https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Asops&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T17:06:59ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_17&diff=2134022021 Fall AMC 12A Problems/Problem 172024-01-27T16:35:18Z<p>Asops: </p>
<hr />
<div>{{duplicate|[[2021 Fall AMC 10A Problems/Problem 20|2021 Fall AMC 10A #20]] and [[2021 Fall AMC 12A Problems/Problem 17|2021 Fall AMC 12A #17]]}}<br />
<br />
== Problem ==<br />
<br />
For how many ordered pairs <math>(b,c)</math> of positive integers does neither <math>x^2+bx+c=0</math> nor <math>x^2+cx+b=0</math> have two distinct real solutions?<br />
<br />
<math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad</math><br />
<br />
== Solution 1 (Casework) ==<br />
A quadratic equation does not have two distinct real solutions if and only if the discriminant is nonpositive. We conclude that:<br />
<ol style="margin-left: 1.5em;"><br />
<li>Since <math>x^2+bx+c=0</math> does not have real solutions, we have <math>b^2\leq 4c.</math></li><p><br />
<li>Since <math>x^2+cx+b=0</math> does not have real solutions, we have <math>c^2\leq 4b.</math></li><p><br />
</ol><br />
Squaring the first inequality, we get <math>b^4\leq 16c^2.</math> Multiplying the second inequality by <math>16,</math> we get <math>16c^2\leq 64b.</math> Combining these results, we get <cmath>b^4\leq 16c^2\leq 64b.</cmath><br />
We apply casework to the value of <math>b:</math><br />
<br />
* If <math>b=1,</math> then <math>1\leq 16c^2\leq 64,</math> from which <math>c=1,2.</math><br />
<br />
* If <math>b=2,</math> then <math>16\leq 16c^2\leq 128,</math> from which <math>c=1,2.</math><br />
<br />
* If <math>b=3,</math> then <math>81\leq 16c^2\leq 192,</math> from which <math>c=3.</math><br />
<br />
* If <math>b=4,</math> then <math>256\leq 16c^2\leq 256,</math> from which <math>c=4.</math><br />
<br />
Together, there are <math>\boxed{\textbf{(B) } 6}</math> ordered pairs <math>(b,c),</math> namely <math>(1,1),(1,2),(2,1),(2,2),(3,3),</math> and <math>(4,4).</math><br />
<br />
~MRENTHUSIASM<br />
<br />
== Solution 2 (Graphing) ==<br />
Similar to Solution 1, use the discriminant to get <math>b^2\leq 4c</math> and <math>c^2\leq 4b</math>. These can be rearranged to <math>c\geq \frac{1}{4}b^2</math> and <math>b\geq \frac{1}{4}c^2</math>. Now, we can roughly graph these two inequalities, letting one of them be the <math>x</math> axis and the other be <math>y</math>. <br />
The graph of solutions should be above the parabola and under its inverse, meaning we want points on the graph or in the first area enclosed by the two graphs:<br />
<asy><br />
unitsize(2);<br />
Label f; <br />
f.p=fontsize(6); <br />
xaxis("$x$",0,5,Ticks(f, 1.0)); <br />
yaxis("$y$",0,5,Ticks(f, 1.0)); <br />
real f(real x) <br />
{ <br />
return 0.25x^2; <br />
} <br />
real g(real x) <br />
{ <br />
return 2*sqrt(x); <br />
} <br />
dot((1,1));<br />
dot((2,1));<br />
dot((1,2));<br />
dot((2,2));<br />
dot((3,3));<br />
dot((4,4));<br />
draw(graph(f,0,sqrt(20)));<br />
draw(graph(g,0,5));<br />
</asy><br />
We are looking for lattice points (since <math>b</math> and <math>c</math> are positive integers), of which we can count <math>\boxed{\textbf{(B) } 6}</math>.<br />
<br />
~aop2014<br />
<br />
== Solution 3 (Graphing) ==<br />
We need to solve the following system of inequalities:<br />
<cmath><br />
\[<br />
\left\{<br />
\begin{array}{ll}<br />
b^2 - 4 c \leq 0 \\<br />
c^2 - 4 b \leq 0<br />
\end{array}<br />
\right..<br />
\]<br />
</cmath><br />
Feasible solutions are in the region formed between two parabolas <math>b^2 - 4 c = 0</math> and <math>c^2 - 4 b = 0</math>.<br />
<br />
Define <math>f \left( b \right) = \frac{b^2}{4}</math> and <math>g \left( b \right) = 2 \sqrt{b}</math>.<br />
Therefore, all feasible solutions are in the region formed between the graphs of these two functions.<br />
<br />
For <math>b = 1</math>, we have <math>f(b) = \frac{1}{4}</math> and <math>g(b) = 2</math>.<br />
Hence, the feasible <math>c</math> are <math>1, 2</math>.<br />
<br />
For <math>b = 2</math>, we have <math>f(b) = 1</math> and <math>g(b) = 2 \sqrt{2}</math>.<br />
Hence, the feasible <math>c</math> are <math>1, 2</math>.<br />
<br />
For <math>b = 3</math>, we have <math>f(b) = \frac{9}{4}</math> and <math>g(b) = 2 \sqrt{3}</math>.<br />
Hence, the feasible <math>c</math> is <math>3</math>.<br />
<br />
For <math>b = 4</math>, we have <math>f(b) = 4</math> and <math>g(b) = 4</math>.<br />
Hence, the feasible <math>c</math> is <math>4</math>.<br />
<br />
For <math>b > 4</math>, we have <math>f(b) > g(b)</math>. Hence, there is no feasible <math>c</math>.<br />
<br />
Putting all cases together, the correct answer is <math>\boxed{\textbf{(B) } 6}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Solution 4 (Oversimplified but Risky)==<br />
A quadratic equation <math>Ax^2+Bx+C=0</math> has one real solution if and only if <math>\sqrt{B^2-4AC}=0.</math> Similarly, it has imaginary solutions if and only if <math>\sqrt{B^2-4AC}<0.</math> We proceed as following:<br />
<br />
We want both <math>x^2+bx+c</math> to be <math>1</math> value or imaginary and <math>x^2+cx+b</math> to be <math>1</math> value or imaginary. <math>x^2+4x+4</math> is one such case since <math>\sqrt {b^2-4ac}</math> is <math>0.</math> Also, <math>x^2+3x+3, x^2+2x+2, x^2+x+1</math> are always imaginary for both <math>b</math> and <math>c.</math> We also have <math>x^2+x+2</math> along with <math>x^2+2x+1</math> since the latter has one solution, while the first one is imaginary. Therefore, we have <math>\boxed{\textbf{(B) } 6}</math> total ordered pairs of integers.<br />
<br />
~Arcticturn<br />
<br />
==Solution 5 (Quick and Easy)==<br />
We see that <math>b^2 \leq 4c</math> and <math>c^2 \leq 4b.</math> WLOG, assume that <math>b \geq c.</math> Then we have that <math>b^2 \leq 4c \leq 4b</math>, so <math>b^2 \leq 4b</math> and therefore <math>b \leq 4</math>, also meaning that <math>c \leq 4.</math> This means that we only need to try 16 cases. Now we can get rid of the assumption that <math>b \geq c</math>, because we want ordered pairs. For <math>b = 1</math> and <math>b = 2</math>, <math>c = 1</math> and <math>c = 2</math> work. When <math>b = 3</math>, <math>c</math> can only be <math>3</math>, and when <math>b = 4</math>, only <math>c = 4</math> works, for a total of <math>\boxed{\textbf{(B) } 6}</math> ordered pairs of integers.<br />
<br />
~littlefox_amc<br />
<br />
==Solution 6 (Fastest)==<br />
We need both <math>b^2\leq 4c</math> and <math>c^2\leq 4b</math>. <br />
<br />
If <math>b=c</math> then the above become <math>b^2\leq 4b\iff b\leq 4</math>, so we have four solutions <math>(k,k)</math>, where <math>k=1</math>,<math>2</math>,<math>3</math>,<math>4</math>.<br />
<br />
If <math>b<c</math> then we only need <math>c^2\leq 4b</math> since it implies <math>b^2< 4c</math>. Now<br />
<math>c^2\leq 4b\leq 4(c-1) \implies (c-2)^2\leq 0 \implies c=2</math>, so <math>b=1</math>. We plug <math>b=1</math>, <math>c=2</math> back into <math>c^2\leq 4b</math> and it works. So there is another solution <math>(1,2)</math>.<br />
<br />
By symmetry, if <math>b>c</math> then <math>(b,c)=(2,1)</math>.<br />
<br />
Therefore the total number of solutions is <math>\boxed{\textbf{(B) } 6}</math>.<br />
<br />
~asops<br />
<br />
== Video Solution by OmegaLearn ==<br />
https://youtu.be/zfChnbMGLVQ?t=4254<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=ef-W3l94k00<br />
<br />
~MathProblemSolvingSkills.com<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=EkaKfkQgFbI<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/RPnfZKv4DVA<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021 Fall|ab=A|num-b=16|num-a=18}}<br />
{{AMC10 box|year=2021 Fall|ab=A|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_23&diff=2054422023 AMC 12B Problems/Problem 232023-11-24T02:16:57Z<p>Asops: /* Solution 4 (Easy computation) */</p>
<hr />
<div>==Problem==<br />
When <math>n</math> standard six-sided dice are rolled, the product of the numbers rolled can be any of <math>936</math> possible values. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9</math><br />
<br />
<br />
==Solution 1==<br />
<br />
We start by trying to prove a function of <math>n</math>, and then we can apply the function and equate it to <math>936</math> to find the value of <math>n</math>.<br />
<br />
It is helpful to think of this problem in the format <math>(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots</math>. Note that if we represent the scenario in this manner, we can think of picking a <math>1</math> for one factor and then a <math>5</math> for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for <math>n=2</math>, <math>4</math> can be reached by picking <math>1</math> and <math>4</math> or <math>2</math> and <math>2</math>. However, this form gives us insights that will be useful later in the problem.<br />
<br />
Note that there are only <math>3</math> primes in the set <math>\{1,2,3,4,5,6\}</math>: <math>2,3,</math> and <math>5</math>. Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form <math>2^h \cdot 3^i \cdot 5^j</math> (the choice of variables will become clear later), for integer nonnegative values <math>h,i,j</math>. So now the problem boils down to how many distinct triplets <math>(h,i,j)</math> can be formed by taking the product of <math>n</math> dice values. <br />
<br />
We start our work on representing <math>j</math>: the powers of <math>5</math>, because it is the simplest in this scenario because there is only one factor of <math>5</math> in the set. Because of this, having <math>j</math> fives in our prime factorization of the product is equivalent to picking <math>j</math> factors from the polynomial <math>(1+\dots + 6) \cdots</math> and choosing each factor to be a <math>5</math>. Now that we've selected <math>j</math> factors, there are <math>n-j</math> factors remaining to choose our powers of <math>3</math> and <math>2</math>. <br />
<br />
Suppose our prime factorization of this product contains <math>i</math> powers of <math>3</math>. These powers of <math>3</math> can either come from a <math>3</math> factor or a <math>6</math> factor, but since both <math>3</math> and <math>6</math> contain only one power of <math>3</math>, this means that a product with <math>i</math> powers of <math>3</math> corresponds directly to picking <math>i</math> factors from the polynomial, each of which is either <math>3</math> or <math>6</math> (but this distinction doesn't matter when we consider only the powers of <math>3</math>. <br />
<br />
Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair <math>(i,j)</math> that match the requirements, corresponding to the number of <math>3</math>'s and the number of <math>5</math>'s our product will have. Then how many different <math>h</math> values for the powers of <math>2</math> are possible?<br />
<br />
In the <math>i+j</math> factors we have already chosen, we obviously can't have any factors of <math>2</math> in the <math>j</math> factors with <math>5</math>. However, we can have a factor of <math>2</math> pairing with factors of <math>3</math>, if we choose a <math>6</math>. The maximal possible power of <math>2</math> in these <math>i</math> factors is thus <math>2^i</math>, which occurs when we pick every factor to be <math>6</math>. <br />
<br />
We now have <math>n-i-j</math> factors remaining, and we want to allocate these to solely powers of <math>2</math>. For each of these factors, we can choose either a <math>1,2,</math> or <math>4</math>. Therefore the maximal power of <math>2</math> achieved in these factors is when we pick <math>4</math> for all of them, which is equivalent to <math>2^{2\cdot (n-i-j)}</math>.<br />
<br />
Now if we multiply this across the total <math>n</math> factors (or <math>n</math> dice) we have a total of <math>2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}</math>, which is the maximal power of <math>2</math> attainable in the product for a pair <math>(i,j)</math>. Now note that every power of <math>2</math> below this power is attainable: we can simply just take away a power of <math>2</math> from an existing factor by dividing by <math>2</math>. Therefore the powers of <math>2</math>, and thus the <math>h</math> value ranges from <math>h=0</math> to <math>h=2n-i-2j</math>, so there are a total of <math>2n+1-i-2j</math> distinct values for <math>h</math> for a given pair <math>(i,j)</math>. <br />
<br />
Now to find the total number of distinct triplets, we must sum this across all possible <math>i</math>s and <math>j</math>s. Lets take note of our restrictions on <math>i,j</math>: the only restriction is that <math>i+j \leq n</math>, since we're picking factors from <math>n</math> dice.<br />
<br />
<cmath> \sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j</cmath><br />
<br />
We start by calculating the first term. <math>2n+1</math> is constant, so we just need to find out how many pairs there are such that <math>i+j \leq n</math>. Set <math>i</math> to <math>0</math>: <math>j</math> can range from <math>0</math> to <math>n</math>, then set <math>i</math> to <math>1</math>: <math>j</math> can range from <math>0</math> to <math>n-1</math>, etc. The total number of pairs is thus <math>n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}</math>. Therefore the left summation evaluates to <cmath>\frac{(2n+1)(n+1)(n+2)}{2}</cmath><br />
<br />
Now we calculate <math>\sum_{i+j \leq n}^{} i+2j</math>. This simplifies to <math>\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j</math>. Note that because <math>i+j = n</math> is symmetric with respect to <math>i,j</math>, the sum of <math>i</math> in all of the pairs will be equal to the sum of <math>j</math> in all of the pairs. Thus this is equal to calculating <math>3 \cdot \sum_{i+j \leq n}^{} i</math>.<br />
<br />
In the pairs, <math>i=1</math> appears for <math>j</math> ranging between <math>0</math> and <math>n-1</math> so the sum here is <math>1 \cdot (n)</math>. Similarly <math>i=2</math> appears for <math>j</math> ranging from <math>0</math> to <math>n-2</math>, so the sum is <math>2 \cdot (n-1)</math>. If we continue the pattern, the sum overall is <math>(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1</math>. We can rearrange this as <math>((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)</math><br />
<br />
<cmath> = \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1</cmath><br />
<br />
We can write this in easier terms as <math>\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k</math><br />
<cmath>=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}</cmath><br />
<cmath> = \frac{n(n+1)(n+2)}{6}</cmath><br />
<br />
We multiply this by <math>3</math> to obtain that <cmath>\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}</cmath><br />
<br />
Thus our final answer for the number of distinct triplets <math>(h,i,j)</math> is:<br />
<cmath>\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}</cmath><br />
<cmath> = \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)</cmath><br />
<cmath> = \frac{(n+1)^2(n+2)}{2}</cmath><br />
<br />
Now most of the work is done. We set this equal to <math>936</math> and prime factorize. <math>936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13</math>, so <math>(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13</math>. Clearly <math>13</math> cannot be anything squared and <math>2^4 \cdot 3^2</math> is a perfect square, so <math>n+2 = 13</math> and <math>n = 11 = \boxed{A}</math><br />
<br />
<br />
<br />
~KingRavi<br />
<br />
==Solution 2==<br />
<br />
The product can be written as <br />
<cmath><br />
\begin{align*}<br />
2^a 3^b 4^c 5^d 6^e<br />
& = 2^{a + 2c + e} 3^{b + e} 5^d .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, we need to find the number of ordered tuples <math>\left( a + 2c + e, b+e, d \right)</math> where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math> are non-negative integers satisfying <math>a+b+c+d+e \leq n</math>.<br />
We denote this number as <math>f(n)</math>.<br />
<br />
Denote by <math>g \left( k \right)</math> the number of ordered tuples <math>\left( a + 2c + e, b+e \right)</math> where <math>\left( a, b, c, e \right) \in \Delta_k</math> with <math>\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}</math>.<br />
<br />
Thus,<br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{d = 0}^n g \left( n - d \right) \\<br />
& = \sum_{k = 0}^n g \left( k \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Next, we compute <math>g \left( k \right)</math>.<br />
<br />
Denote <math>i = b + e</math>. Thus, for each given <math>i</math>, the range of <math>a + 2c + e</math> is from 0 to <math>2 k - i</math>.<br />
Thus, the number of <math>\left( a + 2c + e, b + e \right)</math> is<br />
<cmath><br />
\begin{align*}<br />
g \left( k \right)<br />
& = \sum_{i=0}^k \left( 2 k - i + 1 \right) \\<br />
& = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, <br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{k = 0}^n g \left( k \right) \\<br />
& = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \\<br />
& = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2<br />
- \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \\<br />
& = \frac{3}{2} \cdot<br />
\frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right)<br />
- \frac{1}{2} \cdot<br />
\frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \\<br />
& = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
By solving <math>f \left( n \right) = 936</math>, we get <math>n = \boxed{\textbf{(A) 11}}</math>.<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 3 (Cheese)==<br />
<br />
The product can be written as<br />
<cmath><br />
\begin{align*}<br />
2^x 3^y 5^z<br />
\end{align*}<br />
</cmath><br />
<br />
Letting <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math>, 6 possible values. But if the only restriction of the product if that <math>2x\le n,y\le n,z\le n</math>, we can get <math>(2+1)(1+1)(1+1)=12</math> possible values. We calculate the ratio<br />
<cmath>r = \frac{\text{possible values of real situation}}{\text{possible values of ideal situation}} = \frac{6}{12}=0.5.</cmath><br />
<br />
Letting <math>n=2</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),</math><br /><br />
<math>(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)</math>, 17 possible values.<br />
The number of possibilities in the ideal situation is <math>5*3*3=45</math>, making <math>r = 17/45 \approx 0.378</math>.<br />
<br />
Now we can predict the trend of <math>r</math>: as <math>n</math> increases, <math>r</math> decreases.<br />
Letting<br />
<math>n=3</math>, you get possible values of ideal situation=<math>7*4*4=112</math>.<br />
<math>n=4</math>, the number=<math>9*5*5=225</math>.<br /><br />
<math>n=5</math>, the number=<math>11*6*6=396</math>.<br /><br />
<math>n=6</math>, the number=<math>13*7*7=637,637<936</math> so 6 is not the answer.<br /><br />
<math>n=7</math>, the number=<math>15*8*8=960</math>.<br /><br />
<math>n=8</math>, the number=<math>17*9*9=1377</math>,but <math>1377*0.378</math>≈<math>521</math> still much smaller than 936.<br /><br />
<math>n=9</math>, the number=<math>19*10*10=1900</math>,but <math>1900*0.378</math>≈<math>718</math> still smaller than 936.<br /><br />
<math>n=10</math>, the number=<math>21*11*11=2541</math>, <math>2541*0.378</math>≈<math>960</math> is a little bigger 936, but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is <math>\boxed{\textbf{(A) 11}}</math>.<br />
<br />
Check calculation:<br />
<math>n=11</math>,the number=<math>23*12*12=3312</math>,<math>3312*0.378</math>≈<math>1252</math> is much bigger than 936.<br />
<br />
~Troublemaker<br />
<br />
==Solution 4 (Easy computation)==<br />
<br />
The key observation is if <math>P=2^a\cdot 3^b \cdot 5^c</math>, then given <math>b</math> and <math>c</math>, <math>a</math> can take any value from <math>0</math> to <math>b+2d</math> where <math>d=n-b-c</math> is the number of rolls which is neither divisible by <math>3</math> nor <math>5</math>. We are left to calculate<br />
<cmath><br />
\sum_{b+c\leq n} (b+2d+1)=\sum_{b+c+d=n} (b+2d+1).<br />
</cmath><br />
By symmetry, <math> \sum_{b+c+d=n} d = \sum_{b+c+d=n} c</math>. Therefore,<br />
<cmath><br />
\sum_{b+c\leq n}(b+2d+1)=\sum_{b+c\leq n}(b+c+d+1)=\sum_{b+c\leq n}(n+1)=(n+1)\binom{n+2}{2}.<br />
</cmath><br />
<br />
The rest is the same as above.<br />
<br />
~asops<br />
<br />
==Video Solution 1 by OmegaLearn==<br />
https://youtu.be/FZG1j95owTo<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/BWb1dS4Jba0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2023|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_23&diff=2054392023 AMC 12B Problems/Problem 232023-11-24T02:04:57Z<p>Asops: </p>
<hr />
<div>==Problem==<br />
When <math>n</math> standard six-sided dice are rolled, the product of the numbers rolled can be any of <math>936</math> possible values. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9</math><br />
<br />
<br />
==Solution 1==<br />
<br />
We start by trying to prove a function of <math>n</math>, and then we can apply the function and equate it to <math>936</math> to find the value of <math>n</math>.<br />
<br />
It is helpful to think of this problem in the format <math>(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots</math>. Note that if we represent the scenario in this manner, we can think of picking a <math>1</math> for one factor and then a <math>5</math> for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for <math>n=2</math>, <math>4</math> can be reached by picking <math>1</math> and <math>4</math> or <math>2</math> and <math>2</math>. However, this form gives us insights that will be useful later in the problem.<br />
<br />
Note that there are only <math>3</math> primes in the set <math>\{1,2,3,4,5,6\}</math>: <math>2,3,</math> and <math>5</math>. Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form <math>2^h \cdot 3^i \cdot 5^j</math> (the choice of variables will become clear later), for integer nonnegative values <math>h,i,j</math>. So now the problem boils down to how many distinct triplets <math>(h,i,j)</math> can be formed by taking the product of <math>n</math> dice values. <br />
<br />
We start our work on representing <math>j</math>: the powers of <math>5</math>, because it is the simplest in this scenario because there is only one factor of <math>5</math> in the set. Because of this, having <math>j</math> fives in our prime factorization of the product is equivalent to picking <math>j</math> factors from the polynomial <math>(1+\dots + 6) \cdots</math> and choosing each factor to be a <math>5</math>. Now that we've selected <math>j</math> factors, there are <math>n-j</math> factors remaining to choose our powers of <math>3</math> and <math>2</math>. <br />
<br />
Suppose our prime factorization of this product contains <math>i</math> powers of <math>3</math>. These powers of <math>3</math> can either come from a <math>3</math> factor or a <math>6</math> factor, but since both <math>3</math> and <math>6</math> contain only one power of <math>3</math>, this means that a product with <math>i</math> powers of <math>3</math> corresponds directly to picking <math>i</math> factors from the polynomial, each of which is either <math>3</math> or <math>6</math> (but this distinction doesn't matter when we consider only the powers of <math>3</math>. <br />
<br />
Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair <math>(i,j)</math> that match the requirements, corresponding to the number of <math>3</math>'s and the number of <math>5</math>'s our product will have. Then how many different <math>h</math> values for the powers of <math>2</math> are possible?<br />
<br />
In the <math>i+j</math> factors we have already chosen, we obviously can't have any factors of <math>2</math> in the <math>j</math> factors with <math>5</math>. However, we can have a factor of <math>2</math> pairing with factors of <math>3</math>, if we choose a <math>6</math>. The maximal possible power of <math>2</math> in these <math>i</math> factors is thus <math>2^i</math>, which occurs when we pick every factor to be <math>6</math>. <br />
<br />
We now have <math>n-i-j</math> factors remaining, and we want to allocate these to solely powers of <math>2</math>. For each of these factors, we can choose either a <math>1,2,</math> or <math>4</math>. Therefore the maximal power of <math>2</math> achieved in these factors is when we pick <math>4</math> for all of them, which is equivalent to <math>2^{2\cdot (n-i-j)}</math>.<br />
<br />
Now if we multiply this across the total <math>n</math> factors (or <math>n</math> dice) we have a total of <math>2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}</math>, which is the maximal power of <math>2</math> attainable in the product for a pair <math>(i,j)</math>. Now note that every power of <math>2</math> below this power is attainable: we can simply just take away a power of <math>2</math> from an existing factor by dividing by <math>2</math>. Therefore the powers of <math>2</math>, and thus the <math>h</math> value ranges from <math>h=0</math> to <math>h=2n-i-2j</math>, so there are a total of <math>2n+1-i-2j</math> distinct values for <math>h</math> for a given pair <math>(i,j)</math>. <br />
<br />
Now to find the total number of distinct triplets, we must sum this across all possible <math>i</math>s and <math>j</math>s. Lets take note of our restrictions on <math>i,j</math>: the only restriction is that <math>i+j \leq n</math>, since we're picking factors from <math>n</math> dice.<br />
<br />
<cmath> \sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j</cmath><br />
<br />
We start by calculating the first term. <math>2n+1</math> is constant, so we just need to find out how many pairs there are such that <math>i+j \leq n</math>. Set <math>i</math> to <math>0</math>: <math>j</math> can range from <math>0</math> to <math>n</math>, then set <math>i</math> to <math>1</math>: <math>j</math> can range from <math>0</math> to <math>n-1</math>, etc. The total number of pairs is thus <math>n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}</math>. Therefore the left summation evaluates to <cmath>\frac{(2n+1)(n+1)(n+2)}{2}</cmath><br />
<br />
Now we calculate <math>\sum_{i+j \leq n}^{} i+2j</math>. This simplifies to <math>\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j</math>. Note that because <math>i+j = n</math> is symmetric with respect to <math>i,j</math>, the sum of <math>i</math> in all of the pairs will be equal to the sum of <math>j</math> in all of the pairs. Thus this is equal to calculating <math>3 \cdot \sum_{i+j \leq n}^{} i</math>.<br />
<br />
In the pairs, <math>i=1</math> appears for <math>j</math> ranging between <math>0</math> and <math>n-1</math> so the sum here is <math>1 \cdot (n)</math>. Similarly <math>i=2</math> appears for <math>j</math> ranging from <math>0</math> to <math>n-2</math>, so the sum is <math>2 \cdot (n-1)</math>. If we continue the pattern, the sum overall is <math>(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1</math>. We can rearrange this as <math>((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)</math><br />
