https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Awesomediabrine&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T20:03:38ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_15&diff=1687661998 AIME Problems/Problem 152021-12-28T22:52:50Z<p>Awesomediabrine: </p>
<hr />
<div>== Problem ==<br />
Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which <math>(i,j)</math> and <math>(j,i)</math> do not both appear for any <math>i</math> and <math>j</math>. Let <math>D_{40}</math> be the set of all dominos whose coordinates are no larger than 40. Find the length of the longest proper sequence of dominos that can be formed using the dominos of <math>D_{40}.</math><br />
<br />
== Solution 1 ==<br />
We can draw a comparison between the domino a set of 40 points (labeled 1 through 40) in which every point is connected with every other point. The connections represent the dominoes.<br />
<br />
You need to have all even number of [[segment]]s coming from each point except 0 or 2 which have an odd number of segments coming from the point. (Reasoning for this: Everytime you go to a vertex, you have to leave the vertex, so every vertex reached is equivalent to adding 2 more segments. So the degree of each vertex must be even, with the exception of endpoints) Since there are 39 segments coming from each point it is impossible to touch every segment.<br />
<br />
But you can get up to 38 on each segment because you go in to the point then out on a different segment. Counting going out from the starting and ending at the ending point we have:<br />
<br />
<math>\frac{38\cdot 38 + 2\cdot 39}2 = \boxed{761}</math><br />
<br />
''Clarification'' : To clarify the above solution, every time you move to a new vertex, you take one path in and must leave by another path. Therefore every vertex needs to have an even number of segments leaving it (with the exception of the first and last), because the "in" and "out" segments must make a pair.<br />
<br />
== Solution 2 ==<br />
A proper sequence can be represented by writing the common coordinates of adjacent ordered pairs once. For example, represent (4,7),(7,3),(3,5) as <math>4,7,3,5 .</math> Label the vertices of a regular <math>n</math> -gon <math>1,2,3, \ldots, n .</math> Each domino is thereby represented by a directed segment from one vertex of the <math>n</math> -gon to another, and a proper sequence is represented as a path that retraces no segment. Each time that such a path reaches a non-terminal vertex, it must leave it. <br />
<br />
<br />
Thus, when <math>n</math> is even, it is not possible for such a path to trace every segment, for an odd number of segments emanate from each vertex. By removing <math>\frac{1}{2}(n-2)</math> suitable segments, however, it can be arranged that <math>n-2</math> segments will emanate from <math>n-2</math> of the vertices and that an odd number of segments will emanate from exactly two of the vertices. <br />
<br />
<br />
In this situation, a path can be found that traces every remaining segment exactly once, starting at one of the two exceptional vertices and finishing at the other. This path will have length <math>\left(\begin{array}{c}{n-1} \\ 2\end{array}\right)-\frac{1}{2}(n-2),</math> which is 761 when <math>n=40</math>.<br />
<br />
~phoenixfire<br />
=== Note 1 ===<br />
When <math>n</math> is odd, a proper sequence of length <math>\left(\begin{array}{c}{n-1} \\ 2\end{array}\right)</math> can be found using the dominos of <math>D_{n}</math>. In this case, the second coordinate of the final domino equals the first coordinate of the first domino. <br />
<br />
~phoenixfire<br />
<br />
== Solution 3 ==<br />
Let <math>A_{n}=\{1,2,3, \ldots, n\}</math> and <math>D_{n}</math> be the set of dominos that can be formed using integers in <math>A_{n} .</math> Each <math>k</math> in <math>A_{n}</math> appears in <math>2(n-1)</math> dominos in <math>D_{n},</math> hence appears at most <math>n-1</math> times in a proper sequence from <math>D_{n}.</math> Except possibly for the integers <math>i</math> and <math>j</math> that begin and end a proper sequence, every integer appears an even number of times in the sequence. <br />
<br />
<br />
Thus, if <math>n</math> is even, each integer different from <math>i</math> and <math>j</math> appears on at most <math>n-2</math> dominos in the sequence, because <math>n-2</math> is even, and <math>i</math> and <math>j</math> themselves appear on at most <math>n-1</math> dominos each. This gives an upper bound of<br />
<cmath><br />
\frac{1}{2}\left[(n-2)^{2}+2(n-1)\right]=\frac{n^{2}-2 n+2}{2}<br />
</cmath><br />
dominos in the longest proper sequence in <math>D_{n}.</math> This bound is in fact attained for every even <math>n.</math> It is easy to verify this for <math>n=2</math>, so assume inductively that a sequence of this length has been found for a particular value of <math>n</math>. <br />
<br />
<br />
Without loss of generality, assume <math>i=1</math> and <math>j=2,</math> and let <math>_{p} X_{p+2}</math> denote a four-domino sequence of the form <math>(p, n+1)(n+1, p+1)(p+1, n+2)(n+2, p+2) .</math> By appending<br />
<cmath><br />
{ }_{2} X_{4},{ }_{4} X_{6}, \ldots,{ }_{n-2} X_{n},(n, n+1)(n+1,1)(1, n+2)(n+2,2)<br />
</cmath><br />
to the given proper sequence, a proper sequence of length<br />
<cmath><br />
\frac{n^{2}-2 n+2}{2}+4 \cdot \frac{n-2}{2}+4=\frac{n^{2}+2 n+2}{2}=\frac{(n+2)^{2}-2(n+2)+2}{2}<br />
</cmath><br />
is obtained that starts at <math>i=1</math> and ends at <math>j=2 .</math> This completes the inductive proof. <br />
<br />
<br />
In particular, the longest proper sequence when <math>n=40</math> is 761.<br />
<br />
~phoenixfire<br />
<br />
=== Note 2 ===<br />
In the language of graph theory, this is an example of an Eulerian circuit.<br />
<br />
~phoenixfire<br />
<br />
==Solution 4== <br />
Consider the segments joining the vertices of a regular <math>n</math>-gon. For odd <math>n</math>, we see that the number of segments is quite easily <math>\binom{n-1}{2}</math>. This is because every vertex touches every other vertex the same number of times. (<math>\frac{n-1}{2}</math> times to be exact). Hence the answer for odd cases is <math>n\frac{n-1}{2}=\binom{n-1}{2}</math>. (This is because a segment that starts at the first vertex also ends at the first vertex).<br />
For even <math>n</math> however, every vertex touches <math>\frac{n-2}{2}</math> vertices. However, one may be motivated to say that the answer (as in the odd case) is <math>n\frac{n-2}{2}</math>. But this is incorrect, because for the even case, it never ends at the vertex (the first vertex) you started at. So it must end at another vertex. But that vertex has already <math>\frac{n-2}{2}</math> other segments touching it. So we have that the final answer is <math>1</math> plus <math>n\frac{n-2}{2}</math>. The case for <math>n=40</math>, is <math>761</math>.<br />
<br />
~th1nq3r<br />
<br />
==Note==<br />
The reason it touches every single other vertex <math>\frac{n-1}{2}</math> for odd <math>n</math> is because there are a total of <math>\binom{n-1}{2}</math> segments, and once dividing <math>\binom{n-1}{2}</math> by <math>n</math>, you will then have the number of segments that are connected to each vertex.<br />
For the even case every vertex has at least <math>\frac{n-2}{2}</math> other segments touching it. This is because (you CAN convince yourself through a painful induction/observation argument, but you probably shouldn't) <math>n</math> is even, and if you try to apply the odd case to the even case, (namely that there is <math>\frac{n-1}{2}</math> segments touching each vertex), there would have to then be <math>n\frac{n-1}{2}</math> total segments, which never works since it never loops back to the vertex you started on). So at LEAST, there should be <math>\frac{n-2}{2}</math> segments touching each vertex. However, there is also at most <math>\frac{n-2}{2}</math> segments touching each vertex. Hence by the (what I call) the "less-than-or-greater-than" argument, there must be <math>\frac{n-2}{2}</math> segments out of each vertex. (Mind the plus one extra vertex since once again, the vertex you started on, it doesn't loop around, so it must end at another vertex). Hence the answer is <math>n\frac{n-2}{2}+1</math>. <br />
<br />
(I am not very good at explaining things. Sorry if it didn't make sense. Maybe if you find some way to contact me on aops, I could try and help). (For instance, I could see possible confusion at the part where I claim that the minimum lines be <math>\frac{n-2}{2}</math>. You might think, "Oh why not <math>\frac{n-1}{2}-1</math>, or even <math>\frac{n-1}{2}-2</math> or even subtract <math>3</math>?" Well if you actually start off with the fact that there should be at MOST <math>\frac{n-2}{2}</math> segments from each vertex, then it is obvious, since then there must be a minimum of <math>\frac{n-2}{2}</math> segments from each vertex).<br />
<br />
~th1nq3r<br />
<br />
== Solution 5 ==<br />
<br />
We can see that <math>|D_{40}| = 780</math>, since we are choosing 2 integers <math>[1,40]</math>, and order doesn't matter because <math>(i,j)</math> and <math>(j,i)</math> aren't both in the set. Then from doing a smaller example of <math>D_4</math>, we can note that non-endpoints must have an even number of pairs in <math>D</math> in order for one domino's end to match another's beginning. Then, in order to maximize the length we want to minimize the number of dominoes we remove to make all pairs be even (except endpoints). Then we can see that for every two pairs with an odd number of pairs, we can connect single ends to form a circular chain (i.e. if we want to ensure that 2,3 have even pairs then we can imagine (3,1)(1,2)). Thus we can divide 40 by 4 to get the number of numbers to remove. Thus we need to remove 10 numbers. But each number is connected to another, so we remove 20 dominoes. Thus there are 780-20 = 760 dominoes not including endpoints. Finally, we can add one more domino at the end since it doesn't need to match up with the first of another to make it 761.<br />
<br />
== See also ==<br />
{{AIME box|year=1998|num-b=14|after=Last question}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_6&diff=1684722000 AIME I Problems/Problem 62021-12-25T22:28:31Z<p>Awesomediabrine: /* Solutions */</p>
