https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Addictedtomath&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T12:24:20ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_25&diff=503042006 AMC 12B Problems/Problem 252012-12-25T03:10:13Z<p>Addictedtomath: </p>
<hr />
<div>== Problem ==<br />
A sequence <math>a_1,a_2,\dots</math> of non-negative integers is defined by the rule <math>a_{n+2}=|a_{n+1}-a_n|</math> for <math>n\geq 1</math>. If <math>a_1=999</math>, <math>a_2<999</math> and <math>a_{2006}=1</math>, how many different values of <math>a_2</math> are possible?<br />
<br />
<math><br />
\mathrm{(A)}\ 165<br />
\qquad<br />
\mathrm{(B)}\ 324<br />
\qquad<br />
\mathrm{(C)}\ 495<br />
\qquad<br />
\mathrm{(D)}\ 499<br />
\qquad<br />
\mathrm{(E)}\ 660<br />
</math><br />
<br />
== Solution ==<br />
We say the sequence <math>(a_n)</math> completes at <math>i</math> if <math>i</math> is the minimal positive integer such that <math>a_i = a_{i + 1} = 1</math>. Otherwise, we say <math>(a_n)</math> does not complete.<br />
<br />
<br />
Note that if <math>d = \gcd(999, a_2) \neq 1</math>, then <math>d|a_n</math> for all <math>n \geq 1</math>, and <math>d</math> does not divide <math>1</math>, so if <math>\gcd(999, a_2) \neq 1</math>, then <math>(a_n)</math> does not complete. (Also, <math>a_{2006}</math> cannot be 1 in this case since <math>d</math> does not divide <math>1</math>, so we do not care about these <math>a_2</math> at all.)<br />
<br />
From now on, suppose <math>\gcd(999, a_2) = 1</math>.<br />
<br />
<br />
We will now show that <math>(a_n)</math> completes at <math>i</math> for some <math>i \leq 2006</math>. We will do this with 3 lemmas.<br />
<br />
<br />
'''Lemma:''' If <math>a_j \neq a_{j + 1}</math>, and neither value is <math>0</math>, then <math>\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})</math>.<br />
<br />
'''Proof:''' There are 2 cases to consider.<br />
<br />
If <math>a_j > a_{j + 1}</math>, then <math>a_{j + 2} = a_j - a_{j + 1}</math>, and <math>a_{j + 3} = |a_j - 2a_{j + 1}|</math>. So <math>a_j > a_{j + 2}</math> and <math>a_j > a_{j + 3}</math>.<br />
<br />
If <math>a_j < a_{j + 1}</math>, then <math>a_{j + 2} = a_{j + 1} - a_j</math>, and <math>a_{j + 3} = a_j</math>. So <math>a_{j + 1} > a_{j + 2}</math> and <math>a_{j + 1} > a_{j + 3}</math>.<br />
<br />
In both cases, <math>\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})</math>, as desired.<br />
----------------------<br />
<br />
'''Lemma:''' If <math>a_i = a_{i + 1}</math>, then <math>a_i = 1</math>. Moreover, if instead we have <math>a_i = 0</math> for some <math>i > 2</math>, then <math>a_{i - 1} = a_{i - 2} = 1</math>.<br />
<br />
'''Proof:''' By the way <math>(a_n)</math> is constructed in the problem statement, having two equal consecutive terms <math>a_i = a_{i + 1}</math> implies that <math>a_i</math> divides every term in the sequence. So <math>a_i | 999</math> and <math>a_i | a_2</math>, so <math>a_i | \gcd(999, a_2) = 1</math>, so <math>a_i = 1</math>. For the proof of the second result, note that if <math>a_i = 0</math>, then <math>a_{i - 1} = a_{i - 2}</math>, so by the first result we just proved, <math>a_{i - 2} = a_{i - 1} = 1</math>.<br />
----------------------<br />
<br />
'''Lemma:''' <math>(a_n)</math> completes at <math>i</math> for some <math>i \leq 2000</math>.<br />
<br />
'''Proof:''' Suppose <math>(a_n)</math> completed at some <math>i > 2000</math> or not at all. Then by the second lemma and the fact that neither <math>999</math> nor <math>a_2</math> are <math>0</math>, none of the pairs <math>(a_1, a_2), ..., (a_{1999}, a_{2000})</math> can have a <math>0</math> or be equal to <math>(1, 1)</math>. So the first lemma implies<br />
<cmath>\max(a_1, a_2) > \max(a_3, a_4) > \cdots > \max(a_{1999}, a_{2000}) > 0,</cmath><br />
so <math>999 = \max(a_1, a_2) \geq 1000</math>, a contradiction. Hence <math>(a_n)</math> completes at <math>i</math> for some <math>i \leq 2000</math>.<br />
----------------------<br />
<br />
<br />
Now we're ready to find exactly which values of <math>a_2</math> we want to count.<br />
<br />
Let's keep in mind that <math>2006 \equiv 2 \pmod 3</math> and that <math>a_1 = 999</math> is odd. We have two cases to consider.<br />
<br />
<br />
'''Case 1:''' If <math>a_2</math> is odd, then <math>a_3</math> is even, so <math>a_4</math> is odd, so <math>a_5</math> is odd, so <math>a_6</math> is even, and this pattern must repeat every three terms because of the recursive definition of <math>(a_n)</math>, so the terms of <math>(a_n)</math> reduced modulo 2 are<br />
<cmath>1, 1, 0, 1, 1, 0, ...,</cmath> <br />