<br />
<cmath> = \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1</cmath><br />
<br />
We can write this in easier terms as <math>\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k</math><br />
<cmath>=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}</cmath><br />
<cmath> = \frac{n(n+1)(n+2)}{6}</cmath><br />
<br />
We multiply this by <math>3</math> to obtain that <cmath>\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}</cmath><br />
<br />
Thus our final answer for the number of distinct triplets <math>(h,i,j)</math> is:<br />
<cmath>\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}</cmath><br />
<cmath> = \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)</cmath><br />
<cmath> = \frac{(n+1)^2(n+2)}{2}</cmath><br />
<br />
Now most of the work is done. We set this equal to <math>936</math> and prime factorize. <math>936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13</math>, so <math>(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13</math>. Clearly <math>13</math> cannot be anything squared and <math>2^4 \cdot 3^2</math> is a perfect square, so <math>n+2 = 13</math> and <math>n = 11 = \boxed{A}</math><br />
<br />
<br />
<br />
~KingRavi<br />
<br />
==Solution 2==<br />
<br />
The product can be written as <br />
<cmath><br />
\begin{align*}<br />
2^a 3^b 4^c 5^d 6^e<br />
& = 2^{a + 2c + e} 3^{b + e} 5^d .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, we need to find the number of ordered tuples <math>\left( a + 2c + e, b+e, d \right)</math> where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math> are non-negative integers satisfying <math>a+b+c+d+e \leq n</math>.<br />
We denote this number as <math>f(n)</math>.<br />
<br />
Denote by <math>g \left( k \right)</math> the number of ordered tuples <math>\left( a + 2c + e, b+e \right)</math> where <math>\left( a, b, c, e \right) \in \Delta_k</math> with <math>\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}</math>.<br />
<br />
Thus,<br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{d = 0}^n g \left( n - d \right) \\<br />
& = \sum_{k = 0}^n g \left( k \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Next, we compute <math>g \left( k \right)</math>.<br />
<br />
Denote <math>i = b + e</math>. Thus, for each given <math>i</math>, the range of <math>a + 2c + e</math> is from 0 to <math>2 k - i</math>.<br />
Thus, the number of <math>\left( a + 2c + e, b + e \right)</math> is<br />
<cmath><br />
\begin{align*}<br />
g \left( k \right)<br />
& = \sum_{i=0}^k \left( 2 k - i + 1 \right) \\<br />
& = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, <br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{k = 0}^n g \left( k \right) \\<br />
& = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \\<br />
& = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2<br />
- \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \\<br />
& = \frac{3}{2} \cdot<br />
\frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right)<br />
- \frac{1}{2} \cdot<br />
\frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \\<br />
& = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
By solving <math>f \left( n \right) = 936</math>, we get <math>n = \boxed{\textbf{(A) 11}}</math>.<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 3 (Cheese)==<br />
<br />
The product can be written as<br />
<cmath><br />
\begin{align*}<br />
2^x 3^y 5^z<br />
\end{align*}<br />
</cmath><br />
<br />
Letting <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math>, 6 possible values. But if the only restriction of the product if that <math>2x\le n,y\le n,z\le n</math>, we can get <math>(2+1)(1+1)(1+1)=12</math> possible values. We calculate the ratio<br />
<cmath>r = \frac{\text{possible values of real situation}}{\text{possible values of ideal situation}} = \frac{6}{12}=0.5.</cmath><br />
<br />
Letting <math>n=2</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),</math><br /><br />
<math>(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)</math>, 17 possible values.<br />
The number of possibilities in the ideal situation is <math>5*3*3=45</math>, making <math>r = 17/45 \approx 0.378</math>.<br />
<br />
Now we can predict the trend of <math>r</math>: as <math>n</math> increases, <math>r</math> decreases.<br />
Letting<br />
<math>n=3</math>, you get possible values of ideal situation=<math>7*4*4=112</math>.<br />
<math>n=4</math>, the number=<math>9*5*5=225</math>.<br /><br />
<math>n=5</math>, the number=<math>11*6*6=396</math>.<br /><br />
<math>n=6</math>, the number=<math>13*7*7=637,637<936</math> so 6 is not the answer.<br /><br />
<math>n=7</math>, the number=<math>15*8*8=960</math>.<br /><br />
<math>n=8</math>, the number=<math>17*9*9=1377</math>,but <math>1377*0.378</math>≈<math>521</math> still much smaller than 936.<br /><br />
<math>n=9</math>, the number=<math>19*10*10=1900</math>,but <math>1900*0.378</math>≈<math>718</math> still smaller than 936.<br /><br />
<math>n=10</math>, the number=<math>21*11*11=2541</math>, <math>2541*0.378</math>≈<math>960</math> is a little bigger 936, but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is <math>\boxed{\textbf{(A) 11}}</math>.<br />
<br />
Check calculation:<br />
<math>n=11</math>,the number=<math>23*12*12=3312</math>,<math>3312*0.378</math>≈<math>1252</math> is much bigger than 936.<br />
<br />
~Troublemaker<br />
<br />
==Solution 4 (Easy computation)==<br />
<br />
The key observation is if <math>P=2^a\cdot 3^b \cdot 5^c</math>, then given <math>b</math> and <math>c</math>, <math>a</math> can take any value from <math>0</math> to <math>2(n-b-c)+b=2n-b-2c</math>. We are left to calculate<br />
<cmath><br />
\sum_{b+c\leq n} (2n-b-2c+1).<br />
</cmath><br />
<br />
Let <math>d=n-b-c</math> be the number of rolls which is neither divisible by <math>3</math> nor <math>5</math>, then<br />
<cmath><br />
\begin{align*}<br />
&\sum_{b+c\leq n}(2n-b-2c+1)\\<br />
&=\sum_{b+c+d= n}(2n-b-2c+1)\\<br />
&=\sum_{b+c+d= n}(2n-b-c-d+1) \\<br />
&=\sum_{b+c+d= n} (n+1) = (n+1) \binom{n+2}{2}.<br />
\end{align*}<br />
</cmath><br />
<br />
Notice that we used symmetry argument. We also applied Stars & Bars to obtain <math>\sum_{b+c+d=n} 1 =\binom{n+2}{2}</math>.<br />
<br />
~asops<br />
<br />
==Video Solution 1 by OmegaLearn==<br />
https://youtu.be/FZG1j95owTo<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/BWb1dS4Jba0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2023|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_23&diff=2054382023 AMC 12B Problems/Problem 232023-11-24T02:03:51Z<p>Asops: </p>
<hr />
<div>==Problem==<br />
When <math>n</math> standard six-sided dice are rolled, the product of the numbers rolled can be any of <math>936</math> possible values. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9</math><br />
<br />
<br />
==Solution 1==<br />
<br />
We start by trying to prove a function of <math>n</math>, and then we can apply the function and equate it to <math>936</math> to find the value of <math>n</math>.<br />
<br />
It is helpful to think of this problem in the format <math>(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots</math>. Note that if we represent the scenario in this manner, we can think of picking a <math>1</math> for one factor and then a <math>5</math> for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for <math>n=2</math>, <math>4</math> can be reached by picking <math>1</math> and <math>4</math> or <math>2</math> and <math>2</math>. However, this form gives us insights that will be useful later in the problem.<br />
<br />
Note that there are only <math>3</math> primes in the set <math>\{1,2,3,4,5,6\}</math>: <math>2,3,</math> and <math>5</math>. Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form <math>2^h \cdot 3^i \cdot 5^j</math> (the choice of variables will become clear later), for integer nonnegative values <math>h,i,j</math>. So now the problem boils down to how many distinct triplets <math>(h,i,j)</math> can be formed by taking the product of <math>n</math> dice values. <br />
<br />
We start our work on representing <math>j</math>: the powers of <math>5</math>, because it is the simplest in this scenario because there is only one factor of <math>5</math> in the set. Because of this, having <math>j</math> fives in our prime factorization of the product is equivalent to picking <math>j</math> factors from the polynomial <math>(1+\dots + 6) \cdots</math> and choosing each factor to be a <math>5</math>. Now that we've selected <math>j</math> factors, there are <math>n-j</math> factors remaining to choose our powers of <math>3</math> and <math>2</math>. <br />
<br />
Suppose our prime factorization of this product contains <math>i</math> powers of <math>3</math>. These powers of <math>3</math> can either come from a <math>3</math> factor or a <math>6</math> factor, but since both <math>3</math> and <math>6</math> contain only one power of <math>3</math>, this means that a product with <math>i</math> powers of <math>3</math> corresponds directly to picking <math>i</math> factors from the polynomial, each of which is either <math>3</math> or <math>6</math> (but this distinction doesn't matter when we consider only the powers of <math>3</math>. <br />
<br />
Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair <math>(i,j)</math> that match the requirements, corresponding to the number of <math>3</math>'s and the number of <math>5</math>'s our product will have. Then how many different <math>h</math> values for the powers of <math>2</math> are possible?<br />
<br />
In the <math>i+j</math> factors we have already chosen, we obviously can't have any factors of <math>2</math> in the <math>j</math> factors with <math>5</math>. However, we can have a factor of <math>2</math> pairing with factors of <math>3</math>, if we choose a <math>6</math>. The maximal possible power of <math>2</math> in these <math>i</math> factors is thus <math>2^i</math>, which occurs when we pick every factor to be <math>6</math>. <br />
<br />
We now have <math>n-i-j</math> factors remaining, and we want to allocate these to solely powers of <math>2</math>. For each of these factors, we can choose either a <math>1,2,</math> or <math>4</math>. Therefore the maximal power of <math>2</math> achieved in these factors is when we pick <math>4</math> for all of them, which is equivalent to <math>2^{2\cdot (n-i-j)}</math>.<br />
<br />
Now if we multiply this across the total <math>n</math> factors (or <math>n</math> dice) we have a total of <math>2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}</math>, which is the maximal power of <math>2</math> attainable in the product for a pair <math>(i,j)</math>. Now note that every power of <math>2</math> below this power is attainable: we can simply just take away a power of <math>2</math> from an existing factor by dividing by <math>2</math>. Therefore the powers of <math>2</math>, and thus the <math>h</math> value ranges from <math>h=0</math> to <math>h=2n-i-2j</math>, so there are a total of <math>2n+1-i-2j</math> distinct values for <math>h</math> for a given pair <math>(i,j)</math>. <br />
<br />
Now to find the total number of distinct triplets, we must sum this across all possible <math>i</math>s and <math>j</math>s. Lets take note of our restrictions on <math>i,j</math>: the only restriction is that <math>i+j \leq n</math>, since we're picking factors from <math>n</math> dice.<br />
<br />
<cmath> \sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j</cmath><br />
<br />
We start by calculating the first term. <math>2n+1</math> is constant, so we just need to find out how many pairs there are such that <math>i+j \leq n</math>. Set <math>i</math> to <math>0</math>: <math>j</math> can range from <math>0</math> to <math>n</math>, then set <math>i</math> to <math>1</math>: <math>j</math> can range from <math>0</math> to <math>n-1</math>, etc. The total number of pairs is thus <math>n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}</math>. Therefore the left summation evaluates to <cmath>\frac{(2n+1)(n+1)(n+2)}{2}</cmath><br />
<br />
Now we calculate <math>\sum_{i+j \leq n}^{} i+2j</math>. This simplifies to <math>\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j</math>. Note that because <math>i+j = n</math> is symmetric with respect to <math>i,j</math>, the sum of <math>i</math> in all of the pairs will be equal to the sum of <math>j</math> in all of the pairs. Thus this is equal to calculating <math>3 \cdot \sum_{i+j \leq n}^{} i</math>.<br />
<br />
In the pairs, <math>i=1</math> appears for <math>j</math> ranging between <math>0</math> and <math>n-1</math> so the sum here is <math>1 \cdot (n)</math>. Similarly <math>i=2</math> appears for <math>j</math> ranging from <math>0</math> to <math>n-2</math>, so the sum is <math>2 \cdot (n-1)</math>. If we continue the pattern, the sum overall is <math>(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1</math>. We can rearrange this as <math>((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)</math><br />
<br />
<cmath> = \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1</cmath><br />
<br />
We can write this in easier terms as <math>\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k</math><br />
<cmath>=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}</cmath><br />
<cmath> = \frac{n(n+1)(n+2)}{6}</cmath><br />
<br />
We multiply this by <math>3</math> to obtain that <cmath>\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}</cmath><br />
<br />
Thus our final answer for the number of distinct triplets <math>(h,i,j)</math> is:<br />
<cmath>\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}</cmath><br />
<cmath> = \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)</cmath><br />
<cmath> = \frac{(n+1)^2(n+2)}{2}</cmath><br />
<br />
Now most of the work is done. We set this equal to <math>936</math> and prime factorize. <math>936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13</math>, so <math>(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13</math>. Clearly <math>13</math> cannot be anything squared and <math>2^4 \cdot 3^2</math> is a perfect square, so <math>n+2 = 13</math> and <math>n = 11 = \boxed{A}</math><br />
<br />
<br />
<br />
~KingRavi<br />
<br />
==Solution 2==<br />
<br />
The product can be written as <br />
<cmath><br />
\begin{align*}<br />
2^a 3^b 4^c 5^d 6^e<br />
& = 2^{a + 2c + e} 3^{b + e} 5^d .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, we need to find the number of ordered tuples <math>\left( a + 2c + e, b+e, d \right)</math> where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math> are non-negative integers satisfying <math>a+b+c+d+e \leq n</math>.<br />
We denote this number as <math>f(n)</math>.<br />
<br />
Denote by <math>g \left( k \right)</math> the number of ordered tuples <math>\left( a + 2c + e, b+e \right)</math> where <math>\left( a, b, c, e \right) \in \Delta_k</math> with <math>\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}</math>.<br />
<br />
Thus,<br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{d = 0}^n g \left( n - d \right) \\<br />
& = \sum_{k = 0}^n g \left( k \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Next, we compute <math>g \left( k \right)</math>.<br />
<br />
Denote <math>i = b + e</math>. Thus, for each given <math>i</math>, the range of <math>a + 2c + e</math> is from 0 to <math>2 k - i</math>.<br />
Thus, the number of <math>\left( a + 2c + e, b + e \right)</math> is<br />
<cmath><br />
\begin{align*}<br />
g \left( k \right)<br />
& = \sum_{i=0}^k \left( 2 k - i + 1 \right) \\<br />
& = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, <br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{k = 0}^n g \left( k \right) \\<br />
& = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \\<br />
& = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2<br />
- \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \\<br />
& = \frac{3}{2} \cdot<br />
\frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right)<br />
- \frac{1}{2} \cdot<br />
\frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \\<br />
& = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
By solving <math>f \left( n \right) = 936</math>, we get <math>n = \boxed{\textbf{(A) 11}}</math>.<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 3 (Cheese)==<br />
<br />
The product can be written as<br />
<cmath><br />
\begin{align*}<br />
2^x 3^y 5^z<br />
\end{align*}<br />
</cmath><br />
<br />
Letting <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math>, 6 possible values. But if the only restriction of the product if that <math>2x\le n,y\le n,z\le n</math>, we can get <math>(2+1)(1+1)(1+1)=12</math> possible values. We calculate the ratio<br />
<cmath>r = \frac{\text{possible values of real situation}}{\text{possible values of ideal situation}} = \frac{6}{12}=0.5.</cmath><br />
<br />
Letting <math>n=2</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),</math><br /><br />
<math>(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)</math>, 17 possible values.<br />
The number of possibilities in the ideal situation is <math>5*3*3=45</math>, making <math>r = 17/45 \approx 0.378</math>.<br />
<br />
Now we can predict the trend of <math>r</math>: as <math>n</math> increases, <math>r</math> decreases.<br />
Letting<br />
<math>n=3</math>, you get possible values of ideal situation=<math>7*4*4=112</math>.<br />
<math>n=4</math>, the number=<math>9*5*5=225</math>.<br /><br />
<math>n=5</math>, the number=<math>11*6*6=396</math>.<br /><br />
<math>n=6</math>, the number=<math>13*7*7=637,637<936</math> so 6 is not the answer.<br /><br />
<math>n=7</math>, the number=<math>15*8*8=960</math>.<br /><br />
<math>n=8</math>, the number=<math>17*9*9=1377</math>,but <math>1377*0.378</math>≈<math>521</math> still much smaller than 936.<br /><br />
<math>n=9</math>, the number=<math>19*10*10=1900</math>,but <math>1900*0.378</math>≈<math>718</math> still smaller than 936.<br /><br />
<math>n=10</math>, the number=<math>21*11*11=2541</math>, <math>2541*0.378</math>≈<math>960</math> is a little bigger 936, but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is <math>\boxed{\textbf{(A) 11}}</math>.<br />
<br />
Check calculation:<br />
<math>n=11</math>,the number=<math>23*12*12=3312</math>,<math>3312*0.378</math>≈<math>1252</math> is much bigger than 936.<br />
<br />
~Troublemaker<br />
<br />
==Solution 4 (Easy computation)==<br />
<br />
The key observation is if <math>P=2^a\cdot 3^b \cdot 5^c</math>, then given <math>b</math> and <math>c</math>, <math>a</math> can take any value from <math>0</math> to <math>2(n-b-c)+b=2n-b-2c</math>. We are left to calculate<br />
\[<br />
\sum_{b+c\leq n} (2n-b-2c+1).<br />
\]<br />
<br />
Let <math>d=n-b-c</math> be the number of rolls which is neither divisible by <math>3</math> nor <math>5</math>, then<br />
<cmath><br />
\begin{align*}<br />
&\sum_{b+c\leq n}(2n-b-2c+1)\\<br />
&=\sum_{b+c+d= n}(2n-b-2c+1)\\<br />
&=\sum_{b+c+d= n}(2n-b-c-d+1) \\<br />
&=\sum_{b+c+d= n} (n+1) = (n+1) \binom{n+2}{2}.<br />
\end{align*}<br />
</cmath><br />
<br />
Notice that we used symmetry argument twice. We also applied Stars & Bars to obtain <math>\sum_{b+c+d=n} 1 =\binom{n+2}{2}</math>.<br />
<br />
~asops<br />
<br />
==Video Solution 1 by OmegaLearn==<br />
https://youtu.be/FZG1j95owTo<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/BWb1dS4Jba0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2023|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_23&diff=2054372023 AMC 12B Problems/Problem 232023-11-24T01:54:56Z<p>Asops: </p>
<hr />
<div>==Problem==<br />
When <math>n</math> standard six-sided dice are rolled, the product of the numbers rolled can be any of <math>936</math> possible values. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9</math><br />
<br />
<br />
==Solution 1==<br />
<br />
We start by trying to prove a function of <math>n</math>, and then we can apply the function and equate it to <math>936</math> to find the value of <math>n</math>.<br />
<br />
It is helpful to think of this problem in the format <math>(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots</math>. Note that if we represent the scenario in this manner, we can think of picking a <math>1</math> for one factor and then a <math>5</math> for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for <math>n=2</math>, <math>4</math> can be reached by picking <math>1</math> and <math>4</math> or <math>2</math> and <math>2</math>. However, this form gives us insights that will be useful later in the problem.<br />
<br />
Note that there are only <math>3</math> primes in the set <math>\{1,2,3,4,5,6\}</math>: <math>2,3,</math> and <math>5</math>. Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form <math>2^h \cdot 3^i \cdot 5^j</math> (the choice of variables will become clear later), for integer nonnegative values <math>h,i,j</math>. So now the problem boils down to how many distinct triplets <math>(h,i,j)</math> can be formed by taking the product of <math>n</math> dice values. <br />
<br />
We start our work on representing <math>j</math>: the powers of <math>5</math>, because it is the simplest in this scenario because there is only one factor of <math>5</math> in the set. Because of this, having <math>j</math> fives in our prime factorization of the product is equivalent to picking <math>j</math> factors from the polynomial <math>(1+\dots + 6) \cdots</math> and choosing each factor to be a <math>5</math>. Now that we've selected <math>j</math> factors, there are <math>n-j</math> factors remaining to choose our powers of <math>3</math> and <math>2</math>. <br />
<br />
Suppose our prime factorization of this product contains <math>i</math> powers of <math>3</math>. These powers of <math>3</math> can either come from a <math>3</math> factor or a <math>6</math> factor, but since both <math>3</math> and <math>6</math> contain only one power of <math>3</math>, this means that a product with <math>i</math> powers of <math>3</math> corresponds directly to picking <math>i</math> factors from the polynomial, each of which is either <math>3</math> or <math>6</math> (but this distinction doesn't matter when we consider only the powers of <math>3</math>. <br />
<br />
Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair <math>(i,j)</math> that match the requirements, corresponding to the number of <math>3</math>'s and the number of <math>5</math>'s our product will have. Then how many different <math>h</math> values for the powers of <math>2</math> are possible?<br />
<br />
In the <math>i+j</math> factors we have already chosen, we obviously can't have any factors of <math>2</math> in the <math>j</math> factors with <math>5</math>. However, we can have a factor of <math>2</math> pairing with factors of <math>3</math>, if we choose a <math>6</math>. The maximal possible power of <math>2</math> in these <math>i</math> factors is thus <math>2^i</math>, which occurs when we pick every factor to be <math>6</math>. <br />
<br />
We now have <math>n-i-j</math> factors remaining, and we want to allocate these to solely powers of <math>2</math>. For each of these factors, we can choose either a <math>1,2,</math> or <math>4</math>. Therefore the maximal power of <math>2</math> achieved in these factors is when we pick <math>4</math> for all of them, which is equivalent to <math>2^{2\cdot (n-i-j)}</math>.<br />
<br />
Now if we multiply this across the total <math>n</math> factors (or <math>n</math> dice) we have a total of <math>2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}</math>, which is the maximal power of <math>2</math> attainable in the product for a pair <math>(i,j)</math>. Now note that every power of <math>2</math> below this power is attainable: we can simply just take away a power of <math>2</math> from an existing factor by dividing by <math>2</math>. Therefore the powers of <math>2</math>, and thus the <math>h</math> value ranges from <math>h=0</math> to <math>h=2n-i-2j</math>, so there are a total of <math>2n+1-i-2j</math> distinct values for <math>h</math> for a given pair <math>(i,j)</math>. <br />
<br />
Now to find the total number of distinct triplets, we must sum this across all possible <math>i</math>s and <math>j</math>s. Lets take note of our restrictions on <math>i,j</math>: the only restriction is that <math>i+j \leq n</math>, since we're picking factors from <math>n</math> dice.<br />
<br />
<cmath> \sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j</cmath><br />
<br />
We start by calculating the first term. <math>2n+1</math> is constant, so we just need to find out how many pairs there are such that <math>i+j \leq n</math>. Set <math>i</math> to <math>0</math>: <math>j</math> can range from <math>0</math> to <math>n</math>, then set <math>i</math> to <math>1</math>: <math>j</math> can range from <math>0</math> to <math>n-1</math>, etc. The total number of pairs is thus <math>n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}</math>. Therefore the left summation evaluates to <cmath>\frac{(2n+1)(n+1)(n+2)}{2}</cmath><br />
<br />
Now we calculate <math>\sum_{i+j \leq n}^{} i+2j</math>. This simplifies to <math>\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j</math>. Note that because <math>i+j = n</math> is symmetric with respect to <math>i,j</math>, the sum of <math>i</math> in all of the pairs will be equal to the sum of <math>j</math> in all of the pairs. Thus this is equal to calculating <math>3 \cdot \sum_{i+j \leq n}^{} i</math>.<br />
<br />
In the pairs, <math>i=1</math> appears for <math>j</math> ranging between <math>0</math> and <math>n-1</math> so the sum here is <math>1 \cdot (n)</math>. Similarly <math>i=2</math> appears for <math>j</math> ranging from <math>0</math> to <math>n-2</math>, so the sum is <math>2 \cdot (n-1)</math>. If we continue the pattern, the sum overall is <math>(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1</math>. We can rearrange this as <math>((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)</math><br />
<br />
<cmath> = \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1</cmath><br />
<br />
We can write this in easier terms as <math>\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k</math><br />
<cmath>=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}</cmath><br />
<cmath> = \frac{n(n+1)(n+2)}{6}</cmath><br />
<br />
We multiply this by <math>3</math> to obtain that <cmath>\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}</cmath><br />
<br />
Thus our final answer for the number of distinct triplets <math>(h,i,j)</math> is:<br />
<cmath>\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}</cmath><br />
<cmath> = \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)</cmath><br />
<cmath> = \frac{(n+1)^2(n+2)}{2}</cmath><br />
<br />
