac{OC}{DO}=\frac{1}{3}\Rightarrow \text{(C)}</math><br />
<br />
==Solution 3==<br />
<br />
Say the center of the circle is point <math>O</math>;<br />
Without loss of generality, assume <math>AC=2</math>, so <math>CB=4</math> and the diameter and radius are <math>6</math> and <math>3</math>, respectively. Therefore, <math>CO=1</math>, and <math>DO=3</math>.<br />
The area of <math>\triangle DCE</math> can be expressed as <math>\frac{1}{2}(CD)(6)\text{sin }(CDE).</math> <math>\frac{1}{2}(CD)(6)</math> happens to be the area of <math>\triangle ABD</math>. Furthermore, <math>\text{sin } CDE = \frac{CO}{DO},</math> or <math>\frac{1}{3}.</math> Therefore, the ratio is <math>\frac{1}{3}.</math><br />
<br />
== Solution 4 ==<br />
<br />
WLOG, let <math>AC=1</math>, <math>BC=2</math>, so radius of the circle is <math>\frac{3}{2}</math> and <math>OC=\frac{1}{2}</math>. As in solution 1, By same altitude, the ratio <math>[DCE]/[ABD]=PE/AB</math>, where <math>P</math> is the point where <math>DC</math> extended meets circle <math>O</math>. Note that angle P = 90 deg, so DCO ~ DPE with ratio 1:2, so PE = 1. Thus, our ratio is <math>\frac{1}{3}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Ratio Problems]]<br />
{{MAA Notice}}</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_22&diff=1001312008 AMC 12A Problems/Problem 222019-01-06T00:58:39Z<p>Therealchocolatelover: /* Solution 2 (without trigonometry) */</p>
<hr />
<div>{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #22]] and [[2008 AMC 10A Problems/Problem 25|2008 AMC 10A #25]]}}<br />
==Problem==<br />
A round table has radius <math>4</math>. Six rectangular place mats are placed on the table. Each place mat has width <math>1</math> and length <math>x</math> as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length <math>x</math>. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is <math>x</math>?<br />
<br />
<asy>unitsize(4mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(1\)",(-0.5,3.8),S);</asy><br />
<br />
<math>\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}</math><br />
<br />
==Solution==<br />
=== Solution 1 (trigonometry) ===<br />
Let one of the mats be <math>ABCD</math>, and the center be <math>O</math> as shown: <br />
<br />
<asy>unitsize(8mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(x\)",(0.03,1.5),E);<br />
label("\(A\)",(-3.6,2.5513),E);<br />
label("\(B\)",(-3.15,1.35),E);<br />
label("\(C\)",(0.05,3.20),E);<br />
label("\(D\)",(-0.75,4.15),E);<br />
label("\(O\)",(0.00,-0.10),E);<br />
label("\(1\)",(-0.1,3.8),S);<br />
label("\(4\)",(-0.4,2.2),S);<br />
draw((0,0)--(0,3.103));<br />
draw((0,0)--(-2.687,1.5513));<br />
draw((0,0)--(-0.5,3.9686));</asy><br />
<br />
Since there are <math>6</math> mats, <math>\Delta BOC</math> is [[equilateral]]. So, <math>BC=CO=x</math>. Also, <math>\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ</math>. <br />
<br />
By the [[Law of Cosines]]: <math>4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}</math>. <br />
<br />
Since <math>x</math> must be positive, <math>x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C</math>.<br />
<br />
=== Solution 2 (without trigonometry) ===<br />
Draw <math>OD</math> and <math>OC</math> as in the diagram. Draw the altitude from <math>O</math> to <math>DC</math> and call the intersection <math>E</math><br />
<br />
<br />
<asy>unitsize(8mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(x\)",(0.03,1.5),E);<br />
label("\(A\)",(-3.6,2.5513),E);<br />
label("\(B\)",(-3.15,1.35),E);<br />
label("\(C\)",(0.05,3.20),E);<br />
label("\(D\)",(-0.75,4.15),E);<br />
label("\(O\)",(0.00,-0.10),E);<br />
label("\(1\)",(-0.1,3.8),S);<br />
label("\(4\)",(-0.4,2.2),S);<br />
draw((0,0)--(0,3.103));<br />
draw((0,0)--(-2.687,1.5513));<br />
draw((0,0)--(-0.5,3.9686));</asy><br />
<br />
As proved in the first solution, <math> \angle OCD = 150^\circ</math>. <br />
