https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Apsid&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T00:40:51ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_15&diff=1615612006 AMC 10A Problems/Problem 152021-09-04T21:12:01Z<p>Apsid: Added a solution using the angle between the starting point and first meeting point</p>
<hr />
<div>== Problem ==<br />
Odell and Kershaw run for 30 minutes on a [[circle|circular]] track. Odell runs clockwise at 250 m/min and uses the inner lane with a [[radius]] of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial [[line]] as Odell. How many times after the start do they pass each other? <br />
<br />
<math>\mathrm{(A) \ } 29\qquad\mathrm{(B) \ } 42\qquad\mathrm{(C) \ } 45\qquad\mathrm{(D) \ } 47\qquad\mathrm{(E) \ } 50\qquad</math><br />
<br />
== Solution ==<br />
<center><asy><br />
draw((5,0){up}..{left}(0,5),red);<br />
draw((-5,0){up}..{right}(0,5),red);<br />
draw((5,0){down}..{left}(0,-5),red);<br />
draw((-5,0){down}..{right}(0,-5),red);<br />
draw((6,0){up}..{left}(0,6),blue);<br />
draw((-6,0){up}..{right}(0,6),blue);<br />
draw((6,0){down}..{left}(0,-6),blue);<br />
draw((-6,0){down}..{right}(0,-6),blue);<br />
</asy></center><br />
<br />
Since <math>d = rt</math>, we note that Odell runs one lap in <math>\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}</math> minutes, while Kershaw also runs one lap in <math>\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}</math> minutes. They take the same amount of [[time]] to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are <math>\frac{30}{\frac{2\pi}{5}} \approx 23.8</math> laps run by both, or <math>\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 = 47</math> meeting points <math> \Longrightarrow \mathrm{(D)}</math>.<br />
== Solution 2 ==<br />
We first find the amount of minutes, k, until Odell and Kershaw's next meeting. Let <math>a</math> be the angle in radians between their starting point and the point where they first meet, measured counterclockwise. Since Kershaw has traveled 300k meters at this point and the circumference of his track is <math>120\pi</math>, <math>a=\frac{300k}{120\pi}\cdot 2\pi</math>. Similarly, <math>2\pi-a=\frac{250k}{100\pi}\cdot{2\pi}</math> since Odell has traveled 250k meters in the opposite direction and the circumference of his track is <math>100\pi</math>. Solving for a in the second equation, <math>a=2\pi-\frac{250k}{100\pi}\cdot 2\pi</math>. Then, from the first equation, <math>\frac{300k}{120\pi}\cdot 2\pi=2\pi-\frac{250k}{100\pi}\cdot 2\pi</math>. Solving for k, we get <math>k=\frac{\pi}{5}</math>. After k minutes, they are back at the same position, except rotated, so they will meet again in k minutes. So the total amount of meetings is <math>\lfloor\frac{30}{k}\rfloor=\lfloor\frac{150}{\pi}\rfloor=47</math>. So, the answer is <math>\boxed{(D)}</math>.<br />
<br />
~apsid<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2006|ab=A|num-b=14|num-a=16}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Apsidhttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_6&diff=1615552005 AMC 12B Problems/Problem 62021-09-04T17:16:28Z<p>Apsid: Added a solution using Stewart's Theorem</p>
<hr />
<div>{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #6]] and [[2005 AMC 10B Problems|2005 AMC 10B #10]]}}<br />
== Problem ==<br />
In <math>\triangle ABC</math>, we have <math>AC=BC=7</math> and <math>AB=2</math>. Suppose that <math>D</math> is a point on line <math>AB</math> such that <math>B</math> lies between <math>A</math> and <math>D</math> and <math>CD=8</math>. What is <math>BD</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 3 \qquad<br />