Now most of the work is done. We set this equal to <math>936</math> and prime factorize. <math>936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13</math>, so <math>(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13</math>. Clearly <math>13</math> cannot be anything squared and <math>2^4 \cdot 3^2</math> is a perfect square, so <math>n+2 = 13</math> and <math>n = 11 = \boxed{A}</math><br />
<br />
<br />
<br />
~KingRavi<br />
<br />
==Solution 2==<br />
<br />
The product can be written as <br />
<cmath><br />
\begin{align*}<br />
2^a 3^b 4^c 5^d 6^e<br />
& = 2^{a + 2c + e} 3^{b + e} 5^d .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, we need to find the number of ordered tuples <math>\left( a + 2c + e, b+e, d \right)</math> where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math> are non-negative integers satisfying <math>a+b+c+d+e \leq n</math>.<br />
We denote this number as <math>f(n)</math>.<br />
<br />
Denote by <math>g \left( k \right)</math> the number of ordered tuples <math>\left( a + 2c + e, b+e \right)</math> where <math>\left( a, b, c, e \right) \in \Delta_k</math> with <math>\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}</math>.<br />
<br />
Thus,<br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{d = 0}^n g \left( n - d \right) \\<br />
& = \sum_{k = 0}^n g \left( k \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Next, we compute <math>g \left( k \right)</math>.<br />
<br />
Denote <math>i = b + e</math>. Thus, for each given <math>i</math>, the range of <math>a + 2c + e</math> is from 0 to <math>2 k - i</math>.<br />
Thus, the number of <math>\left( a + 2c + e, b + e \right)</math> is<br />
<cmath><br />
\begin{align*}<br />
g \left( k \right)<br />
& = \sum_{i=0}^k \left( 2 k - i + 1 \right) \\<br />
& = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, <br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{k = 0}^n g \left( k \right) \\<br />
& = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \\<br />
& = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2<br />
- \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \\<br />
& = \frac{3}{2} \cdot<br />
\frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right)<br />
- \frac{1}{2} \cdot<br />
\frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \\<br />
& = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
By solving <math>f \left( n \right) = 936</math>, we get <math>n = \boxed{\textbf{(A) 11}}</math>.<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 3 (Cheese)==<br />
<br />
The product can be written as<br />
<cmath><br />
\begin{align*}<br />
2^x 3^y 5^z<br />
\end{align*}<br />
</cmath><br />
<br />
Letting <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math>, 6 possible values. But if the only restriction of the product if that <math>2x\le n,y\le n,z\le n</math>, we can get <math>(2+1)(1+1)(1+1)=12</math> possible values. We calculate the ratio<br />
<cmath>r = \frac{\text{possible values of real situation}}{\text{possible values of ideal situation}} = \frac{6}{12}=0.5.</cmath><br />
<br />
Letting <math>n=2</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),</math><br /><br />
<math>(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)</math>, 17 possible values.<br />
The number of possibilities in the ideal situation is <math>5*3*3=45</math>, making <math>r = 17/45 \approx 0.378</math>.<br />
<br />
Now we can predict the trend of <math>r</math>: as <math>n</math> increases, <math>r</math> decreases.<br />
Letting<br />
<math>n=3</math>, you get possible values of ideal situation=<math>7*4*4=112</math>.<br />
<math>n=4</math>, the number=<math>9*5*5=225</math>.<br /><br />
<math>n=5</math>, the number=<math>11*6*6=396</math>.<br /><br />
<math>n=6</math>, the number=<math>13*7*7=637,637<936</math> so 6 is not the answer.<br /><br />
<math>n=7</math>, the number=<math>15*8*8=960</math>.<br /><br />
<math>n=8</math>, the number=<math>17*9*9=1377</math>,but <math>1377*0.378</math>≈<math>521</math> still much smaller than 936.<br /><br />
<math>n=9</math>, the number=<math>19*10*10=1900</math>,but <math>1900*0.378</math>≈<math>718</math> still smaller than 936.<br /><br />
<math>n=10</math>, the number=<math>21*11*11=2541</math>, <math>2541*0.378</math>≈<math>960</math> is a little bigger 936, but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is <math>\boxed{\textbf{(A) 11}}</math>.<br />
<br />
Check calculation:<br />
<math>n=11</math>,the number=<math>23*12*12=3312</math>,<math>3312*0.378</math>≈<math>1252</math> is much bigger than 936.<br />
<br />
~Troublemaker<br />
<br />
==Solution 4 (Easy computation)==<br />
<br />
The key observation is if <math>P=2^a\cdot 3^b \cdot 5^c</math>, then given <math>b</math> and <math>c</math>, <math>a</math> can take any value from <math>0</math> to <math>2(n-b-c)+b=2n-b-2c</math>. We are left to calculate<br />
\[<br />
\sum_{b+c\leq n} (2n-b-2c+1).<br />
\]<br />
<br />
Let <math>d=n-b-c</math> be the number of rolls which is neither divisible by <math>3</math> nor <math>5</math>, then<br />
<cmath><br />
\begin{align*}<br />
&\sum_{b+c\leq n}(2n-b-2c+1)\\<br />
&=\sum_{b+c+d= n}(2n-b-2c+1)\\<br />
&= \frac 12 \sum_{b+c+d= n}(2n-b-2c+2n-c-2b+2) \\<br />
&= \frac 12 \sum_{b+c+d= n} (n+2+3d) \\<br />
&= \frac 12 \sum_{b+c+d= n}(n+2+b+c+d) \\<br />
&= \frac 12 \sum_{b+c+d= n} (2n+2) = (n+1) \binom{n+2}{2}.<br />
\end{align*}<br />
</cmath><br />
<br />
Notice that we used symmetry argument twice. We also applied Stars & Bars to obtain <math>\sum_{b+c+d=n} 1 =\binom{n+2}{2}</math>.<br />
<br />
~asops<br />
<br />
==Video Solution 1 by OmegaLearn==<br />
https://youtu.be/FZG1j95owTo<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/BWb1dS4Jba0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2023|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_23&diff=2054362023 AMC 12B Problems/Problem 232023-11-24T01:47:59Z<p>Asops: </p>
<hr />
<div>==Problem==<br />
When <math>n</math> standard six-sided dice are rolled, the product of the numbers rolled can be any of <math>936</math> possible values. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9</math><br />
<br />
<br />
==Solution 1==<br />
<br />
We start by trying to prove a function of <math>n</math>, and then we can apply the function and equate it to <math>936</math> to find the value of <math>n</math>.<br />
<br />
It is helpful to think of this problem in the format <math>(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots</math>. Note that if we represent the scenario in this manner, we can think of picking a <math>1</math> for one factor and then a <math>5</math> for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for <math>n=2</math>, <math>4</math> can be reached by picking <math>1</math> and <math>4</math> or <math>2</math> and <math>2</math>. However, this form gives us insights that will be useful later in the problem.<br />
<br />
Note that there are only <math>3</math> primes in the set <math>\{1,2,3,4,5,6\}</math>: <math>2,3,</math> and <math>5</math>. Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form <math>2^h \cdot 3^i \cdot 5^j</math> (the choice of variables will become clear later), for integer nonnegative values <math>h,i,j</math>. So now the problem boils down to how many distinct triplets <math>(h,i,j)</math> can be formed by taking the product of <math>n</math> dice values. <br />
<br />
We start our work on representing <math>j</math>: the powers of <math>5</math>, because it is the simplest in this scenario because there is only one factor of <math>5</math> in the set. Because of this, having <math>j</math> fives in our prime factorization of the product is equivalent to picking <math>j</math> factors from the polynomial <math>(1+\dots + 6) \cdots</math> and choosing each factor to be a <math>5</math>. Now that we've selected <math>j</math> factors, there are <math>n-j</math> factors remaining to choose our powers of <math>3</math> and <math>2</math>. <br />
<br />
Suppose our prime factorization of this product contains <math>i</math> powers of <math>3</math>. These powers of <math>3</math> can either come from a <math>3</math> factor or a <math>6</math> factor, but since both <math>3</math> and <math>6</math> contain only one power of <math>3</math>, this means that a product with <math>i</math> powers of <math>3</math> corresponds directly to picking <math>i</math> factors from the polynomial, each of which is either <math>3</math> or <math>6</math> (but this distinction doesn't matter when we consider only the powers of <math>3</math>. <br />
<br />
Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair <math>(i,j)</math> that match the requirements, corresponding to the number of <math>3</math>'s and the number of <math>5</math>'s our product will have. Then how many different <math>h</math> values for the powers of <math>2</math> are possible?<br />
<br />
In the <math>i+j</math> factors we have already chosen, we obviously can't have any factors of <math>2</math> in the <math>j</math> factors with <math>5</math>. However, we can have a factor of <math>2</math> pairing with factors of <math>3</math>, if we choose a <math>6</math>. The maximal possible power of <math>2</math> in these <math>i</math> factors is thus <math>2^i</math>, which occurs when we pick every factor to be <math>6</math>. <br />
<br />
We now have <math>n-i-j</math> factors remaining, and we want to allocate these to solely powers of <math>2</math>. For each of these factors, we can choose either a <math>1,2,</math> or <math>4</math>. Therefore the maximal power of <math>2</math> achieved in these factors is when we pick <math>4</math> for all of them, which is equivalent to <math>2^{2\cdot (n-i-j)}</math>.<br />
<br />
Now if we multiply this across the total <math>n</math> factors (or <math>n</math> dice) we have a total of <math>2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}</math>, which is the maximal power of <math>2</math> attainable in the product for a pair <math>(i,j)</math>. Now note that every power of <math>2</math> below this power is attainable: we can simply just take away a power of <math>2</math> from an existing factor by dividing by <math>2</math>. Therefore the powers of <math>2</math>, and thus the <math>h</math> value ranges from <math>h=0</math> to <math>h=2n-i-2j</math>, so there are a total of <math>2n+1-i-2j</math> distinct values for <math>h</math> for a given pair <math>(i,j)</math>. <br />
<br />
Now to find the total number of distinct triplets, we must sum this across all possible <math>i</math>s and <math>j</math>s. Lets take note of our restrictions on <math>i,j</math>: the only restriction is that <math>i+j \leq n</math>, since we're picking factors from <math>n</math> dice.<br />
<br />
<cmath> \sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j</cmath><br />
<br />
We start by calculating the first term. <math>2n+1</math> is constant, so we just need to find out how many pairs there are such that <math>i+j \leq n</math>. Set <math>i</math> to <math>0</math>: <math>j</math> can range from <math>0</math> to <math>n</math>, then set <math>i</math> to <math>1</math>: <math>j</math> can range from <math>0</math> to <math>n-1</math>, etc. The total number of pairs is thus <math>n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}</math>. Therefore the left summation evaluates to <cmath>\frac{(2n+1)(n+1)(n+2)}{2}</cmath><br />
<br />
Now we calculate <math>\sum_{i+j \leq n}^{} i+2j</math>. This simplifies to <math>\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j</math>. Note that because <math>i+j = n</math> is symmetric with respect to <math>i,j</math>, the sum of <math>i</math> in all of the pairs will be equal to the sum of <math>j</math> in all of the pairs. Thus this is equal to calculating <math>3 \cdot \sum_{i+j \leq n}^{} i</math>.<br />
<br />
In the pairs, <math>i=1</math> appears for <math>j</math> ranging between <math>0</math> and <math>n-1</math> so the sum here is <math>1 \cdot (n)</math>. Similarly <math>i=2</math> appears for <math>j</math> ranging from <math>0</math> to <math>n-2</math>, so the sum is <math>2 \cdot (n-1)</math>. If we continue the pattern, the sum overall is <math>(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1</math>. We can rearrange this as <math>((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)</math><br />
<br />
<cmath> = \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1</cmath><br />
<br />
We can write this in easier terms as <math>\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k</math><br />
<cmath>=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}</cmath><br />
<cmath> = \frac{n(n+1)(n+2)}{6}</cmath><br />
<br />
We multiply this by <math>3</math> to obtain that <cmath>\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}</cmath><br />
<br />
Thus our final answer for the number of distinct triplets <math>(h,i,j)</math> is:<br />
<cmath>\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}</cmath><br />
<cmath> = \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)</cmath><br />
<cmath> = \frac{(n+1)^2(n+2)}{2}</cmath><br />
<br />
Now most of the work is done. We set this equal to <math>936</math> and prime factorize. <math>936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13</math>, so <math>(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13</math>. Clearly <math>13</math> cannot be anything squared and <math>2^4 \cdot 3^2</math> is a perfect square, so <math>n+2 = 13</math> and <math>n = 11 = \boxed{A}</math><br />
<br />
<br />
<br />
~KingRavi<br />
<br />
==Solution 2==<br />
<br />
The product can be written as <br />
<cmath><br />
\begin{align*}<br />
2^a 3^b 4^c 5^d 6^e<br />
& = 2^{a + 2c + e} 3^{b + e} 5^d .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, we need to find the number of ordered tuples <math>\left( a + 2c + e, b+e, d \right)</math> where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math> are non-negative integers satisfying <math>a+b+c+d+e \leq n</math>.<br />
We denote this number as <math>f(n)</math>.<br />
<br />
Denote by <math>g \left( k \right)</math> the number of ordered tuples <math>\left( a + 2c + e, b+e \right)</math> where <math>\left( a, b, c, e \right) \in \Delta_k</math> with <math>\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}</math>.<br />
<br />
Thus,<br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{d = 0}^n g \left( n - d \right) \\<br />
& = \sum_{k = 0}^n g \left( k \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Next, we compute <math>g \left( k \right)</math>.<br />
<br />
Denote <math>i = b + e</math>. Thus, for each given <math>i</math>, the range of <math>a + 2c + e</math> is from 0 to <math>2 k - i</math>.<br />
Thus, the number of <math>\left( a + 2c + e, b + e \right)</math> is<br />
<cmath><br />
\begin{align*}<br />
g \left( k \right)<br />
& = \sum_{i=0}^k \left( 2 k - i + 1 \right) \\<br />
& = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, <br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{k = 0}^n g \left( k \right) \\<br />
& = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \\<br />
& = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2<br />
- \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \\<br />
& = \frac{3}{2} \cdot<br />
\frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right)<br />
- \frac{1}{2} \cdot<br />
\frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \\<br />
& = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
By solving <math>f \left( n \right) = 936</math>, we get <math>n = \boxed{\textbf{(A) 11}}</math>.<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 3 (Cheese)==<br />
<br />
The product can be written as<br />
<cmath><br />
\begin{align*}<br />
2^x 3^y 5^z<br />
\end{align*}<br />
</cmath><br />
<br />
Letting <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math>, 6 possible values. But if the only restriction of the product if that <math>2x\le n,y\le n,z\le n</math>, we can get <math>(2+1)(1+1)(1+1)=12</math> possible values. We calculate the ratio<br />
<cmath>r = \frac{\text{possible values of real situation}}{\text{possible values of ideal situation}} = \frac{6}{12}=0.5.</cmath><br />
<br />
Letting <math>n=2</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),</math><br /><br />
<math>(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)</math>, 17 possible values.<br />
The number of possibilities in the ideal situation is <math>5*3*3=45</math>, making <math>r = 17/45 \approx 0.378</math>.<br />
<br />
Now we can predict the trend of <math>r</math>: as <math>n</math> increases, <math>r</math> decreases.<br />
Letting<br />
<math>n=3</math>, you get possible values of ideal situation=<math>7*4*4=112</math>.<br />
<math>n=4</math>, the number=<math>9*5*5=225</math>.<br /><br />
<math>n=5</math>, the number=<math>11*6*6=396</math>.<br /><br />
<math>n=6</math>, the number=<math>13*7*7=637,637<936</math> so 6 is not the answer.<br /><br />
<math>n=7</math>, the number=<math>15*8*8=960</math>.<br /><br />
<math>n=8</math>, the number=<math>17*9*9=1377</math>,but <math>1377*0.378</math>≈<math>521</math> still much smaller than 936.<br /><br />
<math>n=9</math>, the number=<math>19*10*10=1900</math>,but <math>1900*0.378</math>≈<math>718</math> still smaller than 936.<br /><br />
<math>n=10</math>, the number=<math>21*11*11=2541</math>, <math>2541*0.378</math>≈<math>960</math> is a little bigger 936, but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is <math>\boxed{\textbf{(A) 11}}</math>.<br />
<br />
Check calculation:<br />
<math>n=11</math>,the number=<math>23*12*12=3312</math>,<math>3312*0.378</math>≈<math>1252</math> is much bigger than 936.<br />
<br />
~Troublemaker<br />
<br />
==Solution 4 (Easy computation)==<br />
<br />
The key observation is if <math>P=2^a\cdot 3^b \cdot 5^c</math>, then given <math>b</math> and <math>c</math>, <math>b</math> can take any value from <math>0</math> to <math>2(n-b-c)+b=2n-b-2c</math>. We are left to calculate<br />
\[<br />
\sum_{b+c\leq n} (2n-b-2c+1).<br />
\]Let <math>d=n-b-c</math> be the number of rolls which is neither divisible by <math>3</math> nor <math>5</math>, then<br />
<cmath>\sum_{b+c\leq n}(2n-b-2c+1)=\sum_{b+c+d= n}(2n-b-2c+1)</cmath><cmath>= \frac 12 \sum_{b+c+d= n}(2n-b-2c+2n-c-2b+2) = \frac 12 \sum_{b+c+d= n} (n+2+3d) </cmath><cmath>= \frac 12 \sum_{b+c+d= n}(n+2+b+c+d) = \frac 12 \sum_{b+c+d= n} (2n+2) = (n+1) \binom{n+2}{2}.</cmath><br />
The rest is the same as above.<br />
<br />
Notice that we used symmetry argument twice. We also applied Stars & Bars to obtain <math>\sum_{b+c+d=n} 1 =\binom{n+2}{2}</math>.<br />
<br />
~asops<br />
<br />
==Video Solution 1 by OmegaLearn==<br />
https://youtu.be/FZG1j95owTo<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/BWb1dS4Jba0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2023|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_23&diff=2054352023 AMC 12B Problems/Problem 232023-11-24T01:46:36Z<p>Asops: </p>
<hr />
<div>==Problem==<br />
When <math>n</math> standard six-sided dice are rolled, the product of the numbers rolled can be any of <math>936</math> possible values. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9</math><br />
<br />
<br />
==Solution 1==<br />
<br />
We start by trying to prove a function of <math>n</math>, and then we can apply the function and equate it to <math>936</math> to find the value of <math>n</math>.<br />
<br />
It is helpful to think of this problem in the format <math>(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots</math>. Note that if we represent the scenario in this manner, we can think of picking a <math>1</math> for one factor and then a <math>5</math> for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for <math>n=2</math>, <math>4</math> can be reached by picking <math>1</math> and <math>4</math> or <math>2</math> and <math>2</math>. However, this form gives us insights that will be useful later in the problem.<br />
<br />
Note that there are only <math>3</math> primes in the set <math>\{1,2,3,4,5,6\}</math>: <math>2,3,</math> and <math>5</math>. Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form <math>2^h \cdot 3^i \cdot 5^j</math> (the choice of variables will become clear later), for integer nonnegative values <math>h,i,j</math>. So now the problem boils down to how many distinct triplets <math>(h,i,j)</math> can be formed by taking the product of <math>n</math> dice values. <br />
<br />
We start our work on representing <math>j</math>: the powers of <math>5</math>, because it is the simplest in this scenario because there is only one factor of <math>5</math> in the set. Because of this, having <math>j</math> fives in our prime factorization of the product is equivalent to picking <math>j</math> factors from the polynomial <math>(1+\dots + 6) \cdots</math> and choosing each factor to be a <math>5</math>. Now that we've selected <math>j</math> factors, there are <math>n-j</math> factors remaining to choose our powers of <math>3</math> and <math>2</math>. <br />
<br />
Suppose our prime factorization of this product contains <math>i</math> powers of <math>3</math>. These powers of <math>3</math> can either come from a <math>3</math> factor or a <math>6</math> factor, but since both <math>3</math> and <math>6</math> contain only one power of <math>3</math>, this means that a product with <math>i</math> powers of <math>3</math> corresponds directly to picking <math>i</math> factors from the polynomial, each of which is either <math>3</math> or <math>6</math> (but this distinction doesn't matter when we consider only the powers of <math>3</math>. <br />
<br />
Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair <math>(i,j)</math> that match the requirements, corresponding to the number of <math>3</math>'s and the number of <math>5</math>'s our product will have. Then how many different <math>h</math> values for the powers of <math>2</math> are possible?<br />
<br />
In the <math>i+j</math> factors we have already chosen, we obviously can't have any factors of <math>2</math> in the <math>j</math> factors with <math>5</math>. However, we can have a factor of <math>2</math> pairing with factors of <math>3</math>, if we choose a <math>6</math>. The maximal possible power of <math>2</math> in these <math>i</math> factors is thus <math>2^i</math>, which occurs when we pick every factor to be <math>6</math>. <br />
<br />
We now have <math>n-i-j</math> factors remaining, and we want to allocate these to solely powers of <math>2</math>. For each of these factors, we can choose either a <math>1,2,</math> or <math>4</math>. Therefore the maximal power of <math>2</math> achieved in these factors is when we pick <math>4</math> for all of them, which is equivalent to <math>2^{2\cdot (n-i-j)}</math>.<br />
<br />
Now if we multiply this across the total <math>n</math> factors (or <math>n</math> dice) we have a total of <math>2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}</math>, which is the maximal power of <math>2</math> attainable in the product for a pair <math>(i,j)</math>. Now note that every power of <math>2</math> below this power is attainable: we can simply just take away a power of <math>2</math> from an existing factor by dividing by <math>2</math>. Therefore the powers of <math>2</math>, and thus the <math>h</math> value ranges from <math>h=0</math> to <math>h=2n-i-2j</math>, so there are a total of <math>2n+1-i-2j</math> distinct values for <math>h</math> for a given pair <math>(i,j)</math>. <br />
<br />
Now to find the total number of distinct triplets, we must sum this across all possible <math>i</math>s and <math>j</math>s. Lets take note of our restrictions on <math>i,j</math>: the only restriction is that <math>i+j \leq n</math>, since we're picking factors from <math>n</math> dice.<br />
<br />
<cmath> \sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j</cmath><br />
<br />
We start by calculating the first term. <math>2n+1</math> is constant, so we just need to find out how many pairs there are such that <math>i+j \leq n</math>. Set <math>i</math> to <math>0</math>: <math>j</math> can range from <math>0</math> to <math>n</math>, then set <math>i</math> to <math>1</math>: <math>j</math> can range from <math>0</math> to <math>n-1</math>, etc. The total number of pairs is thus <math>n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}</math>. Therefore the left summation evaluates to <cmath>\frac{(2n+1)(n+1)(n+2)}{2}</cmath><br />
<br />
Now we calculate <math>\sum_{i+j \leq n}^{} i+2j</math>. This simplifies to <math>\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j</math>. Note that because <math>i+j = n</math> is symmetric with respect to <math>i,j</math>, the sum of <math>i</math> in all of the pairs will be equal to the sum of <math>j</math> in all of the pairs. Thus this is equal to calculating <math>3 \cdot \sum_{i+j \leq n}^{} i</math>.<br />
<br />
In the pairs, <math>i=1</math> appears for <math>j</math> ranging between <math>0</math> and <math>n-1</math> so the sum here is <math>1 \cdot (n)</math>. Similarly <math>i=2</math> appears for <math>j</math> ranging from <math>0</math> to <math>n-2</math>, so the sum is <math>2 \cdot (n-1)</math>. If we continue the pattern, the sum overall is <math>(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1</math>. We can rearrange this as <math>((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)</math><br />
<br />
<cmath> = \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1</cmath><br />
<br />
We can write this in easier terms as <math>\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k</math><br />
<cmath>=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})</cmath><br />
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}</cmath><br />
<cmath> = \frac{n(n+1)(n+2)}{6}</cmath><br />
<br />
We multiply this by <math>3</math> to obtain that <cmath>\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}</cmath><br />
<br />
Thus our final answer for the number of distinct triplets <math>(h,i,j)</math> is:<br />
<cmath>\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}</cmath><br />
<cmath> = \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)</cmath><br />
<cmath> = \frac{(n+1)^2(n+2)}{2}</cmath><br />
<br />
Now most of the work is done. We set this equal to <math>936</math> and prime factorize. <math>936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13</math>, so <math>(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13</math>. Clearly <math>13</math> cannot be anything squared and <math>2^4 \cdot 3^2</math> is a perfect square, so <math>n+2 = 13</math> and <math>n = 11 = \boxed{A}</math><br />
<br />
<br />
<br />
~KingRavi<br />
<br />
==Solution 2==<br />
<br />
The product can be written as <br />
<cmath><br />
\begin{align*}<br />
2^a 3^b 4^c 5^d 6^e<br />
& = 2^{a + 2c + e} 3^{b + e} 5^d .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, we need to find the number of ordered tuples <math>\left( a + 2c + e, b+e, d \right)</math> where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math> are non-negative integers satisfying <math>a+b+c+d+e \leq n</math>.<br />