<hr />
<div>== Problem ==<br />
For how many [[ordered pair]]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^6</math> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the [[geometric mean]] of <math>x</math> and <math>y</math>?<br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
<cmath>\begin{eqnarray*}<br />
\frac{x+y}{2} &=& \sqrt{xy} + 2\\<br />
x+y-4 &=& 2\sqrt{xy}\\<br />
y - 2\sqrt{xy} + x &=& 4\\<br />
\sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}</cmath><br />
<br />
Because <math>y > x</math>, we only consider <math>+2</math>.<br />
<br />
For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation.<br />
<br />
The maximum that <math>\sqrt{y}</math> can be is <math>\sqrt{10^6} - 1 = 999</math> because <math>\sqrt{y}</math> must be an integer (this is because <math>\sqrt{y} - \sqrt{x} = 2</math>, an integer). Then <math>\sqrt{x} = 997</math>, and we continue this downward until <math>\sqrt{y} = 3</math>, in which case <math>\sqrt{x} = 1</math>. The number of pairs of <math>(\sqrt{x},\sqrt{y})</math>, and so <math>(x,y)</math> is then <math>\boxed{997}</math>.<br />
<!-- solution lost in edit conflict - azjps -<br />
Since <math>y>x</math>, it follows that each ordered pair <math>(x,y) = (n^2, (n+2)^2)</math> satisfies this equation. The minimum value of <math>x</math> is <math>1</math> and the maximum value of <math>y = 999^2</math> which would make <math>x = 997^2</math>. Thus <math>x</math> can be any of the squares between <math>1</math> and <math>997^2</math> inclusive and the answer is <math>\boxed{997}</math>.<br />
--><br />
<br />
=== Solution 2 ===<br />
<br />
<br />
Let <math>a^2</math> = <math>x</math> and <math>b^2</math> = <math>y</math>, where <math>a</math> and <math>b</math> are positive.<br />
<br />
Then <cmath>\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2</cmath><br />
<cmath>a^2 + b^2 = 2ab + 4</cmath><br />
<cmath>(a-b)^2 = 4</cmath><br />
<cmath>(a-b) = \pm 2</cmath><br />
<br />
This makes counting a lot easier since now we just have to find all pairs <math>(a,b)</math> that differ by 2.<br />
<br />
<br />
Because <math>\sqrt{10^6} = 10^3</math>, then we can use all positive integers less than 1000 for <math>a</math> and <math>b</math>.<br />
<br />
<br />
We know that because <math>x < y</math>, we get <math>a < b</math>.<br />
<br />
<br />
We can count even and odd pairs separately to make things easier*:<br />
<br />
<br />
Odd: <cmath>(1,3) , (3,5) , (5,7) . . . (997,999)</cmath><br />
<br />
<br />
Even: <cmath>(2,4) , (4,6) , (6,8) . . . (996,998)</cmath><br />
<br />
<br />
This makes 499 odd pairs and 498 even pairs, for a total of <math>\boxed{997}</math> pairs.<br />
<br />
<br />
<br />
<math>*</math>Note: We are counting the pairs for the values of <math>a</math> and <math>b</math>, which, when squared, translate to the pairs of <math>(x,y)</math> we are trying to find.<br />
<br />
=== Solution 3 ===<br />
Since the arithmetic mean is 2 more than the geometric mean, <math>\frac{x+y}{2} = 2 + \sqrt{xy}</math>. We can multiply by 2 to get <math>x + y = 4 + 2\sqrt{xy}</math>. Subtracting 4 and squaring gives <br />
<cmath>((x+y)-4)^2 = 4xy</cmath><br />
<cmath>((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy</cmath><br />
<cmath>x^2 - 2xy + y^2 + 16 - 8x - 8y = 0</cmath><br />
<br />
Notice that <math>((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y</math>, so the problem asks for solutions of <br />
<cmath>(x-y-4)^2 = 16y</cmath><br />
Since the left hand side is a perfect square, and 16 is a perfect square, <math>y</math> must also be a perfect square. Since <math>0 < y < (1000)^2</math>, <math>y</math> must be from <math>1^2</math> to <math>999^2</math>, giving at most 999 options for <math>y</math>.<br />
<br />
However if <math>y = 1^2</math>, you get <math>(x-5)^2 = 16</math>, which has solutions <math>x = 9</math> and <math>x = 1</math>. Both of those solutions are not less than <math>y</math>, so <math>y</math> cannot be equal to 1. If <math>y = 2^2 = 4</math>, you get <math>(x - 8)^2 = 64</math>, which has 2 solutions, <math>x = 16</math>, and <math>x = 0</math>. 16 is not less than 4, and <math>x</math> cannot be 0, so <math>y</math> cannot be 4. However, for all other <math>y</math>, you get exactly 1 solution for <math>x</math>, and that gives a total of <math>999 - 2 = \boxed{997}</math> pairs.<br />
<br />
- asbodke<br />
<br />
<br />
=== Solution 4 (Similar to Solution 3) ===<br />
Rearranging our conditions to <br />
<br />
<cmath>x^2-2xy+y^2+16-8x-8y=0 \implies</cmath><br />
<cmath>(y-x)^2=8(x+y-2).</cmath><br />
<br />
Thus, <math>4|y-x.</math><br />
<br />
Now, let <math>y = 4k+x.</math> Plugging this back into our expression, we get<br />
<br />
<cmath>(k-1)^2=x-1.</cmath><br />
<br />
There, a unique value of <math>x, y</math> is formed for every value of <math>k</math>. However, we must have <br />
<br />
<cmath>y<10^6 \implies (k+1)^2< 10^6-1</cmath><br />
<br />
and <br />
<br />
<cmath>x=(k-1)^2+1>0.</cmath><br />
<br />
Therefore, there are only <math>997</math> pairs of <math>(x,y).</math><br />
<br />
Solution by Williamgolly<br />
<br />
=== Solution 5 ===<br />
<br />
First we see that our condition is <math>\frac{x+y}{2} = 2 + \sqrt{xy}</math>. Then we can see that <math>x+y = 4 + 2\sqrt{xy}</math>. From trying a simple example to figure out conditions for <math>x,y</math>, we want to find <math>x-y</math> so we can isolate for <math>x</math>. From doing the example we can note that we can square both sides and subtract <math>4xy</math>: <math>(x-y)^2 = 16 + 16\sqrt{xy} \implies x-y = -2(<br />
\sqrt{1+\sqrt{xy}})</math> (note it is negative because <math>y > x</math>. Clearly the square root must be an integer, so now let <math>\sqrt{xy} = a^2-1</math>. Thus <math>x-y = -2a</math>. Thus <math>x = 2 + \sqrt{xy} - a = 2 + a^2 - 1 -2a</math>. We can then find <math>y</math>, and use the quadratic formula on <math>x,y</math> to ensure they are <math>>0</math> and <math><10^6</math> respectively. Thus we get that <math>y</math> can go up to 999 and <math>x</math> can go down to <math>3</math>, leaving <math>997</math> possibilities for <math>x,y</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=I|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_19&diff=1648482015 AMC 12B Problems/Problem 192021-11-08T01:05:46Z<p>Awesomediabrine: </p>
<hr />
<div>==Problem==<br />
In <math>\triangle ABC</math>, <math>\angle C = 90^\circ</math> and <math>AB = 12</math>. Squares <math>ABXY</math> and <math>CBWZ</math> are constructed outside of the triangle. The points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle. What is the perimeter of the triangle?<br />
<br />
<math>\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32</math><br />
<br />
==Solution 1==<br />
<asy><br />
pair A,B,C,M,E,W,Z,X,Y;<br />
A=(2,0);<br />
B=(0,2);<br />
C=(0,0);<br />
M=(A+B)/2;<br />
W=(-2,2);<br />
Z=(-2,-0);<br />
X=(2,4);<br />
Y=(4,2);<br />
E=(W+Z)/2;<br />
draw(A--B--C--cycle);<br />
draw(W--B--C--Z--cycle);<br />
draw(A--B--X--Y--cycle);<br />
dot(M);<br />
dot(E);<br />
label("W",W,NW);<br />
label("Z",Z,SW);<br />
label("C",C,S);<br />
label("A",A,S);<br />
label("B",B,N);<br />
label("X",X,NE);<br />
label("Y",Y,SE);<br />
label("E",E,1.5*plain.W);<br />
label("M",M,NE);<br />
draw(circle(M,sqrt(10)));<br />
</asy><br />
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of <math>WZ</math> and <math>XY</math> and finding their intersection point. This point happens to be the midpoint of <math>AB</math>, the hypotenuse. Let this point be <math>M</math>. To find the radius, determine <math>MY</math>, where <math>MY^{2} = MA^2 + AY^2</math>, <math>MA = \frac{12}{2} = 6</math>, and <math>AY = AB = 12</math>. Thus, the radius <math>=r =MY = 6\sqrt5</math>.<br />
<br />
Next we let <math>AC = b</math> and <math>BC = a</math>. Consider the right triangle <math>ACB</math> first. Using the Pythagorean theorem, we find that <math>a^2 + b^2 = 12^2 = 144</math>. <br />
<asy><br />
pair A,B,C,M,E,W,Z,X,Y;<br />
A=(2,0);<br />
B=(0,2);<br />
C=(0,0);<br />
M=(A+B)/2;<br />
W=(-2,2);<br />
Z=(-2,-0);<br />
X=(2,4);<br />
Y=(4,2);<br />
E=(W+Z)/2;<br />
draw(A--B--C--cycle);<br />
draw(W--B--C--Z--cycle);<br />
draw(A--B--X--Y--cycle);<br />
dot(M);<br />
dot(E);<br />
label("W",W,NW);<br />
label("Z",Z,SW);<br />
label("C",C,S);<br />
label("A",A,S);<br />
label("B",B,N);<br />
label("X",X,NE);<br />
label("Y",Y,SE);<br />
label("E",E,1.5*plain.W);<br />
label("M",M,NE);<br />
draw(circle(M,sqrt(10)));<br />
draw(E--Z--M--cycle,dashed);<br />
</asy><br />
Now, we let <math>E</math> be the midpoint of <math>WZ</math>, and we consider right triangle <math>ZEM</math>. By the Pythagorean theorem, we have that <math>\left(\frac{a}{2}\right)^2 + \left(a + \frac{b}{2}\right)^2 = r^2 = 180</math>. Expanding this equation, we get that<br />
<br />
<cmath>\frac{1}{4}(a^2+b^2) + a^2 + ab = 180</cmath><br />
<cmath>\frac{144}{4} + a^2 + ab = 180</cmath><br />
<cmath>a^2 + ab = 144 = a^2 + b^2</cmath><br />
<cmath>ab = b^2</cmath><br />
<cmath>b = a</cmath><br />
<br />
This means that <math>ABC</math> is a 45-45-90 triangle, so <math>a = b = \frac{12}{\sqrt2} = 6\sqrt2</math>. Thus the perimeter is <math>a + b + AB = 12\sqrt2 + 12</math> which is answer <math>\boxed{\textbf{(C)}\; 12 + 12\sqrt2}</math>.<br />
<br />
==Solution 2==<br />
The center of the circle on which <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie must be equidistant from each of these four points. Draw the perpendicular bisectors of <math>\overline{XY}</math> and of <math>\overline{WZ}</math>. Note that the perpendicular bisector of <math>\overline{WZ}</math> is parallel to <math>\overline{CW}</math> and passes through the midpoint of <math>\overline{AC}</math>. Therefore, the triangle that is formed by <math>A</math>, the midpoint of <math>\overline{AC}</math>, and the point at which this perpendicular bisector intersects <math>\overline{AB}</math> must be similar to <math>\triangle ABC</math>, and the ratio of a side of the smaller triangle to a side of <math>\triangle ABC</math> is 1:2. Consequently, the perpendicular bisector of <math>\overline{XY}</math> passes through the midpoint of <math>\overline{AB}</math>. The perpendicular bisector of <math>\overline{WZ}</math> must include the midpoint of <math>\overline{AB}</math> as well. Since all points on a perpendicular bisector of any two points <math>M</math> and <math>N</math> are equidistant from <math>M</math> and <math>N</math>, the center of the circle must be the midpoint of <math>\overline{AB}</math>.<br />
