https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=MooMoo&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T14:16:37ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Vieta%27s_Formulas&diff=73992Vieta's Formulas2015-12-26T21:04:58Z<p>MooMoo: /* Introduction */</p>
<hr />
<div>'''Vieta's Formulas''', otherwise called Viète's Laws, are a set of [[equation]]s relating the [[root]]s and the [[coefficient]]s of [[polynomial]]s.<br />
<br />
== Introduction ==<br />
<br />
Vieta's Formulas were discovered by the French mathematician [[François Viète]].<br />
<br />
Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as<br />
<br />
<center><math>x^2+ax+b=(x-p)(x-q)</math></center><br />
<br />
(Note that the first term is <math>x^2</math>, not <math>ax^2</math>.) Using the distributive property to expand the right side we get<br />
<br />
<center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center> <br />
<br />
We know that two Polynomials are Equal if and only if their coefficieNts are equal, so <math>x^2+ax+b=x^2-(p+q)x+pq</math> means that <math>a=-(p+q)</math> and <math>b=pq</math>. In other wordS, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the <math>x</math> term.<br />
<br />
A similar set of relations for cubics can be found by expanding <math>x^3+ax^2+bx+c=(x-p)(x-q)(x-r)</math>.<br />
<br />
We can state Vieta's formula's more rigorously and generally. Let <math>P(x)</math> be a polynomial of degree <math>n</math>, so <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,<br />
where the coefficient of <math>x^{i}</math> is <math>{a}_i</math> and <math>a_n \neq 0</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>. We thus have that<br />
<br />
<center><math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math></center><br />
<br />
Expanding out the right hand side gives us<br />
<br />
<math> a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.</math><br />
<br />
The coefficient of <math> x^k </math> in this expression will be the <math>k </math>th [[symmetric sum]] of the <math>r_i</math>. <br />
<br />
We now have two different expressions for <math>P(x)</math>. These must be equal. However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal. So, starting with the coefficient of <math> x^n </math>, we see that<br />
<br />
<center><math>a_n = a_n</math></center><br />
<center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center><br />
<center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center><br />
<center><math>\vdots</math></center><br />
<center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center><br />
<br />
More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>).<br />
<br />
If we denote <math>\sigma_k</math> as the <math>k</math>th symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>1\le k\le {n}</math>.<br />
<br />
==Problems==<br />
<br />
*Let <math>r_1,r_2,</math> and <math>r_3</math> be the three roots of the cubic <math>x^3 + 3x^2 + 4x - 4</math>. Find the value of <math>r_1r_2+r_1r_3+r_2r_3</math>. Yeah<br />
*Suppose the polynomial <math>5x^3 + 4x^2 - 8x + 6</math> has three real roots <math>a,b</math>, and <math>c</math>. Find the value of <math>a(1+b+c)+b(1+a+c)+c(1+a+b)</math>.<br />
*Let <math>m</math> and <math>n</math> be the roots of the quadratic equation <math>4x^2 + 5x + 3 = 0</math>. Find <math>(m + 7)(n + 7)</math>.<br />
<br />
===Intermediate===<br />
*Let <math>a</math>, <math>b</math>, and <math>c</math> be positive real numbers with <math>a<b<c</math> such that <math>a+b+c=12</math>, <math>a^2+b^2+c^2=50</math>, and <math>a^3+b^3+c^3=216</math>. Find <math>a+2b+3c</math>.<br />
*(USAMTS 2010) Find <math>c>0</math> such that if <math>r</math>, <math>s</math>, and <math>t</math> are the roots of the cubic <cmath>f(x)=x^3-4x^2+6x-c,</cmath> then <cmath>1=\dfrac1{r^2+s^2}+\dfrac1{s^2+t^2}+\dfrac1{t^2+r^2}.</cmath><br />
