https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathfanatic&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T10:04:32ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=140212007 AIME I Problems/Problem 102007-03-15T03:43:53Z<p>Mathfanatic: fixed 6\choose3 LaTeX error in my solution</p>
<hr />
<div>== Problem ==<br />
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let <math>N</math> be the number of shadings with this property. Find the remainder when <math>N</math> is divided by 1000.<br />
<br />
== Solution ==<br />
<br />
Consider the first column. There are <math>{6\choose3} = 20</math> ways that the balls could be chosen, but WLOG let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br />
<br />
Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br />
<br />
*All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br />
<br />
*Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have <math>3 \cdot 2 \cdot 3 \cdot 3 = 54</math>.<br />
<br />
*All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br />
<br />
So there are <math>20(3+54+36) = 1860</math> different shadings, and the solution is '''860'''.<br />
<br />
== See also ==<br />
{{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Mathfanatichttps://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=140202007 AIME I Problems/Problem 102007-03-15T03:41:10Z<p>Mathfanatic: deleted old partial solution and wrote new one</p>
<hr />
<div>== Problem ==<br />
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let <math>N</math> be the number of shadings with this property. Find the remainder when <math>N</math> is divided by 1000.<br />
<br />
== Solution ==<br />
<br />
Consider the first column. There are <math>\binom{6}{3} = 20</math> ways that the balls could be chosen, but WLOG let them be the first three rows. (So change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get an answer.<br />
<br />
Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br />
<br />
*All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br />
<br />
*Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have <math>3 \cdot 2 \cdot 3 \cdot 3 = 54</math>.<br />
<br />
*All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br />
[[Casework]] shows that:<br />
<br />
So there are <math>20(3+54+36) = 1860</math> different shadings, and the solution is '''860'''.<br />
<br />
== See also ==<br />
{{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Mathfanatichttps://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems&diff=140042007 AIME I Problems2007-03-14T23:57:41Z<p>Mathfanatic: Typed up problems</p>
<hr />
<div>== Problem 1 ==<br />
How many positive perfect squares less than <math>10^6</math> are multiples of 24?<br />
<br />
[[2007 AIME I Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person.<br />
<br />
[[2007 AIME I Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The complex number <math>z</math> is equal to <math>9+bi</math>, where <math>b</math> is a positive real number and <math>i^{2}=-1</math>. Given that the imaginary parts of <math>z^{2}</math> and <math>z^{3}</math> are the same, what is <math>b</math> equal to?<br />
<br />
[[2007 AIME I Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are <math>60</math>,<math>84</math>, and <math>140</math>. The three planets and the star are currently collinear. What is the fewest number of years from now that they will all be collinear again?<br />
<br />
[[2007 AIME I Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
The formula for converting a Fahrenheit temperature <math>F</math> to the corresponding Celsius temperature <math>C</math> is <math>C = \frac{5}{9}(F-32).</math> An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer.<br />
<br />
For how many integer Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature?<br />
<br />
[[2007 AIME I Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
A frog moves from 0 to 39 on an integral number line in the following way - on a given move, it jumps either to the next highest multiple of 3 or the next highest multiple of 13. Find the number of distinct possible paths the frog can take.<br />
<br />
[[2007 AIME I Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Let <br />
<math>N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor )</math><br />
<br />
Find the remainder when <math>N</math> is divided by 1000. (<math>\lfloor{k}\rfloor</math> is the greatest integer less than or equal to <math>k</math>, and <math>\lceil{k}\rceil</math> is the least integer greater than or equal to <math>k</math>.)<br />
<br />
[[2007 AIME I Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