We denote this number as <math>f(n)</math>.<br />
<br />
Denote by <math>g \left( k \right)</math> the number of ordered tuples <math>\left( a + 2c + e, b+e \right)</math> where <math>\left( a, b, c, e \right) \in \Delta_k</math> with <math>\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}</math>.<br />
<br />
Thus,<br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{d = 0}^n g \left( n - d \right) \\<br />
& = \sum_{k = 0}^n g \left( k \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Next, we compute <math>g \left( k \right)</math>.<br />
<br />
Denote <math>i = b + e</math>. Thus, for each given <math>i</math>, the range of <math>a + 2c + e</math> is from 0 to <math>2 k - i</math>.<br />
Thus, the number of <math>\left( a + 2c + e, b + e \right)</math> is<br />
<cmath><br />
\begin{align*}<br />
g \left( k \right)<br />
& = \sum_{i=0}^k \left( 2 k - i + 1 \right) \\<br />
& = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, <br />
<cmath><br />
\begin{align*}<br />
f \left( n \right)<br />
& = \sum_{k = 0}^n g \left( k \right) \\<br />
& = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \\<br />
& = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2<br />
- \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \\<br />
& = \frac{3}{2} \cdot<br />
\frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right)<br />
- \frac{1}{2} \cdot<br />
\frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \\<br />
& = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
By solving <math>f \left( n \right) = 936</math>, we get <math>n = \boxed{\textbf{(A) 11}}</math>.<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 3 (Cheese)==<br />
<br />
The product can be written as<br />
<cmath><br />
\begin{align*}<br />
2^x 3^y 5^z<br />
\end{align*}<br />
</cmath><br />
<br />
Letting <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math>, 6 possible values. But if the only restriction of the product if that <math>2x\le n,y\le n,z\le n</math>, we can get <math>(2+1)(1+1)(1+1)=12</math> possible values. We calculate the ratio<br />
<cmath>r = \frac{\text{possible values of real situation}}{\text{possible values of ideal situation}} = \frac{6}{12}=0.5.</cmath><br />
<br />
Letting <math>n=2</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),</math><br /><br />
<math>(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)</math>, 17 possible values.<br />
The number of possibilities in the ideal situation is <math>5*3*3=45</math>, making <math>r = 17/45 \approx 0.378</math>.<br />
<br />
Now we can predict the trend of <math>r</math>: as <math>n</math> increases, <math>r</math> decreases.<br />
Letting<br />
<math>n=3</math>, you get possible values of ideal situation=<math>7*4*4=112</math>.<br />
<math>n=4</math>, the number=<math>9*5*5=225</math>.<br /><br />
<math>n=5</math>, the number=<math>11*6*6=396</math>.<br /><br />
<math>n=6</math>, the number=<math>13*7*7=637,637<936</math> so 6 is not the answer.<br /><br />
<math>n=7</math>, the number=<math>15*8*8=960</math>.<br /><br />
<math>n=8</math>, the number=<math>17*9*9=1377</math>,but <math>1377*0.378</math>≈<math>521</math> still much smaller than 936.<br /><br />
<math>n=9</math>, the number=<math>19*10*10=1900</math>,but <math>1900*0.378</math>≈<math>718</math> still smaller than 936.<br /><br />
<math>n=10</math>, the number=<math>21*11*11=2541</math>, <math>2541*0.378</math>≈<math>960</math> is a little bigger 936, but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is <math>\boxed{\textbf{(A) 11}}</math>.<br />
<br />
Check calculation:<br />
<math>n=11</math>,the number=<math>23*12*12=3312</math>,<math>3312*0.378</math>≈<math>1252</math> is much bigger than 936.<br />
<br />
~Troublemaker<br />
<br />
==Solution 4 (Easy computation)<br />
<br />
The key observation is if <math>P=2^a\cdot 3^b \cdot 5^c</math>, then given <math>b</math> and <math>c</math>, <math>b</math> can take any value from <math>0</math> to <math>2(n-b-c)+b=2n-b-2c</math>. We are left to calculate<br />
\[<br />
\sum_{b+c\leq n} (2n-b-2c+1).<br />
\]Let <math>d=n-b-c</math> be the number of rolls which is neither divisible by <math>3</math> nor <math>5</math>, then<br />
<cmath>\sum_{b+c\leq n}(2n-b-2c+1)=\sum_{b+c+d= n}(2n-b-2c+1)</cmath><cmath>= \frac 12 \sum_{b+c+d= n}(2n-b-2c+2n-c-2b+2) = \frac 12 \sum_{b+c+d= n} (n+2+3d) </cmath><cmath>= \frac 12 \sum_{b+c+d= n}(n+2+b+c+d) = \frac 12 \sum_{b+c+d= n} (2n+2) = (n+1) \binom{n+2}{2}.</cmath><br />
The rest is commentary.<br />
<br />
Notice that we used symmetry argument twice. We also applied Stars & Bars to obtain <math>\sum_{b+c+d=n} 1 =\binom{n+2}{2}</math>.<br />
<br />
~asops<br />
<br />
==Video Solution 1 by OmegaLearn==<br />
https://youtu.be/FZG1j95owTo<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/BWb1dS4Jba0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2023|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_20&diff=1895022022 AMC 10B Problems/Problem 202023-02-11T15:58:06Z<p>Asops: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCD</math> be a rhombus with <math>\angle ADC = 46^\circ</math>. Let <math>E</math> be the midpoint of <math>\overline{CD}</math>, and let <math>F</math> be the point<br />
on <math>\overline{BE}</math> such that <math>\overline{AF}</math> is perpendicular to <math>\overline{BE}</math>. What is the degree measure of <math>\angle BFC</math>?<br />
<br />
<math>\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114</math><br />
<br />
==Diagram==<br />
<asy><br />
/* Made by MRENTHUSIASM */<br />
size(300);<br />
pair A, B, C, D, E, F;<br />
D = origin;<br />
A = 6*dir(46);<br />
C = (6,0);<br />
B = C + (A-D);<br />
E = midpoint(C--D);<br />
F = foot(A,B,E);<br />
dot("$A$",A,1.5*NW,linewidth(5));<br />
dot("$B$",B,1.5*NE,linewidth(5));<br />
dot("$C$",C,1.5*SE,linewidth(5));<br />
dot("$D$",D,1.5*SW,linewidth(5));<br />
dot("$E$",E,1.5*S,linewidth(5));<br />
dot("$F$",F,1.5*dir(-20),linewidth(5));<br />
markscalefactor=0.04;<br />
draw(rightanglemark(A,F,B),red);<br />
draw(A--B--C--D--cycle^^A--F--C^^B--E);<br />
label("$46^{\circ}$",D,3*dir(26),red);<br />
</asy><br />
~MRENTHUSIASM<br />
<br />
==Solution 1 (Law of Sines and Law of Cosines)==<br />
<br />
Without loss of generality, we assume the length of each side of <math>ABCD</math> is <math>2</math>.<br />
Because <math>E</math> is the midpoint of <math>CD</math>, <math>CE = 1</math>.<br />
<br />
Because <math>ABCD</math> is a rhombus, <math>\angle BCE = 180^\circ - \angle D</math>.<br />
<br />
In <math>\triangle BCE</math>, following from the law of sines,<br />
<cmath><br />
\[<br />
\frac{CE}{\sin \angle FBC} = \frac{BC}{\sin \angle BEC} .<br />
\]<br />
</cmath><br />
<br />
We have <math>\angle BCE = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC</math>.<br />
<br />
Hence,<br />
<cmath><br />
\[<br />
\frac{1}{\sin \angle FBC} = \frac{2}{\sin \left( 46^\circ - \angle FBC \right)} .<br />
\]<br />
</cmath><br />
<br />
By solving this equation, we get <math>\tan \angle FBC = \frac{\sin 46^\circ}{2 + \cos 46^\circ}</math>.<br />
<br />
Because <math>AF \perp BF</math>,<br />
<cmath><br />
\begin{align*}<br />
BF & = AB \cos \angle ABF \\<br />
& = 2 \cos \left( 46^\circ - \angle FBC \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
In <math>\triangle BFC</math>, following from the law of sines,<br />
<cmath><br />
\[<br />
\frac{BF}{\sin \angle BCF} = \frac{BC}{\sin \angle BFC} .<br />
\]<br />
</cmath><br />
<br />
Because <math>\angle BCF = 180^\circ - \angle BFC - \angle FBC</math>, the equation above can be converted as<br />
<cmath><br />
\[<br />
\frac{BF}{\sin \left( \angle BFC + \angle FBC \right)} = \frac{BC}{\sin \angle BFC} .<br />
\]<br />
</cmath><br />
<br />
Therefore,<br />
<cmath><br />
\begin{align*}<br />
\tan \angle BFC & = \frac{\sin \angle FBC}{\cos \left( 46^\circ - \angle FBC \right) - \cos \angle FBC} \\<br />
& = \frac{1}{\sin 46^\circ - \left( 1 - \cos 46^\circ \right) \cot \angle FBC} \\<br />
& = \frac{\sin 46^\circ}{\cos 46^\circ - 1} \\<br />
& = - \frac{\sin 134^\circ}{1 + \cos 134^\circ} \\<br />
& = - \tan \frac{134^\circ}{2} \\<br />
& = - \tan 67^\circ \\<br />
& = \tan \left( 180^\circ - 67^\circ \right) \\<br />
& = \tan 113^\circ .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, <math>\angle BFC =<br />
\boxed{\textbf{(D)} \ 113}</math>.<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 2==<br />
Extend segments <math>\overline{AD}</math> and <math>\overline{BE}</math> until they meet at point <math>G</math>.<br />
<br />
Because <math>\overline{AB} \parallel \overline{ED}</math>, we have <math>\angle ABG = \angle DEG</math> and <math>\angle GDE = \angle GAB</math>, so <math>\triangle ABG \sim \triangle DEG</math> by AA.<br />
<br />
Because <math>ABCD</math> is a rhombus, <math>AB = CD = 2DE</math>, so <math>AG = 2GD</math>, meaning that <math>D</math> is a midpoint of segment <math>\overline{AG}</math>.<br />
<br />
Now, <math>\overline{AF} \perp \overline{BE}</math>, so <math>\triangle GFA</math> is right and median <math>FD = AD</math>.<br />
<br />
So now, because <math>ABCD</math> is a rhombus, <math>FD = AD = CD</math>. This means that there exists a circle from <math>D</math> with radius <math>AD</math> that passes through <math>F</math>, <math>A</math>, and <math>C</math>.<br />
<br />
AG is a diameter of this circle because <math>\angle AFG=90^\circ</math>. This means that <math>\angle GFC = \angle GAC = \frac{1}{2} \angle GDC</math>, so <math>\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ</math>, which means that <math>\angle BFC = \boxed{\textbf{(D)} \ 113}</math><br />
<br />
~popop614<br />
<br />
==Solution 3==<br />
Let <math>\overline{AC}</math> meet <math>\overline{BD}</math> at <math>O</math>, then <math>AOFB</math> is cyclic and <math>\angle FBO = \angle FAO</math>. Also, <math>AC \cdot BO = [ABCD] = 2 \cdot [ABE] = AF \cdot BE</math>, so <math>\frac{AF}{BO} = \frac{AC}{BE}</math>, thus <math>\triangle AFC \sim \triangle BOE</math> by SAS, and <math>\angle OEB = \angle ACF</math>, then <math>\angle CFE = \angle EOC = \angle DAC = 67^\circ</math>, and <math>\angle BFC = \boxed{\textbf{(D)} \ 113}</math><br />
<br />
~mathfan2020<br />
<br />
A little bit faster: <math>AOFB</math> is cyclic <math>\implies \angle OFE = \angle BAO</math>. <br />
<br />
<math>AB \parallel CD \implies \angle BAO = \angle OCE</math>.<br />
<br />
Therefore <math>\angle OFE=\angle OCE \implies OECF</math> is cyclic.<br />
<br />
Hence <math>\angle CFE=\angle COE=\angle CAD = 67^\circ</math>.<br />
<br />
~asops<br />
<br />
==Solution 4==<br />
Observe that all answer choices are close to <math>112.5 = 90+\frac{45}{2}</math>. A quick solve shows that having <math>\angle D = 90^\circ</math> yields <math>\angle BFC = 135^\circ = 90 + \frac{90}{2}</math>, meaning that <math>\angle BFC</math> increases with <math>\angle D</math>. <br />
Substituting, <math>\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}</math><br />
<br />
~mathfan2020<br />
<br />
==Solution 5 (Similarity and Circle Geometry)==<br />
Let's make a diagram, but extend <math>AD</math> and <math>BE</math> to point <math>G</math>. <br />
<asy><br />
/*<br />
Made by ghfhgvghj10<br />
Edited by MRENTHUSIASM<br />
*/<br />
size(300);<br />
pair A, B, C, D, E, F, G;<br />
D = origin;<br />
A = 6*dir(46);<br />
C = (6,0);<br />
B = C + (A-D);<br />
E = midpoint(C--D);<br />
F = foot(A,B,E);<br />
G = 6*dir(226);<br />
dot("$A$",A,1.5*NW,linewidth(5));<br />
dot("$B$",B,1.5*NE,linewidth(5));<br />
dot("$C$",C,1.5*SE,linewidth(5));<br />
dot("$D$",D,1.5*NW,linewidth(5));<br />
dot("$E$",E,1.5*S,linewidth(5));<br />
dot("$F$",F,1.5*dir(-20),linewidth(5));<br />
dot("$G$",G,1.5*SW,linewidth(5));<br />
markscalefactor=0.04;<br />
draw(rightanglemark(A,F,B),red);<br />
draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G);<br />
label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));<br />
</asy><br />
We know that <math>AB=AD=2</math> and <math>CE=DE=1</math>. <br />
<br />
By AA Similarity, <math>\triangle ABG \sim \triangle DEG</math> with a ratio of <math>2:1</math>. This implies that <math>2AD=AG</math> and <math>AD \cong DG</math>, so <math>AG=2AD=2\cdot2=4</math>. That is, <math>D</math> is the midpoint of <math>AG</math>.<br />
<br />
Now, let's redraw our previous diagram, but construct a circle with radius <math>AD</math> or <math>2</math> centered at <math>D</math> and by extending <math>CD</math> to point <math>H</math>, which is on the circle. <br />
<asy><br />
/*<br />
Made by ghfhgvghj10<br />
Edited by MRENTHUSIASM<br />
*/<br />
size(300);<br />
pair A, B, C, D, E, F, G;<br />
D = origin;<br />
A = 6*dir(46);<br />
C = (6,0);<br />
B = C + (A-D);<br />
E = midpoint(C--D);<br />
F = foot(A,B,E);<br />
G = 6*dir(226);<br />
dot("$A$",A,1.5*NE,linewidth(5));<br />
dot("$B$",B,1.5*NE,linewidth(5));<br />
dot("$C$",C,1.5*SE,linewidth(5));<br />
dot("$D$",D,1.5*NW,linewidth(5));<br />
dot("$E$",E,1.5*S,linewidth(5));<br />
dot("$F$",F,1.5*dir(-20),linewidth(5));<br />
dot("$G$",G,1.5*SW,linewidth(5));<br />
markscalefactor=0.04;<br />
draw(rightanglemark(A,F,B),red);<br />
draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G);<br />
label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));<br />
draw(Circle(D,6),dashed);<br />
</asy><br />
Notice how <math>F</math> and <math>C</math> are on the circle and that <math>\angle CFE</math> intercepts with <math>\overset{\Large\frown} {CG}</math>. <br />
<br />
Let's call <math>\angle CFE = \theta</math>.<br />
<br />
Note that <math>\angle CDG</math> also intercepts <math>\overset{\Large\frown} {CG}</math>, So <math>\angle CDG = 2\angle CFE</math>. <br />
<br />
Let <math>\angle CDG = 2\theta</math>. Notice how <math>\angle CDG</math> and <math>\angle ADC</math> are supplementary to each other. We conclude that <cmath>\begin{align*}<br />
2\theta &= 180-\angle ADC \\<br />
2\theta &= 180-46 \\<br />
2\theta &= 134 \\<br />
\theta &= 67. <br />
\end{align*}</cmath><br />
Since <math>\angle BFC=180-\theta</math>, we have <math>\angle BFC=180-67=\boxed{\textbf{(D)} \ 113}</math>.<br />
<br />
~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits).<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=HWJe96s_ugs&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=6<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/Ysb1EK_5B2g<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
== Video Solution by OmegaLearn Using Clever Similar Triangles and Angle Chasing ==<br />
https://youtu.be/lEmCprb20n4<br />
<br />
~ pi_is_3.14<br />
<br />
== Video Solution, best solution (family friendly, no circles drawn) ==<br />
https://www.youtube.com/watch?v=vwI3I7dxw0Q<br />
<br />
== Video Solution, by Challenge 25 ==<br />
https://youtu.be/W1jbMaO8BIQ (cyclic quads)<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2022|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_9&diff=1786612019 AIME I Problems/Problem 92022-10-04T13:21:19Z<p>Asops: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>\tau(n)</math> denote the number of positive integer divisors of <math>n</math>. Find the sum of the six least positive integers <math>n</math> that are solutions to <math>\tau (n) + \tau (n+1) = 7</math>.<br />
<br />
==Solution==<br />
In order to obtain a sum of <math>7</math>, we must have:<br />
* either a number with <math>5</math> divisors (''a fourth power of a prime'') and a number with <math>2</math> divisors (''a prime''), or<br />
* a number with <math>4</math> divisors (''a semiprime or a cube of a prime'') and a number with <math>3</math> divisors (''a square of a prime''). (No integer greater than <math>1</math> can have fewer than <math>2</math> divisors.)<br />
Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square-like <math>3^2</math> with <math>3</math> divisors, or a fourth power like <math>2^4</math> with <math>5</math> divisors. We then find the smallest such values by hand.<br />
* <math>2^2</math> has two possibilities: <math>3</math> and <math>4</math> or <math>4</math> and <math>5</math>. Neither works.<br />
* <math>3^2</math> has two possibilities: <math>8</math> and <math>9</math> or <math>9</math> and <math>10</math>. <math>\boxed{(8,9)}</math> and <math>\boxed{(9,10)}</math> both work.<br />
* <math>2^4</math> has two possibilities: <math>15</math> and <math>16</math> or <math>16</math> and <math>17</math>. Only <math>\boxed{(16,17)}</math> works.<br />
* <math>5^2</math> has two possibilities: <math>24</math> and <math>25</math> or <math>25</math> and <math>26</math>. Only <math>\boxed{(25,26)}</math> works.<br />
* <math>7^2</math> has two possibilities: <math>48</math> and <math>49</math> or <math>49</math> and <math>50</math>. Neither works.<br />
* <math>3^4</math> has two possibilities: <math>80</math> and <math>81</math> or <math>81</math> and <math>82</math>. Neither works.<br />
* <math>11^2</math> has two possibilities: <math>120</math> and <math>121</math> or <math>121</math> and <math>122</math>. Only <math>\boxed{(121,122)}</math> works.<br />
* <math>13^2</math> has two possibilities: <math>168</math> and <math>169</math> or <math>169</math> and <math>170</math>. Neither works.<br />
* <math>17^2</math> has two possibilities: <math>288</math> and <math>289</math> or <math>289</math> and <math>290</math>. Neither works.<br />
* <math>19^2</math> has two possibilities: <math>360</math> and <math>361</math> or <math>361</math> and <math>362</math>. Only <math>\boxed{(361,362)}</math> works.<br />
Having computed the working possibilities, we take the sum of the corresponding values of <math>n</math>: <math>8+9+16+25+121+361 = \boxed{\textbf{540}}</math>. ~Kepy.<br />
<br />
<br />
Possible improvement: since all primes <math>>2</math> are odd, their fourth powers are odd as well, which cannot be adjacent to any primes because both of the adjacent numbers will be even. Thus, we only need to check <math>16</math> for the fourth power case. - mathleticguyyy<br />
<br />
==Solution 2==<br />
Let the ordered pair <math>(a,b)</math> represent the number of divisors of <math>n</math> and <math>n+1</math> respectively.<br />
We see that to obtain a sum of <math>7</math>, we can have <math>(2,5), (3,4), (4,3),</math> and <math>(5,2)</math>.<br />
<br />
<br />
Case 1: When we have <math>(2,5)</math><br />
For <math>n</math> to have 2 divisors, it must be a prime number.<br />
For <math>n+1</math> to have 5 divisors, it must be in the form <math>a^4</math>.<br />
If <math>n+1</math> is in the form <math>a^4</math>, then <math>n = a^4-1 = (a^2+1)(a-1)(a+1)</math>. This means that <math>n</math>, or <math>a^4-1</math> has factors other than 1 and itself; <math>n</math> is not prime.<br />
No cases work in this case<br />
<br />
<br />
Case 2: When we have <math>(4,3)</math><br />
For <math>n</math> to have 4 divisors, it must be in the form <math>a^3</math> or <math>ab</math>, where <math>a</math> and <math>b</math> are distinct prime numbers .<br />
For <math>n+1</math> to have 3 divisors, it must be a square number. <br />
Let <math>n+1 = A^2</math> (<math>A</math> is a prime number). When <math>n = a^3, a^3+1 = A^2, (A-1)(A+1)=a^3</math>.<br />
We see that the only case when it works is when <math>a=2, A=3</math>, so <math>n=8</math> works.<br />
<br />
<br />
Case 3: When we have <math>(5,2)</math><br />
For <math>n</math> to have 5 divisors, it must be in the form <math>a^4</math>, where <math>a</math> is a prime number.<br />
For <math>n+1</math> to have 2 divisors, it must be a prime number.<br />
Notice that <math>a</math> and <math>a^4</math> have the same parity (even/odd). Since every prime greater than 2 are odd, <math>n = a^4</math> must be even. Since <math>a^4</math> is even, <math>a</math> must be even as well, and the only prime number that is even is 2. When <math>a=2, n=16</math>.<br />
<br />
<br />
Case 4: When we have <math>(3,4)</math><br />
For <math>n</math> to have 3 divisors, it must be a square number. <br />
For <math>n+1</math> to have 4 divisors, it must be in the form <math>a^3</math> or <math>ab</math>, where <math>a</math> and <math>b</math> are distinct prime numbers. <br />
Similar to Case 2, let <math>n = A^2</math> (<math>A</math> is a prime number).<br />
<br />
* When <math>n+1 = a^3, A^2+1 = a^3</math>.<br />
There are no cases that satisfy this equation.<br />
<br />
* When <math>n+1=ab, A^2+1 = ab</math>.<br />
We test squares of primes to find values of n that work.<br />
* <math>A=2</math>, <math>4+1=5</math>. Doesn't work.<br />
* <math>A=3</math>, <math>9+1=10=2*5</math>. It works. <math>n=9</math><br />
* <math>A=5</math>, <math>25+1=26=2*13</math>. It works. <math>n=25</math><br />
* <math>A=7</math>, <math>49+1=50=2*5^2</math>. Doesn't work.<br />
* <math>A=11</math>, <math>121+1=122=2*61</math>. It works. <math>n=121</math><br />
* <math>A=13</math>, <math>169+1=170=2*5*17</math>. Doesn't work<br />
* <math>A=17</math>, <math>289+1=290=2*5*29</math>. Doesn't work<br />
* <math>A=19</math>, <math>361+1=362=2*181</math>. It works. <math>n=361</math><br />
<br />
Now we add up the values of <math>n</math> to get the answer: <math>8+16+9+25+121+361 = \boxed{540}</math>.<br />
~toastybaker<br />
==Solution 3 (Official MAA)==<br />
Let <math>p,\,q,</math> and <math>r</math> represent primes. Because <math>\tau(n)=1</math> only for <math>n=1,</math> there is no <math>n</math> for which <math>\{\tau(n),\tau(n+1)\}=\{1,6\}</math>. If <math>\{\tau(n),\tau(n+1)\}=\{2,5\},</math> then <math>\{n,n+1\}=\{p,q^4\},</math> so <math>|p-q^4|=1.</math> Checking <math>q=2</math> and <math>p=17</math> yields the solution <math>n=16.</math> If <math>q>2,</math> then <math>q</math> is odd, and <math>p=q^4\pm 1</math> is even, so <math>p</math> cannot be prime.<br />
<br />
If <math>\{\tau(n),\tau(n+1)\}=\{3,4\},</math> then <math>\{n,n+1\}=\{p^2,q^3\}</math> or <math>\{p^2,qr\}.</math> Consider <math>|p^2-q^3|=1.</math> If <math>p^2-1=(p-1)(p+1)=q^3,</math> Then <math>q=2.</math> This yields the solution <math>p=3</math> and <math>q=2,</math> so <math>n=8.</math> If <math>q^3-1=(q-1)(q^2+q+1)=p^2,</math> then <math>q-1=1,</math> which does not give a solution. Consider <math>|p^2-qr|=1.</math> If <math>p^2-1=(p-1)(p+1)=qr,</math> then if <math>p>2,</math> the left side is divisible by 8, so there are no solutions. Finding the smallest four primes such that <math>p^2+1=qr</math> gives <math>3^2+1=10,\,5^2+1=26,\,11^2+1=122,</math> and <math>19^2+1=362.</math> The six least values of <math>n</math> are <math>8,9,16,25,121,</math> and <math>361,</math> whose sum is <math>540.</math><br />
<br />
==Video Solution ==<br />
https://www.youtube.com/watch?v=2ouOexOnG1A<br />
<br />
~ North America Math COntest Go Go Go<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_8&diff=1696712014 AIME I Problems/Problem 82022-01-11T02:27:46Z<p>Asops: /* Problem 8 */</p>
<hr />
<div>== Problem 8 ==<br />
The positive integers <math>N</math> and <math>N^2</math> both end in the same sequence of four digits <math>abcd</math> when written in base <math>10</math>, where digit <math>a</math> is not zero. Find the three-digit number <math>abc</math>.<br />
<br />
==Solution 1 (similar to Solution 3)== <br />
<br />
We have that <math>N^2 - N = N(N - 1)\equiv 0\mod{10000}</math><br />
<br />
Thus, <math>N(N-1)</math> must be divisible by both <math>5^4</math> and <math>2^4</math>. Note, however, that if either <math>N</math> or <math>N-1</math> has both a <math>5</math> and a <math>2</math> in its factorization, the other must end in either <math>1</math> or <math>9</math>, which is impossible for a number that is divisible by either <math>2</math> or <math>5</math>. Thus, one of them is divisible by <math>2^4 = 16</math>, and the other is divisible by <math>5^4 = 625</math>. Noting that <math>625 \equiv 1\mod{16}</math>, we see that <math>625</math> would work for <math>N</math>, except the thousands digit is <math>0</math>. The other possibility is that <math>N</math> is a multiple of <math>16</math> and <math>N-1</math> is a multiple of <math>625</math>. In order for this to happen, <cmath>N-1 \equiv -1 \pmod {16}.</cmath> Since <math>625 \equiv 1 \pmod{16}</math>, we know that <math>15 \cdot 625 = 9375 \equiv 15 \equiv -1 \mod{16}</math>. Thus, <math>N-1 = 9375</math>, so <math>N = 9376</math>, and our answer is <math>\boxed{937}</math>.<br />
<br />
== Solution 2 (bashing)==<br />
let <math>N= 10000t+1000a+100b+10c+d</math> for positive integer values <math>t,a,b,c,d</math>.<br />
When we square <math>N</math> we get that <br />
<cmath>\begin{align*}<br />
N^2 &=(10000t+1000a+100b+10c+d)^2\\<br />
&=10^8t^2+10^6a^2+10^4b^2+10^2c^2+d^2+2(10^7ta+10^6tb+10^5tc+10^4td+10^5ab+10^4ac+10^3bc+10^ad+10^2bd+10cd)<br />
\end{align*}</cmath><br />
<br />
However, we don't have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only:<br />
<cmath>2000ad+2000bc+100c^2+200bd+20cd+d^2.</cmath><br />
Now we need to compare each decimal digit with <math>1000a+100b+10c+d</math> and see whether the digits are congruent in base 10.<br />
we first consider the ones digits:<br />
<br />
<math>d^2\equiv d \pmod{10}.</math><br />
<br />
This can happen for only 3 values : 1, 5 and 6.<br />
<br />
We can try to solve each case:<br />
<br />
*Case 1 <math>(d=1)</math><br />
Considering the tenths place, <br />
we have that:<br />
<br />
<math>20cd=20c\equiv 10c \pmod {100}</math><br />
so <math>c= 0</math>.<br />
<br />
Considering the hundreds place we have that <br />
<br />
<math>200bd+100c^2= 200b \equiv 100b \pmod{1000}</math><br />
so again <math>b=0</math><br />
<br />
now considering the thousands place we have that <br />
<br />
<math>2000ad+2000bc = 2000a \equiv 1000a \pmod {10000}</math><br />
so we get <math>a=0</math> but <math>a</math> cannot be equal to <math>0</math> so we consider <math>d=5.</math><br />
<br />
*Case 2 <math>(d=5)</math><br />
considering the tenths place<br />
we have that:<br />
<br />
<math>20cd+20=100c+20\equiv 20 \equiv 10c \mod {100}</math><br />
( the extra <math>20</math> is carried from <math>d^2</math> which is equal to <math>25</math>)<br />
so <math>c=2</math><br />
<br />
considering the hundreds place we have that <br />
<br />
<math>200bd+100c^2+100c= 1000b+600 \equiv600\equiv 100b \pmod{1000}</math><br />
( the extra <math>100c</math> is carried from the tenths place)<br />
so <math>b=6</math><br />
<br />
now considering the thousands place we have that <br />
<br />
<math>2000ad+2000bc +1000b= 10000a+24000+ 6000\equiv0\equiv 1000a \pmod {10000}</math><br />
( the extra <math>1000b</math> is carried from the hundreds place)<br />
so a is equal 0 again<br />
<br />
*Case 3<math>(d=6)</math><br />
considering the tenths place<br />
we have that:<br />
<br />
<math>20cd+30=120c+30\equiv 30+20c \equiv 10c \pmod {100}</math><br />
( the extra <math>20</math> is carried from <math>d^2</math> which is equal to <math>25</math>)<br />
if <math>c=7</math> then we have <br />