*(HMMT 2007) The complex numbers <math>\alpha_1</math>, <math>\alpha_2</math>, <math>\alpha_3</math>, and <math>\alpha_4</math> are the four distinct roots of the equation <math>x^4+2x^3+2=0</math>. Determine the unordered set <cmath>\{\alpha_1\alpha_2+\alpha_3\alpha_4,\,\alpha_1\alpha_3+\alpha_2\alpha_4,\,\alpha_1\alpha_4+\alpha_2\alpha_3\}.</cmath><br />
<br />
===Olympiad===<br />
[2008 AIME II] Let <math>r</math>, <math>s</math>, and <math>t</math> be the three roots of the equation <math>8x^3 + 1001x + 2008 = 0</math>. Find <math>(r + s)^3 + (s + t)^3 + (t + r)^3</math>.<br />
<br />
== See Also ==<br />
<br />
* [[Algebra]]<br />
* [[Polynomials]]<br />
* [[Newton's Sums]]<br />
<br />
== External Links ==<br />
*[http://mathworld.wolfram.com/VietasFormulas.html Mathworld's Article]<br />
<br />
[[Category:Elementary algebra]]</div>MooMoohttps://artofproblemsolving.com/wiki/index.php?title=Vieta%27s_Formulas&diff=73991Vieta's Formulas2015-12-26T21:04:06Z<p>MooMoo: /* Problems */</p>
<hr />
<div>'''Vieta's Formulas''', otherwise called Viète's Laws, are a set of [[equation]]s relating the [[root]]s and the [[coefficient]]s of [[polynomial]]s.<br />
<br />
== Introduction ==<br />
<br />
Vieta's Formulas were discovered by the French mathematician [[François Viète]].<br />
<br />
Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as<br />
<br />
<center><math>x^2+ax+b=(x-p)(x-q)</math></center><br />
<br />
(Note that the first term is <math>x^2</math>, not <math>ax^2</math>.) Using the distributive property to expand the right side we get<br />
<br />
<center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center> <br />
<br />
We know that two polynomials are equal if and only if their coefficients are equal, so <math>x^2+ax+b=x^2-(p+q)x+pq</math> means that <math>a=-(p+q)</math> and <math>b=pq</math>. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the <math>x</math> term.<br />
<br />
A similar set of relations for cubics can be found by expanding <math>x^3+ax^2+bx+c=(x-p)(x-q)(x-r)</math>.<br />
<br />
We can state Vieta's formula's more rigorously and generally. Let <math>P(x)</math> be a polynomial of degree <math>n</math>, so <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,<br />
where the coefficient of <math>x^{i}</math> is <math>{a}_i</math> and <math>a_n \neq 0</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>. We thus have that<br />
<br />
<center><math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math></center><br />
<br />
Expanding out the right hand side gives us<br />
<br />
<math> a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.</math><br />
<br />
The coefficient of <math> x^k </math> in this expression will be the <math>k </math>th [[symmetric sum]] of the <math>r_i</math>. <br />
<br />
We now have two different expressions for <math>P(x)</math>. These must be equal. However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal. So, starting with the coefficient of <math> x^n </math>, we see that<br />
<br />
<center><math>a_n = a_n</math></center><br />
<center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center><br />
<center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center><br />
<center><math>\vdots</math></center><br />
<center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center><br />
<br />
More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>).<br />
<br />
If we denote <math>\sigma_k</math> as the <math>k</math>th symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>1\le k\le {n}</math>.<br />
<br />
==Problems==<br />
<br />
*Let <math>r_1,r_2,</math> and <math>r_3</math> be the three roots of the cubic <math>x^3 + 3x^2 + 4x - 4</math>. Find the value of <math>r_1r_2+r_1r_3+r_2r_3</math>. Yeah<br />