<br />
Now the distance between the midpoint of <math>\overline{AB}</math> and <math>Z</math>, which is equal to the radius of this circle, is <math>\sqrt{12^2 + 6^2} = \sqrt{180}</math>. Let <math>a=AC</math>. Then the distance between the midpoint of <math>\overline{AB}</math> and <math>Y</math>, also equal to the radius of the circle, is given by <math>\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}</math> (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have<br />
<br />
<cmath>\left(\frac{a}{2}\right)^2+\left(a+\frac{\sqrt{144-a^2}}{2}\right)^{2} = 180</cmath><br />
<cmath>144 - a^2 = a\sqrt{144-a^2}</cmath><br />
<cmath>(144-a^2)^2 = a^2(144-a^2)</cmath><br />
<br />
Since <math>a</math> cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by <math>(144-a^2)</math>, and arrive at <math>a = 6\sqrt{2}</math>. The length of other leg of the triangle must be <math>\sqrt{144-72} = 6\sqrt{2}</math>. Thus, the perimeter of the triangle is <math>12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.<br />
<br />
==Solution 3==<br />
In order to solve this problem, we can search for similar triangles. Begin by drawing triangle <math>ABC</math> and squares <math>ABXY</math> and <math>ACWZ</math>. Draw segments <math>\overline{YZ}</math> and <math>\overline{WX}</math>. Because we are given points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle, we can conclude that <math>WXYZ</math> forms a cyclic quadrilateral. Take <math>\overline{AC}</math> and extend it through a point <math>P</math> on <math>\overline{YZ}</math>. Now, we must do some angle chasing to prove that <math>\triangle WBX</math> is similar to <math>\triangle YAZ</math>. <br />
<br />
Let <math>\alpha</math> denote the measure of <math>\angle ABC</math>. Following this, <math>\angle BAC</math> measures <math>90 - \alpha</math>. By our construction, <math>\overline{CAP}</math> is a straight line, and we know <math>\angle YAB</math> is a right angle. Therefore, <math>\angle PAY</math> measures <math>\alpha</math>. Also, <math>\angle CAZ</math> is a right angle and thus, <math>\angle ZAP</math> is a right angle. Sum <math>\angle ZAP</math> and <math>\angle PAY</math> to find <math>\angle ZAY</math>, which measures <math>90 + \alpha</math>. We also know that <math>\angle WBY</math> measures <math>90 + \alpha</math>. Therefore, <math>\angle ZAY = \angle WBX</math>. <br />
<br />
Let <math>\beta</math> denote the measure of <math>\angle AZY</math>. It follows that <math>\angle WZY</math> measures <math>90 + \beta^\circ</math>. Because <math>WXYZ</math> is a cyclic quadrilateral, <math>\angle WZY + \angle YXW = 180^\circ</math>. Therefore, <math>\angle YXW</math> must measure <math>90 - \beta</math>, and <math>\angle BXW</math> must measure <math>\beta</math>. Therefore, <math>\angle AZY = \angle BXW</math>. <br />
<br />
<math>\angle ZAY = \angle WBX</math> and <math>\angle AZY = \angle BXW</math>, so <math>\triangle AZY \sim \triangle BXW</math>! Let <math>x = AC = WC</math>. By Pythagorean theorem, <math>BC = \sqrt{144-x^2}</math>. Now we have <math>WB = WC + BC = x + \sqrt{144-x^2}</math>, <math>BX = 12</math>, <math>YA = 12</math>, and <math>AZ = x</math>. <br />
We can set up an equation: <br />
<br />
<cmath>\frac{YA}{AZ} = \frac{WB}{BX}</cmath><br />
<cmath>\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}</cmath><br />
<cmath>144 = x^2 + x\sqrt{144-x^2}</cmath><br />
<cmath>12^2 - x^2 = x\sqrt{144-x^2}</cmath><br />
<cmath>12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4</cmath><br />
<cmath>2x^4 - 3(12^2)x^2 + 12^4 = 0</cmath><br />
<cmath>(2x^2 - 144)(x^2 - 144) = 0</cmath><br />
<br />
Solving for <math>x</math>, we find that <math>x = 6\sqrt{2}</math> or <math>x = 12</math>, which we omit. The perimeter of the triangle is <math>12 + x + \sqrt{144-x^2}</math>. Plugging in <math>x = 6\sqrt{2}</math>, we get <math>\boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.<br />
<br />
<br />
Alternatively, let <math>BC = S_1, AB = S_2</math> and <math>AC = x</math>. Because <math>\frac{WB}{BX} = \frac{AY}{AZ}</math>, we get that <math>S_2^2 = S_1^2 + S_1x</math>. <math>S_1 = x</math> satisfies the equation because of Pythagorean theorem, so <math>\triangle ABC</math> is right isosceles.<br />
<br />
==Solution 4==<br />
We claim that <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles right triangle.<br />
<br />
''Proof:'' If <math>\triangle ACB</math> is an isosceles right triangle, then <math>\angle WAY=180º</math>. Therefore, <math>W</math>, <math>A</math>, and <math>Y</math> are collinear. Since <math>WY</math> and <math>YX</math> form a right angle, <math>WX</math> is the diameter of the circumcircle of <math>\triangle WYX</math>. Similarly, <math>Z</math>, <math>A</math>, and <math>X</math> are collinear, and <math>ZX</math> forms a right angle with <math>ZW</math>. Thus, <math>WX</math> is also the diameter of the circumcircle of <math>\triangle WZX</math>. Therefore, since <math>\triangle WYX</math> and <math>\triangle WZX</math> share a circumcircle, <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles triangle.<br />
<br />
If <math>\triangle ACB</math> is isosceles, then its legs have length <math>6\sqrt{2}</math>. The perimeter of <math>\triangle ACB</math> is <math>\boxed{\textbf{(C) }12+12\sqrt{2}}</math>.<br />
<br />
==Solution 5==<br />
Note that because <math>\angle WZA=\angle AYX=90^\circ</math>, these two angles inscribe the semicircle defined by diameter <math>WX</math>. Since <math>A</math> lies on line <math>ZA</math>, and <math>\angle XAY=45^\circ</math> (because <math>ABXY</math> is a square), we can find that <math>\angle ZAY= 180^\circ - 45^\circ = 135^\circ</math>.<br />
<br />
Now, we can see that <math>\angle BAC=45^\circ</math> to complete the full <math>360^\circ</math>. Therefore, <math>\triangle ABC</math> is and isosceles right triangle, and <math>AC=BC=6\sqrt2</math>. So Our answer is <math>6\sqrt2 + 6\sqrt 2 + 12 = \boxed{\textbf{(C) }12+12\sqrt{2}}</math>.<br />
.<br />
<br />
==Solution 5==<br />
<br />
We see a circle and little information of position of the shapes inside the triangle, so we think of things associated with circles and think of cyclic quads. We then notice that quadrilateral <math>WXYZ</math> is cyclic. Then we see that <math>\angle W = \angle X</math> making extensions of <math>\overline{WA}</math> and <math>\overline{XZ}</math> diagonals of <math>WXYZ</math>. Let <math>x = \overline{AC}=\overline{AW}</math>. We can see that <math>\overline{AY} = 12\sqrt{2}</math> and <math>\overline{AZ} = x\sqrt{2}</math>. Thus <math>\overline{YZ}^2 = x^2 + (x+12\sqrt{2})^2 = 2x^2 + 24x\sqrt{2} + 288</math>. Let <math>\angle ZAY = \alpha</math>. Then we can see that <math>\angle ZAC</math> and <math>\angle BAY</math> are <math>45</math> degrees, making <math>\angle ZAY</math> <math>\alpha + 90</math> degrees.<br />
<br />
We can verify with the cosine addition identity that <math>-\cos{\alpha + 90 degree} = \sin{\alpha}</math> (knowing that <math>\sin{\theta + 90 degree} = \cos{\theta}</math> motivates this). By law of cosine's, <math>\overline{YZ}^2 = (12\sqrt{2})^2 + x\sqrt{2})^2 - 2\cdot 12\sqrt{2} \cdot x\sqrt{2}cos{\alpha + 90 degree} = 288 + x^2 + 48cos{\alpha + 90 degree} = 288 + 2x^2 + 48x\sin{\alpha}</math>. Since <math>\angle{A}</math> is opposite <math>\overline{BC}</math>, <math>\sin{\alpha} = \frac{\sqrt{144-x^2}}{12}</math>. Thus <math>\overline{YZ}^2 = 288 + 2x^2 + 4x\sqrt{144-x^2}</math>.<br />
<br />
Setting the first paragraph's <math>\overline{YZ}^2</math> equal to the second:<br />
<cmath>288 + x^2 + 4x\sqrt{144-x^2}2x^2 + 24x\sqrt{2} + 288</cmath><br />
<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=1571471987 AIME Problems/Problem 122021-07-01T18:06:17Z<p>Awesomediabrine: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Let <math>m</math> be the smallest [[integer]] whose [[cube root]] is of the form <math>n+r</math>, where <math>n</math> is a [[positive integer]] and <math>r</math> is a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>.<br />
== Solution 1 ==<br />
In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.<br />
<br />
<math>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>. Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>. This means that the smallest possible <math>n</math> should be quite a bit smaller than 1000. In particular, <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>. <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>. Since we want to minimize <math>n</math>, we take <math>n = 19</math>. Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>. However, we still have to make sure a sufficiently small <math>r</math> exists. <br />
<br />
In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>. The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>. Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set. The answer is <math>\boxed{019}</math>.<br />
<br />
== Solution 2 ==<br />
To minimize <math>m</math>, we should minimize <math>n</math>. We have that <math>(n + \frac{1}{1000})^3 = n^3 + \frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9}</math>. For a given value of <math>n</math>, if <math>(n + \frac{1}{1000})^3 - n^3 > 1</math>, there exists an integer between <math>(n + \frac{1}{1000})^3</math> and <math>n^3</math>, and the cube root of this integer would be between <math>n</math> and <math>n + \frac{1}{1000}</math> as desired. We seek the smallest <math>n</math> such that <math>(n + \frac{1}{1000})^3 - n^3 > 1</math>. <br />
<br />
<cmath>(n + \frac{1}{1000})^3 - n^3 > 1</cmath><br />
<cmath>\frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9} > 1</cmath><br />
<cmath>3n^2 + \frac{3}{10^3} n + \frac{1}{10^6} > 10^3</cmath><br />
<br />
Trying values of <math>n</math>, we see that the smallest value of <math>n</math> that works is <math>\boxed{019}</math>.<br />
<br />
Why is it <math>(n + \frac{1}{1000})^3 - n^3 > 1</math> and not greater than or equal to? - awesomediabrine<br />
<br />
== Solution 3 (Similar to Solution 2) ==<br />
Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < n + \frac{1}{1000}</math>. Notice that since we want <math>m</math> minimized, <math>n</math> should also be minimized. Also, <math>n^3</math> should be as close as possible, but not exceeding <math>m</math>. This means <math>m</math> should be set to <math>n^3+1</math>. Substituting and simplifying, we get <cmath>\sqrt[3]{n^3+1} < n + \frac{1}{1000}</cmath> <cmath>n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}</cmath><br />
The last two terms in the right side can be ignored in the calculation because they are too small. This results in <math>1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}</math>. The minimum positive integer <math>n</math> that satisfies this is <math>\boxed{019}</math>. ~ Hb10<br />
<br />
== See also ==<br />
{{AIME box|year=1987|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=1571461987 AIME Problems/Problem 122021-07-01T18:05:48Z<p>Awesomediabrine: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Let <math>m</math> be the smallest [[integer]] whose [[cube root]] is of the form <math>n+r</math>, where <math>n</math> is a [[positive integer]] and <math>r</math> is a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>.<br />
== Solution 1 ==<br />
In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.<br />
<br />
<math>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>. Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>. This means that the smallest possible <math>n</math> should be quite a bit smaller than 1000. In particular, <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>. <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>. Since we want to minimize <math>n</math>, we take <math>n = 19</math>. Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>. However, we still have to make sure a sufficiently small <math>r</math> exists. <br />
<br />
In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>. The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>. Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set. The answer is <math>\boxed{019}</math>.<br />
<br />
== Solution 2 ==<br />
To minimize <math>m</math>, we should minimize <math>n</math>. We have that <math>(n + \frac{1}{1000})^3 = n^3 + \frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9}</math>. For a given value of <math>n</math>, if <math>(n + \frac{1}{1000})^3 - n^3 > 1</math>, there exists an integer between <math>(n + \frac{1}{1000})^3</math> and <math>n^3</math>, and the cube root of this integer would be between <math>n</math> and <math>n + \frac{1}{1000}</math> as desired. We seek the smallest <math>n</math> such that <math>(n + \frac{1}{1000})^3 - n^3 > 1</math>. <br />
<br />
<cmath>(n + \frac{1}{1000})^3 - n^3 > 1</cmath><br />
<cmath>\frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9} > 1</cmath><br />
<cmath>3n^2 + \frac{3}{10^3} n + \frac{1}{10^6} > 10^3</cmath><br />
<br />
Trying values of <math>n</math>, we see that the smallest value of <math>n</math> that works is <math>\boxed{019}</math>.<br />
<br />
Why is it <math>(n + \frac{1}{1000})^3 - n^3 > 1</math> and not greator than or equal to? - awesomediabrine<br />
<br />
== Solution 3 (Similar to Solution 2) ==<br />
Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < n + \frac{1}{1000}</math>. Notice that since we want <math>m</math> minimized, <math>n</math> should also be minimized. Also, <math>n^3</math> should be as close as possible, but not exceeding <math>m</math>. This means <math>m</math> should be set to <math>n^3+1</math>. Substituting and simplifying, we get <cmath>\sqrt[3]{n^3+1} < n + \frac{1}{1000}</cmath> <cmath>n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}</cmath><br />
The last two terms in the right side can be ignored in the calculation because they are too small. This results in <math>1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}</math>. The minimum positive integer <math>n</math> that satisfies this is <math>\boxed{019}</math>. ~ Hb10<br />
<br />
== See also ==<br />
{{AIME box|year=1987|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_16&diff=1537842004 AMC 12B Problems/Problem 162021-05-16T21:57:10Z<p>Awesomediabrine: /* Solutions */</p>
<hr />
<div>== Problem ==<br />
A [[function]] <math>f</math> is defined by <math>f(z) = i\overline{z}</math>, where <math>i=\sqrt{-1}</math> and <math>\overline{z}</math> is the [[complex conjugate]] of <math>z</math>. How many values of <math>z</math> satisfy both <math>|z| = 5</math> and <math>f(z) = z</math>?<br />
<br />
<math>\mathrm{(A)}\ 0<br />
\qquad\mathrm{(B)}\ 1<br />
\qquad\mathrm{(C)}\ 2 <br />
\qquad\mathrm{(D)}\ 4<br />
\qquad\mathrm{(E)}\ 8</math><br />
<br />
== Solutions==<br />
<br />
===Solution 1===<br />
Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>, which implies that all solutions to <math>f(z) = z</math> lie on the line <math>y=x</math> on the complex plane. The graph of <math>|z| = 5</math> is a [[circle]] centered at the origin, and there are <math>2 \Rightarrow \mathrm{(C)}</math> intersections.<br />
<br />
===Solution 2===<br />
We start the same as the above solution: Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>. Since we are given <math>|z| = 5</math>, this implies that <math>a^2+b^2=25</math>. We recognize the Pythagorean triple <math>3,4,5</math> so we see that <math>(a,b)=(3,4)</math> or <math>(4,3)</math>. So the answer is <math>2 \Rightarrow \mathrm{(C)}</math>.<br />
<br />
Solution by franzliszt<br />
<br />
===Solution 3===<br />
Let <math>z=a+bi</math>, like above. Therefore, <math>z = a+bi = i\overline{z} = i(a-bi) = ai+b</math>. We move some terms around to get <math>bi-b = ai-a</math>. We factor: <math>b(i-1) = a(i-1)</math>. We divide out the common factor to see that <math>b = a</math>. Next we put this into the definition of <math>|z| = a^2 + b^2 = a^2 + a^2 = 2a^2 = 25</math>. Finally, <math>a = \pm\sqrt{\frac{25}{2}}</math>, and <math>a</math> has two solutions.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2004|ab=B|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=Physics_competitions&diff=150758Physics competitions2021-04-01T16:25:54Z<p>Awesomediabrine: /* International Physics competitions */</p>
<hr />
<div>This is a resources page for students interested in '''Physics competitions'''. Please add quality resources.<br />
<br />
<br />
== International Physics competitions ==<br />
* [[International Physics Olympiad]] [http://www.jyu.fi/ipho website]<br />
* [[Online Physics Olympiad]] [https://physoly.tech/opho/]<br />
* [[International Physics Online Olympiad (IPhOO)]] [http://onlinescienceolympiads.org website]<br />
* [[Rudolf Ortvay Problem Solving Contest in Physics]] [http://ortvay.elte.hu/ website]<br />
* [[The University Physics Competition]] [http://www.uphysicsc.com website]<br />
* [[Online Physics Brawl]] – next on '''28th November 2018''' [http://physicsbrawl.org website]<br />
* [[FYKOS – The Internet Physics Competition]] [http://fykos.org website]<br />
* [[Princeton University Physics Competition]] [http://physics.princeton.edu/pupc/ website]<br />
<br />
== National Physics competitions ==<br />
=== Canada ===<br />
* [[Canadian Association of Physicists High School Competition]] [http://physics.usask.ca/~pywell/HighSchool/index.html website]<br />
<br />
<br />
=== United States ===<br />
* [[Unites States Physics Olympiad]] [http://www.aapt.org/Contests/olympiad.cfm website] <br />
<br><br />
* [[American Association of PhysicsTeachers Physics Bowl]] [http://www.aapt.org/Contests/physicsbowl.cfm website]<br />
* [[Auburn Physics Invitation]] [http://www.physics.auburn.edu/~msimon/InvBro99-2K.html website]<br />
* [[Iowa State Physics Olympics]] [http://www.educ.drake.edu/gerlovich/physics%20olympics/physics_olympics.html website]<br />
* [[NJAAPT Physics Olympics]] [http://www.njaapt.org/PhysicsOlympics/2005-2006/NJ%20Physics%20Olympics%202005-2006.htm website]<br />
* [[Princeton University Physics Competition]] [http://physics.princeton.edu/pupc/ website]<br />
* [[The Raytheon/TAPT Annual High School Physics Competition]]<br />
* [[St. Louis Area Physics Teachers Physics Competition]] [http://www.slapt.org/ website]<br />
* [[San Diego COE/SDSA Physics Team Competition]] [http://www.sdsa.org/Physics/about.html website]<br />
* [[Unites States Physics Olympiad]] [http://www.aapt.org/Contests/olympiad.cfm website] <br />
* [[University of Alabama High School Physics contest]] [http://bama.ua.edu/~physics/Contest2006.html website]<br />
* [[University of Miami Physics Competition]] [http://www.physics.miami.edu/competition/ website]<br />
* [[University of Michigan Physics Olympiad]] [http://phys-advlab.physics.lsa.umich.edu/ website]<br />
* [[UNC-Charlotte Physics Competition]] [http://education.uncc.edu/aewickli/supercomp1.htm#Physics website]<br />
* [[University of Northern Iowa Regional Physics Olympics]] [http://www.physics.uni.edu/uni_aea267_olympics.shtml website]<br />
* [[Utah Physics Olympiad]] [http://departments.weber.edu/sciencecenter/ScienceOlympiad/Olympiadpage.htm website]<br />
* [[Western Kentucky Physics Olympics]] [http://physics.wku.edu/olympics/events.html website]<br />
* [[Yale Physics Olympics]] [http://wnsl.physics.yale.edu/events/olympics/ website]<br />
<br />
== Regional Physics competitions ==<br />
* [[Asian Physics Olympiad]] [http://www.apho.org/en/home/index.php?charencode=en website]<br />
*[[Ibero-American Physics Olympiad]] - [http://oc.uan.edu.co/oibf/oibf.htm website]<br />
<br />
== Resources ==<br />
Khan Academy: AP Physics 1/2<br />
<br />
==Past Contests==<br />
*[[Boston Area Undergraduate Physics Competition]] [http://liquids.deas.harvard.edu/oleg/competition/ website]<br />
=== Books ===<br />
<br />
<br />
=== Online Resources ===<br />
* [[AoPS]] hosts both an [http://www.artofproblemsolving.com/Forum/index.php?f=405 Introductory Physics Forum] as well as an [http://www.artofproblemsolving.com/Forum/index.php?f=332 Advanced Physics Forum].<br />
* [[AoPS]] also teaches [https://artofproblemsolving.com/school/woot-physics PhysicsWOOT]—like WOOT, but for physics.<br />
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<br />
== See also ==<br />
* [[Physics books]]<br />
* [[Physics scholarships]]<br />
* [[Science competitions]]<br />