The polynomial <math>P(x)</math> is cubic. What is the largest value of <math>k</math> for which the polynomials <math>Q_1(x) = x^2 + (k-29)x - k</math> and <math>Q_2(x) = 2x^2+ (2k-43)x + k</math> are both factors of <math>P(x)</math>?<br />
<br />
[[2007 AIME I Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
In right triangle <math>ABC</math> with right angle <math>C</math>, <math>CA = 30</math> and <math>CB = 16</math>. Its legs <math>CA</math> and <math>CB</math> are extended beyond <math>A</math> and <math>B</math>. Points <math>O_1</math> and <math>O_2</math> lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center <math>O_1</math> is tangent to the hypotenuse and to the extension of leg <math>CA</math>, the circle with center <math>O_2</math> is tangent to the hypotenuse and to the extension of leg <math>CB</math>, and the circles are externally tangent to each other. The length of the radius either circle can be expressed as <math>p/q</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
[[2007 AIME I Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let <math>N</math> be the number of shadings with this property. Find the remainder when <math>N</math> is divided by 1000.<br />
<br />
[[2007 AIME I Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
For each positive integer <math>p</math>, let <math>b(p)</math> denote the unique positive integer <math>k</math> such that <math>|k-\sqrt{p}| < \frac{1}{2}</math>. For example, <math>b(6) = 2</math> and <math>b(23) = 5</math>. If <math>S = \Sigma_{p=1}^{2007} b(p),</math> find the remainder when <math>S</math> is divided by 1000.<br />
<br />
[[2007 AIME I Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
In isosceles triangle <math>ABC</math>, <math>A</math> is located at the origin and <math>B</math> is located at (20,0). Point <math>C</math> is in the first quadrant with <math>AC = BC</math> and angle <math>BAC = 75^{\circ}</math>. If triangle <math>ABC</math> is rotated counterclockwise about point <math>A</math> until the image of <math>C</math> lies on the positive <math>y</math>-axis, the area of the region common to the original and the rotated triangle is in the form <math>p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s</math>, where <math>p,q,r,s</math> are integers. Find <math>(p-q+r-s)/2</math>.<br />
<br />
[[2007 AIME I Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
A square pyramid with base <math>ABCD</math> and vertex <math>E</math> has eight edges of length 4. A plane passes through the midpoints of <math>AE</math>, <math>BC</math>, and <math>CD</math>. The plane's intersection with the pyramid has an area that can be expressed as <math>\sqrt{p}</math>. Find <math>p</math>.<br />
<br />
[[2007 AIME I Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A sequence is defined over non-negative integral indexes in the following way: <math>a_{0}=a_{1}=3</math>, <math>a_{n+1}a_{n-1}=a_{n}^{2}+2007</math>.<br />
<br />
Find the greatest integer that does not exceed <math>\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}</math><br />
<br />
[[2007 AIME I Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Let <math>ABC</math> be an equilateral triangle, and let <math>D</math> and <math>F</math> be points on sides <math>BC</math> and <math>AB</math>, respectively, with <math>FA = 5</math> and <math>CD = 2</math>. Point <math>E</math> lies on side <math>CA</math> such that angle <math>DEF = 60^{\circ}</math>. The area of triangle <math>DEF</math> is <math>14\sqrt{3}</math>. The two possible values of the length of side <math>AB</math> are <math>p \pm q \sqrt{r}</math>, where <math>p</math> and <math>q</math> are rational, and <math>r</math> is an integer not divisible by the square of a prime. Find <math>r</math>.<br />
<br />
[[2007 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>Mathfanatichttps://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems&diff=140032007 AIME I Problems2007-03-14T23:57:16Z<p>Mathfanatic: </p>
<hr />
<div>== Problem 1 ==<br />
How many positive perfect squares less than <math>10^6</math> are multiples of 24?<br />
<br />
[[2007 AIME I Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person.<br />
[[2007 AIME I Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The complex number <math>z</math> is equal to <math>9+bi</math>, where <math>b</math> is a positive real number and <math>i^{2}=-1</math>. Given that the imaginary parts of <math>z^{2}</math> and <math>z^{3}</math> are the same, what is <math>b</math> equal to?<br />
<br />