*Suppose the polynomial <math>5x^3 + 4x^2 - 8x + 6</math> has three real roots <math>a,b</math>, and <math>c</math>. Find the value of <math>a(1+b+c)+b(1+a+c)+c(1+a+b)</math>.<br />
*Let <math>m</math> and <math>n</math> be the roots of the quadratic equation <math>4x^2 + 5x + 3 = 0</math>. Find <math>(m + 7)(n + 7)</math>.<br />
<br />
===Intermediate===<br />
*Let <math>a</math>, <math>b</math>, and <math>c</math> be positive real numbers with <math>a<b<c</math> such that <math>a+b+c=12</math>, <math>a^2+b^2+c^2=50</math>, and <math>a^3+b^3+c^3=216</math>. Find <math>a+2b+3c</math>.<br />
*(USAMTS 2010) Find <math>c>0</math> such that if <math>r</math>, <math>s</math>, and <math>t</math> are the roots of the cubic <cmath>f(x)=x^3-4x^2+6x-c,</cmath> then <cmath>1=\dfrac1{r^2+s^2}+\dfrac1{s^2+t^2}+\dfrac1{t^2+r^2}.</cmath><br />
*(HMMT 2007) The complex numbers <math>\alpha_1</math>, <math>\alpha_2</math>, <math>\alpha_3</math>, and <math>\alpha_4</math> are the four distinct roots of the equation <math>x^4+2x^3+2=0</math>. Determine the unordered set <cmath>\{\alpha_1\alpha_2+\alpha_3\alpha_4,\,\alpha_1\alpha_3+\alpha_2\alpha_4,\,\alpha_1\alpha_4+\alpha_2\alpha_3\}.</cmath><br />
<br />
===Olympiad===<br />
[2008 AIME II] Let <math>r</math>, <math>s</math>, and <math>t</math> be the three roots of the equation <math>8x^3 + 1001x + 2008 = 0</math>. Find <math>(r + s)^3 + (s + t)^3 + (t + r)^3</math>.<br />
<br />
== See Also ==<br />
<br />
* [[Algebra]]<br />
* [[Polynomials]]<br />
* [[Newton's Sums]]<br />
<br />
== External Links ==<br />
*[http://mathworld.wolfram.com/VietasFormulas.html Mathworld's Article]<br />
<br />
[[Category:Elementary algebra]]</div>MooMoohttps://artofproblemsolving.com/wiki/index.php?title=User:MooMoo&diff=73990User:MooMoo2015-12-26T20:52:33Z<p>MooMoo: Blanked the page</p>
<hr />
<div></div>MooMoohttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems&diff=584182001 AMC 10 Problems2013-12-28T16:45:45Z<p>MooMoo: /* Problem 20 */</p>
<hr />
<div>==Problem 1==<br />
<br />
The median of the list <math>n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15</math><br />
is <math>10</math>. What is the mean?<br />
<br />
<math>\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 11</math><br />
<br />
[[2001 AMC 10 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
A number <math>x</math> is <math>2</math> more than the product of its reciprocal and its additive inverse. In which interval does the number lie?<br />
<br />
<math>\mathrm{(A)}\ -4\leq x\leq -2 \qquad\mathrm{(B)}\ -2<x\leq 0 \qquad\mathrm{(C)}\ 0<x\leq 2</math><br />
<br />
<math>\mathrm{(D)}\ 2<x\leq 4 \qquad\mathrm{(E)}\ 4<x\leq 6</math><br />
<br />
[[2001 AMC 10 Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
The sum of two numbers is <math>S</math>. Suppose <math>3</math> is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?<br />
<br />
<math>\mathrm{(A)}\ 2S+3 \qquad\mathrm{(B)}\ 3S+2 \qquad\mathrm{(C)}\ 3S+6 \qquad\mathrm{(D)}\ 2S+6 \qquad\mathrm{(E)}\ 2S+12</math><br />
<br />
[[2001 AMC 12 Problems/Problem 1|Solution]]<br />
<br />
==Problem 4==<br />
<br />
What is the maximum number for the possible points of intersection of a circle and a triangle?<br />
<br />
<math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 3 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 5 \qquad\mathrm{(E)}\ 6</math><br />
<br />
[[2001 AMC 10 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
How many of the twelve pentominoes pictured below have at least one line of symmetry?<br />
<br />
<asy><br />
unitsize(5mm);<br />
defaultpen(linewidth(1pt));<br />
draw(shift(2,0)*unitsquare);<br />
draw(shift(2,1)*unitsquare);<br />
draw(shift(2,2)*unitsquare);<br />
draw(shift(1,2)*unitsquare);<br />
draw(shift(0,2)*unitsquare);<br />
draw(shift(2,4)*unitsquare);<br />
draw(shift(2,5)*unitsquare);<br />
draw(shift(2,6)*unitsquare);<br />