<br />
<math>30+20 \cdot 7 \equiv 70\equiv7 \cdot 10 \pmod{100}</math><br />
<br />
so <math>c=7</math><br />
<br />
considering the hundreds place we have that <br />
<br />
<math>200bd+100c^2+100c+100= 1200b+4900+800 \equiv200b+700\equiv 100b \pmod{1000}</math><br />
( the extra <math>100c+100</math> is carried from the tenths place)<br />
<br />
if <math>b=3</math> then we have <br />
<br />
<math>700+200 \cdot 3 \equiv 300\equiv3 \cdot 100 \pmod {1000}</math><br />
<br />
so <math>b=3</math><br />
<br />
now considering the thousands place we have that <br />
<br />
<math>2000ad+2000bc +1000b+5000+1000= 12000a+42000+ 3000+6000\equiv0\equiv 2000a+1000\equiv 1000a \pmod {10000}</math><br />
( the extra <math>1000b+6000</math> is carried from the hundreds place)<br />
<br />
if <math>a=9</math> then we have <br />
<br />
<math>2000 \cdot 9+1000 \equiv 9000\equiv9 \cdot 1000 \pmod {1000}</math><br />
<br />
so <math>a=9</math><br />
<br />
so we have that the last 4 digits of <math>N</math> are <math>9376</math><br />
and <math>abc</math> is equal to <math>\boxed{937}</math><br />
<br />
== Solution 3 (general) ==<br />
By the Chinese Remainder Theorem, the equation <math>N(N-1)\equiv 0\pmod{10000}</math> is equivalent to the two equations:<br />
<cmath>\begin{align*}<br />
N(N-1)&\equiv 0\pmod{16},\\<br />
N(N-1)&\equiv 0\pmod{625}.<br />
\end{align*}</cmath><br />
Since <math>N</math> and <math>N-1</math> are coprime, the only solutions are when <math>(N\mod{16},N\mod{625})\in\{(0,0),(0,1),(1,0),(1,1)\}</math>.<br />
<br />
Let <cmath>\varphi:\mathbb Z/10000\mathbb Z\to\mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,</cmath> <cmath>x\mapsto (x\mod{16},x\mod{625}).</cmath> The statement of the Chinese Remainder theorem is that <math>\varphi</math> is an isomorphism between the two rings. In this language, the solutions are <math>\varphi^{-1}(0,0)</math>, <math>\varphi^{-1}(0,1)</math>, <math>\varphi^{-1}(1,0)</math>, and <math>\varphi^{-1}(1,1)</math>. Now we easily see that <cmath>\varphi^{-1}(0,0)=0</cmath> and <cmath>\varphi^{-1}(1,1)=1.</cmath> Noting that <math>625\equiv 1\pmod{16}</math>, it follows that <cmath>\varphi^{-1}(1,0)=625.</cmath> To compute <math>\varphi^{-1}(0,1)</math>, note that <cmath>(0,1)=15(1,0)+(1,1)</cmath> in <cmath>\mathbb \mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,</cmath> so since <math>\varphi^{-1}</math> is linear in its arguments (by virtue of being an isomorphism), <cmath>\varphi^{-1}(0,1)=15\varphi^{-1}(1,0)+\varphi^{-1}(1,1)=15\times 625+1=9376.</cmath><br />
<br />
The four candidate digit strings <math>abcd</math> are then <math>0000,0001,0625,9376</math>. Of those, only <math>9376</math> has nonzero first digit, and therefore the answer is <math>\boxed{937}</math>.<br />
<br />
== Solution 4 (semi-bashing) ==<br />
<br />
*Note - <math>\overline{abcd}</math> means the number formed when the digits represented by <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are substituted in. <math>\overline{abcd}\ne a\times b\times c\times d</math>.<br />
<br />
WLOG, we can assume that <math>N</math> is a 4-digit integer <math>\overline{abcd}</math>. Note that the only <math>d</math> that will satisfy <math>N</math> will also satisfy <math>d^2\equiv d\pmod{10}</math>, as the units digit of <math>\overline{abcd}^2</math> is affected only by <math>d</math>, regardless of <math>a</math>, <math>b</math>, or <math>c</math>.<br />
<br />
By checking the numbers 0-9, we see that the only possible values of <math>d</math> are <math>d=0, 1, 5, 6</math>.<br />
<br />
Now, we seek to find <math>c</math>. Note that the only <math>\overline{cd}</math> that will satisfy <math>N</math> will also satisfy <math>\overline{cd}^2 \equiv \overline{cd}\pmod{100}</math>, by the same reasoning as above - the last two digits of <math>\overline{abcd}^2</math> are only affected by <math>c</math> and <math>d</math>. As we already have narrowed choices for <math>d</math>, we start reasoning out.<br />
<br />
First, we note that if <math>d=0</math>, then <math>c=0</math>, as a number ending in 0, and therefore divisible by 10, is squared, the result is divisible by 100, meaning it ends in two 0's. However, if <math>N</math> ends in <math>00</math>, then recursively, <math>a</math> and <math>b</math> must be <math>0</math>, as a number divisible by 100 squared ends in four zeros. As <math>a</math> cannot be 0, we throw out this possibility, as the only solution in this case is <math>0</math>.<br />
<br />
Now, let's assume that <math>d=1</math>. <math>\overline{cd}</math> is equal to <math>10c + d = 10c + 1</math>. Squaring this gives <math>100c^2 + 20c + 1</math>, and when modulo 100 is taken, it must equal <math>10c + 1</math>. As <math>c</math> is an integer, <math>100c^2</math> must be divisible by 100, so <math>100c^2+20c+1 \equiv 20c + 1\pmod{100}</math>, which must be equivalent to <math>10c + 1</math>. Note that this is really <math>\overline{(2c)1}</math> and <math>\overline{c1}</math>, and comparing the 10's digits. So really, we're just looking for when the units digit of <math>2c</math> and <math>c</math> are equal, and a quick check reveals that this is only true when <math>c=0</math>.However, if we extend this process to find <math>b</math> and <math>a</math>, we'd find that they are also 0. The only solution in this case is <math>1</math>, and since <math>a=0</math> here, this is not our solution. Therefore, there are no valid solutions in this case.<br />
<br />
Let's assume that <math>d=5</math>. Note that <math>(10c + 5)^2 = 100c^2 + 100c + 25</math>, and when modulo <math>100</math> is taken, <math>25</math> is the remainder. So all cases here have squares that end in 25, so <math>\overline{cd}=25</math> is our only case here. A quick check reveals that <math>25^2=625</math>, which works for now.<br />
<br />
Now, let <math>d=6</math>. Note that <math>(10c + 6)^2 = 100c^2 + 120c + 36</math>. Taking modulo 100, this reduces to <math>20c+36</math>, which must be equivalent to <math>10c+6</math>. Again, this is similar to <math>\overline{(2c+3)6}</math> and <math>\overline{c6}</math>, so we see when the units digits of <math>2c+3</math> and <math>c</math> are equal. To make checking faster, note that <math>2c</math> is necessarily even, so <math>2c+3</math> is necessarily odd, so <math>c</math> must be odd. Checking all the odds reveals that only <math>c=3</math> works, so this case gives <math>76</math>. Checking quickly <math>76^2 = 5776</math>, which works for now.<br />
<br />
Now, we find <math>b</math>, given two possibilities for <math>\overline{cd}</math>. <br />
<br />
Start with <math>\overline{cd} = 25</math>. <math>\overline{bcd} = 100b + \overline{cd} = 100b + 25.</math> Note that if we square this, we get <math>10000b^2 + 5000b + 625</math>, which should be equivalent to <math>100b + 25</math> modulo 1000. Note that, since <math>b</math> is an integer, <math>10000b^2 + 5000 + 625</math> simplifies modulo 1000 to <math>625</math>. Therefore, the only <math>\overline{bcd}</math> that works here is <math>625</math>. <math>625^2 = 390625</math>.<br />
<br />
Now, assume that <math>\overline{cd}=76</math>. We have <math>100b + 76</math>, and when squared, becomes <math>10000b^2 + 15200b + 5776</math>, which, modulo 1000, should be equivalent to <math>100b+76</math>. Reducing <math>10000b^2 + 15200b + 5776</math> modulo <math>1000</math> gives <math>200b + 776</math>. Using the same technique as before, we must equate the hundreds digit of <math>\overline{(2b+7)76}</math> to <math>\overline{b76}</math>, or equate the units digit of <math>2b+7</math> and <math>b</math>. Since <math>2b+7</math> is necessarily odd, any possible <math>b</math>'s must be odd. A quick check reveals that <math>b=3</math> is the only solution, so we get a solution of <math>376</math>. <math>376^2 = 141376</math>.<br />
<br />
Finally, we solve for <math>a</math>. Start with <math>\overline{bcd}=625</math>. We have <math>1000a + 625</math>, which, squared, gives <cmath>1000000a^2 + 1250000a + 390625,</cmath> and reducing modulo 10000 gives simply 625. So <math>\overline{abcd}=625</math>. However, that makes <math>a=0</math>. Therefore, no solutions exist in this case.<br />
<br />
We turn to our last case, <math>\overline{bcd}=376</math>. We have <cmath>1000a + 376^2 = 1000000a^2 + 752000a + 141376,</cmath> and reducing modulo <math>10000</math> gives <math>2000a + 1376</math>, which must be equivalent to <math>1000a + 376</math>. So we must have <math>\overline{(2a+1)376}</math> being equivalent to <math>\overline{a376}</math> modulo 1000. So, the units digit of <math>2a+1</math> must be equal to <math>a</math>. Since <math>2a+1</math> is odd, <math>a</math> must be odd. Lo and behold, the only possibility for <math>a</math> is <math>a=3</math>. Therefore, <math>\overline{abcd}=9376</math>, so our answer is <math>\boxed{937}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=7|num-a=9}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_10&diff=1694152021 Fall AMC 10B Problems/Problem 102022-01-06T01:33:30Z<p>Asops: </p>
<hr />
<div>==Problem==<br />
Forty slips of paper numbered <math>1</math> to <math>40</math> are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by <math>100</math> and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?<br />
<br />
<math>\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67</math><br />
<br />
==Solution 1==<br />
Because Alice doesn't know who has the larger number, she doesn't have <math>1.</math> Because Alice says that she doesn't know who has the larger number, Bob knows that she doesn't have <math>1.</math> But Bob knows who has the larger number, this implies that Bob has the smallest possible number. Because Bob's number is prime, Bob's number is <math>2</math>. Thus, the perfect square is in the <math>200's.</math> The only perfect square is <math>225.</math> Thus, Alice's number is <math>25.</math> The sum of Alice's and Bob's number is <math>25+2 = 27.</math> Thus the answer is <math>\boxed{(\textbf{A}.)}.</math><br />
<br />
~NH14<br />
<br />
== Solution 2 ==<br />
Denote by <math>A</math> and <math>B</math> the numbers drawn by Alice and Bob, respectively.<br />
<br />
Alice's sentence ``I can't tell who has the larger number.'' implies <math>A \in \left\{ 2 , \cdots , 39 \right\}</math>.<br />
<br />
Bob's sentence ``I know who has the larger number.'' implies <math>B \in \left\{ 1 , 2 , 39, 40 \right\}</math>.<br />
<br />
Their subsequent conversation that <math>B</math> is prime implies <math>B = 2</math>.<br />
<br />
Then, Alice's next sentence ``In that case, if I multiply your number by 100 and add my number, the result is a perfect square.'' implies <math>200 + A</math> is a perfect square.<br />
Hence, <math>A = 25</math>.<br />
<br />
Therefore, the answer is <math>\boxed{\textbf{(A) }27}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Video Solution by Interstigation==<br />
https://youtu.be/p9_RH4s-kBA?t=1524<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=11|num-b=9}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_6&diff=1675162014 AIME I Problems/Problem 62021-12-09T00:50:32Z<p>Asops: /* Solution 6 (Vieta's solution) */</p>
<hr />
<div>== Problem 6 ==<br />
<br />
The graphs <math>y=3(x-h)^2+j</math> and <math>y=2(x-h)^2+k</math> have y-intercepts of <math>2013</math> and <math>2014</math>, respectively, and each graph has two positive integer x-intercepts. Find <math>h</math>.<br />
<br />
== Solution 1 ==<br />
Begin by setting <math>x</math> to 0, then set both equations to <math>h^2=\frac{2013-j}{3}</math> and <math>h^2=\frac{2014-k}{2}</math>, respectively. Notice that because the two parabolas have to have positive x-intercepts, <math>h\ge32</math>.<br />
<br />
We see that <math>h^2=\frac{2014-k}{2}</math>, so we now need to find a positive integer <math>h</math> which has positive integer x-intercepts for both equations.<br />
<br />
Notice that if <math>k=2014-2h^2</math> is -2 times a square number, then you have found a value of <math>h</math> for which the second equation has positive x-intercepts. We guess and check <math>h=36</math> to obtain <math>k=-578=-2(17^2)</math>.<br />
<br />
Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is <math>\boxed{036}</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>x=0</math> and <math>y=2013</math> for the first equation, resulting in <math>j=2013-3h^2</math>. Substituting back in to the original equation, we get <math>y=3(x-h)^2+2013-3h^2</math>.<br />
<br />
Now we set <math>y</math> equal to zero, since there are two distinct positive integer roots. Rearranging, we get <math>2013=3h^2-3(x-h)^2</math>, which simplifies to <math>671=h^2-(x-h)^2</math>. Applying difference of squares, we get <math>671=(2h-x)(x)</math>.<br />
<br />
Now, we know that <math>x</math> and <math>h</math> are both integers, so we can use the fact that <math>671=61\times11</math>, and set <math>2h-x=11</math> and <math>x=61</math> (note that letting <math>x=11</math> gets the same result). Therefore, <math>h=\boxed{036}</math>.<br />
<br />
Note that we did not use the second equation since we took advantage of the fact that AIME answers must be integers. However, one can enter <math>h=36</math> into the second equation to verify the validity of the answer. <br />
<br />
Note on the previous note: we still must use the second equation since we could also use <math>671=671\times1</math>, yielding <math>h=336.</math> This answer however does not check out with the second equation which is why it is invalid.<br />
<br />
==Solution 3==<br />
Similar to the first two solutions, we deduce that <math>\text{(-)}j</math> and <math>\text{(-)}k</math> are of the form <math>3a^2</math> and <math>2b^2</math>, respectively, because the roots are integers and so is the <math>y</math>-intercept of both equations. So the <math>x</math>-intercepts should be integers also.<br />
<br />
The first parabola gives<br />
<cmath>3h^2+j=3\left(h^2-a^2\right)=2013</cmath><br />
<cmath>h^2-a^2=671</cmath><br />
And the second parabola gives<br />
<cmath>2h^2+k=2\left(h^2-b^2\right)=2014</cmath><br />
<cmath>h^2-b^2=1007</cmath><br />
<br />
We know that <math>671=11\cdot 61</math> and that <math>1007=19\cdot 53</math>. It is just a fitting coincidence that the average of <math>11</math> and <math>61</math> is the same as the average of <math>19</math> and <math>53</math>. That is <math>\boxed{036}</math>.<br />
<br />
To check, we have<br />
<cmath>(h-a)(h+a)=671=11\cdot 61</cmath><br />
<cmath>(h-b)(h+b)=1007=19\cdot 53</cmath><br />
Those are the only two prime factors of <math>671</math> and <math>1007</math>, respectively. So we don't need any new factorizations for those numbers.<br />
<br />
<math>h+a=61,h-a=11\implies (h,a)=\{36,25\}</math><br />
<br />
<math>h+b=53,h-b=19\implies (h,b)=\{36,17\}</math><br />
<br />
Thus the common integer value for <math>h</math> is <math>\boxed{036}</math>.<br />
<br />
<cmath>y=3(x-h)^2+j\rightarrow y=3(x-11)(x-61)=3x^2-216x+2013</cmath><br />
<cmath>y=2(x-h)^2+k\rightarrow y=2(x-19)(x-53)=2x^2-144x+2014</cmath><br />
<br />
==Solution 4==<br />
First, we expand both equations to get <math>y=3x^2-6hx+3h^2+j</math> and <math>y=2x^2-4hx+2h^2+k</math>. The <math>y</math>-intercept for the first equation can be expressed as <math>3h^2+j</math>. From this, the x-intercepts for the first equation can be written as <br />
<br />
<cmath>x=h \pm \sqrt{(-6h)^2-4*3(3h^2+j)}=h \pm \sqrt{36h^2-12(2013)}=h \pm \sqrt{36h^2-24156}</cmath><br />
<br />
Since the <math>x</math>-intercepts must be integers, <math>\sqrt{36h^2-24156}</math> must also be an integer. From solution 1, we know <math>h</math> must be greater than or equal to 32. We can substitute increasing integer values for <math>h</math> starting from 32; we find that <math>h=36</math>. <br />
<br />
We can test this result using the second equation, whose <math>x</math>-intercepts are <cmath>x=h \pm \sqrt{(-4h)^2-4*2(2h^2+k)}=h \pm \sqrt{16h^2-8(2014)}=h \pm \sqrt{16h^2-16112}</cmath> Substituting 36 in for <math>h</math>, we get <math>h=36 \pm 68</math>, which satisfies the requirement that all x-intercepts must be (positive) integers.<br />
<br />
Thus, <math>h=\boxed{036}</math>.<br />
<br />
==Solution 5==<br />
We have the equation <math>y=3(x-h)^2 + j.</math><br />
<br />
We know: <math>(x,y):(0,2013)</math>, so <math>h^2=2013/3 - j/3</math> after plugging in the values and isolating <math>h^2</math>. Therefore, <math>h^2=671-j/3</math>.<br />
<br />
Lets call the x-intercepts <math>x_1</math>, <math>x_2</math>. Since both <math>x_1</math> and <math>x_2</math> are positive there is a relationship between <math>x_1</math>, <math>x_2</math> and <math>h</math>. Namely, <math>x_1+x_2=2h</math>. The is because: <math>x_1-h=-(x_2-h)</math>, <br />
<br />
Similarly, we know: <math>(x,y):(x_1,0)</math>, so <math>j=-3(x_1-h)^2</math>. Combining the two equations gives us <br />
<cmath>h^2=671+(x_1-h)^2</cmath><br />
<cmath>h^2=671+x_1^2-2x_1h+h^2</cmath><br />
<cmath>h=(671+x_1^2)/2x_1.</cmath><br />
<br />
Now since we have this relationship, <math>2h=x_1+x_2</math>, we can just multiply the last equation by 2(so that we get <math>2h</math> on the left side) which gives us <br />
<cmath>2h=671/x_1+x_1^2/x^1</cmath> <cmath>2h=671/x_1+x_1</cmath> <cmath>x_1+x_2=671/x_1+x_1</cmath> <cmath>x_2=671/x_1</cmath> <cmath>x_1x_2=671.</cmath> Prime factorization of 671 gives 11 and 61. So now we know <math>x_1=11</math> and <math>x_2=61</math>. Lastly, we plug in the numbers,11 and 61, into <math>x_1+x_2=2h</math>, so <math>\boxed{h=36}</math>.<br />
<br />
==Solution 6 (Vieta's solution)==<br />
First, we start of exactly like solutions above and we find out that <math>j=2013-3h^2</math> and <math>k=2014-2h^2</math> We then plug j and k into <math>3(x-h)^2+j</math> and <math>y=2(x-h)^2+k</math> respectively. After that, we get two equations, <math>y=3x^2-6xh+2013</math> and <math>y=2x^2-4xh+2014</math>. We can apply Vieta's. Let the roots of the first equation be <math>a, b</math> and the roots of the second equation be <math>c, d</math>. Thus, we have that <math>a\cdot b=1007</math>, <math>a+b=2h</math> and <math>c\cdot d=671</math>, <math>c+d=2h</math>. Simple evaluations finds that <math>\boxed{h=36}</math><br />
<br />
~Jske25<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems&diff=1671152014 AIME I Problems2021-11-30T17:59:45Z<p>Asops: /* Problem 4 */</p>
<hr />
<div>{{AIME Problems|year=2014|n=I}}<br />
<br />
==Problem 1==<br />
The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters. <br />
<br />
<asy><br />
size(200);<br />
defaultpen(linewidth(0.7));<br />
path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin;<br />
path laceR=reflect((75,0),(75,-240))*laceL;<br />
draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray);<br />
for(int i=0;i<=3;i=i+1)<br />
{<br />
path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5);<br />
unfill(circ1); draw(circ1);<br />
unfill(circ2); draw(circ2);<br />
}<br />
draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));</asy><br />
[[2014 AIME I Problems/Problem 1|Solution]]<br />
<br />
==Problem 2== <br />
<br />
An urn contains <math>4</math> green balls and <math>6</math> blue balls. A second urn contains <math>16</math> green balls and <math>N</math> blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is <math>0.58</math>. Find <math>N</math>.<br />
<br />
[[2014 AIME I Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
Find the number of rational numbers <math>r</math>, <math>0<r<1,</math> such that when <math>r</math> is written as a fraction in lowest terms, the numerator and the denominator have a sum of <math>1000</math>.<br />
<br />
[[2014 AIME I Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at <math>20</math> miles per hour, and Steve rides west at <math>20</math> miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly <math>1</math> minute to go past Jon. The westbound train takes <math>10</math> times as long as the eastbound train to go past Steve. The length of each train is <math>\tfrac{m}{n}</math> miles, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
[[2014 AIME I Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Let the set <math>S = \{P_1, P_2, \dots, P_{12}\}</math> consist of the twelve vertices of a regular <math>12</math>-gon. A subset <math>Q</math> of <math>S</math> is called communal if there is a circle such that all points of <math>Q</math> are inside the circle, and all points of <math>S</math> not in <math>Q</math> are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)<br />
<br />
[[2014 AIME I Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
The graphs <math>y = 3(x-h)^2 + j</math> and <math>y = 2(x-h)^2 + k</math> have y-intercepts of <math>2013</math> and <math>2014</math>, respectively, and each graph has two positive integer x-intercepts. Find <math>h</math>.<br />
<br />
[[2014 AIME I Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
Let <math>w</math> and <math>z</math> be complex numbers such that <math>|w| = 1</math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. (Note that <math>\arg(w)</math>, for <math>w \neq 0</math>, denotes the measure of the angle that the ray from <math>0</math> to <math>w</math> makes with the positive real axis in the complex plane.)<br />
<br />
[[2014 AIME I Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
The positive integers <math>N</math> and <math>N^2</math> both end in the same sequence of four digits <math>abcd</math> when written in base 10, where digit <math>a</math> is not zero. Find the three-digit number <math>abc</math>.<br />
<br />
[[2014 AIME I Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
Let <math>x_1< x_2 < x_3</math> be the three real roots of the equation <math>\sqrt{2014} x^3 - 4029x^2 + 2 = 0</math>. Find <math>x_2(x_1+x_3)</math>.<br />
<br />
[[2014 AIME I Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
A disk with radius <math>1</math> is externally tangent to a disk with radius <math>5</math>. Let <math>A</math> be the point where the disks are tangent, <math>C</math> be the center of the smaller disk, and <math>E</math> be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of <math>360^\circ</math>. That is, if the center of the smaller disk has moved to the point <math>D</math>, and the point on the smaller disk that began at <math>A</math> has now moved to point <math>B</math>, then <math>\overline{AC}</math> is parallel to <math>\overline{BD}</math>. Then <math>\sin^2(\angle BEA)=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
[[2014 AIME I Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
A token starts at the point <math>(0,0)</math> of an <math>xy</math>-coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of <math>|y|=|x|</math> is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
[[2014 AIME I Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
Let <math>A=\{1,2,3,4\}</math>, and <math>f</math> and <math>g</math> be randomly chosen (not necessarily distinct) functions from <math>A</math> to <math>A</math>. The probability that the range of <math>f</math> and the range of <math>g</math> are disjoint is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m</math>.<br />
<br />
[[2014 AIME I Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
On square <math>ABCD</math>, points <math>E,F,G</math>, and <math>H</math> lie on sides <math>\overline{AB},\overline{BC},\overline{CD},</math> and <math>\overline{DA},</math> respectively, so that <math>\overline{EG} \perp \overline{FH}</math> and <math>EG=FH = 34</math>. Segments <math>\overline{EG}</math> and <math>\overline{FH}</math> intersect at a point <math>P</math>, and the areas of the quadrilaterals <math>AEPH, BFPE, CGPF,</math> and <math>DHPG</math> are in the ratio <math>269:275:405:411.</math> Find the area of square <math>ABCD</math>.<br />
<br />
<asy><br />
pair A = (0,sqrt(850));<br />
pair B = (0,0);<br />
pair C = (sqrt(850),0);<br />
pair D = (sqrt(850),sqrt(850));<br />
draw(A--B--C--D--cycle);<br />
dotfactor = 3;<br />
dot("$A$",A,dir(135));<br />
dot("$B$",B,dir(215));<br />
dot("$C$",C,dir(305));<br />
dot("$D$",D,dir(45));<br />
pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850));<br />
pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);<br />
dot("$H$",H,dir(90));<br />
dot("$F$",F,dir(270));<br />
draw(H--F);<br />
pair E = (0,(sqrt(850)-6)/2);<br />
pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);<br />
dot("$E$",E,dir(180));<br />
dot("$G$",G,dir(0));<br />
draw(E--G);<br />
pair P = extension(H,F,E,G);<br />
dot("$P$",P,dir(60));<br />
label("$w$", intersectionpoint( A--P, E--H ));<br />
label("$x$", intersectionpoint( B--P, E--F ));<br />
label("$y$", intersectionpoint( C--P, G--F ));<br />
label("$z$", intersectionpoint( D--P, G--H ));</asy><br />
<br />
[[2014 AIME I Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
Let <math>m</math> be the largest real solution to the equation<br />
<br />
<cmath> \dfrac{3}{x-3} + \dfrac{5}{x-5} + \dfrac{17}{x-17} + \dfrac{19}{x-19} = x^2 - 11x - 4</cmath><br />
<br />
There are positive integers <math>a, b,</math> and <math>c</math> such that <math>m = a + \sqrt{b + \sqrt{c}}</math>. Find <math>a+b+c</math>.<br />
<br />
[[2014 AIME I Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
In <math>\triangle ABC, AB = 3, BC = 4,</math> and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B, \overline{BC}</math> at <math>B</math> and <math>D,</math> and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4},</math> length <math>DE=\frac{a\sqrt{b}}{c},</math> where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. Find <math>a+b+c</math>.<br />
<br />
[[2014 AIME I Problems/Problem 15|Solution]]<br />
<br />
{{AIME box|year=2014|n=I|before=[[2013 AIME II Problems]]|after=[[2014 AIME II Problems]]}}<br />
<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_4&diff=1671142014 AIME I Problems/Problem 42021-11-30T17:59:02Z<p>Asops: /* Problem 4 */</p>
<hr />
<div>== Problem 4 ==<br />
Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at <math>20</math> miles per hour, and Steve rides west at <math>20</math> miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly <math>1</math> minute to go past Jon. The westbound train takes <math>10</math> times as long as the eastbound train to go past Steve. The length of each train is <math>\tfrac{m}{n}</math> miles, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
== Solution 1 ==<br />
For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of <math>\dfrac{1}{3}</math> mi/min.<br />
<br />
Let <math>d</math> be the length of the trains, <math>r_1</math> be the speed of train 1 (the faster train), and <math>r_2</math> be the speed of train 2.<br />
<br />
Consider the problem from the bikers' moving frame of reference. In order to pass Jon, the first train has to cover a distance equal to its own length, at a rate of <math>r_1 - \dfrac{1}{3}</math>. Similarly, the second train has to cover a distance equal to its own length, at a rate of <math>r_2 + \dfrac{1}{3}</math>. Since the times are equal and <math>d = rt</math>, we have that <math>\dfrac{d}{r_1 - \dfrac{1}{3}} = \dfrac{d}{r_2 + \dfrac{1}{3}}</math>. Solving for <math>r_1</math> in terms of <math>r_2</math>, we get that <math>r_1 = r_2 + \dfrac{2}{3}</math>.<br />