so <math>a_{2006}</math> is odd and hence <math>1</math> (since if <math>(a_n)</math> completes at <math>i</math>, then <math>a_k</math> must be <math>0</math> or <math>1</math> for all <math>k \geq i</math>).<br />
---------------------<br />
<br />
'''Case 2:''' If <math>a_2</math> is even, then <math>a_3</math> is odd, so <math>a_4</math> is odd, so <math>a_5</math> is even, so <math>a_6</math> is odd, and this pattern must repeat every three terms, so the terms of <math>(a_n)</math> reduced modulo 2 are <br />
<cmath>1, 0, 1, 1, 0, 1, ...,</cmath><br />
so <math>a_{2006}</math> is even, and hence <math>0</math>.<br />
---------------------<br />
<br />
<br />
We have found that <math>a_{2006} = 1</math> is true precisely when <math>\gcd(999, a_2) = 1</math> and <math>a_2</math> is odd. This tells us what we need to count.<br />
<br />
<br />
There are <math>\phi(999) = 648</math> numbers less than <math>999</math> and relatively prime to it (<math>\phi</math> is the Euler totient function). We want to count how many of these are even. Note that<br />
<cmath>t \mapsto 999 - t</cmath><br />
is a 1-1 correspondence between the odd and even numbers less than and relatively prime to <math>999</math>. So our final answer is <math>648/2 = 324</math>, or <math>\boxed{\text{B}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=24|after=Last Question}}</div>Addictedtomathhttps://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems&diff=449912010 AIME I Problems2012-02-23T01:38:40Z<p>Addictedtomath: /* Problem 9 */</p>
<hr />
<div>{{AIME Problems|year=2010|n=I}}<br />
<br />
== Problem 1 ==<br />
Maya lists all the positive divisors of <math>2010^2</math>. She then randomly selects two distinct divisors from this list. Let <math>p</math> be the probability that exactly one of the selected divisors is a perfect square. The probability <math>p</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Find the remainder when <math>9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}</math> is divided by <math>1000</math>.<br />
<br />
[[2010 AIME I Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Suppose that <math>y = \frac34x</math> and <math>x^y = y^x</math>. The quantity <math>x + y</math> can be expressed as a rational number <math>\frac {r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r + s</math>.<br />
<br />
[[2010 AIME I Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Jackie and Phil have two fair coins and a third coin that comes up heads with probability <math>\frac47</math>. Jackie flips the three coins, and then Phil flips the three coins. Let <math>\frac {m}{n}</math> be the probability that Jackie gets the same number of heads as Phil, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> satisfy <math>a > b > c > d</math>, <math>a + b + c + d = 2010</math>, and <math>a^2 - b^2 + c^2 - d^2 = 2010</math>. Find the number of possible values of <math>a</math>.<br />
<br />
[[2010 AIME I Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
Let <math>P(x)</math> be a quadratic polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
[[2010 AIME I Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Define an ordered triple <math>(A, B, C)</math> of sets to be minimally intersecting if <math>|A \cap B| = |B \cap C| = |C \cap A| = 1</math> and <math>A \cap B \cap C = \emptyset</math>. For example, <math>(\{1,2\},\{2,3\},\{1,3,4\})</math> is a minimally intersecting triple. Let <math>N</math> be the number of minimally intersecting ordered triples of sets for which each set is a subset of <math>\{1,2,3,4,5,6,7\}</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.<br />
<br />
'''Note''': <math>|S|</math> represents the number of elements in the set <math>S</math>.<br />
<br />
[[2010 AIME I Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
For a real number <math>a</math>, let <math>\lfloor a \rfloor</math> denominate the greatest integer less than or equal to <math>a</math>. Let <math>\mathcal{R}</math> denote the region in the coordinate plane consisting of points <math>(x,y)</math> such that <math>\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25</math>. The region <math>\mathcal{R}</math> is completely contained in a disk of radius <math>r</math> (a disk is the union of a circle and its interior). The minimum value of <math>r</math> can be written as <math>\frac {\sqrt {m}}{n}</math>, where <math>m</math> and <math>n</math> are integers and <math>m</math> is not divisible by the square of any prime. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Let <math>(a,b,c)</math> be a real solution of the system of equations <math>x^3 - xyz = 2</math>, <math>y^3 - xyz = 6</math>, <math>z^3 - xyz = 20</math>. The greatest possible value of <math>a^3 + b^3 + c^3</math> can be written in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Let <math>N</math> be the number of ways to write <math>2010</math> in the form <math>2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0</math>, where the <math>a_i</math>'s are integers, and <math>0 \le a_i \le 99</math>. An example of such a representation is <math>1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0</math>. Find <math>N</math>.<br />