draw(shift(1,5)*unitsquare);<br />
draw(shift(0,5)*unitsquare);<br />
draw(shift(4,8)*unitsquare);<br />
draw(shift(3,8)*unitsquare);<br />
draw(shift(2,8)*unitsquare);<br />
draw(shift(1,8)*unitsquare);<br />
draw(shift(0,8)*unitsquare);<br />
draw(shift(6,8)*unitsquare);<br />
draw(shift(7,8)*unitsquare);<br />
draw(shift(8,8)*unitsquare);<br />
draw(shift(9,8)*unitsquare);<br />
draw(shift(9,9)*unitsquare);<br />
draw(shift(6,5)*unitsquare);<br />
draw(shift(7,5)*unitsquare);<br />
draw(shift(8,5)*unitsquare);<br />
draw(shift(7,6)*unitsquare);<br />
draw(shift(7,4)*unitsquare);<br />
draw(shift(6,1)*unitsquare);<br />
draw(shift(7,1)*unitsquare);<br />
draw(shift(8,1)*unitsquare);<br />
draw(shift(6,0)*unitsquare);<br />
draw(shift(7,2)*unitsquare);<br />
draw(shift(11,8)*unitsquare);<br />
draw(shift(12,8)*unitsquare);<br />
draw(shift(13,8)*unitsquare);<br />
draw(shift(14,8)*unitsquare);<br />
draw(shift(13,9)*unitsquare);<br />
draw(shift(11,5)*unitsquare);<br />
draw(shift(12,5)*unitsquare);<br />
draw(shift(13,5)*unitsquare);<br />
draw(shift(11,6)*unitsquare);<br />
draw(shift(13,4)*unitsquare);<br />
draw(shift(11,1)*unitsquare);<br />
draw(shift(12,1)*unitsquare);<br />
draw(shift(13,1)*unitsquare);<br />
draw(shift(13,2)*unitsquare);<br />
draw(shift(14,2)*unitsquare);<br />
draw(shift(16,8)*unitsquare);<br />
draw(shift(17,8)*unitsquare);<br />
draw(shift(18,8)*unitsquare);<br />
draw(shift(17,9)*unitsquare);<br />
draw(shift(18,9)*unitsquare);<br />
draw(shift(16,5)*unitsquare);<br />
draw(shift(17,6)*unitsquare);<br />
draw(shift(18,5)*unitsquare);<br />
draw(shift(16,6)*unitsquare);<br />
draw(shift(18,6)*unitsquare);<br />
draw(shift(16,0)*unitsquare);<br />
draw(shift(17,0)*unitsquare);<br />
draw(shift(17,1)*unitsquare);<br />
draw(shift(18,1)*unitsquare);<br />
draw(shift(18,2)*unitsquare);</asy><br />
<br />
<math>\mathrm{(A)}\ 3 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7</math><br />
<br />
[[2001 AMC 10 Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
Let <math>P(n)</math> and <math>S(n)</math> denote the product and the sum, respectively, of the digits<br />
of the integer <math>n</math>. For example, <math>P(23) = 6</math> and <math>S(23) = 5</math>. Suppose <math>N</math> is a<br />
two-digit number such that <math>N = P(N)+S(N)</math>. What is the units digit of <math>N</math>?<br />
<br />
<math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 3 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 8 \qquad\mathrm{(E)}\ 9</math><br />
<br />
[[2001 AMC 12 Problems/Problem 2|Solution]]<br />
<br />
==Problem 7==<br />
<br />
When the decimal point of a certain positive decimal number is moved four<br />
places to the right, the new number is four times the reciprocal of the original<br />
number. What is the original number?<br />
<br />
<math>\mathrm{(A)}\ 0.0002 \qquad\mathrm{(B)}\ 0.002 \qquad\mathrm{(C)}\ 0.02 \qquad\mathrm{(D)}\ 0.2 \qquad\mathrm{(E)}\ 2</math><br />
<br />
[[2001 AMC 10 Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their<br />
schedule is as follows: Darren works every third school day, Wanda works<br />
every fourth school day, Beatrice works every sixth school day, and Chi works<br />
every seventh school day. Today they are all working in the math lab. In how<br />
many school days from today will they next be together tutoring in the lab?<br />
<br />
<math>\mathrm{(A)}\ 42 \qquad\mathrm{(B)}\ 84 \qquad\mathrm{(C)}\ 126 \qquad\mathrm{(D)}\ 178 \qquad\mathrm{(E)}\ 252</math><br />
<br />
[[2001 AMC 10 Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
The state income tax where Kristin lives is levied at the rate of <math> p\% </math> of the first <math>\textdollar 28000 </math> of annual income plus <math> (p + 2)\% </math> of any amount above <math>\textdollar 28000 </math>. Kristin noticed that the state income tax she paid amounted to <math> (p + 0.25)\% </math> of her annual income. What was her annual income?<br />