* [[Mathematics competitions]]<br />
* [[Physics]]</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_9&diff=1472421983 AIME Problems/Problem 92021-02-17T04:37:29Z<p>Awesomediabrine: /* Solution 5 */</p>
<hr />
<div>== Problem ==<br />
Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.<br />
<br />
== Solution 1 ==<br />
Let <math>y=x\sin{x}</math>. We can rewrite the expression as <math>\frac{9y^2+4}{y}=9y+\frac{4}{y}</math>.<br />
<br />
Since <math>x>0</math>, and <math>\sin{x}>0</math> because <math>0< x<\pi</math>, we have <math>y>0</math>. So we can apply [[AM-GM]]:<br />
<br />
<cmath>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</cmath><br />
<br />
The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>.<br />
<br />
Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when we have <math>x \sin{x} = \frac{2}{3}</math> in the original equation (since <math>x\sin x</math> is continuous and increasing on the interval <math>0 \le x \le \frac{\pi}{2}</math>, and its range on that interval is from <math>0 \le x\sin x \le \frac{\pi}{2}</math>, this value of <math>\frac{2}{3}</math> is attainable by the [[Intermediate Value Theorem]]).<br />
<br />
== Solution 2 ==<br />
We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x \sin x}{x \sin x}</math>. Thus, we must also add back <math>12</math>.<br />
<br />
This results in <math>\dfrac{(3x \sin x-2)^2}{x \sin x}+12</math>.<br />
<br />
Thus, if <math>3x \sin x-2=0</math>, then the minimum is obviously <math>12</math>. We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>.<br />
<br />
== Solution 3 (uses calculus) ==<br />
<br />
Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>, similar to the previous solution. To minimize <math>f(y)</math>, take the [[derivative]] of <math>f(y)</math> and set it equal to zero. <br />
<br />
The derivative of <math>f(y)</math>, using the Power Rule, is<br />
<br />
<math>f'(y)</math> = <math>9 - 4y^{-2}</math><br />
<br />
<math>f'(y)</math> is zero only when <math>y = \frac{2}{3}</math> or <math>y = -\frac{2}{3}</math>. It can further be verified that <math>\frac{2}{3}</math> and <math>-\frac{2}{3}</math> are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative <math>f''(y)=8y^{-3}</math> is positive. However, since <math>x \sin{x}</math> is always positive in the given domain, <math>y = \frac{2}{3}</math>. Therefore, <math>x\sin{x}</math> = <math>\frac{2}{3}</math>, and the answer is <math>\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}</math>.<br />
<br />
== Solution 4 (also uses calculus) ==<br />
<br />
As above, let <math>y = x\sin{x}</math>. Add <math>\frac{12y}{y}</math> to the expression and subtract <math>12</math>, giving <math>f(x) = \frac{(3y+2)^2}{y} - 12</math>. Taking the [[derivative]] of <math>f(x)</math> using the [[Chain Rule]] and [[Quotient Rule]], we have <math>\frac{\text{d}f(x)}{\text{d}x} = \frac{6y(3y+2)-(3y+2)^2}{y^2}</math>. We find the minimum value by setting this to <math>0</math>. Simplifying, we have <math>6y(3y+2) = (3y+2)^2</math> and <math>y = \pm{\frac{2}{3}} = x\sin{x}</math>. Since both <math>x</math> and <math>\sin{x}</math> are positive on the given interval, we can ignore the negative root. Plugging <math>y = \frac{2}{3}</math> into our expression for <math>f(x)</math>, we have <math>\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}</math>.<br />
<br />
== Solution 5 ==<br />
<br />
Set <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> equal to <math>y</math>. Then multiply by <math>x\sin x</math> on both sides to get <math>9x^2\sin^2 x + 4 = y\cdot x\sin x</math> We then subtract <math>yx\sin x</math> from both sides to get <math>9x^2\sin^2 x + 4 - yx\sin x = 0</math> This looks like a quadratic so I set <math>z= x\sin x</math> and use quadratic equation on <math>9z^2 - yz + 4 = 0</math> to see that <math>z = \frac{y\pm\sqrt{y^2-144}}{18}</math> We know that <math>y</math> must be an integer and as small as it can be, so <math>y</math> = 12. We plug this back in to see that <math>x\sin x = \frac{2}{3}</math> which we can prove works using methods from solution 1. This makes the answer <math>\boxed{012}</math><br />
<br />
-awesomediabrine<br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
[[Category:Intermediate Trigonometry Problems]]</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_9&diff=1472411983 AIME Problems/Problem 92021-02-17T04:37:12Z<p>Awesomediabrine: </p>
<hr />
<div>== Problem ==<br />
Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.<br />
<br />
== Solution 1 ==<br />
Let <math>y=x\sin{x}</math>. We can rewrite the expression as <math>\frac{9y^2+4}{y}=9y+\frac{4}{y}</math>.<br />
<br />
Since <math>x>0</math>, and <math>\sin{x}>0</math> because <math>0< x<\pi</math>, we have <math>y>0</math>. So we can apply [[AM-GM]]:<br />
<br />
<cmath>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</cmath><br />
<br />
The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>.<br />
<br />
Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when we have <math>x \sin{x} = \frac{2}{3}</math> in the original equation (since <math>x\sin x</math> is continuous and increasing on the interval <math>0 \le x \le \frac{\pi}{2}</math>, and its range on that interval is from <math>0 \le x\sin x \le \frac{\pi}{2}</math>, this value of <math>\frac{2}{3}</math> is attainable by the [[Intermediate Value Theorem]]).<br />
<br />
== Solution 2 ==<br />
We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x \sin x}{x \sin x}</math>. Thus, we must also add back <math>12</math>.<br />
<br />
This results in <math>\dfrac{(3x \sin x-2)^2}{x \sin x}+12</math>.<br />
<br />
Thus, if <math>3x \sin x-2=0</math>, then the minimum is obviously <math>12</math>. We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>.<br />
<br />
== Solution 3 (uses calculus) ==<br />
<br />
Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>, similar to the previous solution. To minimize <math>f(y)</math>, take the [[derivative]] of <math>f(y)</math> and set it equal to zero. <br />
<br />
The derivative of <math>f(y)</math>, using the Power Rule, is<br />
<br />
<math>f'(y)</math> = <math>9 - 4y^{-2}</math><br />
<br />
<math>f'(y)</math> is zero only when <math>y = \frac{2}{3}</math> or <math>y = -\frac{2}{3}</math>. It can further be verified that <math>\frac{2}{3}</math> and <math>-\frac{2}{3}</math> are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative <math>f''(y)=8y^{-3}</math> is positive. However, since <math>x \sin{x}</math> is always positive in the given domain, <math>y = \frac{2}{3}</math>. Therefore, <math>x\sin{x}</math> = <math>\frac{2}{3}</math>, and the answer is <math>\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}</math>.<br />
<br />
== Solution 4 (also uses calculus) ==<br />
<br />
As above, let <math>y = x\sin{x}</math>. Add <math>\frac{12y}{y}</math> to the expression and subtract <math>12</math>, giving <math>f(x) = \frac{(3y+2)^2}{y} - 12</math>. Taking the [[derivative]] of <math>f(x)</math> using the [[Chain Rule]] and [[Quotient Rule]], we have <math>\frac{\text{d}f(x)}{\text{d}x} = \frac{6y(3y+2)-(3y+2)^2}{y^2}</math>. We find the minimum value by setting this to <math>0</math>. Simplifying, we have <math>6y(3y+2) = (3y+2)^2</math> and <math>y = \pm{\frac{2}{3}} = x\sin{x}</math>. Since both <math>x</math> and <math>\sin{x}</math> are positive on the given interval, we can ignore the negative root. Plugging <math>y = \frac{2}{3}</math> into our expression for <math>f(x)</math>, we have <math>\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}</math>.<br />
<br />
== Solution 5 ==<br />
<br />
Set <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> equal to <math>y</math>. Then multiply by <math>x\sin x</math> on both sides to get <math>9x^2\sin^2 x + 4 = y\cdot x\sin x</math> We then subtract <math>yx\sin x</math> from both sides to get <math>9x^2\sin^2 x + 4 - yx\sin x = 0</math> This looks like a quadratic so I set <math>z= x\sin x</math> and use quadratic equation on <math>9z^2 - yz + 4 = 0</math> to see that <math>z = \frac{y\pm\sqrt{y^2-144}}{18}</math> We know that <math>y</math> must be an integer and as small as it can be, so <math>y</math> = 12. We plug this back in to see that <math>x\sin x = \frac{2}{3}</math> which we can prove works using methods from solution 1. This makes the answer <math>\boxed{012}</math><br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
[[Category:Intermediate Trigonometry Problems]]</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_22&diff=1470682021 AMC 12B Problems/Problem 222021-02-16T05:24:37Z<p>Awesomediabrine: /* Solution 3 (Nim-values) */</p>
<hr />
<div>==Problem==<br />
Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes <math>4</math> and <math>2</math> can be changed into any of the following by one move: <math>(3,2),(2,1,2),(4),(4,1),(2,2),</math> or <math>(1,1,2).</math><br />
<br />
<asy> unitsize(4mm); real[] boxes = {0,1,2,3,5,6,13,14,15,17,18,21,22,24,26,27,30,31,32,33}; for(real i:boxes){ draw(box((i,0),(i+1,3))); } draw((8,1.5)--(12,1.5),Arrow()); defaultpen(fontsize(20pt)); label(",",(20,0)); label(",",(29,0)); label(",...",(35.5,0)); </asy><br />
<br />
Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?<br />
<br />
<math>\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)</math><br />
<br />
== Solution ==<br />
First we note that symmetrical positions are losing for the player to move. Then we start checking small positions. <math>(n)</math> is always winning for the first player. Furthermore, <math>(3, 2, 1)</math> is losing and so is <math>(4, 1).</math> We look at all the positions created from <math>(6, 2, 1),</math> as <math>(6, 1, 1)</math> is obviously winning by playing <math>(2, 2, 1, 1).</math> There are several different positions that can be played by the first player from <math>(6, 2, 1).</math> They are <math>(2, 2, 2, 1), (1, 3, 2, 1), (4, 2, 1), (6, 1), (5, 2, 1), (4, 1, 2, 1), (3, 2, 2, 1).</math> Now we list refutations for each of these moves:<br />
<br />
<br />
<math>(2, 2, 2, 1) - (2, 1, 2, 1)</math><br />
<br />
<br />
<math>(1, 3, 2, 1) - (3, 2, 1)</math><br />
<br />
<br />
<math>(4, 2, 1) - (4, 1)</math><br />
<br />
<br />
<math>(6, 1) - (4, 1)</math><br />
<br />
<br />
<math>(5, 2, 1) - (3, 2, 1)</math><br />
<br />
<br />
<math>(4, 1, 2, 1) - (2, 1, 2, 1)</math><br />