<br />
[[2010 AIME I Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Let <math>\mathcal{R}</math> be the region consisting of the set of points in the coordinate plane that satisfy both <math>|8 - x| + y \le 10</math> and <math>3y - x \ge 15</math>. When <math>\mathcal{R}</math> is revolved around the line whose equation is <math>3y - x = 15</math>, the volume of the resulting solid is <math>\frac {m\pi}{n\sqrt {p}}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p</math>.<br />
<br />
[[2010 AIME I Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let <math>m \ge 3</math> be an integer and let <math>S = \{3,4,5,\ldots,m\}</math>. Find the smallest value of <math>m</math> such that for every partition of <math>S</math> into two subsets, at least one of the subsets contains integers <math>a</math>, <math>b</math>, and <math>c</math> (not necessarily distinct) such that <math>ab = c</math>.<br />
<br />
'''Note''': a partition of <math>S</math> is a pair of sets <math>A</math>, <math>B</math> such that <math>A \cap B = \emptyset</math>, <math>A \cup B = S</math>.<br />
<br />
[[2010 AIME I Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Rectangle <math>ABCD</math> and a semicircle with diameter <math>AB</math> are coplanar and have nonoverlapping interiors. Let <math>\mathcal{R}</math> denote the region enclosed by the semicircle and the rectangle. Line <math>\ell</math> meets the semicircle, segment <math>AB</math>, and segment <math>CD</math> at distinct points <math>N</math>, <math>U</math>, and <math>T</math>, respectively. Line <math>\ell</math> divides region <math>\mathcal{R}</math> into two regions with areas in the ratio <math>1: 2</math>. Suppose that <math>AU = 84</math>, <math>AN = 126</math>, and <math>UB = 168</math>. Then <math>DA</math> can be represented as <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
For each positive integer n, let <math>f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor</math>. Find the largest value of n for which <math>f(n) \le 300</math>.<br />
<br />
'''Note:''' <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.<br />
<br />
[[2010 AIME I Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the incircles of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal radii. Let <math>p</math> and <math>q</math> be positive relatively prime integers such that <math>\frac {AM}{CM} = \frac {p}{q}</math>. Find <math>p + q</math>.<br />
<br />
[[2010 AIME I Problems/Problem 15|Solution]]<br />
<br />
== See also ==<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [[Mathematics competition resources]]</div>Addictedtomathhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_18&diff=440562003 AMC 12A Problems/Problem 182012-01-01T17:27:16Z<p>Addictedtomath: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>n</math> be a <math>5</math>-digit number, and let <math>q</math> and <math>r</math> be the quotient and the remainder, respectively, when <math>n</math> is divided by <math>100</math>. For how many values of <math>n</math> is <math>q+r</math> divisible by <math>11</math>? <br />
<br />
<math> \mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090 </math><br />
<br />
== Solution ==<br />
<br />
When a <math>5</math>-digit number is divided by <math>100</math>, the first <math>3</math> digits become the quotient, <math>q</math>, and the last <math>2</math> digits become the remainder, <math>r</math>. <br />
<br />
Therefore, <math>q</math> can be any integer from <math>100</math> to <math>999</math> inclusive, and <math>r</math> can be any integer from <math>0</math> to <math>99</math> inclusive. <br />
<br />
For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\lfloor \frac{100}{11} \rfloor = 9</math> possible values of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. <br />
<br />
Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\lfloor \frac{900}{11} \rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. <br />
<br />
Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)} </math>.<br />
<br />
== See Also ==<br />
*[[2003 AMC 12A Problems]]<br />
*[[2003 AMC 12A Problems/Problem 17|Previous Problem]]<br />