That makes <math>\Delta OCE</math> a <math>30-60-90</math> triangle, so <math>OE = \frac{x}{2}</math> and <math>CE= \frac{x\sqrt 3}{2}</math><br />
<br />
Since <math> \Delta OEC</math> is a right triangle, <br />
<math>\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0</math><br />
<br />
Solving for <math>x</math> gives <math> x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow C </math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=A|num-b=21|num-a=23}}<br />
{{AMC10 box|year=2008|ab=A|num-b=24|after=Last Question}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Introductory Trigonometry Problems]]<br />
{{MAA Notice}}</div>Therealchocolateloverhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_22&diff=1001302008 AMC 12A Problems/Problem 222019-01-06T00:58:14Z<p>Therealchocolatelover: /* Solution 2 (without trigonometry) */</p>
<hr />
<div>{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #22]] and [[2008 AMC 10A Problems/Problem 25|2008 AMC 10A #25]]}}<br />
==Problem==<br />
A round table has radius <math>4</math>. Six rectangular place mats are placed on the table. Each place mat has width <math>1</math> and length <math>x</math> as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length <math>x</math>. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is <math>x</math>?<br />
<br />
<asy>unitsize(4mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(1\)",(-0.5,3.8),S);</asy><br />
<br />
<math>\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}</math><br />
<br />
==Solution==<br />
=== Solution 1 (trigonometry) ===<br />
Let one of the mats be <math>ABCD</math>, and the center be <math>O</math> as shown: <br />
<br />
<asy>unitsize(8mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(x\)",(0.03,1.5),E);<br />
label("\(A\)",(-3.6,2.5513),E);<br />
label("\(B\)",(-3.15,1.35),E);<br />
label("\(C\)",(0.05,3.20),E);<br />
label("\(D\)",(-0.75,4.15),E);<br />
label("\(O\)",(0.00,-0.10),E);<br />
label("\(1\)",(-0.1,3.8),S);<br />
label("\(4\)",(-0.4,2.2),S);<br />
draw((0,0)--(0,3.103));<br />
draw((0,0)--(-2.687,1.5513));<br />
draw((0,0)--(-0.5,3.9686));</asy><br />
<br />
Since there are <math>6</math> mats, <math>\Delta BOC</math> is [[equilateral]]. So, <math>BC=CO=x</math>. Also, <math>\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ</math>. <br />
<br />
By the [[Law of Cosines]]: <math>4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}</math>. <br />
<br />
Since <math>x</math> must be positive, <math>x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C</math>.<br />
<br />
=== Solution 2 (without trigonometry) ===<br />
Draw <math>OD</math> and <math>OC</math> as in the diagram. Draw the altitude from <math>O</math> to <math>DC</math> and call the intersection <math>E</math><br />
<br />
<br />
<asy>unitsize(8mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(x\)",(0.03,1.5),E);<br />
label("\(A\)",(-3.6,2.5513),E);<br />
label("\(B\)",(-3.15,1.35),E);<br />
label("\(C\)",(0.05,3.20),E);<br />
label("\(D\)",(-0.75,4.15),E);<br />
label("\(O\)",(0.00,-0.10),E);<br />
label("\(1\)",(-0.1,3.8),S);<br />
label("\(4\)",(-0.4,2.2),S);<br />
draw((0,0)--(0,3.103));<br />
draw((0,0)--(-2.687,1.5513));<br />
draw((0,0)--(-0.5,3.9686));</asy><br />
<br />
As proved in the first solution, <math> \angle OCD = 150^\circ</math>. <br />
That makes <math>\Delta OCE</math> a <math>30-60-90</math> triangle, so <math>OE = \frac{x}{2}</math> and <math>CE= \frac{x\sqrt 3}{2}</math><br />
<br />
Since <math> \Delta OCD</math> is a right triangle, <br />
<math>\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0</math><br />
<br />
Solving for <math>x</math> gives <math> x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow C </math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=A|num-b=21|num-a=23}}<br />
{{AMC10 box|year=2008|ab=A|num-b=24|after=Last Question}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Introductory Trigonometry Problems]]<br />
{{MAA Notice}}</div>Therealchocolatelover