<br />
Now, let's examine the times it takes the trains to pass Steve. This time, we augment train 1's speed by <math>\dfrac{1}{3}</math>, and decrease train 2's speed by <math>\dfrac{1}{3}</math>. Thus, we have that <math>\dfrac{d}{r_2 - \dfrac{1}{3}} = 10\dfrac{d}{r_1 + \dfrac{1}{3}}</math>.<br />
<br />
Multiplying this out and simplifying, we get that <math>r_1 = 10r_2 - \dfrac{11}{3}</math>. Since we now have 2 expressions for <math>r_1</math> in terms of <math>r_2</math>, we can set them equal to each other:<br />
<br />
<math>r_2 + \dfrac{2}{3} = 10r_2 - \dfrac{11}{3}</math>. Solving for <math>r_2</math>, we get that <math>r_2 = \dfrac{13}{27}</math>. Since we know that it took train 2 1 minute to pass Jon, we know that <math>1 = \dfrac{d}{r_2 + \dfrac{1}{3}}</math>. Plugging in <math>\dfrac{13}{27}</math> for <math>r_2</math> and solving for <math>d</math>, we get that <math>d = \dfrac{22}{27}</math>, and our answer is <math>27 + 22 = \boxed{049}</math>.<br />
<br />
==Solution 2==<br />
Using a similar approach to Solution 1, let the speed of the east bound train be <math>a</math> and the speed of the west bound train be <math>b</math>.<br />
<br />
So <math>a-20=b+20</math> and <math>a+20=10(b-20)</math>.<br />
<br />
From the first equation, <math>a=b+40</math>. Substituting into the second equation,<br />
<cmath>b+60=10b-200</cmath><br />
<cmath>260=9b</cmath><br />
<cmath>b=\frac{260}{9}\text{ mph}</cmath><br />
This means that <br />
<cmath>a=\frac{260}{9}+40=\frac{620}{9}\text{ mph}</cmath><br />
Checking, we get that the common difference in Jon's speed and trains' speeds is <math>\frac{440}{9}</math> and the differences for Steve is <math>\frac{800}{9}</math> and <math>\frac{80}{9}</math>.<br />
<br />
This question assumes the trains' lengths in MILES:<br />
<cmath>\frac{440}{9}\cdot \frac{1}{60}=\frac{440}{540}=\frac{22}{27}\text{ miles}</cmath><br />
Adding up, we get <math>22+27=\boxed{049}</math>.<br />
<br />
==Solution 3==<br />
Let the length of the trains be <math>L</math>, let the rate of the westward train be <math>W_{R}</math>, ;et the rate of the eastward train be <math>E_{R}</math>, and let the time it takes for the eastward train to pass Steve be <math>E_{T}</math>.<br />
<br />
We have that<br />
<br />
<math>L=(\frac{1}{60})(W_{R}+20)</math><br />
<br />
<math>L=(\frac{1}{60})(E_{R}-20)</math>.<br />
<br />
Adding both of the equations together, we get that <br />
<br />
<math>2L=\frac{W_{R}}{60}+\frac{E_{R}}{60}\implies 120L=W_{R}+E_{R}</math>.<br />
<br />
Now, from the second part of the problem, we acquire that<br />
<br />
<math>L=(E_{T})(E_{R}+20)</math><br />
<br />
<math>L=(10E_{T})(W_{R}-20)</math><br />
<br />
Dividing the second equation by the first, we get that...<br />
<math>1=\frac{10(W_{R}-20)}{E_{R}+20}\implies E_{R}+20=10W_{R}-200\implies E_{R}+220=10W_{R}\implies E_{R}=10W_{R}-220</math>.<br />
<br />
Now, substituting into the <math>120L=W_{R}+E_{R}</math>.<br />
<br />
<math>120L=W_{R}+(10W_{R}-220)\implies 120L= 11W_{R}-220\implies W_{R}=\frac{120L+220}{11}</math>.<br />
<br />
Finally, plugging this back into our very first equation..<br />
<br />
<math>L=(\frac{1}{60})((\frac{120L+220}{11})+20)\implies 660L=120L+440\implies 540L=440\implies L=\frac{22}{27}</math>.<br />
<br />
Hence, the answer is <math>22+27=\boxed{049}</math>.<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_9&diff=1670062014 AIME I Problems/Problem 92021-11-29T16:45:00Z<p>Asops: /* Solution 5 */</p>
<hr />
<div>== Problem 9 ==<br />
<br />
Let <math>x_1<x_2<x_3</math> be the three real roots of the equation <math>\sqrt{2014}x^3-4029x^2+2=0</math>. Find <math>x_2(x_1+x_3)</math>.<br />
<br />
== Solution 1==<br />
<br />
Substituting <math>n</math> for <math>2014</math>, we get <cmath>\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2</cmath> <cmath>= x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0</cmath> Noting that <math>nx^2 - 1</math> factors as a difference of squares to <cmath>(\sqrt{n}x - 1)(\sqrt{n}x+1)</cmath> we can factor the left side as <cmath>(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))</cmath> This means that <math>\frac{1}{\sqrt{n}}</math> is a root, and the other two roots are the roots of <math>x^2 - 2\sqrt{n}x - 2</math>. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to <math>2\sqrt{n}</math>, so the positive root must be greater than <math>2\sqrt{n}</math> in order to produce this sum when added to a negative value. Since <math>0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}</math> is clearly true, <math>x_2 = \frac{1}{\sqrt{2014}}</math> and <math>x_1 + x_3 = 2\sqrt{2014}</math>. Multiplying these values together, we find that <math>x_2(x_1+x_3) = \boxed{002}</math>.<br />
<br />
==Solution 2==<br />
From Vieta's formulae, we know that <cmath>x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}</cmath> <cmath>x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}}</cmath> and <cmath>x_1x_2 + x_2x_3 + x_1x_3 = 0</cmath> Thus, we know that <cmath>x_2(x_1 + x_3) = -x_1x_3</cmath><br />
<br />
Now consider the polynomial with roots <math>x_1x_2, x_2x_3,</math> and <math>x_1x_3</math>. Expanding the polynomial <cmath>(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)</cmath>we get the polynomial <cmath>x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2</cmath> Substituting the values obtained from Vieta's formulae, we find that this polynomial is <cmath>x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}</cmath> We know <math>x_1x_3</math> is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to <cmath>1007x^3 - 4029x - 2 = 0</cmath><br />
<br />
Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the <math>x^3</math> term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that <math>x = -2</math> is a solution. Factoring it out, we get that <cmath>1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)</cmath> Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, <cmath>x_1x_3 = -2</cmath> so <cmath>-x_1x_3 = \boxed{002}</cmath>and we're done.<br />
<br />
==Solution 3==<br />
Observing the equation, we notice that the coefficient for the middle term <math>-4029</math> is equal to <cmath>-2{\sqrt{2014}}^2-1</cmath>. <br />
<br />
Also notice that the coefficient for the <math>{x^3}</math> term is <math>\sqrt{2014}</math>. Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the <math>x</math> term of the binomial would have a coefficient of <math>\sqrt{2014}</math>. Similarly, the <math>x</math> term of the trinomial would also have a coefficient of <math>\sqrt{2014}</math>. The factored form of the expression would look something like the following:<br />
<cmath>({\sqrt{2014}}x-a)(x^2-n{\sqrt{2014}}x-b)</cmath> where <math>{a, b,c}</math> are all positive integers (because the <math>{x^2}</math> term of the original expression is negative, and the constant term is positive), and <cmath>{ab=2}</cmath><br />
<br />
Multiplying this expression out gives <cmath>{{\sqrt{2014}x^3-(2014n+a)x^2+(an{\sqrt{2014}}-b{\sqrt{2014}})x+ab}}</cmath> Equating this with the original expression gives <cmath>{2014n+a}=-4029</cmath> The only positive integer solutions of this expression is <math>(n, a)=(1, 2015)</math> or <math>(2, 1)</math>. If <math>(n, a)=(1, 2015)</math> then setting <math>{an{\sqrt{2014}}-b{\sqrt{2014}}}=0</math> yields <math>{b=2015}</math> and therefore <math>{ab=2015^2}</math> which clearly isn't equal to <math>2</math> as the constant term. Therefore, <math>(n, a)=(2, 1)</math> and the factored form of the expression is:<br />
<cmath>({\sqrt{2014}}x-1)(x^2-2{\sqrt{2014}}x-2)</cmath> Therefore, one of the three roots of the original expression is <cmath>{x=\dfrac{1}{\sqrt{2014}}}</cmath><br />
Using the quadratic formula yields the other two roots as <cmath>{x={\sqrt{2014}}+{\sqrt{2016}}}</cmath> and <cmath>{x={\sqrt{2014}}-{\sqrt{2016}}}</cmath> Arranging the roots in ascending order (in the order <math>x_1<x_2<x_3</math>), <cmath>{\sqrt{2014}}-{\sqrt{2016}}<\dfrac{1}{\sqrt{2014}}<{\sqrt{2014}}+{\sqrt{2016}}</cmath><br />
Therefore, <cmath>x_2(x_1+x_3)=\dfrac{1}{\sqrt{2014}}{2\sqrt{2014}}=\boxed{002}</cmath><br />
<br />
==Solution 4==<br />
By Vieta's, we are seeking to find <math>x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}</math>. Substitute <math>n=-x_1x_3</math> and <math>x_2=\frac{2}{\sqrt{2014}n}</math>. Substituting this back into the original equation, we have <math>\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0</math>, so <math>2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0</math>. Hence, <math>8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}</math>, and <math>n\equiv 2\pmod{1007}</math>. But since <math>n\le 999</math> because it is our desired answer, the only possible value for <math>n</math> is <math>\boxed{002}</math><br />
BEST PROOOFFFF<br />
Stormersyle & mathleticguyyy<br />
<br />
<br />
== Solution 5 ==<br />
<br />
Let <math>x =\frac{y}{\sqrt{2014}}.</math> The original equation simplifies to <math>\frac{y^3}{2014} -\frac{4029y^2}{2014}+2 = 0 \implies y^3 - 4029y^2 + 4028=0.</math> Here we clearly see that <math>y=1</math> is a root. Dividing <math>y-1</math> from the sum we find that <math>(y-1)(y^2-4028y-4028)=0.</math> From simple bounding we see that <math>y=1</math> is the middle root. Therefore <math>x_{2}(x_{1}+x_{3}) =\frac{1}{\sqrt{2014}} \cdot\frac{4028}{\sqrt{2014}} = \boxed{002}.</math><br />
<br />
== Solution 6 ==<br />
<br />
<math>\sqrt{2014}</math> occurs multiple times, so let k = <math>\sqrt{2014}</math>.<br />
<br />
The equation becomes <math>0 = kx^3 - (2k^2 + 1)x^2 + 2</math>. Since we want to relate k and x, we should solve for one of them. We can't solve for x, since that would require the cubic formula, so we solve for k, and express it in terms of a quadratic, and apply the quadratic formula.<br />
<br />
We get the roots are:<br />
<br />
<math>y = \frac{1}{x}</math>, and <math>y = \frac{x}{2} - \frac{1}{x}</math>.<br />
<br />
In the first case, <math>x = \frac{1}{y} = \frac{1}{\sqrt{2014}}</math>.<br />
<br />
In the second case, <math>x^2 - 2\sqrt{2014} - 2 = 0</math>. The solutions are <math>\sqrt{2014} \pm \sqrt{2016}</math>. The sum of these 2 solutions is <math>2 \sqrt{2014}</math>, and <math>\frac{1}{\sqrt{2014}}</math> is the middle solution, and thus, <math>(x_1 + x_3) \cdot x_2 = 2</math><br />
<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1643852020 AMC 10B Problems/Problem 222021-10-31T15:53:04Z<p>Asops: /* Solution 6 (Modular Arithmetic) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the denominator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2==<br />
Similar to Solution 1, let <math>x=2^{50}</math>. It suffices to find remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. Dividing polynomials results in a remainder of <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 3 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 4==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 5==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 6 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1643842020 AMC 10B Problems/Problem 222021-10-31T15:52:48Z<p>Asops: /* Solution 6 (Modular Arithmetic) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the denominator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2==<br />
Similar to Solution 1, let <math>x=2^{50}</math>. It suffices to find remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. Dividing polynomials results in a remainder of <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 3 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 4==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 5==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 6 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
~ (edited by asops)<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_15&diff=1639852010 AIME I Problems/Problem 152021-10-23T21:41:21Z<p>Asops: /* Solution 6 (Similar to Solution 1 with easier computation) */</p>
<hr />
<div>__TOC__<br />
== Problem ==<br />
In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the [[incircle]]s of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal [[inradius|radii]]. Then <math>\frac{AM}{CM} = \frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p + q</math>.<br />
<br />
== Solution ==<br />
<center><asy> /* from geogebra: see azjps, userscripts.org/scripts/show/72997 */<br />
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200);<br />
<br />
/* segments and figures */<br />
draw((0,0)--(15,0));<br />
draw((15,0)--(6.66667,9.97775));<br />
draw((6.66667,9.97775)--(0,0));<br />
draw((7.33333,0)--(6.66667,9.97775));<br />
draw(circle((4.66667,2.49444),2.49444));<br />
draw(circle((9.66667,2.49444),2.49444));<br />
draw((4.66667,0)--(4.66667,2.49444));<br />
draw((9.66667,2.49444)--(9.66667,0));<br />
<br />
/* points and labels */<br />
label("r",(10.19662,1.92704),SE);<br />
label("r",(5.02391,1.8773),SE);<br />
dot((0,0));<br />
label("$A$",(-1.04408,-0.60958),NE);<br />
dot((15,0));<br />
label("$C$",(15.41907,-0.46037),NE);<br />
dot((6.66667,9.97775));<br />
label("$B$",(6.66525,10.23322),NE);<br />
label("$15$",(6.01866,-1.15669),NE);<br />
label("$13$",(11.44006,5.50815),NE);<br />
label("$12$",(2.28834,5.75684),NE);<br />
dot((7.33333,0));<br />
label("$M$",(7.56053,-1.000),NE);<br />
label("$H_1$",(3.97942,-1.200),NE);<br />
label("$H_2$",(9.54741,-1.200),NE);<br />
dot((4.66667,2.49444));<br />
label("$I_1$",(3.97942,2.92179),NE);<br />
dot((9.66667,2.49444));<br />
label("$I_2$",(9.54741,2.92179),NE);<br />
clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle);<br />
</asy></center><br />
=== Solution 1 ===<br />
Let <math>AM = x</math>, then <math>CM = 15 - x</math>. Also let <math>BM = d</math> Clearly, <math>\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}</math>. We can also express each area by the rs formula. Then <math>\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}</math>. Equating and cross-multiplying yields <math>25x + 2dx = 15d + 180</math> or <math>d = \frac {25x - 180}{15 - 2x}.</math> Note that for <math>d</math> to be positive, we must have <math>7.2 < x < 7.5</math>.<br />
<br />
By [[Stewart's Theorem]], we have <math>12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)</math> or <math>432 = 3d^2 + 40x - 3x^2.</math> Brute forcing by plugging in our previous result for <math>d</math>, we have <math>432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.</math> Clearing the fraction and gathering like terms, we get <math>0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.</math><br />
<br />
''Aside: Since <math>x</math> must be rational in order for our answer to be in the desired form, we can use the [[Rational Root Theorem]] to reveal that <math>12x</math> is an integer. The only such <math>x</math> in the above-stated range is <math>\frac {22}3</math>.''<br />
<br />
Legitimately solving that quartic, note that <math>x = 0</math> and <math>x = 15</math> should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get <math>0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).</math> The only solution in the desired range is thus <math>\frac {22}3</math>. Then <math>CM = \frac {23}3</math>, and our desired ratio <math>\frac {AM}{CM} = \frac {22}{23}</math>, giving us an answer of <math>\boxed{045}</math>.<br />
<br />
=== Solution 2 ===<br />
Let <math>AM = 2x</math> and <math>BM = 2y</math> so <math>CM = 15 - 2x</math>. Let the [[incenter]]s of <math>\triangle ABM</math> and <math>\triangle BCM</math> be <math>I_1</math> and <math>I_2</math> respectively, and their equal inradii be <math>r</math>. From <math>r = \sqrt {(s - a)(s - b)(s - c)/s}</math>, we find that<br />
<br />
<cmath>\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\<br />
& = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}</cmath><br />
<br />
Let the incircle of <math>\triangle ABM</math> meet <math>AM</math> at <math>P</math> and the incircle of <math>\triangle BCM</math> meet <math>CM</math> at <math>Q</math>. Then note that <math>I_1 P Q I_2</math> is a rectangle. Also, <math>\angle I_1 M I_2</math> is right because <math>MI_1</math> and <math>MI_2</math> are the angle bisectors of <math>\angle AMB</math> and <math>\angle CMB</math> respectively and <math>\angle AMB + \angle CMB = 180^\circ</math>. By properties of [[tangent (geometry)|tangents]] to [[circle]]s <math>MP = (MA + MB - AB)/2 = x + y - 6</math> and <math>MQ = (MB + MC - BC)/2 = - x + y + 1</math>. Now notice that the altitude of <math>M</math> to <math>I_1 I_2</math> is of length <math>r</math>, so by similar triangles we find that <math>r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)</math> (3). Equating (3) with (1) and (2) separately yields<br />
<br />
<cmath>\begin{align*}<br />
2y^2 - 30 = 2xy + 5x - 7y \\<br />
2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}<br />
</cmath><br />
<br />
and adding these we have<br />
<br />
<cmath><br />
4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\<br />
\implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.<br />
</cmath><br />
<br />
=== Solution 3 ===<br />
Let the incircle of <math>ABM</math> hit <math>AM</math>, <math>AB</math>, <math>BM</math> at <math>X_{1},Y_{1},Z_{1}</math>, and let the incircle of <math>CBM</math> hit <math>MC</math>, <math>BC</math>, <math>BM</math> at <math>X_{2},Y_{2},Z_{2}</math>. Draw the incircle of <math>ABC</math>, and let it be tangent to <math>AC</math> at <math>X</math>. Observe that we have a homothety centered at A sending the incircle of <math>ABM</math> to that of <math>ABC</math>, and one centered at <math>C</math> taking the incircle of <math>BCM</math> to that of <math>ABC</math>. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is <math>AX_{1}/AX=CX_{2}/CX</math>.<br />
<br />
By standard computations, <math>AX=\dfrac{AB+AC-BC}{2}=7</math> and <math>CX=\dfrac{BC+AC-AB}{2}=8</math>. Now, let <math>AX_{1}=7x</math> and <math>CX_{2}=8x</math>. We will now go around and chase lengths. Observe that <math>BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x</math>. Then, <math>BZ_{1}=12-7x</math>. We also have <math>CY_{2}=CX_{2}=8x</math>, so <math>BY_{2}=13-8x</math> and <math>BZ_{2}=13-8x</math>.<br />
<br />
Observe now that <math>X_{1}M+MX_{2}=AC-15x=15(1-x)</math>. Also,<math>X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)</math>. Solving, we get <math>X_{1}M=8-8x</math> and <math>MX_{2}=7-7x</math> (as a side note, note that <math>AX_{1}+MX_{2}=X_{1}M+X_{2}C</math>, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).<br />
<br />
Now, we get <math>BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x</math>. To finish, we will compute area ratios. <math>\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}</math>. Also, since their inradii are equal, we get <math>\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}</math>. Equating and cross multiplying yields the quadratic <math>3x^{2}-8x+4=0</math>, so <math>x=2/3,2</math>. However, observe that <math>AX_{1}+CX_{2}=15x<15</math>, so we take <math>x=2/3</math>. Our ratio is therefore <math>\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}</math>, giving the answer <math>\boxed{045}</math>.<br />
<br />
Note: Once we have <math>MX_1=8-8x</math> and <math>MX_2=7-7x</math>, it's bit easier to use use the right triangle of <math>O_1MO_2</math> than chasing the area ratio. The inradius of <math>\triangle{ABC}</math> can be calculated to be <math>r=\sqrt{14}</math>, and the inradius of <math>ABM</math> and <math>ACM</math> are <math>r_1=r_2= xr</math>, so, <br />
<cmath> O_1O_2^2 = O_1M^2 + O_2M^2 = r_1^2+X_1M^2 + r_2^2 + X_2M^2</cmath><br />
or,<br />
<cmath> (15(1-x))^2 = 2(\sqrt{14}x)^2 + (7(1-x))^2 + (8(1-x))^2 </cmath><br />
<cmath>112(1-x)^2 = 28x^2</cmath><br />
<cmath>4(1-x)^2 = x^2</cmath><br />
We get <math>x=\frac{2}{3}</math> or <math>x=2</math>.<br />
<br />
=== Solution 4 ===<br />
Suppose the incircle of <math>ABM</math> touches <math>AM</math> at <math>X</math>, and the incircle of <math>CBM</math> touches <math>CM</math> at <math>Y</math>. Then<br />
<br />
<cmath>r = AX \tan(A/2) = CY \tan(C/2)</cmath><br />
<br />
We have <math>\cos A = \frac{12^2+15^2-13^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}</math>, <math>\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}{\sqrt{14}}</math><br />
<br />
<math>\cos C = \frac{13^2+15^2-12^2}{2\cdot 13\cdot 15} = \frac{250}{30\cdot 13} = \frac{25}{39}</math>, <math>\tan(C/2) = \sqrt{\frac{39-25}{39+25}}=\frac{\sqrt{14}}{8}</math>,<br />
<br />
Therefore <math>AX/CY = \tan(C/2)/\tan(A/2) = \frac{14}{2\cdot 8}= \frac{7}{8}.</math><br />
<br />
And since <math>AX=\frac{1}{2}(12+AM-BM)</math>, <math>CY = \frac{1}{2}(13+CM-BM)</math>, <br />
<br />
<cmath> \frac{12+AM-BM}{13+CM-BM} = \frac{7}{8}</cmath><br />
<br />
<cmath> 96+8AM-8BM = 91 +7CM-7BM</cmath><br />
<br />
<cmath>BM= 5 + 8AM-7CM = 5 + 15AM - 7(CM+AM) = 5+15(AM-7) \dots\dots (1)</cmath><br />
<br />
Now,<br />
<br />
<math>\frac{AM}{CM} = \frac{[ABM]}{[CBM]} = \frac{\frac{1}{2}(12+AM+BM)r}{\frac{1}{2}(13+CM+BM)r}=\frac{12+AM+BM}{13+CM+BM}= \frac{12+BM}{13+BM} = \frac{17+15(AM-7)}{18+15(AM-7)}</math><br />
<br />
<cmath>\frac{AM}{15} = \frac{17+15(AM-7)}{35+30(AM-7)} = \frac{15AM-88}{30AM-175}</cmath><br />
<cmath>6AM^2 - 35AM = 45AM-264</cmath><br />
<cmath>3AM^2 -40AM+132=0</cmath><br />
<cmath>(3AM-22)(AM-6)=0</cmath><br />
<br />
So <math>AM=22/3</math> or <math>6</math>. But from (1) we know that <math>5+15(AM-7)>0</math>, or <math>AM>7-1/3>6</math>, so <math>AM=22/3</math>, <math>CM=15-22/3=23/3</math>, <math>AM/CM=22/23</math>.<br />
<br />
=== Solution 5 ===<br />
<br />
Let the common inradius equal r, <math>BM = x</math>, <math>AM = y</math>, <math>MC = z</math><br />
<br />
From the prespective of <math>\triangle{ABM}</math> and <math>\triangle{BMC}</math> we get:<br />
<br />
<math>S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})</math> <math>S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})</math><br />
<br />
Add two triangles up, we get <math>\triangle{ABC}</math> :<br />
<br />
<math>S_{ABC} = S_{ABM} + S_{BMC} = r \cdot \frac{25+2x+y+z}{2}</math><br />
<br />
Since <math>y + z = 15</math>, we get:<br />
<br />
<math>r = \frac{S_{ABC}}{20 + x}</math><br />
<br />
By drawing an altitude from <math>I_1</math> down to a point <math>H_1</math> and from <math>I_2</math> to <math>H_2</math>, we can get:<br />
<br />
<math>r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2} </math> and<br />
<br />
<math>r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C = r \cdot \frac{13+z-x}{2}</math><br />
<br />
Adding these up, we get:<br />
<br />
<math>r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x</math><br />
<br />
<math>r = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}</math><br />
<br />
Now, we have 2 values equal to r, we can set them equal to each other:<br />
<br />
<math>\frac{S_{ABC}}{20 + x} = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}</math> <br />
<br />
If we let R denote the incircle of ABC, note:<br />
<br />
AC = <math>(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15</math> and<br />
<br />
<math>S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R</math>. <br />
<br />
By cross multiplying the equation above, we get:<br />
<br />
<math>400 - x^2 = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot S_{ABC} = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R \cdot 20 = 15 \cdot 20 = 300</math><br />
<br />
We can find out x:<br />
<br />
<math>x = 10</math>.<br />
<br />
Now, we can find ratio of y and z:<br />
<br />
<math>\frac{AM}{CM} = \frac{y}{z} = \frac{S_{ABM}}{S_{BCM}} = \frac{r \cdot \frac{22+y}{2} }{r \cdot \frac{23+z}{2}} = \frac{22+y}{23+z} = \frac{22}{23}</math> <br />
<br />
The answer is <math>\boxed{045}</math>. <br />
<br />
-Alexlikemath<br />
<br />
=== Solution 6 (Similar to Solution 1 with easier computation) ===<br />
<br />
Let <math>CM=x, AM=rx, BM=d</math>. <math>x+rx=15\Rightarrow x=\frac{15}{1+r}</math>.<br />
<br />
Similar to Solution 1, we have <br />
<cmath><br />
r=\frac{[AMB]}{[CMB]}=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r}<br />
</cmath><br />
as well as<br />
<cmath><br />
12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2 (\text{via Stewart's Theorem})<br />
</cmath><br />
<cmath><br />
\frac{(12^2 + 13^2r) \cdot 15}{1+r} - \frac{15r\cdot 15^2}{(1+r)^2}=\frac{15(13r-12)^2}{(1-r)^2}<br />
</cmath><br />
<cmath><br />
\frac{169r^2+88r+144}{(1+r)^2}=\frac{(13r-12)^2}{(1-r)^2}<br />
</cmath><br />
<cmath><br />
(169r^2+88r+144)((r^2+1)-2r)=(169r^2-312r+144)((r^2+1)+2r)<br />
</cmath><br />
<cmath><br />
(r^2+1)(400r)=2r(338r^2-224r+288)<br />
</cmath><br />
<cmath><br />
100(r^2+1)=169r^2-112r+144 \Rightarrow 69r^2-112r+44=(23r-22)(3r-2)=0<br />
</cmath><br />
<br />
Since <math>d=\frac{13r-12}{1-r}>0</math>, we have <math>r=\frac{22}{23} \longrightarrow \boxed{045}</math>.<br />
<br />
~ asops<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=UQVI0Q2PWZw&feature=youtu.be&fbclid=IwAR338IdppnZVuze4rzT0gm8G2NB_uIsmj175WgD6sa43gg3EhFAGm5bAvV0<br />
<br />
== Sidenote ==<br />
In the problem, <math>BM=10</math> and the equal inradius of the two triangles happens to be <math> \frac {2\sqrt{14}}{3}</math>.<br />
<br />
== See Also ==<br />
<br />
<br />
{{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_15&diff=1639842010 AIME I Problems/Problem 152021-10-23T21:32:10Z<p>Asops: </p>
<hr />
<div>__TOC__<br />
== Problem ==<br />
In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the [[incircle]]s of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal [[inradius|radii]]. Then <math>\frac{AM}{CM} = \frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p + q</math>.<br />
<br />
== Solution ==<br />
<center><asy> /* from geogebra: see azjps, userscripts.org/scripts/show/72997 */<br />
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200);<br />
<br />
/* segments and figures */<br />
draw((0,0)--(15,0));<br />
draw((15,0)--(6.66667,9.97775));<br />
draw((6.66667,9.97775)--(0,0));<br />
draw((7.33333,0)--(6.66667,9.97775));<br />
draw(circle((4.66667,2.49444),2.49444));<br />
draw(circle((9.66667,2.49444),2.49444));<br />
draw((4.66667,0)--(4.66667,2.49444));<br />
draw((9.66667,2.49444)--(9.66667,0));<br />
<br />
/* points and labels */<br />
label("r",(10.19662,1.92704),SE);<br />
label("r",(5.02391,1.8773),SE);<br />
dot((0,0));<br />
label("$A$",(-1.04408,-0.60958),NE);<br />
dot((15,0));<br />
label("$C$",(15.41907,-0.46037),NE);<br />
dot((6.66667,9.97775));<br />
label("$B$",(6.66525,10.23322),NE);<br />
label("$15$",(6.01866,-1.15669),NE);<br />
label("$13$",(11.44006,5.50815),NE);<br />
label("$12$",(2.28834,5.75684),NE);<br />
dot((7.33333,0));<br />