\mathrm{(B)}\ 2\sqrt{3} \qquad<br />
\mathrm{(C)}\ 4 \qquad<br />
\mathrm{(D)}\ 5 \qquad<br />
\mathrm{(E)}\ 4\sqrt{2}<br />
</math><br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 ===<br />
Draw height <math>CH</math> (Perpendicular line from point C to line AD). We have that <math>BH=1</math>. From the [[Pythagorean Theorem]], <math>CH=\sqrt{48}</math>. Since <math>CD=8</math>, <math>HD=\sqrt{8^2-48}=\sqrt{16}=4</math>, and <math>BD=HD-1</math>, so <math>BD=\boxed{\text{(A)}3}</math>.<br />
<br />
=== Solution 2 (Trig) ===<br />
<br />
After drawing out a diagram, let <math>\angle{ABC}=\theta</math>. By the Law of Cosines, <math>7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}</math>. In <math>\triangle CBD</math>, we have <math>\angle{CBD}=(180-\theta)</math>, and using the identity <math>\cos(180-\theta)=-\cos{\theta}</math> and Law of Cosines one more time: <math>8^2=7^2+x^2-2(7)(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0</math>. The only positive value for <math>x</math> is <math>3</math>, which gives the length of <math>\overline{BD}</math>. Thus the answer is <math>\boxed{\text{A}}</math>.<br />
<br />
~Bowser498<br />
<br />
== Solution 3 (Stewart's Theorem) ==<br />
Let <math>BD=k</math>. Then, by Stewart's Theorem, <br />
<br />
<math>2k(2+k)+7^2(2+k)=7^2k+8^2\cdot 2<br />
\implies k^2-2k-15=0<br />
\implies k=\boxed{3}</math><br />
<br />
~apsid<br />
<br />
<br />
== See also ==<br />
{{AMC10 box|year=2005|ab=B|num-b=9|num-a=11}}<br />
{{AMC12 box|year=2005|ab=B|num-b=5|num-a=7}}<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Apsidhttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_24&diff=1614922005 AMC 10A Problems/Problem 242021-09-03T15:41:59Z<p>Apsid: /* Note: Solution 1 */</p>
<hr />
<div>==Problem==<br />
For each positive integer <math> n > 1 </math>, let <math>P(n)</math> denote the greatest prime factor of <math>n</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 />
==Solution 1==<br />
If <math> P(n) = \sqrt{n} </math>, then <math> n = p_{1}^{2} </math>, where <math> p_{1} </math> is a [[prime number]].<br />
<br />
If <math> P(n+48) = \sqrt{n+48} </math>, then <math>n + 48</math> is a square, but we know that n is <math>p_{1}^{2} </math>.<br />
<br />
<br />
This means we just have to check for squares of primes, add 48 and look whether the root is a prime number.<br />
We can easily see that the difference between two consecutive square after 576 is greater than or equal to 49,<br />
Hence we have to consider only the prime numbers till 23.<br />
<br />
<br />
Squaring prime numbers below 23 including 23 we get the following list.<br />
<br />
<math>4 , 9 , 25 , 49 , 121, 169 , 289 , 361 , 529</math><br />
<br />
<br />
But adding 48 to a number ending with 9 will result in a number ending with 7, but we know that a perfect square does not end in 7, so we can eliminate those cases to get the new list.<br />
<br />
<math>4 , 25 , 121 , 361</math><br />
<br />
<br />
Adding 48, we get 121 as the only possible solution.<br />
Hence the answer is (B).<br />
<br />
The only positive integer that satisfies both requirements is 11.<br />
<br />
edited by mobius247<br />
<br />
==Note: Solution 1==<br />