[[2007 AIME I Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are <math>60</math>,<math>84</math>, and <math>140</math>. The three planets and the star are currently collinear. What is the fewest number of years from now that they will all be collinear again?<br />
<br />
[[2007 AIME I Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
The formula for converting a Fahrenheit temperature <math>F</math> to the corresponding Celsius temperature <math>C</math> is <math>C = \frac{5}{9}(F-32).</math> An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer.<br />
<br />
For how many integer Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature?<br />
<br />
[[2007 AIME I Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
A frog moves from 0 to 39 on an integral number line in the following way - on a given move, it jumps either to the next highest multiple of 3 or the next highest multiple of 13. Find the number of distinct possible paths the frog can take.<br />
<br />
[[2007 AIME I Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Let <br />
<math>N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor )</math><br />
<br />
Find the remainder when <math>N</math> is divided by 1000. (<math>\lfloor{k}\rfloor</math> is the greatest integer less than or equal to <math>k</math>, and <math>\lceil{k}\rceil</math> is the least integer greater than or equal to <math>k</math>.)<br />
<br />
[[2007 AIME I Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
The polynomial <math>P(x)</math> is cubic. What is the largest value of <math>k</math> for which the polynomials <math>Q_1(x) = x^2 + (k-29)x - k</math> and <math>Q_2(x) = 2x^2+ (2k-43)x + k</math> are both factors of <math>P(x)</math>?<br />
<br />
[[2007 AIME I Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
In right triangle <math>ABC</math> with right angle <math>C</math>, <math>CA = 30</math> and <math>CB = 16</math>. Its legs <math>CA</math> and <math>CB</math> are extended beyond <math>A</math> and <math>B</math>. Points <math>O_1</math> and <math>O_2</math> lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center <math>O_1</math> is tangent to the hypotenuse and to the extension of leg <math>CA</math>, the circle with center <math>O_2</math> is tangent to the hypotenuse and to the extension of leg <math>CB</math>, and the circles are externally tangent to each other. The length of the radius either circle can be expressed as <math>p/q</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
[[2007 AIME I Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let <math>N</math> be the number of shadings with this property. Find the remainder when <math>N</math> is divided by 1000.<br />
<br />
[[2007 AIME I Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
For each positive integer <math>p</math>, let <math>b(p)</math> denote the unique positive integer <math>k</math> such that <math>|k-\sqrt{p}| < \frac{1}{2}</math>. For example, <math>b(6) = 2</math> and <math>b(23) = 5</math>. If <math>S = \Sigma_{p=1}^{2007} b(p),</math> find the remainder when <math>S</math> is divided by 1000.<br />
<br />
[[2007 AIME I Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
In isosceles triangle <math>ABC</math>, <math>A</math> is located at the origin and <math>B</math> is located at (20,0). Point <math>C</math> is in the first quadrant with <math>AC = BC</math> and angle <math>BAC = 75^{\circ}</math>. If triangle <math>ABC</math> is rotated counterclockwise about point <math>A</math> until the image of <math>C</math> lies on the positive <math>y</math>-axis, the area of the region common to the original and the rotated triangle is in the form <math>p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s</math>, where <math>p,q,r,s</math> are integers. Find <math>(p-q+r-s)/2</math>.<br />
<br />
[[2007 AIME I Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
A square pyramid with base <math>ABCD</math> and vertex <math>E</math> has eight edges of length 4. A plane passes through the midpoints of <math>AE</math>, <math>BC</math>, and <math>CD</math>. The plane's intersection with the pyramid has an area that can be expressed as <math>\sqrt{p}</math>. Find <math>p</math>.<br />
<br />
[[2007 AIME I Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A sequence is defined over non-negative integral indexes in the following way: <math>a_{0}=a_{1}=3</math>, <math>a_{n+1}a_{n-1}=a_{n}^{2}+2007</math>.<br />
<br />
Find the greatest integer that does not exceed <math>\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}</math><br />
<br />