<br />
<br />
<math>(3, 2, 2, 1) - (1, 2, 2, 1)</math><br />
<br />
<br />
This proves that <math>(6, 2, 1)</math> is losing for the first player.<br />
<br />
-Note: In general, this game is very complicated. For example <math>(8, 7, 5, 3, 2)</math> is winning for the first player but good luck showing that.<br />
<br />
== Solution 2 (Process of Elimination)==<br />
<br />
<math>(6,1,1)</math> can be turned into <math>(2,2,1,1)</math> by Arjun, which is symmetric, so Beth will lose.<br />
<br />
<math>(6,3,1)</math> can be turned into <math>(3,1,3,1)</math> by Arjun, which is symmetric, so Beth will lose.<br />
<br />
<math>(6,2,2)</math> can be turned into <math>(2,2,2,2)</math> by Arjun, which is symmetric, so Beth will lose.<br />
<br />
<math>(6,3,2)</math> can be turned into <math>(3,2,3,2)</math> by Arjun, which is symmetric, so Beth will lose.<br />
<br />
That leaves <math>(6,2,1)</math> or <math>\boxed{\textbf{(B)}}</math>.<br />
<br />
== Solution 3 (Nim-values) ==<br />
<br />
Let the nim-value of the ending game state, where someone has just removed the final brick, be <math>0</math>. Then, any game state with a nim-value of <math>0</math> is losing. It is well-known that the nim-value of a supergame (a combination of two or more individual games) is the binary xor function on the nim-values of the individual games that compose the supergame. Therefore, we calculate the nim-values of the states with a single wall up to <math>6</math> bricks long (since the answer choices only go up to <math>6</math>). <br />
<br />
First, the game with <math>1</math> brick has a nim-value of <math>1</math>. <br />
<br />
Similarly, the game with <math>2</math> bricks has a nim-value of <math>2</math>.<br />
<br />
Next, we consider a <math>3</math> brick wall. After the next move, the possible resulting game states are <math>1</math> brick, a <math>2</math> brick wall, or <math>2</math> separate bricks. The first two options have nim-values of <math>1</math> and <math>2</math>. The final option has a nim-value of <math>1\oplus 1 = 0</math>, so the nim-value of this game state is <math>3</math>. <br />
<br />
Next, the <math>4</math> brick wall. The possible states are a <math>2</math> brick wall, a <math>3</math> brick wall, a <math>2</math> brick wall and a <math>1</math> brick wall, or a <math>1</math> brick wall and a <math>1</math> brick wall. The nim-values of these states are <math>2</math>, <math>3</math>, <math>3</math>, and <math>0</math>, respectively, and hence the nim-value of this game state is <math>1</math>.<br />
(Wait why is the nim-value of it <math>1</math>? - awesomediabrine) <br />
<br />
The possible game states after the <math>5</math> brick wall are the following: a <math>3</math> brick wall, a <math>4</math> brick wall, a <math>3</math> brick wall and a <math>1</math> brick wall, a two <math>2</math> brick walls, and a <math>2</math> brick wall plus a <math>1</math> brick wall. The nim-values of these are <math>3</math>, <math>1</math>, <math>2</math>, <math>0</math>, and <math>3</math>, respectively, meaning the nim-value of a <math>5</math> brick wall is <math>4</math>. <br />
<br />
Finally, we find the nim-value of a <math>6</math> brick wall. The possible states are a <math>5</math> brick wall, a <math>4</math> brick wall and a <math>1</math> brick wall, a <math>3</math> brick wall and a <math>2</math> brick wall, a <math>4</math> brick wall, a <math>3</math> brick wall and a <math>1</math> brick wall, and finally two <math>2</math> brick walls. The nim-values of these game states are <math>4</math>, <math>0</math>, <math>1</math>, <math>1</math>, <math>2</math>, and <math>0</math>, respectively. This means the <math>6</math> brick wall has a nim-value of <math>3</math>. <br />
<br />
The problem is asking which of the answer choices is losing, or has a nim-value of <math>0</math>. We see that option <math>\textbf{(A)}</math> has a nim-value of <math>3\oplus1\oplus1=3</math>, option <math>\textbf{(B)}</math> has a nim-value of <math>3\oplus2\oplus1=0</math>, option <math>\textbf{(C)}</math> has a nim-value of <math>3\oplus2\oplus2=3</math>, option <math>\textbf{(D)}</math> has a nim-value of <math>3\oplus3\oplus1=1</math>, and option <math>\textbf{(E)}</math> has a nim-value of <math>3\oplus3\oplus2=2</math>, so the answer is <math>\boxed{\textbf{(B) }(6, 2, 1)}</math>.<br />
<br />
This method can also be extended to solve the note after the first solution. The nim-values of the <math>7</math> brick wall and the <math>8</math> brick wall are <math>2</math> and <math>1</math>, using the same method as above. The nim-value of <math>(8, 7, 5, 3, 2)</math> is therefore <math>1\oplus2\oplus4\oplus3\oplus2 = 5</math>, which is winning.<br />
<br />
== Video Solution by OmegaLearn (Game Theory) ==<br />
https://youtu.be/zkSBMVAfYLo<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=21|num-a=23}}<br />
{{AMC10 box|year=2021|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_25&diff=1467232021 AMC 10A Problems/Problem 252021-02-15T02:54:06Z<p>Awesomediabrine: /* Solution 2 (Casework) */</p>
<hr />
<div>==Problem==<br />
How many ways are there to place <math>3</math> indistinguishable red chips, <math>3</math> indistinguishable blue chips, and <math>3</math> indistinguishable green chips in the squares of a <math>3 \times 3</math> grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally.<br />
<br />
<math>\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36</math><br />
==Solution 1==<br />
Call the different colors A,B,C. There are <math>3!=6</math> ways to rearrange these colors to these three letters, so <math>6</math> must be multiplied after the letters are permuted in the grid. <br />
WLOG assume that A is in the center. <br />
<cmath>\begin{tabular}{ c c c }<br />
? & ? & ? \\ <br />
? & A & ? \\ <br />
? & ? & ? <br />
\end{tabular}</cmath><br />
In this configuration, there are two cases, either all the A's lie on the same diagonal:<br />
<cmath>\begin{tabular}{ c c c }<br />
? & ? & A \\ <br />
? & A & ? \\ <br />
A & ? & ? <br />
\end{tabular}</cmath><br />
or all the other two A's are on adjacent corners:<br />
<cmath>\begin{tabular}{ c c c }<br />
A & ? & A \\ <br />
? & A & ? \\ <br />
? & ? & ? <br />
\end{tabular}</cmath><br />
In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.<br />
<br />
In each case there is only one way to put the three B's and the three C's as shown in the diagrams. <br />
<cmath>\begin{tabular}{ c c c }<br />
C & B & A \\ <br />
B & A & C \\ <br />
A & C & B <br />
\end{tabular}</cmath><br />
<cmath>\begin{tabular}{ c c c }<br />
A & B & A \\ <br />
C & A & C \\ <br />
B & C & B <br />
\end{tabular}</cmath><br />
This means that there are 4+2=6 ways to arrange A,B, and C in the grid, and there are 6 ways to rearrange the colors. Therefore, there are <math>6\cdot6=36</math> ways in total, which is <br />
<math>\boxed{\text{E}}</math>.<br />
<br />
-happykeeper<br />
<br />
==Solution 2 (Casework)==<br />
Without the loss of generality, we place a red ball in the top-left square. There are two cases:<br />
<br />
<b>Case (1)</b>: The two balls adjacent to the top-left red ball have different colors.<br />
<cmath>\begin{tabular}{ c c c }<br />
R & B & ? \\ <br />
G & R & ? \\ <br />
? & ? & ? <br />
\end{tabular}</cmath><br />
Each placement has <math>6</math> permutations, as there are <math>3!=6</math> ways to permute RBG.<br />
<br />
There are three sub-cases for Case (1):<br />
<cmath>\begin{tabular}{ c c c }<br />
R & B & R \\ <br />
G & R & G \\ <br />
B & G & B <br />
\end{tabular}</cmath><br />
<cmath>\begin{tabular}{ c c c }<br />
R & B & G \\ <br />
G & R & B \\ <br />
R & B & G <br />
\end{tabular}</cmath><br />
<cmath>\begin{tabular}{ c c c }<br />
R & B & G \\ <br />
G & R & B \\ <br />
R & G & R <br />
\end{tabular}</cmath><br />
[why does this have 4 R's? - awesomediabrine]<br />
So, Case (1) has <math>3\cdot6=18</math> ways.<br />
<br />
<br />
<b>Case (2)</b>: The two balls adjacent to the top-left red ball have the same color.<br />
<cmath>\begin{tabular}{ c c c }<br />
R & B & ? \\ <br />
B & ? & ? \\ <br />
? & ? & ? <br />
\end{tabular}</cmath><br />
Each placement has <math>6</math> permutations, as there are <math>\binom32\binom21=6</math> ways to choose three balls consisting of exactly two colors (RBB, RGG, BRR, BGG, GRR, GBB).<br />
There are three sub-cases for Case (2):<br />
<cmath>\begin{tabular}{ c c c }<br />
R & B & R \\ <br />
B & G & B \\ <br />
G & R & G <br />
\end{tabular}</cmath><br />
<cmath>\begin{tabular}{ c c c }<br />
R & B & G \\ <br />
B & G & R \\ <br />
R & B & G <br />
\end{tabular}</cmath><br />
<cmath>\begin{tabular}{ c c c }<br />
R & B & G \\ <br />
B & G & R \\ <br />
G & R & B <br />
\end{tabular}</cmath><br />
So, Case (2) has <math>3\cdot6=18</math> ways.<br />
<br />
<br />
Together, the answer is <math>18+18=\boxed{\textbf{(E)} ~36}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
== Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations) ==<br />
https://youtu.be/wKJ9ppI-8Ew<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021|ab=A|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_24&diff=1467212021 AMC 10A Problems/Problem 242021-02-15T02:22:34Z<p>Awesomediabrine: /* Solution 3 (Geometry) */</p>
<hr />
<div>==Problem==<br />
The interior of a quadrilateral is bounded by the graphs of <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math>, where <math>a</math> a positive real number. What is the area of this region in terms of <math>a</math>, valid for all <math>a > 0</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}</math><br />
<br />
==Solution 1==<br />
The conditions <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math> give <math>|x+ay| = |2a|</math> and <math>|ax-y| = |a|</math> or <math>x+ay = \pm 2a</math> and <math>ax-y = \pm a</math>. The slopes here are perpendicular, so the quadrilateral is a rectangle.<br />