<br />
<math> \textbf{(A) } \textdollar 28,000\qquad\textbf{(B) } \textdollar 32,000\qquad\textbf{(C) }\textdollar 35,000\qquad\textbf{(D) } \textdollar 42,000\qquad\textbf{(E) } \textdollar 56,000 </math><br />
<br />
[[2001 AMC 10 Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
If <math>x</math>, <math>y</math>, and <math>z</math> are positive with <math>xy = 24</math>, <math>xz = 48</math>, and <math>yz = 72</math>, then <math>x + y + z</math> is<br />
<br />
<math>\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24</math><br />
<br />
[[2001 AMC 10 Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains <math> 8 </math> unit squares. The second ring contains <math> 16 </math> unit squares. If we continue this process, the number of unit squares in the <math> 100^\text{th} </math> ring is<br />
<br />
<math>\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404</math><br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(1pt));<br />
fill((2,2)--(2,7)--(7,7)--(7,2)--cycle, mediumgray);<br />
fill((3,3)--(6,3)--(6,6)--(3,6)--cycle, gray);<br />
fill((4,4)--(5,4)--(5,5)--(4,5)--cycle, black);<br />
for(real i=0; i<=9; ++i)<br />
{<br />
draw((i,0)--(i,9));<br />
draw((0,i)--(9,i));<br />
}</asy><br />
<br />
[[2001 AMC 10 Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
Suppose that <math> n </math> is the product of three consecutive integers and that <math> n </math> is divisible by <math> 7 </math>. Which of the following is not necessarily a divisor of <math> n </math>?<br />
<br />
<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42 </math><br />
<br />
[[2001 AMC 10 Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
A telephone number has the form <math> ABC - DEF - GHIJ </math>, where each letter represents a different digit. The digits in each part of the numbers are in decreasing order; that is, <math> A > B > C </math>, <math> D > E > F </math>, and <math> G > H > I > J </math>. Furthermore, <math> D </math>, <math> E </math>, and <math> F </math> are consecutive even digits; <math>G</math>, <math>H</math>, <math>I</math>, and <math>J</math> are consecutive odd digits; and <math>A + B + C = 9</math>. Find <math>A</math>.<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 />
[[2001 AMC 12 Problems/Problem 6|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
A charity sells <math> 140 </math> benefit tickets for a total of <math> \textdollar 2001 </math>. Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?<br />
<br />
<math> \textbf{(A)} \textdollar 782 \qquad \textbf{(B)} \textdollar 986 \qquad \textbf{(C)} \textdollar 1158 \qquad \textbf{(D)} \textdollar 1219 \qquad \textbf{(E)} \textdollar 1449 </math><br />
<br />
[[2001 AMC 12 Problems/Problem 7|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
A street has parallel curbs <math> 40 </math> feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is <math> 15 </math> feet and each stripe is <math> 50 </math> feet long. Find the distance, in feet, between the stripes.<br />
<br />
<math> \textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25 </math><br />
<br />
[[2001 AMC 10 Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
The mean of three numbers is <math> 10 </math> more than the least of the numbers and <math> 15 </math> less than the greatest. The median of the three numbers is <math> 5 </math>. What is their sum?<br />
<br />
<math> \textbf{(A)} \ 5 \qquad \textbf{(B)} \ 20 \qquad \textbf{(C)} \ 25 \qquad \textbf{(D)} \ 30 \qquad \textbf{(E)} \ 36 </math><br />
<br />
[[2001 AMC 12 Problems/Problem 4|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
Which of the cones listed below can be formed from a <math> 252^\circ </math> sector of a circle of radius <math> 10 </math> by aligning the two straight sides?<br />