*[[2003 AMC 12A Problems/Problem 19|Next Problem]]<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Addictedtomathhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_18&diff=440552003 AMC 12A Problems/Problem 182012-01-01T17:19:59Z<p>Addictedtomath: /* See Also */</p>
<hr />
<div>== Problem ==<br />
Let <math>n</math> be a <math>5</math>-digit number, and let <math>q</math> and <math>r</math> be the quotient and the remainder, respectively, when <math>n</math> is divided by <math>100</math>. For how many values of <math>n</math> is <math>q+r</math> divisible by <math>11</math>? <br />
<br />
<math> \mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090 </math><br />
<br />
== Solution ==<br />
<br />
Solution 1:<br />
<br />
When a <math>5</math>-digit number is divided by <math>100</math>, the first <math>3</math> digits become the quotient, <math>q</math>, and the last <math>2</math> digits become the remainder, <math>r</math>. <br />
<br />
Therefore, <math>q</math> can be any integer from <math>100</math> to <math>999</math> inclusive, and <math>r</math> can be any integer from <math>0</math> to <math>99</math> inclusive. <br />
<br />
For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\lfloor \frac{100}{11} \rfloor = 9</math> possible values of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. <br />
<br />
Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\lfloor \frac{900}{11} \rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. <br />
<br />
Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow B</math>.<br />
<br />
Solution 2:<br />
<br />
Let the 5 digit number be abcde, where a, b, c, d, and e are digits. Then q = abc, and r = de. Then the digits of q +r theoretically are a, then b+d, then c + e. If this is to be divisible by 11, we can use the divisibility check for 11 to say that a + c + e - b -d is divisible by 11. If we revert this again, but we want all the digits to be single digit, then we get abcde again. So if q + r is to be divisible by 11, then the number we started with, abcde, must also be divisible by 11. There are 90000 total possible values of abcde, of which 8181 of them are divisible by 11. So for those numbers q + r is divisible by 11, so we have 8181 (B).<br />
<br />
== See Also ==<br />
*[[2003 AMC 12A Problems]]<br />
*[[2003 AMC 12A Problems/Problem 17|Previous Problem]]<br />
*[[2003 AMC 12A Problems/Problem 19|Next Problem]]<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Addictedtomathhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_17&diff=440542003 AMC 12A Problems/Problem 172012-01-01T17:18:51Z<p>Addictedtomath: /* See Also */</p>
<hr />
<div>== Problem ==<br />
Square <math>ABCD</math> has sides of length <math>4</math>, and <math>M</math> is the midpoint of <math>\overline{CD}</math>. A circle with radius <math>2</math> and center <math>M</math> intersects a circle with radius <math>4</math> and center <math>A</math> at points <math>P</math> and <math>D</math>. What is the distance from <math>P</math> to <math>\overline{AD}</math>?<br />
<br />
[[Image:5d50417537c6cddfb70810403c62787b889cdcb1.png]]<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math><br />
<br />
== Solution ==<br />
<br />
Let <math>D</math> be the origin. <math>A</math> is the point <math>(0,4)</math> and <math>M</math> is the point <math>(2,0)</math>. We are given the radius of the quarter circle and semicircle as <math>4</math> and <math>2</math>, respectively, so their equations, respectively, are:<br />
<br />
<math>x^2 + (y-4)^2 = 4^2</math><br />
<br />
<math>(x-2)^2 + y^2 = 2^2</math><br />
<br />
<br />
Algebraically manipulating the second equation gives:<br />
<br />
<math>y^2 = 2^2 - (x-2)^2</math><br />
<br />
<math>y^2 = (2-(x-2)(2+(x-2))</math><br />
<br />
<math>y^2 = (4-x)(x)</math><br />
<br />
<math>y = \sqrt{4x - x^2}</math><br />
<br />
<br />
Substituting this back into the first equation:<br />
<br />
<math>x^2 + (\sqrt{4x - x^2} - 4)^2 = 4^2</math><br />
<br />
<math>x^2 + 4x - x^2 - 8\sqrt{4x - x^2} + 16 = 16</math><br />
<br />
<math>4x - 8\sqrt{4x - x^2} = 0</math><br />
<br />
<math>4x = 8\sqrt{4x - x^2}</math><br />
<br />
<math>16x^2 = 64(4x - x^2)</math><br />
<br />
<math>16x^2 = 256x - 64x^2</math><br />
<br />
<math>80x^2 - 256x = 0</math><br />
<br />
<math>x(80x - 256) = 0</math><br />
<br />
<br />
Solving each factor for 0 yields <math>x = 0 , \frac{16}{5}</math>. The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow B</math><br />
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== See Also ==<br />
*[[2003 AMC 12A Problems]]<br />
*[[2003 AMC 12A Problems/Problem 16|Previous Problem]]<br />
*[[2003 AMC 12A Problems/Problem 18|Next Problem]]</div>Addictedtomath