Plug in <math>a=1</math> and graph it. We quickly see that the area is <math>2\sqrt{2} \cdot \sqrt{2} = 4</math>, so the answer can't be <math>A</math> or <math>B</math> by testing the values they give (test it!). Now plug in <math>a=2</math>. We see using a ruler that the sides of the rectangle are about <math>\frac74</math> and <math>\frac72</math>. So the area is about <math>\frac{49}8 = 6.125</math>. Testing <math>C</math> we get <math>\frac{16}3</math> which is clearly less than <math>6</math>, so it is out. Testing <math>D</math> we get <math>\frac{32}5</math> which is near our answer, so we leave it. Testing <math>E</math> we get <math>\frac{16}5</math>, way less than <math>6</math>, so it is out. So, the only plausible answer is <math>\boxed{D}</math> ~firebolt360<br />
<br />
==Solution 2 (Casework)==<br />
For the equation <math>(x+ay)^2 = 4a^2,</math> the cases are <br />
<br />
(1) <math>x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a,y</math>-intercept <math>2,</math> and slope <math>-\frac 1a.</math><br />
<br />
(2) <math>x+ay=-2a.</math> This is a line with <math>x</math>-intercept <math>-2a,y</math>-intercept <math>-2,</math> and slope <math>-\frac 1a.</math><br />
<br />
For the equation <math>(ax-y)^2 = a^2,</math> the cases are<br />
<br />
(1)* <math>ax-y=a.</math> This is a line with <math>x</math>-intercept <math>1,y</math>-intercept <math>-a,</math> and slope <math>a.</math><br />
<br />
(2)* <math>ax-y=-a.</math> This is a line with <math>x</math>-intercept <math>-1,y</math>-intercept <math>a,</math> and slope <math>a.</math><br />
<br />
Since the slopes of the intersecting lines (from the four above equations) are negative reciprocals, the quadrilateral is a rectangle.<br />
<br />
Plugging <math>a=2</math> into the four above equations and solving systems of equations for intersecting lines [(1) and (1)*, (1) and (2)*, (2) and (1)*, (2) and (2)*], we get the respective solutions <cmath>(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).</cmath> <br />
Finally, by the Distance Formula, the length and width of the rectangle are <math>\frac{8\sqrt5}{5}</math> and <math>\frac{4\sqrt5}{5}.</math> The area we seek is <cmath>\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.</cmath><br />
Plugging <math>a=2</math> into the choices gives <br />
<br />
<math>\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}</math><br />
<br />
The answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3 (Geometry)==<br />
Similar to Solution 2, we will use the equations of the four cases:<br />
<br />
(1) <math>x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a</math>, <math>y</math>-intercept <math>2</math>, and slope <math>-\frac 1a.</math><br />
<br />
(2) <math>x+ay=-2a.</math> This is a line with <math>x</math>-intercept <math>-2a</math>, <math>y</math>-intercept <math>-2</math>, and slope <math>-\frac 1a.</math><br />
<br />
(3)* <math>ax-y=a.</math> This is a line with <math>x</math>-intercept <math>1</math>, <math>y</math>-intercept <math>-a</math>, and slope <math>a.</math><br />
<br />
(4)* <math>ax-y=-a.</math> This is a line with <math>x</math>-intercept <math>-1</math>, <math>y</math>-intercept <math>a</math>, and slope <math>a.</math><br />
<br />
The area of the rectangle created by the four equations can be written as <math>2a\cdot \sin A\cdot4\sin A</math><br />
<br />
= <math>8a(\sin A)^2</math><br />
<br />
= <math>8a(~\frac{1}{\sqrt{a^2+1}})^2</math><br />
<br />
= <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math><br />
<br />
(Note: <math>\tan A=</math> slope <math>a</math>)<br />
<br />
-fnothing4994<br />
<br />
Where did the <math>a^2</math> and <math>\sin A = \frac{1}{\sqrt{a^2+1}}</math> come from?<br />
<br />
==Solution 4 (bruh moment solution)==<br />
<br />
Trying <math>a = 1</math> narrows down the choices to options <math>\textbf{(C)}</math>, <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math>. Trying <math>a = 2</math> and <math>a = 3</math> eliminates <math>\textbf{(C)}</math> and <math>\textbf{(E)}</math>, to obtain <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> as our answer. -¢<br />
<br />
== Video Solution by OmegaLearn (System of Equations and Shoelace Formula) ==<br />
https://youtu.be/2iohPYkZpkQ<br />
<br />
~ pi_is_3.14<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_24&diff=1467192021 AMC 10A Problems/Problem 242021-02-15T02:20:33Z<p>Awesomediabrine: /* Solution 3 (Geometry) */</p>
<hr />
<div>==Problem==<br />
The interior of a quadrilateral is bounded by the graphs of <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math>, where <math>a</math> a positive real number. What is the area of this region in terms of <math>a</math>, valid for all <math>a > 0</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}</math><br />
<br />
==Solution 1==<br />
The conditions <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math> give <math>|x+ay| = |2a|</math> and <math>|ax-y| = |a|</math> or <math>x+ay = \pm 2a</math> and <math>ax-y = \pm a</math>. The slopes here are perpendicular, so the quadrilateral is a rectangle.<br />
Plug in <math>a=1</math> and graph it. We quickly see that the area is <math>2\sqrt{2} \cdot \sqrt{2} = 4</math>, so the answer can't be <math>A</math> or <math>B</math> by testing the values they give (test it!). Now plug in <math>a=2</math>. We see using a ruler that the sides of the rectangle are about <math>\frac74</math> and <math>\frac72</math>. So the area is about <math>\frac{49}8 = 6.125</math>. Testing <math>C</math> we get <math>\frac{16}3</math> which is clearly less than <math>6</math>, so it is out. Testing <math>D</math> we get <math>\frac{32}5</math> which is near our answer, so we leave it. Testing <math>E</math> we get <math>\frac{16}5</math>, way less than <math>6</math>, so it is out. So, the only plausible answer is <math>\boxed{D}</math> ~firebolt360<br />
<br />
==Solution 2 (Casework)==<br />
For the equation <math>(x+ay)^2 = 4a^2,</math> the cases are <br />
<br />
(1) <math>x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a,y</math>-intercept <math>2,</math> and slope <math>-\frac 1a.</math><br />
<br />
(2) <math>x+ay=-2a.</math> This is a line with <math>x</math>-intercept <math>-2a,y</math>-intercept <math>-2,</math> and slope <math>-\frac 1a.</math><br />
<br />
For the equation <math>(ax-y)^2 = a^2,</math> the cases are<br />
<br />
(1)* <math>ax-y=a.</math> This is a line with <math>x</math>-intercept <math>1,y</math>-intercept <math>-a,</math> and slope <math>a.</math><br />
<br />
(2)* <math>ax-y=-a.</math> This is a line with <math>x</math>-intercept <math>-1,y</math>-intercept <math>a,</math> and slope <math>a.</math><br />
<br />
Since the slopes of the intersecting lines (from the four above equations) are negative reciprocals, the quadrilateral is a rectangle.<br />
<br />
Plugging <math>a=2</math> into the four above equations and solving systems of equations for intersecting lines [(1) and (1)*, (1) and (2)*, (2) and (1)*, (2) and (2)*], we get the respective solutions <cmath>(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).</cmath> <br />
Finally, by the Distance Formula, the length and width of the rectangle are <math>\frac{8\sqrt5}{5}</math> and <math>\frac{4\sqrt5}{5}.</math> The area we seek is <cmath>\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.</cmath><br />
Plugging <math>a=2</math> into the choices gives <br />
<br />
<math>\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}</math><br />
<br />
The answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3 (Geometry)==<br />
Similar to Solution 2, we will use the equations of the four cases:<br />
<br />
(1) <math>x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a</math>, <math>y</math>-intercept <math>2</math>, and slope <math>-\frac 1a.</math><br />
<br />
(2) <math>x+ay=-2a.</math> This is a line with <math>x</math>-intercept <math>-2a</math>, <math>y</math>-intercept <math>-2</math>, and slope <math>-\frac 1a.</math><br />
<br />
(3)* <math>ax-y=a.</math> This is a line with <math>x</math>-intercept <math>1</math>, <math>y</math>-intercept <math>-a</math>, and slope <math>a.</math><br />
<br />
(4)* <math>ax-y=-a.</math> This is a line with <math>x</math>-intercept <math>-1</math>, <math>y</math>-intercept <math>a</math>, and slope <math>a.</math><br />
<br />
The area of the rectangle created by the four equations can be written as <math>2a\cdot \sin A\cdot4\sin A</math><br />
<br />
= <math>8a(\sin A)^2</math><br />
<br />
= <math>8a(~\frac{1}{\sqrt{a^2+1}})^2</math><br />
<br />
= <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math><br />
<br />
(Note: <math>\tan A=</math> slope <math>a</math>)<br />
<br />
-fnothing4994<br />
<br />
Where did the <math>a^2</math> come from?<br />
<br />
==Solution 4 (bruh moment solution)==<br />
<br />
Trying <math>a = 1</math> narrows down the choices to options <math>\textbf{(C)}</math>, <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math>. Trying <math>a = 2</math> and <math>a = 3</math> eliminates <math>\textbf{(C)}</math> and <math>\textbf{(E)}</math>, to obtain <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> as our answer. -¢<br />
<br />
== Video Solution by OmegaLearn (System of Equations and Shoelace Formula) ==<br />
https://youtu.be/2iohPYkZpkQ<br />
<br />
~ pi_is_3.14<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_18&diff=1466582021 AMC 12A Problems/Problem 182021-02-14T18:59:58Z<p>Awesomediabrine: /* Solution 1 (but where do you get 10=11+f(\frac{25}{11}) */</p>
<hr />
<div>{{duplicate|[[2021 AMC 10A Problems#Problem 18|2021 AMC 10A #18]] and [[2021 AMC 12A Problems#Problem 18|2021 AMC 12A #18]]}}<br />
<br />
==Problem==<br />
Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a\cdot b) = f(a)+f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. Furthermore, suppose that <math>f</math> also has the property that <math>f(p)=p</math> for every prime number <math>p</math>. For which of the following numbers <math>x</math> is <math>f(x) < 0</math>?<br />
<br />
<math>\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad</math><br />
<br />
==Solution 1 (but where do you get <math>10=11+f(\frac{25}{11})</math> ==<br />
Looking through the solutions we can see that <math>f(\frac{25}{11})</math> can be expressed as <math>f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})</math> so using the prime numbers to piece together what we have we can get <math>10=11+f(\frac{25}{11})</math>, so <math>f(\frac{25}{11})=-1</math> or <math>\boxed{E}</math>.<br />
<br />
-Lemonie<br />
<br />
<math>f(\frac{25}{11} \cdot 11) = f(25) = f(5) + f(5) = 10</math><br />
<br />
- awesomediabrine<br />