<br />
<asy>import graph;unitsize(1.5cm);defaultpen(fontsize(8pt));draw(Arc((0,0),1,-72,180),linewidth(.8pt));draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt));label("$10$",(-0.5,0),S);draw(Arc((0,0),0.1,-72,180));label("$252^{\circ}$",(0.05,0.05),NE);</asy><br />
<br />
<math> \textbf{(A)} \text{A cone with slant height of 10 and radius 6} </math><br />
<math> \textbf{(B)} \text{A cone with height of 10 and radius 6} </math><br />
<math> \textbf{(C)} \text{A cone with slant height of 10 and radius 7} </math><br />
<math> \textbf{(D)} \text{A cone with height of 10 and radius 7} </math><br />
<math> \textbf{(E)} \text{A cone with slant height of 10 and radius 8} </math><br />
<br />
<br />
[[2001 AMC 12 Problems/Problem 8|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to<br />
<br />
<math> \textbf{(A)} \ 50 \qquad \textbf{(B)} \ 52 \qquad \textbf{(C)} \ 54 \qquad \textbf{(D)} \ 56 \qquad \textbf{(E)} \ 58 \qquad </math><br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(0.8pt));<br />
path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0);<br />
path p2=(0,1)--(1,1)--(1,0);<br />
path p3=(2,0)--(2,1)--(3,1);<br />
path p4=(3,2)--(2,2)--(2,3);<br />
path p5=(1,3)--(1,2)--(0,2);<br />
path p6=(1,1)--(2,2);<br />
path p7=(2,1)--(1,2);<br />
path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7;<br />
for(int i=0; i<3; ++i)<br />
{<br />
for(int j=0; j<3; ++j)<br />
{<br />
draw(shift(3*i,3*j)*p);<br />
}<br />
}</asy><br />
<br />
[[2001 AMC 12 Problems/Problem 10|Solution]]<br />
<br />
== Problem 19==<br />
<br />
Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?<br />
<br />
<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 </math><br />
<br />
[[2001 AMC 10 Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length <math> 2000 </math>. What is the length of each side of the octagon?<br />
<br />
<math> \textbf{(A)} \frac{1}{3}(2000) \qquad \textbf{(B)} 2000(\sqrt{2}-1) \qquad \textbf{(C)} 2000(2-\sqrt{2})<br />
\qquad \textbf{(D)} 1000 \qquad \textbf{(E)} 1000\sqrt{2} </math><br />
<br />
[[2001 AMC 10 Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter <math> 10 </math> and altitude <math> 12 </math>, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.<br />
<br />
<math> \textbf{(A)}\ \frac{8}3\qquad\textbf{(B)}\ \frac{30}{11}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{25}{8}\qquad\textbf{(E)}\ \frac{7}{2} </math><br />
<br />
[[2001 AMC 10 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by <math> v </math>, <math> w </math>, <math> x </math>, <math> y </math>, and <math> z </math>. Find <math> y + z </math>.<br />
<br />
<math> \textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47 </math><br />
<br />
<asy><br />
unitsize(10mm);<br />
defaultpen(linewidth(1pt));<br />
for(int i=0; i<=3; ++i)<br />
{<br />
draw((0,i)--(3,i));<br />
draw((i,0)--(i,3));<br />
}<br />
label("$25$",(0.5,0.5));<br />
label("$z$",(1.5,0.5));<br />
label("$21$",(2.5,0.5));<br />
label("$18$",(0.5,1.5));<br />
label("$x$",(1.5,1.5));<br />
label("$y$",(2.5,1.5));<br />
label("$v$",(0.5,2.5));<br />
label("$24$",(1.5,2.5));<br />
label("$w$",(2.5,2.5));</asy><br />
<br />
[[2001 AMC 10 Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?<br />
<br />
<math> \textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{7}{10} </math><br />
<br />
[[2001 AMC 12 Problems/Problem 11|Solution]]<br />
<br />
==Problem 24==<br />
<br />
In trapezoid <math> ABCD </math>, <math> \overline{AB} </math> and <math> \overline{CD} </math> are perpendicular to <math> \overline{AD} </math>, with <math> AB+CD=BC </math>, <math> AB<CD </math>, and <math> AD=7 </math>. What is <math> AB\cdot CD </math>?<br />
<br />
<math> \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13 </math><br />