Since all primes greater than 2 are odd, we know that the difference between the squares of any two consecutive primes greater than 2 is at least <math>(p+2)^2-p^2=4p+4</math>, where p is the smaller of the consecutive primes. For <math>p>11</math>, <math>4p+4>48</math>. This means that the difference between the squares of any two consecutive primes both greater than 11 is greater than 48, so <math>n</math> and <math>n+48</math> can't both be the squares of primes if <math>n=p^2</math> and p>11. So, we only need to check <math>n=2^2, 3^2, 5^2, 7^2, 11^2</math>.<br />
<br />
~apsid<br />
<br />
==Video Solution==<br />
CHECK OUT Video Solution:https://youtu.be/IsqrsMkR-mA<br />
<br />
~<i>rudolf1279<i><br />
<br />
==Solution 2==<br />
If <math> P(n) = \sqrt{n} </math>, then <math> n = p_{1}^{2} </math>, where <math> p_{1} </math> is a [[prime number]]. <br />
<br />
If <math> P(n+48) = \sqrt{n+48} </math>, then <math> n+48 = p_{2}^{2} </math>, where <math> p_{2} </math> is a different prime number. <br />
<br />
So: <br />
<br />
<math> p_{2}^{2} = n+48 </math><br />
<br />
<math> p_{1}^{2} = n </math><br />
<br />
<math> p_{2}^{2} - p_{1}^{2} = 48 </math><br />
<br />
<math> (p_{2}+p_{1})(p_{2}-p_{1})=48 </math><br />
<br />
Since <math> p_{1} > 0 </math> : <math> (p_{2}+p_{1}) > (p_{2}-p_{1}) </math>. <br />
<br />
Looking at pairs of [[divisor]]s of <math>48</math>, we have several possibilities to solve for <math>p_{1}</math> and <math>p_{2}</math>:<br />
<br />
<br />
<math> (p_{2}+p_{1}) = 48 </math><br />
<br />
<math> (p_{2}-p_{1}) = 1 </math><br />
<br />
<math> p_{1} = \frac{47}{2} </math><br />
<br />
<math> p_{2} = \frac{49}{2} </math><br />
<br />
<br />
<math> (p_{2}+p_{1}) = 24 </math><br />
<br />
<math> (p_{2}-p_{1}) = 2 </math><br />
<br />
<math> p_{1} = 11 </math><br />
<br />
<math> p_{2} = 13 </math><br />
<br />
<br />
<math> (p_{2}+p_{1}) = 16 </math><br />
<br />
<math> (p_{2}-p_{1}) = 3 </math><br />
<br />
<math> p_{1} = \frac{13}{2} </math><br />
<br />
<math> p_{2} = \frac{19}{2} </math><br />
<br />
<br />
<math> (p_{2}+p_{1}) = 12 </math><br />
<br />
<math> (p_{2}-p_{1}) = 4 </math><br />
<br />
<math> p_{1} = 4 </math><br />
<br />
<math> p_{2} = 8 </math><br />
<br />
<br />
<math> (p_{2}+p_{1}) = 8 </math><br />
<br />
<math> (p_{2}-p_{1}) = 6 </math><br />
<br />
<math> p_{1} = 1 </math><br />
<br />
<math> p_{2} = 7 </math><br />
<br />
<br />
The only solution <math> (p_{1} , p_{2}) </math> where both numbers are primes is <math>(11,13)</math>. <br />
<br />
Therefore the number of [[positive integer]]s <math>n</math> that satisfy both statements is <math>1\Rightarrow \mathrm{(B)}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2005|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Apsidhttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_24&diff=1614912005 AMC 10A Problems/Problem 242021-09-03T15:41:21Z<p>Apsid: A note about a quicker solution using similar methods to Solution 1</p>
<hr />
<div>==Problem==<br />
For each positive integer <math> n > 1 </math>, let <math>P(n)</math> denote the greatest prime factor of <math>n</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 />
==Solution 1==<br />
If <math> P(n) = \sqrt{n} </math>, then <math> n = p_{1}^{2} </math>, where <math> p_{1} </math> is a [[prime number]].<br />
<br />
If <math> P(n+48) = \sqrt{n+48} </math>, then <math>n + 48</math> is a square, but we know that n is <math>p_{1}^{2} </math>.<br />
<br />
<br />
This means we just have to check for squares of primes, add 48 and look whether the root is a prime number.<br />
We can easily see that the difference between two consecutive square after 576 is greater than or equal to 49,<br />