[[2007 AIME I Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Let <math>ABC</math> be an equilateral triangle, and let <math>D</math> and <math>F</math> be points on sides <math>BC</math> and <math>AB</math>, respectively, with <math>FA = 5</math> and <math>CD = 2</math>. Point <math>E</math> lies on side <math>CA</math> such that angle <math>DEF = 60^{\circ}</math>. The area of triangle <math>DEF</math> is <math>14\sqrt{3}</math>. The two possible values of the length of side <math>AB</math> are <math>p \pm q \sqrt{r}</math>, where <math>p</math> and <math>q</math> are rational, and <math>r</math> is an integer not divisible by the square of a prime. Find <math>r</math>.<br />
<br />
[[2007 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>Mathfanatichttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_15&diff=140022005 AIME II Problems/Problem 152007-03-14T18:06:46Z<p>Mathfanatic: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math> w_1 </math> and <math> w_2 </math> denote the circles <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest possible value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally tangent to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math><br />
<br />
== Solution ==<br />
<br />
Rewrite the given equations as<br />
<math>(x+5)^2 + (y-12)^2 = 256</math><br />
<math>(x-5)^2 + (y-12)^2 = 16</math><br />
<br />
Let <math>w_3</math> have center <math>(x,y)</math> and radius <math>r</math>. Now, if two circles with radii <math>r_1</math> and <math>r_2</math> are externally tangent, then the distance between their centers is <math>r_1 + r_2</math>, and if they are internally tangent, it is <math>|r_1 - r_2|</math>. So we have<br />
<br />
<math>r + 4 = \sqrt{(x-5)^2 + (y-12)^2}</math><br />
<br />
<math>16 - r = \sqrt{(x+5)^2 + (y-12)^2}.</math><br />
<br />
Solving for <math>r</math> in both equations and setting them equal yields<br />
<math>20 - \sqrt{(x+5)^2 + (y-12)^2} = \sqrt{(x-5)^2 + (y-12)^2}</math><br />
<br />
Squaring both sides, canceling common terms, and rearranging yields<br />
<br />
<math>20+x = 2\sqrt{(x+5)^2 + (y-12)^2}</math><br />
<br />
Squaring again and canceling yields<br />
<br />
<math>1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.</math><br />
<br />
So the locus of points that can be the center of the circle with the desired properties is an ellipse.<br />
<br />
Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math>:<br />
<br />
<math>1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75}</math><br />
<br />
Expanding yields<br />
<br />
<math>(3+4a^2)x^2 - 96ax + 276 = 0</math><br />
<br />
We want the value of <math>a</math> that makes the line <math>y=ax</math> tangent to the ellipse, which will mean that for that choice of <math>a</math> there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is 0. So<br />
<br />
<math>(-96a)^2 - 4(3+4a^2)(276) = 0</math><br />
<br />
Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is 169.<br />
<br />
== See also ==<br />
<br />
* [[2005 AIME II Problems/Problem 14| Previous problem]]<br />
* [[2005 AIME II Problems]]<br />
<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Mathfanatichttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_15&diff=140002005 AIME II Problems/Problem 152007-03-14T07:51:23Z<p>Mathfanatic: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math> w_1 </math> and <math> w_2 </math> denote the circles <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest possible value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally tangent to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math><br />
<br />
== Solution ==<br />
<br />
Rewrite the given equations as<br />
<math>(x+5)^2 + (y-12)^2 = 256</math><br />
<math>(x-5)^2 + (y-12)^2 = 16</math><br />
<br />
Let <math>w_3</math> have center <math>(x,y)</math> and radius <math>r</math>. Now, if two circles with radii <math>r_1</math> and <math>r_2</math> are externally tangent, then the distance between their centers is <math>r_1 + r_2</math>, and if they are internally tangent, it is <math>|r_1 - r_2|</math>. So we have<br />
<br />
<math>r + 4 = \sqrt{(x-5)^2 + (y-12)^2}</math><br />
<math>16 - r = \sqrt{(x+5)^2 + (y-12)^2}.</math><br />
<br />
Solving for <math>r</math> in both equations and setting them equal yields<br />
<math>20 - \sqrt{(x+5)^2 + (y-12)^2} = \sqrt{(x-5)^2 + (y-12)^2}</math><br />
<br />
Squaring both sides, canceling common terms, and rearranging yields<br />
<br />
<math>20+x = 2\sqrt{(x+5)^2 + (y-12)^2}</math><br />
<br />
Squaring again and canceling yields<br />
<br />
<math>1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.</math><br />
<br />
So the locus of points that can be the center of the circle with the desired properties is an ellipse.<br />