<br />
[[2001 AMC 10 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
How many positive integers not exceeding <math> 2001 </math> are multiples of <math> 3 </math> or <math> 4 </math> but not <math> 5 </math>?<br />
<br />
<math> \textbf{(A)}\ 768 \qquad \textbf{(B)}\ 801 \qquad \textbf{(C)}\ 934 \qquad \textbf{(D)}\ 1067 \qquad \textbf{(E)}\ 1167 </math><br />
<br />
[[2001 AMC 12 Problems/Problem 12|Solution]]<br />
{{MAA Notice}}</div>MooMoohttps://artofproblemsolving.com/wiki/index.php?title=User:MooMoo&diff=48232User:MooMoo2012-09-06T19:18:59Z<p>MooMoo: Created page with "Hi. I'm MooMoo. Contrary to the belief, I actually don't like cows. That's all I have time for. Bye."</p>
<hr />
<div>Hi. I'm MooMoo. Contrary to the belief, I actually don't like cows. That's all I have time for. Bye.</div>MooMoohttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems&diff=463472005 AMC 10A Problems2012-04-15T15:41:26Z<p>MooMoo: /* Problem 19 */</p>
<hr />
<div>==Problem 1==<br />
While eating out, Mike and Joe each tipped their server <math>2</math> dollars. Mike tipped <math>10\%</math> of his bill and Joe tipped <math>20\%</math> of his bill. What was the difference, in dollars between their bills? <br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 20 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For each pair of real numbers <math>a</math><math>\neq</math><math>b</math>, define the operation <math>\star</math> as<br />
<br />
<math> (a \star b) = \frac{a+b}{a-b} </math>.<br />
<br />
What is the value of <math> ((1 \star 2) \star 3)</math>?<br />
<br />
<math> \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \textrm{This\, value\, is\, not\, defined.} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The equations <math> 2x + 7 = 3 </math> and <math> bx - 10 = -2 </math> have the same solution <math>x</math>. What is the value of <math>b</math>? <br />
<br />
<math> \mathrm{(A) \ } -8\qquad \mathrm{(B) \ } -4\qquad \mathrm{(C) \ } -2\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A rectangle with a diagonal of length <math>x</math> is twice as long as it is wide. What is the area of the rectangle? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
A store normally sells windows at <dollar/>100 each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?<br />
<br />
<math> \mathrm{(A) \ } 100\qquad \mathrm{(B) \ } 200\qquad \mathrm{(C) \ } 300\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The average (mean) of <math>20</math> numbers is <math>30</math>, and the average of <math>30</math> other numbers is <math>20</math>. What is the average of all <math>50</math> numbers?<br />
<br />
<math> \mathrm{(A) \ } 23\qquad \mathrm{(B) \ } 24\qquad \mathrm{(C) \ } 25\qquad \mathrm{(D) \ } 26\qquad \mathrm{(E) \ } 27 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Josh and Mike live <math>13</math> miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met? <br />
<br />
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math> and <math>BE</math>=1. What is the area of the inner square <math>EFGH</math>?<br />
<br />
[[File:AMC102005Aq.png]]<br />
<br />
<math> \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Three tiles are marked <math>X</math> and two other tiles are marked <math>O</math>. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads <math>XOXOX</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{12}\qquad \mathrm{(B) \ } \frac{1}{10}\qquad \mathrm{(C) \ } \frac{1}{6}\qquad \mathrm{(D) \ } \frac{1}{4}\qquad \mathrm{(E) \ } \frac{1}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
There are two values of <math>a</math> for which the equation <math> 4x^2 + ax + 8x + 9 = 0 </math> has only one solution for <math>x</math>. What is the sum of those values of <math>a</math>?<br />
<br />
<math> \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
