https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Yizhou76&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T13:56:26ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_2&diff=1266582007 AIME I Problems/Problem 22020-06-26T22:27:34Z<p>Yizhou76: /* Problem */</p>
<hr />
<div>== Problem ==<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 />
== Solution ==<br />
Clearly we have people ''moving'' at speeds of <math>6,8</math> and <math>10</math> feet/second. Notice that out of the three people, Cy is at the largest disadvantage to begin with and since all speeds are close, it is hardest for him to catch up. Furthermore, Bob is clearly the farthest along. Thus it is reasonable to assume that there is some point when Al is halfway between Cy and Bob. At this time <math>s</math>, we have that<br />
<br />
<math><br />
\frac{8(s-4)+10(s-2)}{2}=6s<br />
</math><br />
After solving, <math>s=\frac{26}{3}</math>. At this time, Al has traveled <math>6\cdot\frac{26}{3}=52</math> feet.<br />
<br />
We could easily check that Al is in the middle by trying all three possible cases. <math>\frac{6s + 8(s-4)}{2} = 10(s-2)</math> yields that <math>s = \frac 43</math>, which can be disregarded since both Bob and Cy hadn't started yet. <math>\frac{6s + 10(s-2)}{2} = 8(s-4)</math> yields that <math>-10=-32</math>, a contradiction. Thus, the answer is <math>\boxed{52}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2007|n=I|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Yizhou76https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems&diff=1266572007 AIME I Problems2020-06-26T22:27:11Z<p>Yizhou76: /* Problem 2 */</p>
<hr />
<div>{{AIME Problems|year=2007|n=I}}<br />
<br />
== 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> years. 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 <math>32</math> and <math>1000</math> 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 is placed at the [[origin]] on the [[number line]], and moves according to the following rule: in a given move, the frog advances to either the closest [[point]] with a greater [[integer]] [[coordinate]] that is a multiple of <math>3</math>, or to the closest point with a greater integer coordinate that is a multiple of <math>13</math>. A ''move sequence'' is a [[sequence]] of coordinates which correspond to valid moves, beginning with <math>0</math>, and ending with <math>39</math>. For example, <math>0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39</math> is a move sequence. How many move sequences are possible for the frog?<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 of 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 />
[[Image:AIME I 2007-9.png]]<br />
<br />
[[2007 AIME I Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
In a <math>6 \times 4</math> grid (<math>6</math> rows, <math>4</math> columns), <math>12</math> of the <math>24</math> 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 <math>1000</math>.<br />
<br />
[[Image:AIME I 2007-10.png]]<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 = \sum_{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 />
[[Image:AIME I 2007-13.png]]<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]]<br />
{{MAA Notice}}</div>Yizhou76https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems&diff=1233482007 AIME I Problems2020-05-31T18:09:29Z<p>Yizhou76: /* Problem 2 */</p>
<hr />
<div>{{AIME Problems|year=2007|n=I}}<br />
<br />
== 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 2 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 4 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> years. 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 <math>32</math> and <math>1000</math> 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 is placed at the [[origin]] on the [[number line]], and moves according to the following rule: in a given move, the frog advances to either the closest [[point]] with a greater [[integer]] [[coordinate]] that is a multiple of <math>3</math>, or to the closest point with a greater integer coordinate that is a multiple of <math>13</math>. A ''move sequence'' is a [[sequence]] of coordinates which correspond to valid moves, beginning with <math>0</math>, and ending with <math>39</math>. For example, <math>0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39</math> is a move sequence. How many move sequences are possible for the frog?<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 of 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 />
[[Image:AIME I 2007-9.png]]<br />
<br />
[[2007 AIME I Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
In a <math>6 \times 4</math> grid (<math>6</math> rows, <math>4</math> columns), <math>12</math> of the <math>24</math> 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 <math>1000</math>.<br />
<br />
[[Image:AIME I 2007-10.png]]<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 = \sum_{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 />
[[Image:AIME I 2007-13.png]]<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]]<br />
{{MAA Notice}}</div>Yizhou76https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_2&diff=1233472007 AIME I Problems/Problem 22020-05-31T17:59:16Z<p>Yizhou76: /* Problem */</p>
<hr />
<div>== Problem ==<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 2 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 4 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 />
== Solution ==<br />
Clearly we have people ''moving'' at speeds of <math>6,8</math> and <math>10</math> feet/second. Notice that out of the three people, Cy is at the largest disadvantage to begin with and since all speeds are close, it is hardest for him to catch up. Furthermore, Bob is clearly the farthest along. Thus it is reasonable to assume that there is some point when Al is halfway between Cy and Bob. At this time <math>s</math>, we have that<br />
<br />
<math><br />
\frac{8(s-4)+10(s-2)}{2}=6s<br />
</math><br />
After solving, <math>s=\frac{26}{3}</math>. At this time, Al has traveled <math>6\cdot\frac{26}{3}=52</math> feet.<br />
<br />
We could easily check that Al is in the middle by trying all three possible cases. <math>\frac{6s + 8(s-4)}{2} = 10(s-2)</math> yields that <math>s = \frac 43</math>, which can be disregarded since both Bob and Cy hadn't started yet. <math>\frac{6s + 10(s-2)}{2} = 8(s-4)</math> yields that <math>-10=-32</math>, a contradiction. Thus, the answer is <math>\boxed{52}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2007|n=I|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Yizhou76