<br />
==Solution 2==<br />
We know that <math>f(2)=2</math>. Adding <math>f(1)</math> to both sides, we get <br />
<cmath>\begin{align*}<br />
f(2)+f(1)&=2+f(1)\\ why<br />
f(2)&=2+f(1)\\<br />
2&=2+f(1)\\<br />
f(1)&=0<br />
\end{align*}</cmath><br />
Also<br />
<cmath>f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2</cmath><br />
<cmath>f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3</cmath><br />
<cmath>f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11</cmath><br />
In <math>\textbf{(A)}</math> we have <math>f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7</math>.<br />
<br />
In <math>\textbf{(B)}</math> we have <math>f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3</math>.<br />
<br />
In <math>\textbf{(C)}</math> we have <math>f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1</math>.<br />
<br />
In <math>\textbf{(D)}</math> we have <math>f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2</math>.<br />
<br />
In <math>\textbf{(E)}</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>.<br />
<br />
Thus, our answer is <math>\boxed{\textbf{(E)} \frac{25}{11}}</math><br />
~JHawk0224<br />
<br />
I found the first part confusing to read, so I am rewriting it:<br />
<cmath>f(2) = f(2 \cdot 1) = f(2) + f(1)</cmath><br />
<cmath>f(2) = f(2) + f(1)</cmath><br />
<cmath>0 = f(1)</cmath><br />
- awesomediabrine<br />
<br />
==Solution 3 (Deeper)==<br />
Consider the rational <math>\frac{a}{b}</math>, for <math>a,b</math> integers. We have <math>f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)</math>. So <math>f\left(\frac{a}{b}\right)=f(a)-f(b)</math>. Let <math>p</math> be a prime. Notice that <math>f(p^k)=kf(p)</math>. And <math>f(p)=p</math>. So if <math>a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}</math>, <math>f(a)=a_1p_1+a_2p_2+....+a_kp_k</math>. We simply need this to be greater than what we have for <math>f(b)</math>. Notice that for answer choices <math>A,B,C, </math> and <math>D</math>, the numerator <math>(a)</math> has less prime factors than the denominator, and so they are less likely to work. We check <math>E</math> first, and it works, therefore the answer is <math>\boxed{\textbf{(E)}}</math>. <br />
<br />
~yofro<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=dvlTA8Ncp58<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtu.be/8gGcj95rlWY<br />
<br />
== Video Solution by OmegaLearn (Using Functions and manipulations) ==<br />
https://youtu.be/aGv99CLzguE<br />
<br />
~ pi_is_3.14<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=17|num-a=19}}<br />
{{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_18&diff=1466572021 AMC 12A Problems/Problem 182021-02-14T18:58:25Z<p>Awesomediabrine: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2021 AMC 10A Problems#Problem 18|2021 AMC 10A #18]] and [[2021 AMC 12A Problems#Problem 18|2021 AMC 12A #18]]}}<br />
<br />
==Problem==<br />
Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a\cdot b) = f(a)+f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. Furthermore, suppose that <math>f</math> also has the property that <math>f(p)=p</math> for every prime number <math>p</math>. For which of the following numbers <math>x</math> is <math>f(x) < 0</math>?<br />
<br />
<math>\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad</math><br />
<br />
==Solution 1 (but where do you get <math>10=11+f(\frac{25}{11})</math> ==<br />
Looking through the solutions we can see that <math>f(\frac{25}{11})</math> can be expressed as <math>f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})</math> so using the prime numbers to piece together what we have we can get <math>10=11+f(\frac{25}{11})</math>, so <math>f(\frac{25}{11})=-1</math> or <math>\boxed{E}</math>.<br />
<br />
-Lemonie<br />
<br />
<math>f(\frac{25}{11} \cdot 11) = f(25) = f(5) \cdot f(5) = 10</math><br />
<br />
- awesomediabrine<br />
<br />
==Solution 2==<br />
We know that <math>f(2)=2</math>. Adding <math>f(1)</math> to both sides, we get <br />
<cmath>\begin{align*}<br />
f(2)+f(1)&=2+f(1)\\ why<br />
f(2)&=2+f(1)\\<br />
2&=2+f(1)\\<br />
f(1)&=0<br />
\end{align*}</cmath><br />
Also<br />
<cmath>f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2</cmath><br />
<cmath>f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3</cmath><br />
<cmath>f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11</cmath><br />
In <math>\textbf{(A)}</math> we have <math>f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7</math>.<br />
<br />
In <math>\textbf{(B)}</math> we have <math>f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3</math>.<br />
<br />
In <math>\textbf{(C)}</math> we have <math>f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1</math>.<br />
<br />
In <math>\textbf{(D)}</math> we have <math>f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2</math>.<br />
<br />
In <math>\textbf{(E)}</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>.<br />
<br />
Thus, our answer is <math>\boxed{\textbf{(E)} \frac{25}{11}}</math><br />
~JHawk0224<br />
<br />
I found the first part confusing to read, so I am rewriting it:<br />
<cmath>f(2) = f(2 \cdot 1) = f(2) + f(1)</cmath><br />
<cmath>f(2) = f(2) + f(1)</cmath><br />
<cmath>0 = f(1)</cmath><br />
- awesomediabrine<br />
<br />
==Solution 3 (Deeper)==<br />
Consider the rational <math>\frac{a}{b}</math>, for <math>a,b</math> integers. We have <math>f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)</math>. So <math>f\left(\frac{a}{b}\right)=f(a)-f(b)</math>. Let <math>p</math> be a prime. Notice that <math>f(p^k)=kf(p)</math>. And <math>f(p)=p</math>. So if <math>a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}</math>, <math>f(a)=a_1p_1+a_2p_2+....+a_kp_k</math>. We simply need this to be greater than what we have for <math>f(b)</math>. Notice that for answer choices <math>A,B,C, </math> and <math>D</math>, the numerator <math>(a)</math> has less prime factors than the denominator, and so they are less likely to work. We check <math>E</math> first, and it works, therefore the answer is <math>\boxed{\textbf{(E)}}</math>. <br />
<br />
~yofro<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=dvlTA8Ncp58<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtu.be/8gGcj95rlWY<br />
<br />
== Video Solution by OmegaLearn (Using Functions and manipulations) ==<br />
https://youtu.be/aGv99CLzguE<br />
<br />
~ pi_is_3.14<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=17|num-a=19}}<br />
{{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Awesomediabrinehttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_18&diff=1466542021 AMC 12A Problems/Problem 182021-02-14T18:51:38Z<p>Awesomediabrine: /* Solution 1 (but where do you get 10=11+f(\frac{25}{11}) */</p>
<hr />
<div>{{duplicate|[[2021 AMC 10A Problems#Problem 18|2021 AMC 10A #18]] and [[2021 AMC 12A Problems#Problem 18|2021 AMC 12A #18]]}}<br />
<br />
==Problem==<br />
Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a\cdot b) = f(a)+f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. Furthermore, suppose that <math>f</math> also has the property that <math>f(p)=p</math> for every prime number <math>p</math>. For which of the following numbers <math>x</math> is <math>f(x) < 0</math>?<br />
<br />
<math>\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad</math><br />
<br />
==Solution 1 (but where do you get <math>10=11+f(\frac{25}{11})</math> ==<br />
Looking through the solutions we can see that <math>f(\frac{25}{11})</math> can be expressed as <math>f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})</math> so using the prime numbers to piece together what we have we can get <math>10=11+f(\frac{25}{11})</math>, so <math>f(\frac{25}{11})=-1</math> or <math>\boxed{E}</math>.<br />
<br />
-Lemonie<br />
<br />
<math>f(\frac{25}{11} \cdot 11) = f(25) = f(5) \cdot f(5) = 10</math><br />
<br />
- awesomediabrine<br />
<br />
==Solution 2==<br />
We know that <math>f(2)=2</math>. Adding <math>f(1)</math> to both sides, we get <cmath>\begin{align*}<br />
f(2)+f(1)&=2+f(1)\\ why<br />
f(2)&=2+f(1)\\<br />
2&=2+f(1)\\<br />
f(1)&=0<br />
\end{align*}</cmath><br />
Also<br />
<cmath>f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2</cmath><br />
<cmath>f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3</cmath><br />
<cmath>f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11</cmath><br />
In <math>\textbf{(A)}</math> we have <math>f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7</math>.<br />
<br />
In <math>\textbf{(B)}</math> we have <math>f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3</math>.<br />
<br />
In <math>\textbf{(C)}</math> we have <math>f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1</math>.<br />
<br />
In <math>\textbf{(D)}</math> we have <math>f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2</math>.<br />
<br />
In <math>\textbf{(E)}</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>.<br />
<br />
Thus, our answer is <math>\boxed{\textbf{(E)} \frac{25}{11}}</math><br />
~JHawk0224<br />
<br />
==Solution 3 (Deeper)==<br />
Consider the rational <math>\frac{a}{b}</math>, for <math>a,b</math> integers. We have <math>f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)</math>. So <math>f\left(\frac{a}{b}\right)=f(a)-f(b)</math>. Let <math>p</math> be a prime. Notice that <math>f(p^k)=kf(p)</math>. And <math>f(p)=p</math>. So if <math>a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}</math>, <math>f(a)=a_1p_1+a_2p_2+....+a_kp_k</math>. We simply need this to be greater than what we have for <math>f(b)</math>. Notice that for answer choices <math>A,B,C, </math> and <math>D</math>, the numerator <math>(a)</math> has less prime factors than the denominator, and so they are less likely to work. We check <math>E</math> first, and it works, therefore the answer is <math>\boxed{\textbf{(E)}}</math>. <br />
<br />
~yofro<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=dvlTA8Ncp58<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtu.be/8gGcj95rlWY<br />
<br />
== Video Solution by OmegaLearn (Using Functions and manipulations) ==<br />
https://youtu.be/aGv99CLzguE<br />
<br />
~ pi_is_3.14<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=17|num-a=19}}<br />
{{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Awesomediabrine