<br />
Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math>:<br />
<br />
<math>1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75}</math><br />
<br />
Expanding yields<br />
<math>(3+4a^2)x^2 - 96ax + 276 = 0</math><br />
<br />
We want the value of <math>a</math> that makes the line <math>y=ax</math> tangent to the ellipse, which will mean that for that choice of <math>a</math> there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is 0. So<br />
<br />
<math>(-96a)^2 - 4(3+4a^2)(276) = 0</math><br />
<br />
Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is 169.<br />
<br />
== See also ==<br />
<br />
* [[2005 AIME II Problems/Problem 14| Previous problem]]<br />
* [[2005 AIME II Problems]]<br />
<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Mathfanatichttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_15&diff=139992005 AIME II Problems/Problem 152007-03-14T07:50:43Z<p>Mathfanatic: </p>
<hr />
<div>== Problem ==<br />
<br />
Let <math> w_1 </math> and <math> w_2 </math> denote the circles <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest possible value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally tangent to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math><br />
<br />
== Solution ==<br />
<br />
Rewrite the given equations as<br />
<math>(x+5)^2 + (y-12)^2 = 256</math><br />
<math>(x-5)^2 + (y-12)^2 = 16</math><br />
<br />
Let <math>w_3</math> have center <math>(x,y)</math> and radius <math>r</math>. Now, if two circles with radii <math>r_1</math> and <math>r_2</math> are externally tangent, then the distance between their centers is <math>r_1 + r_2</math>, and if they are internally tangent, it is <math>|r_1 - r_2|</math>. So we have<br />
<br />
<math>r + 4 = \sqrt{(x-5)^2 + (y-12)^2}</math><br />
<math>16 - r = \sqrt{(x+5)^2 + (y-12)^2}.</math><br />
<br />
Solving for <math>r</math> in both equations and setting them equal yields<br />
<math>20 - \sqrt{(x+5)^2 + (y-12)^2} = \sqrt{(x-5)^2 + (y-12)^2}</math><br />
<br />
Squaring both sides, canceling common terms, and rearranging yields<br />
<br />
<math>20+x = 2\sqrt{(x+5)^2 + (y-12)^2}</math><br />
<br />
Squaring again and canceling yields<br />
<br />
<math>1 = \frac{x^2}{100} + {(y-12)^2}{75}.</math><br />
<br />
So the locus of points that can be the center of the circle with the desired properties is an ellipse.<br />
<br />
Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math>:<br />
<br />
<math>1 = \frac{x^2}{100} + {(ax-12)^2}{75}</math><br />
<br />
Expanding yields<br />
<math>(3+4a^2)x^2 - 96ax + 276 = 0</math><br />
<br />
We want the value of <math>a</math> that makes the line <math>y=ax</math> tangent to the ellipse, which will mean that for that choice of <math>a</math> there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is 0. So<br />
<br />
<math>(-96a)^2 - 4(3+4a^2)(276) = 0</math><br />
<br />
Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is 169.<br />
<br />
== See also ==<br />
<br />
* [[2005 AIME II Problems/Problem 14| Previous problem]]<br />
* [[2005 AIME II Problems]]<br />
<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Mathfanatichttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_6_Pre_2005&diff=9656Mock AIME 6 Pre 20052006-08-17T23:21:17Z<p>Mathfanatic: /* See also */</p>
<hr />
<div>The '''Mock AIME 6 Pre 2005''' was written by [[Art of Problem Solving]] community member paladin8.<br />
<br />
* [[Mock AIME 6 Pre 2005/Problems|Entire Exam]]<br />
* [[Mock AIME 6 Pre 2005/Answer Key|Answer Key]]<br />
* [[Mock AIME 6 Pre 2005/Problem 1|Problem 1]]<br />
* [[Mock AIME 6 Pre 2005/Problem 2|Problem 2]]<br />
* [[Mock AIME 6 Pre 2005/Problem 3|Problem 3]]<br />
* [[Mock AIME 6 Pre 2005/Problem 4|Problem 4]]<br />
* [[Mock AIME 6 Pre 2005/Problem 5|Problem 5]]<br />
* [[Mock AIME 6 Pre 2005/Problem 6|Problem 6]]<br />
* [[Mock AIME 6 Pre 2005/Problem 7|Problem 7]]<br />
* [[Mock AIME 6 Pre 2005/Problem 8|Problem 8]]<br />
* [[Mock AIME 6 Pre 2005/Problem 9|Problem 9]]<br />
* [[Mock AIME 6 Pre 2005/Problem 10|Problem 10]]<br />
* [[Mock AIME 6 Pre 2005/Problem 11|Problem 11]]<br />
* [[Mock AIME 6 Pre 2005/Problem 12|Problem 12]]<br />
* [[Mock AIME 6 Pre 2005/Problem 13|Problem 13]]<br />
* [[Mock AIME 6 Pre 2005/Problem 14|Problem 14]]<br />
* [[Mock AIME 6 Pre 2005/Problem 15|Problem 15]]<br />
<br />
== See also ==<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mock AIME]] <br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28066 Problems] <-- THIS LINK IS WRONG!<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28368 Solutions]</div>Mathfanatic