label("$M$",(7.56053,-1.000),NE);<br />
label("$H_1$",(3.97942,-1.200),NE);<br />
label("$H_2$",(9.54741,-1.200),NE);<br />
dot((4.66667,2.49444));<br />
label("$I_1$",(3.97942,2.92179),NE);<br />
dot((9.66667,2.49444));<br />
label("$I_2$",(9.54741,2.92179),NE);<br />
clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle);<br />
</asy></center><br />
=== Solution 1 ===<br />
Let <math>AM = x</math>, then <math>CM = 15 - x</math>. Also let <math>BM = d</math> Clearly, <math>\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}</math>. We can also express each area by the rs formula. Then <math>\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}</math>. Equating and cross-multiplying yields <math>25x + 2dx = 15d + 180</math> or <math>d = \frac {25x - 180}{15 - 2x}.</math> Note that for <math>d</math> to be positive, we must have <math>7.2 < x < 7.5</math>.<br />
<br />
By [[Stewart's Theorem]], we have <math>12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)</math> or <math>432 = 3d^2 + 40x - 3x^2.</math> Brute forcing by plugging in our previous result for <math>d</math>, we have <math>432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.</math> Clearing the fraction and gathering like terms, we get <math>0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.</math><br />
<br />
''Aside: Since <math>x</math> must be rational in order for our answer to be in the desired form, we can use the [[Rational Root Theorem]] to reveal that <math>12x</math> is an integer. The only such <math>x</math> in the above-stated range is <math>\frac {22}3</math>.''<br />
<br />
Legitimately solving that quartic, note that <math>x = 0</math> and <math>x = 15</math> should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get <math>0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).</math> The only solution in the desired range is thus <math>\frac {22}3</math>. Then <math>CM = \frac {23}3</math>, and our desired ratio <math>\frac {AM}{CM} = \frac {22}{23}</math>, giving us an answer of <math>\boxed{045}</math>.<br />
<br />
=== Solution 2 ===<br />
Let <math>AM = 2x</math> and <math>BM = 2y</math> so <math>CM = 15 - 2x</math>. Let the [[incenter]]s of <math>\triangle ABM</math> and <math>\triangle BCM</math> be <math>I_1</math> and <math>I_2</math> respectively, and their equal inradii be <math>r</math>. From <math>r = \sqrt {(s - a)(s - b)(s - c)/s}</math>, we find that<br />
<br />
<cmath>\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\<br />
& = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}</cmath><br />
<br />
Let the incircle of <math>\triangle ABM</math> meet <math>AM</math> at <math>P</math> and the incircle of <math>\triangle BCM</math> meet <math>CM</math> at <math>Q</math>. Then note that <math>I_1 P Q I_2</math> is a rectangle. Also, <math>\angle I_1 M I_2</math> is right because <math>MI_1</math> and <math>MI_2</math> are the angle bisectors of <math>\angle AMB</math> and <math>\angle CMB</math> respectively and <math>\angle AMB + \angle CMB = 180^\circ</math>. By properties of [[tangent (geometry)|tangents]] to [[circle]]s <math>MP = (MA + MB - AB)/2 = x + y - 6</math> and <math>MQ = (MB + MC - BC)/2 = - x + y + 1</math>. Now notice that the altitude of <math>M</math> to <math>I_1 I_2</math> is of length <math>r</math>, so by similar triangles we find that <math>r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)</math> (3). Equating (3) with (1) and (2) separately yields<br />
<br />
<cmath>\begin{align*}<br />
2y^2 - 30 = 2xy + 5x - 7y \\<br />
2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}<br />
</cmath><br />
<br />
and adding these we have<br />
<br />
<cmath><br />
4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\<br />
\implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.<br />
</cmath><br />
<br />
=== Solution 3 ===<br />
Let the incircle of <math>ABM</math> hit <math>AM</math>, <math>AB</math>, <math>BM</math> at <math>X_{1},Y_{1},Z_{1}</math>, and let the incircle of <math>CBM</math> hit <math>MC</math>, <math>BC</math>, <math>BM</math> at <math>X_{2},Y_{2},Z_{2}</math>. Draw the incircle of <math>ABC</math>, and let it be tangent to <math>AC</math> at <math>X</math>. Observe that we have a homothety centered at A sending the incircle of <math>ABM</math> to that of <math>ABC</math>, and one centered at <math>C</math> taking the incircle of <math>BCM</math> to that of <math>ABC</math>. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is <math>AX_{1}/AX=CX_{2}/CX</math>.<br />
<br />
By standard computations, <math>AX=\dfrac{AB+AC-BC}{2}=7</math> and <math>CX=\dfrac{BC+AC-AB}{2}=8</math>. Now, let <math>AX_{1}=7x</math> and <math>CX_{2}=8x</math>. We will now go around and chase lengths. Observe that <math>BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x</math>. Then, <math>BZ_{1}=12-7x</math>. We also have <math>CY_{2}=CX_{2}=8x</math>, so <math>BY_{2}=13-8x</math> and <math>BZ_{2}=13-8x</math>.<br />
<br />
Observe now that <math>X_{1}M+MX_{2}=AC-15x=15(1-x)</math>. Also,<math>X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)</math>. Solving, we get <math>X_{1}M=8-8x</math> and <math>MX_{2}=7-7x</math> (as a side note, note that <math>AX_{1}+MX_{2}=X_{1}M+X_{2}C</math>, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).<br />
<br />
Now, we get <math>BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x</math>. To finish, we will compute area ratios. <math>\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}</math>. Also, since their inradii are equal, we get <math>\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}</math>. Equating and cross multiplying yields the quadratic <math>3x^{2}-8x+4=0</math>, so <math>x=2/3,2</math>. However, observe that <math>AX_{1}+CX_{2}=15x<15</math>, so we take <math>x=2/3</math>. Our ratio is therefore <math>\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}</math>, giving the answer <math>\boxed{045}</math>.<br />
<br />
Note: Once we have <math>MX_1=8-8x</math> and <math>MX_2=7-7x</math>, it's bit easier to use use the right triangle of <math>O_1MO_2</math> than chasing the area ratio. The inradius of <math>\triangle{ABC}</math> can be calculated to be <math>r=\sqrt{14}</math>, and the inradius of <math>ABM</math> and <math>ACM</math> are <math>r_1=r_2= xr</math>, so, <br />
<cmath> O_1O_2^2 = O_1M^2 + O_2M^2 = r_1^2+X_1M^2 + r_2^2 + X_2M^2</cmath><br />
or,<br />
<cmath> (15(1-x))^2 = 2(\sqrt{14}x)^2 + (7(1-x))^2 + (8(1-x))^2 </cmath><br />
<cmath>112(1-x)^2 = 28x^2</cmath><br />
<cmath>4(1-x)^2 = x^2</cmath><br />
We get <math>x=\frac{2}{3}</math> or <math>x=2</math>.<br />
<br />
=== Solution 4 ===<br />
Suppose the incircle of <math>ABM</math> touches <math>AM</math> at <math>X</math>, and the incircle of <math>CBM</math> touches <math>CM</math> at <math>Y</math>. Then<br />
<br />
<cmath>r = AX \tan(A/2) = CY \tan(C/2)</cmath><br />
<br />
We have <math>\cos A = \frac{12^2+15^2-13^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}</math>, <math>\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}{\sqrt{14}}</math><br />
<br />
<math>\cos C = \frac{13^2+15^2-12^2}{2\cdot 13\cdot 15} = \frac{250}{30\cdot 13} = \frac{25}{39}</math>, <math>\tan(C/2) = \sqrt{\frac{39-25}{39+25}}=\frac{\sqrt{14}}{8}</math>,<br />
<br />
Therefore <math>AX/CY = \tan(C/2)/\tan(A/2) = \frac{14}{2\cdot 8}= \frac{7}{8}.</math><br />
<br />
And since <math>AX=\frac{1}{2}(12+AM-BM)</math>, <math>CY = \frac{1}{2}(13+CM-BM)</math>, <br />
<br />
<cmath> \frac{12+AM-BM}{13+CM-BM} = \frac{7}{8}</cmath><br />
<br />
<cmath> 96+8AM-8BM = 91 +7CM-7BM</cmath><br />
<br />
<cmath>BM= 5 + 8AM-7CM = 5 + 15AM - 7(CM+AM) = 5+15(AM-7) \dots\dots (1)</cmath><br />
<br />
Now,<br />
<br />
<math>\frac{AM}{CM} = \frac{[ABM]}{[CBM]} = \frac{\frac{1}{2}(12+AM+BM)r}{\frac{1}{2}(13+CM+BM)r}=\frac{12+AM+BM}{13+CM+BM}= \frac{12+BM}{13+BM} = \frac{17+15(AM-7)}{18+15(AM-7)}</math><br />
<br />
<cmath>\frac{AM}{15} = \frac{17+15(AM-7)}{35+30(AM-7)} = \frac{15AM-88}{30AM-175}</cmath><br />
<cmath>6AM^2 - 35AM = 45AM-264</cmath><br />
<cmath>3AM^2 -40AM+132=0</cmath><br />
<cmath>(3AM-22)(AM-6)=0</cmath><br />
<br />
So <math>AM=22/3</math> or <math>6</math>. But from (1) we know that <math>5+15(AM-7)>0</math>, or <math>AM>7-1/3>6</math>, so <math>AM=22/3</math>, <math>CM=15-22/3=23/3</math>, <math>AM/CM=22/23</math>.<br />
<br />
=== Solution 5 ===<br />
<br />
Let the common inradius equal r, <math>BM = x</math>, <math>AM = y</math>, <math>MC = z</math><br />
<br />
From the prespective of <math>\triangle{ABM}</math> and <math>\triangle{BMC}</math> we get:<br />
<br />
<math>S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})</math> <math>S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})</math><br />
<br />
Add two triangles up, we get <math>\triangle{ABC}</math> :<br />
<br />
<math>S_{ABC} = S_{ABM} + S_{BMC} = r \cdot \frac{25+2x+y+z}{2}</math><br />
<br />
Since <math>y + z = 15</math>, we get:<br />
<br />
<math>r = \frac{S_{ABC}}{20 + x}</math><br />
<br />
By drawing an altitude from <math>I_1</math> down to a point <math>H_1</math> and from <math>I_2</math> to <math>H_2</math>, we can get:<br />
<br />
<math>r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2} </math> and<br />
<br />
<math>r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C = r \cdot \frac{13+z-x}{2}</math><br />
<br />
Adding these up, we get:<br />
<br />
<math>r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x</math><br />
<br />
<math>r = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}</math><br />
<br />
Now, we have 2 values equal to r, we can set them equal to each other:<br />
<br />
<math>\frac{S_{ABC}}{20 + x} = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}</math> <br />
<br />
If we let R denote the incircle of ABC, note:<br />
<br />
AC = <math>(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15</math> and<br />
<br />
<math>S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R</math>. <br />
<br />
By cross multiplying the equation above, we get:<br />
<br />
<math>400 - x^2 = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot S_{ABC} = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R \cdot 20 = 15 \cdot 20 = 300</math><br />
<br />
We can find out x:<br />
<br />
<math>x = 10</math>.<br />
<br />
Now, we can find ratio of y and z:<br />
<br />
<math>\frac{AM}{CM} = \frac{y}{z} = \frac{S_{ABM}}{S_{BCM}} = \frac{r \cdot \frac{22+y}{2} }{r \cdot \frac{23+z}{2}} = \frac{22+y}{23+z} = \frac{22}{23}</math> <br />
<br />
The answer is <math>\boxed{045}</math>. <br />
<br />
-Alexlikemath<br />
<br />
=== Solution 6 (Similar to Solution 1 with easier computation) ===<br />
<br />
Let <math>CM=x, AM=rx, BM=d</math>. <math>x+rx=15\Rightarrow x=\frac{15}{1+r}</math>.<br />
<br />
Similar to Solution 1, we have <br />
<cmath><br />
r=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r}<br />
</cmath><br />
as well as<br />
<cmath><br />
12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2<br />
</cmath><br />
<cmath><br />
\frac{(12^2 + 13^2r) \cdot 15}{1+r} - \frac{15r\cdot 15^2}{(1+r)^2}=\frac{15(13r-12)^2}{(1-r)^2}<br />
</cmath><br />
<cmath><br />
\frac{169r^2+88r+144}{(1+r)^2}=\frac{(13r-12)^2}{(1-r)^2}<br />
</cmath><br />
<cmath><br />
(169r^2+88r+144)((r^2+1)-2r)=(169r^2-312r+144)((r^2+1)+2r)<br />
</cmath><br />
<cmath><br />
(r^2+1)(400r)=2r(338r^2-224r+288)<br />
</cmath><br />
<cmath><br />
100(r^2+1)=169r^2-112r+144 \Rightarrow 69r^2-112r+44=(23r-22)(3r-2)=0<br />
</cmath><br />
<br />
Since <math>d=\frac{13r-12}{1-r}>0</math>, we have <math>r=\frac{22}{23} \longrightarrow \boxed{045}</math>.<br />
<br />
~ asops<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=UQVI0Q2PWZw&feature=youtu.be&fbclid=IwAR338IdppnZVuze4rzT0gm8G2NB_uIsmj175WgD6sa43gg3EhFAGm5bAvV0<br />
<br />
== Sidenote ==<br />
In the problem, <math>BM=10</math> and the equal inradius of the two triangles happens to be <math> \frac {2\sqrt{14}}{3}</math>.<br />
<br />
== See Also ==<br />
<br />
<br />
{{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_13&diff=1638292010 AIME I Problems/Problem 132021-10-21T03:05:50Z<p>Asops: </p>
<hr />
<div>__TOC__<br />
== Problem ==<br />
[[Rectangle]] <math>ABCD</math> and a [[semicircle]] with diameter <math>AB</math> are coplanar and have nonoverlapping interiors. Let <math>\mathcal{R}</math> denote the region enclosed by the semicircle and the rectangle. Line <math>\ell</math> meets the semicircle, segment <math>AB</math>, and segment <math>CD</math> at distinct points <math>N</math>, <math>U</math>, and <math>T</math>, respectively. Line <math>\ell</math> divides region <math>\mathcal{R}</math> into two regions with areas in the ratio <math>1: 2</math>. Suppose that <math>AU = 84</math>, <math>AN = 126</math>, and <math>UB = 168</math>. Then <math>DA</math> can be represented as <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>.<br />
<br />
== Solution ==<br />
===Diagram===<br />
<center><asy> /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */<br />
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500);<br />
pen zzttqq = rgb(0.6,0.2,0);<br />
pen xdxdff = rgb(0.4902,0.4902,1);<br />
<br />
/* segments and figures */<br />
draw((0,-154.31785)--(0,0));<br />
draw((0,0)--(252,0));<br />
draw((0,0)--(126,0),zzttqq);<br />
draw((126,0)--(63,109.1192),zzttqq);<br />
draw((63,109.1192)--(0,0),zzttqq);<br />
draw((-71.4052,(+9166.01287-109.1192*-71.4052)/21)--(504.60925,(+9166.01287-109.1192*504.60925)/21));<br />
draw((0,-154.31785)--(252,-154.31785));<br />
draw((252,-154.31785)--(252,0));<br />
draw((0,0)--(84,0));<br />
draw((84,0)--(252,0));<br />
draw((63,109.1192)--(63,0));<br />
draw((84,0)--(84,-154.31785));<br />
draw(arc((126,0),126,0,180));<br />
<br />
/* points and labels */<br />
dot((0,0));<br />
label("$A$",(-16.43287,-9.3374),NE/2);<br />
dot((252,0));<br />
label("$B$",(255.242,5.00321),NE/2);<br />
dot((0,-154.31785));<br />
label("$D$",(3.48464,-149.55669),NE/2);<br />
dot((252,-154.31785));<br />
label("$C$",(255.242,-149.55669),NE/2);<br />
dot((126,0));<br />
label("$O$",(129.36332,5.00321),NE/2);<br />
dot((63,109.1192));<br />
label("$N$",(44.91307,108.57427),NE/2);<br />
label("$126$",(28.18236,40.85473),NE/2);<br />
dot((84,0));<br />
label("$U$",(87.13819,5.00321),NE/2);<br />
dot((113.69848,-154.31785));<br />
label("$T$",(116.61611,-149.55669),NE/2);<br />
dot((63,0));<br />
label("$N'$",(66.42398,5.00321),NE/2);<br />
label("$84$",(41.72627,-12.5242),NE/2);<br />
label("$168$",(167.60494,-12.5242),NE/2);<br />
dot((84,-154.31785));<br />
label("$T'$",(87.13819,-149.55669),NE/2);<br />
dot((252,0));<br />
label("$I$",(255.242,5.00321),NE/2);<br />
clip((-71.4052,-225.24323)--(-71.4052,171.51361)--(504.60925,171.51361)--(504.60925,-225.24323)--cycle);<br />
</asy></center><br />
=== Solution 1 ===<br />
<br />
The center of the semicircle is also the midpoint of <math>AB</math>. Let this point be O. Let <math>h</math> be the length of <math>AD</math>.<br />
<br />
Rescale everything by 42, so <math>AU = 2, AN = 3, UB = 4</math>. Then <math>AB = 6</math> so <math>OA = OB = 3</math>.<br />
<br />
Since <math>ON</math> is a radius of the semicircle, <math>ON = 3</math>. Thus <math>OAN</math> is an equilateral triangle.<br />
<br />
Let <math>X</math>, <math>Y</math>, and <math>Z</math> be the areas of triangle <math>OUN</math>, sector <math>ONB</math>, and trapezoid <math>UBCT</math> respectively.<br />
<br />
<math>X = \frac {1}{2}(UO)(NO)\sin{O} = \frac {1}{2}(1)(3)\sin{60^\circ} = \frac {3}{4}\sqrt {3}</math><br />
<br />
<math>Y = \frac {1}{3}\pi(3)^2 = 3\pi</math><br />
<br />
To find <math>Z</math> we have to find the length of <math>TC</math>. Project <math>T</math> and <math>N</math> onto <math>AB</math> to get points <math>T'</math> and <math>N'</math>. Notice that <math>UNN'</math> and <math>TUT'</math> are similar. Thus:<br />
<br />
<math>\frac {TT'}{UT'} = \frac {UN'}{NN'} \implies \frac {TT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies TT' = \frac {\sqrt {3}}{9}h</math>.<br />
<br />
Then <math>TC = T'C - T'T = UB - TT' = 4 - \frac {\sqrt {3}}{9}h</math>. So:<br />
<br />
<math>Z = \frac {1}{2}(BU + TC)(CB) = \frac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2</math><br />
<br />
Let <math>L</math> be the area of the side of line <math>l</math> containing regions <math>X, Y, Z</math>. Then<br />
<br />
<math>L = X + Y + Z = \frac {3}{4}\sqrt {3} + 3\pi + 4h - \frac {\sqrt {3}}{18}h^2</math><br />
<br />
Obviously, the <math>L</math> is greater than the area on the other side of line <math>l</math>. This other area is equal to the total area minus <math>L</math>. Thus:<br />
<br />
<math>\frac {2}{1} = \frac {L}{6h + \frac {9}{2}{\pi} - L} \implies 12h + 9\pi = 3L</math>.<br />
<br />
Now just solve for <math>h</math>.<br />
<br />
<cmath>\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\<br />
0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\<br />
h^2 & = \frac {9}{4}(6) \\<br />
h & = \frac {3}{2}\sqrt {6} \end{align*}</cmath><br />
<br />
Don't forget to un-rescale at the end to get <math>AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}</math>. <br />
<br />
Finally, the answer is <math>63 + 6 = \boxed{069}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
Let <math>O</math> be the center of the semicircle. It follows that <math>AU + UO = AN = NO = 126</math>, so triangle <math>ANO</math> is [[equilateral]].<br />
<br />
Let <math>Y</math> be the foot of the altitude from <math>N</math>, such that <math>NY = 63\sqrt{3}</math> and <math>NU = 21</math>.<br />
<br />
Finally, denote <math>DT = a</math>, and <math>AD = x</math>. Extend <math>U</math> to point <math>Z</math> so that <math>Z</math> is on <math>CD</math> and <math>UZ</math> is perpendicular to <math>CD</math>. It then follows that <math>ZT = a-84</math>. Since <math>NYU</math> and <math>UZT</math> are [[similar]],<br />
<br />
<math>\frac {x}{a-84} = \frac {63\sqrt{3}}{21} = 3\sqrt{3}</math><br />
<br />
Given that line <math>NT</math> divides <math>R</math> into a ratio of <math>1:2</math>, we can also say that<br />
<br />
<math>(x)(\frac{84+a}{2}) + \frac {126^2\pi}{6} - (63)(21)(\sqrt{3}) = (\frac{1}{3})(252x + \frac{126^2\pi}{2})</math><br />
<br />
where the first term is the area of trapezoid <math>AUTD</math>, the second and third terms denote the areas of <math>\frac{1}{6}</math> a full circle, and the area of <math>NUO</math>, respectively, and the fourth term on the right side of the equation is equal to <math>R</math>. Cancelling out the <math>\frac{126^2\pi}{6}</math> on both sides, we obtain<br />
<br />
<math>(x)(\frac{84+a}{2}) - \frac{252x}{3} = (63)(21)(\sqrt{3})</math><br />
<br />
By adding and collecting like terms,<br />
<math>\frac{3ax - 252x}{6} = (63)(21)(\sqrt{3})</math><br />
<br />
<math>\frac{(3x)(a-84)}{6} = (63)(21)(\sqrt{3})</math>.<br />
<br />
Since <math>a - 84 = \frac{x}{3\sqrt{3}}</math>,<br />
<br />
<math>\frac {(3x)(\frac{x}{3\sqrt{3}})}{6} = (63)(21)(\sqrt{3})</math><br />
<br />
<math>\frac {x^2}{\sqrt{3}} = (63)(126)(\sqrt{3})</math><br />
<br />
<math>x^2 = (63)(126)(3) = (2)(3^5)(7^2)</math><br />
<br />
<math>x = AD = (7)(3^2)(\sqrt{6}) = 63\sqrt{6}</math>, so the answer is <math>\boxed{069}.</math><br />
<br />
<br />
=== Solution 3 ===<br />
<br />
Note that the total area of <math> \mathcal{R} </math> is <math>252DA + \frac {126^2 \pi}{2}</math> and thus one of the regions has area <math>84DA + \frac {126^2 \pi}{6}</math><br />
<br />
As in the above solutions we discover that <math>\angle AON = 60^\circ</math>, thus sector <math>ANO</math> of the semicircle has <math>\frac{1}{3}</math> of the semicircle's area.<br />
<br />
Similarly, dropping the <math>N'T'</math> perpendicular we observe that <math>[AN'T'D] = 84DA</math>, which is <math>\frac{1}{3}</math> of the total rectangle.<br />
<br />
Denoting the region to the left of <math>\overline {NT}</math> as <math>\alpha</math> and to the right as <math>\beta</math>, it becomes clear that if <math>[\triangle UT'T] = [\triangle NUO]</math> then the regions will have the desired ratio.<br />
<br />
Using the 30-60-90 triangle, the slope of <math>NT</math>, is <math>{-3\sqrt{3}}</math>, and thus <math>[\triangle UT'T] = \frac {DA^2}{6\sqrt{3}}</math>.<br />
<br />
<math>[NUO]</math> is most easily found by <math>\frac{absin(c)}{2}</math>: <math>[\triangle NUO] = \frac {126*42 * \frac {\sqrt{3}}{2}}{2}</math><br />
<br />
Equating, <math>\frac {126*42 * \frac {\sqrt{3}}{2}}{2} = \frac {DA^2}{6\sqrt{3}}</math><br />
<br />
Solving, <math> 63 * 21 * 3 * 6 = DA^2</math><br />
<br />
<math>DA = 63 \sqrt{6} \longrightarrow \boxed {069}</math><br />
<br />
=== Solution 4 (Coordinates) ===<br />
Like above solutions, note that <math>ANO</math> is equilateral with side length <math>126,</math> where <math>O</math> is the midpoint of <math>AB.</math> Then, if we let <math>DA=a</math> and set origin at <math>D=(0,0),</math> we get <math>N=(63,a+63\sqrt{3}), U=(84,a).</math> Line <math>NU</math> is then <math>y-a=\sqrt{27}(x-84),</math> so it intersects <math>CA,</math> the <math>x</math>-axis, at <math>x=(a/\sqrt{27}+84),</math> giving us point <math>T.</math> Now the area of region <math>R</math> is <math>252a+\pi(126)^2 / 2,</math> so one third of that is <math>84a+\pi(126)^2 / 6.</math><br />
<br />
The area of the smaller piece of <math>R</math> is <math>[AUTD] + [ANU] + [\text{lune} AN] = \frac{1}{2} \cdot a(84+\frac{a}{\sqrt{27}}+84)+\frac{1}{2} \cdot 84 \cdot 63 \sqrt{3}+ \frac{\pi (126)^2}{6}-\frac{1}{2} \cdot 126 \cdot 63 \sqrt{2}</math><br />
<math>=\frac{a^2}{2\sqrt{27}}+84a-21\cdot 63\sqrt{2} + \frac{\pi(126)^2}{6}.</math><br />
Setting this equal to <math>84a+\pi(126)^2 / 6</math> and canceling the <math>84a + \pi(126)^2</math> yields<br />
<math>\frac{a^2}{2\sqrt{27}}=21 \cdot 63 \sqrt{3},</math> so <math>a = 63 \sqrt{6}</math> and the anser is <math>\boxed{069}.</math><br />
~ rzlng<br />
<br />
=== Solution 5 (Minimal calculation) ===<br />
<br />
Once we establish that <math>\Delta ANO</math> is equilateral, we have <br />
<cmath>[{\rm Sector } BON] = 2[{\rm Sector } AON], [BCT'U]=2[ADT'U]</cmath><br />
<cmath>\Rightarrow [\overset{\large\frown}{NB} CT'UO]=2[\overset{\large\frown}{NA} DT'UO]</cmath><br />
On the other hand,<br />
<cmath>[\overset{\large\frown}{NB} CT]=2[\overset{\large\frown}{NA} DT]</cmath><br />
Therefore, <math>[UT'T]=[NUO]</math>. <br />
<br />
Now, <math>UO=42, NU=21 \Rightarrow [UT'T]=[NUO]=2[NN'U]</math>. Also <math>\Delta UT'T \sim \Delta NN'U</math>. Therefore,<br />
<cmath>DA=UT'=\sqrt{2} NN'=\sqrt{2} \left(\frac{\sqrt{3}}{2}\cdot 126\right)=63\sqrt{6}\longrightarrow \boxed {069}</cmath><br />
<br />
~asops<br />
<br />
== See Also ==<br />
*<url>viewtopic.php?t=338915 Discussion</url>, with a Geogebra diagram.<br />
<br />
{{AIME box|year=2010|num-b=12|num-a=14|n=I}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_13&diff=1638282010 AIME I Problems/Problem 132021-10-21T02:37:19Z<p>Asops: </p>
<hr />
<div>__TOC__<br />
== Problem ==<br />
[[Rectangle]] <math>ABCD</math> and a [[semicircle]] with diameter <math>AB</math> are coplanar and have nonoverlapping interiors. Let <math>\mathcal{R}</math> denote the region enclosed by the semicircle and the rectangle. Line <math>\ell</math> meets the semicircle, segment <math>AB</math>, and segment <math>CD</math> at distinct points <math>N</math>, <math>U</math>, and <math>T</math>, respectively. Line <math>\ell</math> divides region <math>\mathcal{R}</math> into two regions with areas in the ratio <math>1: 2</math>. Suppose that <math>AU = 84</math>, <math>AN = 126</math>, and <math>UB = 168</math>. Then <math>DA</math> can be represented as <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>.<br />
<br />
== Solution ==<br />
===Diagram===<br />
<center><asy> /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */<br />
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500);<br />
pen zzttqq = rgb(0.6,0.2,0);<br />
pen xdxdff = rgb(0.4902,0.4902,1);<br />
<br />
/* segments and figures */<br />
draw((0,-154.31785)--(0,0));<br />
draw((0,0)--(252,0));<br />
draw((0,0)--(126,0),zzttqq);<br />
draw((126,0)--(63,109.1192),zzttqq);<br />
draw((63,109.1192)--(0,0),zzttqq);<br />
draw((-71.4052,(+9166.01287-109.1192*-71.4052)/21)--(504.60925,(+9166.01287-109.1192*504.60925)/21));<br />
draw((0,-154.31785)--(252,-154.31785));<br />
draw((252,-154.31785)--(252,0));<br />
draw((0,0)--(84,0));<br />
draw((84,0)--(252,0));<br />
draw((63,109.1192)--(63,0));<br />
draw((84,0)--(84,-154.31785));<br />
draw(arc((126,0),126,0,180));<br />
<br />
/* points and labels */<br />
dot((0,0));<br />
label("$A$",(-16.43287,-9.3374),NE/2);<br />
dot((252,0));<br />
label("$B$",(255.242,5.00321),NE/2);<br />
dot((0,-154.31785));<br />
label("$D$",(3.48464,-149.55669),NE/2);<br />
dot((252,-154.31785));<br />
label("$C$",(255.242,-149.55669),NE/2);<br />
dot((126,0));<br />
label("$O$",(129.36332,5.00321),NE/2);<br />
dot((63,109.1192));<br />
label("$N$",(44.91307,108.57427),NE/2);<br />
label("$126$",(28.18236,40.85473),NE/2);<br />
dot((84,0));<br />
label("$U$",(87.13819,5.00321),NE/2);<br />
dot((113.69848,-154.31785));<br />
label("$T$",(116.61611,-149.55669),NE/2);<br />
dot((63,0));<br />
label("$N'$",(66.42398,5.00321),NE/2);<br />
label("$84$",(41.72627,-12.5242),NE/2);<br />
label("$168$",(167.60494,-12.5242),NE/2);<br />
dot((84,-154.31785));<br />
label("$T'$",(87.13819,-149.55669),NE/2);<br />
dot((252,0));<br />
label("$I$",(255.242,5.00321),NE/2);<br />
clip((-71.4052,-225.24323)--(-71.4052,171.51361)--(504.60925,171.51361)--(504.60925,-225.24323)--cycle);<br />
</asy></center><br />
=== Solution 1 ===<br />
<br />
The center of the semicircle is also the midpoint of <math>AB</math>. Let this point be O. Let <math>h</math> be the length of <math>AD</math>.<br />
<br />
Rescale everything by 42, so <math>AU = 2, AN = 3, UB = 4</math>. Then <math>AB = 6</math> so <math>OA = OB = 3</math>.<br />
<br />
Since <math>ON</math> is a radius of the semicircle, <math>ON = 3</math>. Thus <math>OAN</math> is an equilateral triangle.<br />
<br />
Let <math>X</math>, <math>Y</math>, and <math>Z</math> be the areas of triangle <math>OUN</math>, sector <math>ONB</math>, and trapezoid <math>UBCT</math> respectively.<br />
<br />
<math>X = \frac {1}{2}(UO)(NO)\sin{O} = \frac {1}{2}(1)(3)\sin{60^\circ} = \frac {3}{4}\sqrt {3}</math><br />
<br />
<math>Y = \frac {1}{3}\pi(3)^2 = 3\pi</math><br />
<br />
To find <math>Z</math> we have to find the length of <math>TC</math>. Project <math>T</math> and <math>N</math> onto <math>AB</math> to get points <math>T'</math> and <math>N'</math>. Notice that <math>UNN'</math> and <math>TUT'</math> are similar. Thus:<br />
<br />
<math>\frac {TT'}{UT'} = \frac {UN'}{NN'} \implies \frac {TT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies TT' = \frac {\sqrt {3}}{9}h</math>.<br />
<br />
Then <math>TC = T'C - T'T = UB - TT' = 4 - \frac {\sqrt {3}}{9}h</math>. So:<br />
<br />
<math>Z = \frac {1}{2}(BU + TC)(CB) = \frac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2</math><br />