A wooden cube <math>n</math> units on a side is painted red on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math>n</math>?<br />
<br />
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 7 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?<br />
<br />
[[Image:2005amc10a12.gif]]<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many positive integers <math>n</math> satisfy the following condition:<br />
<br />
<math> (130n)^{50} > n^{100} > 2^{200} </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 65\qquad \mathrm{(E) \ } 125 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br />
<br />
<math> \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
How many positive cubes divide <math> 3! \cdot 5! \cdot 7! </math> ?<br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is <math>6</math>. How many two-digit numbers have this property? <br />
<br />
<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 19 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are replaced by the numbers <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math>, although not necessarily in this order. The sums of the numbers at the ends of the line segments <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, and <math>EA</math> form an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence? <br />
<br />
[[Image:2005amc10a17.gif]]<br />
<br />
<math> \mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 13 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{5}\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 />
[[2005 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point <math>B</math> from the line on which the bases of the original squares were placed?<br />
<br />
[[File:AMC10A200519.png]]<br />
<br />
[[2005 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
??<br />
<br />
== Problem 20 ==<br />
An equiangular octagon has four sides of length 1 and four sides of length <math>\frac{\sqrt{2}}{2}</math>, arranged so that no two consecutive sides have the same length. What is the area of the octagon?<br />
<br />
<math> \mathrm{(A) \ } \frac72\qquad \mathrm{(B) \ } \frac{7\sqrt2}{2}\qquad \mathrm{(C) \ } \frac{5+4\sqrt2}{2}\qquad \mathrm{(D) \ } \frac{4+5\sqrt2}{2}\qquad \mathrm{(E) \ } 7 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide <math>6n</math>? <br />
<br />
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Let <math>S</math> be the set of the <math>2005</math> smallest positive multiples of <math>4</math>, and let <math>T</math> be the set of the <math>2005</math> smallest positive multiples of <math>6</math>. How many elements are common to <math>S</math> and <math>T</math>?<br />
<br />
<math> \mathrm{(A) \ } 166\qquad \mathrm{(B) \ } 333\qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ } 668\qquad \mathrm{(E) \ } 1001 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<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 />
[[2005 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
For each positive integer <math> m > 1 </math>, let <math>P(m)</math> denote the greatest prime factor of <math>m</math>. For how many positive integers <math>n</math> is it true that both <math> P(n) = \sqrt{n} </math> and <math> P(n+48) = \sqrt{n+48} </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
In <math>ABC</math> we have <math> AB = 25 </math>, <math> BC = 39 </math>, and <math>AC=42</math>. Points <math>D</math> and <math>E</math> are on <math>AB</math> and <math>AC</math> respectively, with <math> AD = 19 </math> and <math> AE = 14 </math>. What is the ratio of the area of triangle <math>ADE</math> to the area of the quadrilateral <math>BCED</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{19}{56}\qquad \mathrm{(E) \ } 1 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
* [[AMC Problems and Solutions]]</div>MooMoo