Hence we have to consider only the prime numbers till 23.<br />
<br />
<br />
Squaring prime numbers below 23 including 23 we get the following list.<br />
<br />
<math>4 , 9 , 25 , 49 , 121, 169 , 289 , 361 , 529</math><br />
<br />
<br />
But adding 48 to a number ending with 9 will result in a number ending with 7, but we know that a perfect square does not end in 7, so we can eliminate those cases to get the new list.<br />
<br />
<math>4 , 25 , 121 , 361</math><br />
<br />
<br />
Adding 48, we get 121 as the only possible solution.<br />
Hence the answer is (B).<br />
<br />
The only positive integer that satisfies both requirements is 11.<br />
<br />
edited by mobius247<br />
<br />
==Note: Solution 1==<br />
Since all primes greater than 2 are odd, we know that the difference between the squares of any two consecutive primes greater than 2 is at least <math>(p+2)^2-p^2=4p+4</math>, where p is the smaller of the consecutive primes. For <math>p>11</math>, <math>4p+4>48</math>. This means that the difference between the squares of any two consecutive primes both greater than 11 is greater than 48, so <math>n</math> and <math>n+48</math> can't both be the squares of primes if <math>n=p^2</math> and p>11. So, we only need to check <math>n=2^2, 3^2, 5^2, 7^2, 11^2</math>.<br />
~apsid<br />
<br />
==Video Solution==<br />
CHECK OUT Video Solution:https://youtu.be/IsqrsMkR-mA<br />
<br />
~<i>rudolf1279<i><br />
<br />
==Solution 2==<br />
If <math> P(n) = \sqrt{n} </math>, then <math> n = p_{1}^{2} </math>, where <math> p_{1} </math> is a [[prime number]]. <br />
<br />
If <math> P(n+48) = \sqrt{n+48} </math>, then <math> n+48 = p_{2}^{2} </math>, where <math> p_{2} </math> is a different prime number. <br />
<br />
So: <br />
<br />
<math> p_{2}^{2} = n+48 </math><br />
<br />
<math> p_{1}^{2} = n </math><br />
<br />
<math> p_{2}^{2} - p_{1}^{2} = 48 </math><br />
<br />
<math> (p_{2}+p_{1})(p_{2}-p_{1})=48 </math><br />
<br />
Since <math> p_{1} > 0 </math> : <math> (p_{2}+p_{1}) > (p_{2}-p_{1}) </math>. <br />
<br />
Looking at pairs of [[divisor]]s of <math>48</math>, we have several possibilities to solve for <math>p_{1}</math> and <math>p_{2}</math>:<br />
<br />
<br />
<math> (p_{2}+p_{1}) = 48 </math><br />
<br />
<math> (p_{2}-p_{1}) = 1 </math><br />
<br />
<math> p_{1} = \frac{47}{2} </math><br />
<br />
<math> p_{2} = \frac{49}{2} </math><br />
<br />
<br />
<math> (p_{2}+p_{1}) = 24 </math><br />
<br />
<math> (p_{2}-p_{1}) = 2 </math><br />
<br />
<math> p_{1} = 11 </math><br />
<br />
<math> p_{2} = 13 </math><br />
<br />
<br />
<math> (p_{2}+p_{1}) = 16 </math><br />
<br />
<math> (p_{2}-p_{1}) = 3 </math><br />
<br />
<math> p_{1} = \frac{13}{2} </math><br />
<br />
<math> p_{2} = \frac{19}{2} </math><br />
<br />
<br />
<math> (p_{2}+p_{1}) = 12 </math><br />
<br />
<math> (p_{2}-p_{1}) = 4 </math><br />
<br />
<math> p_{1} = 4 </math><br />
<br />
<math> p_{2} = 8 </math><br />
<br />
<br />
<math> (p_{2}+p_{1}) = 8 </math><br />
<br />
<math> (p_{2}-p_{1}) = 6 </math><br />
<br />
<math> p_{1} = 1 </math><br />
<br />
<math> p_{2} = 7 </math><br />
<br />
<br />
The only solution <math> (p_{1} , p_{2}) </math> where both numbers are primes is <math>(11,13)</math>. <br />
<br />
Therefore the number of [[positive integer]]s <math>n</math> that satisfy both statements is <math>1\Rightarrow \mathrm{(B)}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2005|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Apsid