<br />
Let <math>L</math> be the area of the side of line <math>l</math> containing regions <math>X, Y, Z</math>. Then<br />
<br />
<math>L = X + Y + Z = \frac {3}{4}\sqrt {3} + 3\pi + 4h - \frac {\sqrt {3}}{18}h^2</math><br />
<br />
Obviously, the <math>L</math> is greater than the area on the other side of line <math>l</math>. This other area is equal to the total area minus <math>L</math>. Thus:<br />
<br />
<math>\frac {2}{1} = \frac {L}{6h + \frac {9}{2}{\pi} - L} \implies 12h + 9\pi = 3L</math>.<br />
<br />
Now just solve for <math>h</math>.<br />
<br />
<cmath>\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\<br />
0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\<br />
h^2 & = \frac {9}{4}(6) \\<br />
h & = \frac {3}{2}\sqrt {6} \end{align*}</cmath><br />
<br />
Don't forget to un-rescale at the end to get <math>AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}</math>. <br />
<br />
Finally, the answer is <math>63 + 6 = \boxed{069}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
Let <math>O</math> be the center of the semicircle. It follows that <math>AU + UO = AN = NO = 126</math>, so triangle <math>ANO</math> is [[equilateral]].<br />
<br />
Let <math>Y</math> be the foot of the altitude from <math>N</math>, such that <math>NY = 63\sqrt{3}</math> and <math>NU = 21</math>.<br />
<br />
Finally, denote <math>DT = a</math>, and <math>AD = x</math>. Extend <math>U</math> to point <math>Z</math> so that <math>Z</math> is on <math>CD</math> and <math>UZ</math> is perpendicular to <math>CD</math>. It then follows that <math>ZT = a-84</math>. Since <math>NYU</math> and <math>UZT</math> are [[similar]],<br />
<br />
<math>\frac {x}{a-84} = \frac {63\sqrt{3}}{21} = 3\sqrt{3}</math><br />
<br />
Given that line <math>NT</math> divides <math>R</math> into a ratio of <math>1:2</math>, we can also say that<br />
<br />
<math>(x)(\frac{84+a}{2}) + \frac {126^2\pi}{6} - (63)(21)(\sqrt{3}) = (\frac{1}{3})(252x + \frac{126^2\pi}{2})</math><br />
<br />
where the first term is the area of trapezoid <math>AUTD</math>, the second and third terms denote the areas of <math>\frac{1}{6}</math> a full circle, and the area of <math>NUO</math>, respectively, and the fourth term on the right side of the equation is equal to <math>R</math>. Cancelling out the <math>\frac{126^2\pi}{6}</math> on both sides, we obtain<br />
<br />
<math>(x)(\frac{84+a}{2}) - \frac{252x}{3} = (63)(21)(\sqrt{3})</math><br />
<br />
By adding and collecting like terms,<br />
<math>\frac{3ax - 252x}{6} = (63)(21)(\sqrt{3})</math><br />
<br />
<math>\frac{(3x)(a-84)}{6} = (63)(21)(\sqrt{3})</math>.<br />
<br />
Since <math>a - 84 = \frac{x}{3\sqrt{3}}</math>,<br />
<br />
<math>\frac {(3x)(\frac{x}{3\sqrt{3}})}{6} = (63)(21)(\sqrt{3})</math><br />
<br />
<math>\frac {x^2}{\sqrt{3}} = (63)(126)(\sqrt{3})</math><br />
<br />
<math>x^2 = (63)(126)(3) = (2)(3^5)(7^2)</math><br />
<br />
<math>x = AD = (7)(3^2)(\sqrt{6}) = 63\sqrt{6}</math>, so the answer is <math>\boxed{069}.</math><br />
<br />
<br />
=== Solution 3 ===<br />
<br />
Note that the total area of <math> \mathcal{R} </math> is <math>252DA + \frac {126^2 \pi}{2}</math> and thus one of the regions has area <math>84DA + \frac {126^2 \pi}{6}</math><br />
<br />
As in the above solutions we discover that <math>\angle AON = 60^\circ</math>, thus sector <math>ANO</math> of the semicircle has <math>\frac{1}{3}</math> of the semicircle's area.<br />
<br />
Similarly, dropping the <math>N'T'</math> perpendicular we observe that <math>[AN'T'D] = 84DA</math>, which is <math>\frac{1}{3}</math> of the total rectangle.<br />
<br />
Denoting the region to the left of <math>\overline {NT}</math> as <math>\alpha</math> and to the right as <math>\beta</math>, it becomes clear that if <math>[\triangle UT'T] = [\triangle NUO]</math> then the regions will have the desired ratio.<br />
<br />
Using the 30-60-90 triangle, the slope of <math>NT</math>, is <math>{-3\sqrt{3}}</math>, and thus <math>[\triangle UT'T] = \frac {DA^2}{6\sqrt{3}}</math>.<br />
<br />
<math>[NUO]</math> is most easily found by <math>\frac{absin(c)}{2}</math>: <math>[\triangle NUO] = \frac {126*42 * \frac {\sqrt{3}}{2}}{2}</math><br />
<br />
Equating, <math>\frac {126*42 * \frac {\sqrt{3}}{2}}{2} = \frac {DA^2}{6\sqrt{3}}</math><br />
<br />
Solving, <math> 63 * 21 * 3 * 6 = DA^2</math><br />
<br />
<math>DA = 63 \sqrt{6} \longrightarrow \boxed {069}</math><br />
<br />
=== Solution 3b (Improvement over Solution 3) ===<br />
To find <math>DA</math> such that <math>[UT'T]=[NUO]</math>, we notice that <br />
<cmath>UO=2NU \Rightarrow [UT'T]=[NUO]=2[NN'U]</cmath> <br />
<br />
We also have <math>\Delta UT'T \sim \Delta NN'U</math>. Therefore,<br />
<cmath>DA=UT'=\sqrt{2} NN'=\sqrt{2} \left(\frac{\sqrt{3}}{2}\cdot 126\right)=63\sqrt{6}\longrightarrow \boxed {069}</cmath><br />
<br />
~asops<br />
=== Solution 4 (Coordinates) ===<br />
Like above solutions, note that <math>ANO</math> is equilateral with side length <math>126,</math> where <math>O</math> is the midpoint of <math>AB.</math> Then, if we let <math>DA=a</math> and set origin at <math>D=(0,0),</math> we get <math>N=(63,a+63\sqrt{3}), U=(84,a).</math> Line <math>NU</math> is then <math>y-a=\sqrt{27}(x-84),</math> so it intersects <math>CA,</math> the <math>x</math>-axis, at <math>x=(a/\sqrt{27}+84),</math> giving us point <math>T.</math> Now the area of region <math>R</math> is <math>252a+\pi(126)^2 / 2,</math> so one third of that is <math>84a+\pi(126)^2 / 6.</math><br />
<br />
The area of the smaller piece of <math>R</math> is <math>[AUTD] + [ANU] + [\text{lune} AN] = \frac{1}{2} \cdot a(84+\frac{a}{\sqrt{27}}+84)+\frac{1}{2} \cdot 84 \cdot 63 \sqrt{3}+ \frac{\pi (126)^2}{6}-\frac{1}{2} \cdot 126 \cdot 63 \sqrt{2}</math><br />
<math>=\frac{a^2}{2\sqrt{27}}+84a-21\cdot 63\sqrt{2} + \frac{\pi(126)^2}{6}.</math><br />
Setting this equal to <math>84a+\pi(126)^2 / 6</math> and canceling the <math>84a + \pi(126)^2</math> yields<br />
<math>\frac{a^2}{2\sqrt{27}}=21 \cdot 63 \sqrt{3},</math> so <math>a = 63 \sqrt{6}</math> and the anser is <math>\boxed{069}.</math><br />
~ rzlng<br />
<br />
== See Also ==<br />
*<url>viewtopic.php?t=338915 Discussion</url>, with a Geogebra diagram.<br />
<br />
{{AIME box|year=2010|num-b=12|num-a=14|n=I}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=1632602020 AMC 10A Problems/Problem 202021-10-07T18:55:09Z<p>Asops: </p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #18]] and [[2020 AMC 10A Problems|2020 AMC 10A #20]]}}<br />
<br />
== Problem ==<br />
Quadrilateral <math>ABCD</math> satisfies <math>\angle ABC = \angle ACD = 90^{\circ}, AC=20,</math> and <math>CD=30.</math> Diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at point <math>E,</math> and <math>AE=5.</math> What is the area of quadrilateral <math>ABCD?</math><br />
<br />
<math>\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370</math><br />
<br />
== Solution 1 (Just Drop An Altitude)==<br />
<br />
<asy><br />
size(15cm,0);<br />
import olympiad;<br />
draw((0,0)--(0,2)--(6,4)--(4,0)--cycle);<br />
label("A", (0,2), NW);<br />
label("B", (0,0), SW);<br />
label("C", (4,0), SE);<br />
label("D", (6,4), NE);<br />
label("E", (1.714,1.143), N);<br />
label("F", (1,1.5), N);<br />
draw((0,2)--(4,0), dashed);<br />
draw((0,0)--(6,4), dashed);<br />
draw((0,0)--(1,1.5), dashed);<br />
label("20", (0,2)--(4,0), SW);<br />
label("30", (4,0)--(6,4), SE);<br />
label("$x$", (1,1.5)--(1.714,1.143), NE);<br />
draw(rightanglemark((0,2),(0,0),(4,0)));<br />
draw(rightanglemark((0,2),(4,0),(6,4)));<br />
draw(rightanglemark((0,0),(1,1.5),(0,2)));<br />
</asy><br />
<br />
It's crucial to draw a good diagram for this one. Since <math>AC=20</math> and <math>CD=30</math>, we get <math>[ACD]=300</math>. Now we need to find <math>[ABC]</math> to get the area of the whole quadrilateral. Drop an altitude from <math>B</math> to <math>AC</math> and call the point of intersection <math>F</math>. Let <math>FE=x</math>. Since <math>AE=5</math>, then <math>AF=5-x</math>. <br />
<br />
By dropping this altitude, we can also see two similar triangles, <math>\triangle BFE \sim \triangle DCE</math>. Since <math>EC</math> is <math>20-5=15</math>, and <math>DC=30</math>, we get that <math>BF=2x</math>. <br />
<br />
Now, if we redraw another diagram just of <math>ABC</math>, we get that <math>(2x)^2=(5-x)(15+x)</math> because of the altitude geometric mean theorem which states that in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into. <br />
<br />
Expanding, simplifying, and dividing by the GCF, we get <math>x^2+2x-15=0</math>. This factors to <math>(x+5)(x-3)</math>. Since lengths cannot be negative, <math>x=3</math>. Since <math>x=3</math>, that means the altitude <math>BF=2\cdot3=6</math>, or <math>[ABC]=60</math>. Thus <math>[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}</math><br />
<br />
~ Solution by Ultraman<br />
<br />
~ Diagram by ciceronii<br />
<br />
~ Formatting by BakedPotato66<br />
<br />
==Solution 2 (Coordinates)==<br />
<asy><br />
size(10cm,0);<br />
draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));<br />
draw((-20,0)--(20,0));<br />
draw((0,-15)--(0,35));<br />
draw((10,30)--(-8,-6));<br />
draw(circle((0,0),10));<br />
label("E",(-4.05,-.25),S);<br />
label("D",(10,30),NE);<br />
label("C",(10,0),NE);<br />
label("B",(-8,-6),SW);<br />
label("A",(-10,0),NW);<br />
label("5",(-10,0)--(-5,0), NE);<br />
label("15",(-5,0)--(10,0), N);<br />
label("30",(10,0)--(10,30), E);<br />
dot((-5,0));<br />
dot((-10,0));<br />
dot((-8,-6));<br />
dot((10,0));<br />
dot((10,30));<br />
</asy><br />
Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,10)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,10)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}</math>.<br />
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==Solution 3 (Trigonometry)==<br />
Let <math>\angle C = \angle{ACB}</math> and <math>\angle{B} = \angle{CBE}.</math> Using Law of Sines on <math>\triangle{BCE}</math> we get <cmath>\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}</cmath> and LoS on <math>\triangle{ABE}</math> yields <cmath>\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.</cmath> Divide the two to get <math>\tan{B} = 3 \tan{C}.</math> Now, <cmath>\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}</cmath> and solve the quadratic, taking the positive solution (C is acute) to get <math>\tan{C} = \frac{1}{3}.</math> So if <math>AB = a,</math> then <math>BC = 3a</math> and <math>[ABC] = \frac{3a^2}{2}.</math> By Pythagorean Theorem, <math>10a^2 = 400 \iff \frac{3a^2}{2} = 60</math> and the answer is <math>300 + 60 \iff \boxed{\textbf{(D)}}.</math><br />
<br />
(This solution is incomplete, can someone complete it please-Lingjun) ok<br />
Latex edited by kc5170<br />
<br />
We could use the famous m-n rule in trigonometry in <math>\triangle ABC</math> with Point <math>E</math> <br />
[Unable to write it here.Could anybody write the expression]<br />
. We will find that <math>\overrightarrow{BD}</math> is an angle bisector of <math>\triangle ABC</math> (because we will get <math>\tan(x) = 1</math>). <br />
Therefore by converse of angle bisector theorem <math>AB:BC = 1:3</math>. By using Pythagorean theorem, we have values of <math>AB</math> and <math>AC</math>. <br />
Computing <math>AB \cdot AC = 120</math>. Adding the areas of <math>ABC</math> and <math>ACD</math>, hence the answer is <math>\boxed{\textbf{(D)}\:360}</math>.<br />
<br />
By: Math-Amaze<br />
Latex: Catoptrics.<br />
<br />
==Solution 4 (Answer Choices)==<br />
We know that the big triangle has area 300. Using the answer choices would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. We guess that the legs are equal to <math>\sqrt{20a}</math> and <math>\sqrt{20b}</math>, and because the hypotenuse is 20 we get <math>a+b=20</math>. Testing small numbers, we get that when <math>a=2</math> and <math>b=18</math>, <math>ab</math> is indeed a square. The area of the triangle is thus <math>60</math>, so the answer is <math>\boxed {\textbf{(D) }360}</math>.<br />
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~tigershark22<br />
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==Solution 5 (Law of Cosines)==<br />
<br />
<asy><br />
import olympiad;<br />
pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798);<br />
dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N);<br />
draw(A--B--C--D--A);<br />
draw(A--C, dotted); draw(B--D, dotted);<br />
</asy><br />
<br />
Denote <math>EB</math> as <math>x</math>. By the Law of Cosine:<br />
<cmath>AB^2 = 25 + x^2 - 10x\cos(\angle DEC)</cmath><br />
<cmath>BC^2 = 225 + x^2 + 30x\cos(\angle DEC)</cmath><br />
<br />
Adding these up yields:<br />
<cmath>400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0</cmath><br />
By the quadratic formula, <math>x = 3\sqrt5</math>.<br />
<br />
Observe:<br />
<cmath>[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60</cmath>.<br />
<br />
Thus the desired area is <math>\frac{1}{2}(30)(20) + 60 = \boxed{\textbf{(D) } 360}</math><br />
<br />
~qwertysri987<br />
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==Solution 6 (Basic Vectors / Coordinates)==<br />
<br />
Let <math>C = (0, 0)</math> and <math>D = (0, 30)</math>. Then <math>E = (-15, 0), A = (-20, 0),</math> and <math>B</math> lies on the line <math>y=2x+30.</math> So the coordinates of <math>B</math> are <cmath>(x, 2x+30).</cmath><br />
<br />
We can make this a vector problem.<br />
<math>\overrightarrow{\mathbf{B}} = \begin{pmatrix}<br />
x \\<br />
2x+30<br />
\end{pmatrix}.</math> We notice that point <math>B</math> forms a right angle, meaning vectors <math>\overrightarrow{\mathbf{BC}}</math> and <math>\overrightarrow{\mathbf{BA}}</math> are orthogonal, and their dot-product is <math>0</math>.<br />
<br />
We determine <math>\overrightarrow{\mathbf{BC}}</math> and <math>\overrightarrow{\mathbf{BA}}</math> to be <math>\begin{pmatrix}<br />
-x \\<br />
-2x-30<br />
\end{pmatrix}</math> and <math>\begin{pmatrix}<br />
-20-x \\<br />
-2x-30<br />
\end{pmatrix}</math> , respectively. (To get this, we use the fact that <math>\overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}</math> and similarly, <math>\overrightarrow{\mathbf{BA}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}.</math> )<br />
<br />
Equating the cross-product to <math>0</math> gets us the quadratic <math>-x(-20-x)+(-2x-30)(-2x-30)=0.</math> The solutions are <math>x=-18, -10.</math> Since <math>B</math> clearly has a more negative x-coordinate than <math>E</math>, we take <math>x=-18</math>. So <math>B = (-18, -6).</math><br />
<br />
From here, there are multiple ways to get the area of <math>\Delta{ABC}</math> to be <math>60</math>, and since the area of <math>\Delta{ACD}</math> is <math>300</math>, we get our final answer to be <cmath>60 + 300 = \boxed{\text{(D) } 360}.</cmath><br />
<br />
-PureSwag<br />
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== Solution 7 (Power of a Point/No quadratics)==<br />
<br />
<asy><br />
import olympiad;<br />
pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798),F=(285,94.5),G=(361.2,361.2);<br />
dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N);dot("$F$",F,N);dot("$G$",G,N);<br />
draw(A--B--C--D--A);<br />
draw(A--C, dotted); draw(B--D, dotted); draw(F--G, dotted);<br />
</asy><br />
<br />
Let <math>F</math> be the midpoint of <math>AC</math>, and draw <math>FG // CD</math> where <math>G</math> is on <math>BD</math>. We have <math>EF=5,FC=10</math>.<br />
<br />
<math>\Delta EFG \sim \Delta ECD \implies FG=10=FA=FC</math>. Therefore <math>ABCG</math> is a cyclic quadrilateral.<br />
<br />
Notice that <math>\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{3} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3\sqrt{5}</math> via Power of a Point.<br />
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The altitude from <math>B</math> to <math>AC</math> is then equal to <math>GF\cdot \frac{BE}{GE}=10\cdot \frac{3\sqrt 5}{5 \sqrt 5}=6</math>. <br />
<br />
Finally, the total area of <math>ABCD</math> is equal to <math>\frac 12 \cdot 20 \left(30+6 \right) =\boxed{\text{(D) } 360}.</math><br />
<br />
~asops<br />
<br />
==Video Solutions==<br />
===Video Solution 1===<br />
Education, The Study of Everything<br />
https://youtu.be/5lb8kk1qbaA<br />
<br />
<br />
===Video Solution 2===<br />
On The Spot STEM<br />
https://www.youtube.com/watch?v=hIdNde2Vln4<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=sHrjx968ZaM&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=2 ~ MathEx<br />
<br />
<br />
===Video Solution 4===<br />
The Beauty Of Math https://www.youtube.com/watch?v=RKlG6oZq9so&ab_channel=TheBeautyofMath<br />
<br />
===Video Solution 5===<br />
https://youtu.be/R220vbM_my8?t=658<br />
<br />
(amritvignesh0719062.0)<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_13&diff=1630872019 AMC 12B Problems/Problem 132021-10-03T13:15:15Z<p>Asops: </p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #17]] and [[2019 AMC 12B Problems|2019 AMC 12B #13]]}}<br />
==Problem==<br />
<br />
A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin <math>k</math> is <math>2^{-k}</math> for <math>k = 1,2,3....</math> What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?<br><br />
<math>\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}</math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin. Clearly, the probability of both landing in the same bin is <math>\sum_{k=1}^{\infty}{2^{-k} \cdot 2^{-k}} = \sum_{k=1}^{\infty}2^{-2k} = \frac{1}{3}</math> (by the geometric series sum formula). Therefore, since the other two probabilities have to both the same, they have to be <math>\frac{1-\frac{1}{3}}{2} = \boxed{\textbf{(C) } \frac{1}{3}}</math>.<br />
<br />
=== Solution 2 ===<br />
Suppose the green ball goes in bin <math>i</math>, for some <math>i \ge 1</math>. The probability of this occurring is <math>\frac{1}{2^i}</math>. Given that this occurs, the probability that the red ball goes in a higher-numbered bin is <math>\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}</math> (by the geometric series sum formula). Thus the probability that the green ball goes in bin <math>i</math>, and the red ball goes in a bin greater than <math>i</math>, is <math>\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}</math>. Summing from <math>i=1</math> to infinity, we get<br />
<br />
<cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath><br />
where we again used the geometric series sum formula. (Alternatively, if this sum equals <math>n</math>, then by writing out the terms and multiplying both sides by <math>4</math>, we see <math>4n = n+1</math>, which gives <math>n = \frac{1}{3}</math>.)<br />
<br />
=== Solution 3 ===<br />
For red ball in bin <math>k</math>, <math>\Pr(\text{Green Below Red})=\sum\limits_{i=1}^{k-1}2^{-i}</math> (GBR) and <math>\Pr(\text{Red in Bin k}=2^{-k}</math> (RB). <br />
<cmath>\Pr(\text{GBR}|\text{RB})=\sum\limits_{k=1}^{\infty}2^{-k}\sum\limits_{i=1}^{k-1}2^{-i}=\sum\limits_{k=1}^{\infty}2^{-k}\cdot\frac{1}{2}(\frac{1-(1/2)^{k-1}}{1-1/2})</cmath><br />
<cmath>\sum\limits_{k=1}^{\infty}\frac{1}{2^{-k}}-2\sum\limits_{k=1}^\infty\frac{1}{(2^2)^{-k}}\implies 1-2/3=\boxed{(\textbf{C}) \frac{1}{3}}</cmath><br />
<br />
=== Solution 4 ===<br />
The probability that the two balls will go into adjacent bins is <math>\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots = \frac{1}{6}</math> by the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of <math>2</math> from each other is <math>\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} + \cdots = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{12}</math> (again recognizing a geometric series). We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is <math>\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots</math>, which, by the geometric series sum formula, is <math>\boxed{\textbf{(C) } \frac{1}{3}}</math>.<br />
-fidgetboss_4000<br />
<br />
=== Solution 5 (quick, conceptual) ===<br />
Define a win as a ball appearing in higher numbered box.<br />
<br />
Start from the first box. <br />
<br />
There are <math>4</math> possible results in the box: Red, Green, Red and Green, or none, with an equal probability of <math>\frac{1}{4}</math> for each. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution, so if <math>p</math> is the probability that Red wins, we can write <math>p = \frac{1}{4} + \frac{1}{4}p</math>: there is a <math>\frac{1}{4}</math> probability that "Red" wins immediately, a <math>0</math> probability in the cases "Green" or "Red and Green", and in the "None" case (which occurs with <math>\frac{1}{4}</math> probability), we then start again, giving the same probability <math>p</math>. Hence, solving the equation, we get <math>p = \boxed{\textbf{(C) } \frac{1}{3}}</math>.<br />
<br />
=== Solution 6 ===<br />
Write out the infinite geometric series as <math>\frac{1}{2}</math>, <math>\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \cdots</math>. To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms (i.e term <math>1</math>, term <math>3</math>, etc.), and then sum the remaining terms - this is in fact precisely equivalent to the method of Solution 2. Writing this out as another infinite geometric sequence, we are left with <math>\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \cdots</math>. Summing, we get <cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath><br />
<br />
=== Solution 7 ===<br />
Fixing the green ball to fall into bin <math>1</math> gives a probability of <math>\frac{1}{2}\left(\frac{1}{2^2}+\frac{1}{2^3} +...\right)</math> for the red ball to fall into a higher bin. Fixing the green ball to fall into bin <math>2</math> gives a probability of <math>\frac{1}{2^2}\left(\frac{1}{2^3}+\frac{1}{2^4} +...\right)</math>. Factoring out the denominator of the first fraction in each probability gives <math>\frac{1}{2^3}\left(1+\frac{1}{2}+\frac{1}{2^2}+...\right)+\frac{1}{2^5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...\right)+...</math> so factoring out <math>\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\right)</math> results in the probability simplifying to <math>\left(\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+...\right)\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\right)</math> and using the formula <math>\frac{a}{1-r}</math> to find both series, we obtain <math>\left(\frac{\frac{1}{2^3}}{1-\frac{1}{4}}\right)\left(\frac{1}{1-\frac{1}{2}}\right)</math> which simplifies to <math>\boxed{\textbf{(C) } \frac{1}{3}}</math> -- OGBooger<br />
<br />
=== Solution 8 ===<br />
We can think of this problem as "what is the probability that the green ball's bin is less than the red ball's bin". We do not consider the case where the red ball goes into bin <math>1</math> because the green ball has no where to go then. The chance that the green one is below the red one if the red one goes to bin <math>2</math> is <math>\frac{1}{4}</math> chance that the red ball even goes in bin <math>2</math> and <math>\frac{1}{2}</math> chance that the green ball goes into any bin less than <math>2</math>. Similarly, if the red goes into bin <math>3</math>, there is a <math>\frac{1}{8} \cdot \left(\frac{1}{4} + \frac{1}{2}\right)</math> chance, or <math>\frac{3}{32}</math>, continuing like this, we get this sequence:<br />
<br />
<math>\frac{1}{8}, \frac{3}{32}, \frac{7}{128}, ...</math><br />
<br />
Let <math>S</math> equal the sum of our series:<br />
<br />
<math>S = \frac{1}{8} + \frac{3}{32} + \frac{7}{128} + ...</math>. That means we can write another equation:<br />
<math>\frac{S}{4} = \frac{1}{32} + \frac{3}{128} + ...</math><br />
<br />
Subtracting <math>\frac{S}{4}</math> from <math>S</math>, yields:<br />
<br />
<math>S - \frac{S}{4} = \frac{1}{8} + \frac{2}{32} + \frac{4}{128} + ...</math><br />
<br />
We see that the above series is a infinite geometric sequence with common ratio <math>\frac{1}{2}</math>. Therefore, the sum of that infinite series is <math>\frac{\frac{1}{8}}{\frac{1}{2}}</math>, which equals <math>\frac{1}{4}</math>. Our equation is now <math>S - \frac{S}{4} = \frac{1}{4}</math>. Solving for <math>S</math> shows that <math>S = \frac{1}{3}</math>.<br />
<br />
Our answer is <math>\boxed{\textbf{(C) }\frac{1}{3}}</math><br />
<br />
~ericshi1685<br />
<br />
=== Solution 9 (quick, symmetry) ===<br />
<br />
Denote <math>G,R</math> the bin numbers of the green and red balls, respectively. The common probability distribution of <math>G,B</math> can be constructed by keep splitting the remaining unassigned probability into two halves: one goes to the smallest number that has not been assigned, and the other goes to the rest. In other words, <math>\Pr(G=k) = \Pr (G>k), \forall k \in \mathbb{N}</math>. Then,<br />
<br />
<cmath><br />
\Pr(G>R)=\sum_{k=1}^\infty \Pr(G>k) \Pr(R=k) = \sum_{k=1}^\infty \Pr(G=k) \Pr(R=k) = \Pr (G=R)<br />
</cmath><br />
<br />
Similarly <math>\Pr(G<R)=\Pr(G=R)</math>. Therefore all three probabilities equal <math>\boxed{\textbf{(C) }\frac{1}{3}}</math>.<br />
<br />
~asops<br />
<br />
== Video Solutions ==<br />
=== Video Solution 1 ===<br />
For those who want a video solution: https://youtu.be/VP7ltu-XEq8<br />
<br />
=== Video Solution 2 ===<br />
https://youtu.be/_0YaCyxiMBo?t=353<br />
<br />
~IceMatrix<br />
<br />
===Video Solution 3===<br />
https://youtu.be/IRyWOZQMTV8?t=2484<br />
<br />
~ pi_is_3.14<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2019|ab=B|num-b=16|num-a=18}}<br />
{{AMC12 box|year=2019|ab=B|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=1630592019 AMC 10B Problems/Problem 252021-10-02T21:35:45Z<p>Asops: </p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br />
<br />
==Problem==<br />
<br />
How many sequences of <math>0</math>s and <math>1</math>s of length <math>19</math> are there that begin with a <math>0</math>, end with a <math>0</math>, contain no two consecutive <math>0</math>s, and contain no three consecutive <math>1</math>s?<br />
<br />
<math>\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75</math><br />
<br />
==Solution 1 (Recursion)==<br />
We can deduce, from the given restrictions, that any valid sequence of length <math>n</math> will start with a <math>0</math> followed by either <math>10</math> or <math>110</math>.<br />
Let <math>f(n)</math> be the number of valid (meaning: the sequence contains 0s and 1s, starts and ends with both 0, and there are no two consecutive 0s and no three consecutive 1s) sequences of length <math>n</math>. <br />
<br />
Then we can define a recursive function <math>f(n) = f(n-3) + f(n-2)</math>, with <math>n \ge 3</math> (because otherwise, the sequence would contain only 0s and this is not allowed due to the give conditions). <br />
<br />
We derived the recursive function, since for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> to return to the starting position, 0, and the resulting sequence will still satisfy the given conditions.<br />
<br />
It is easy to find <math>f(3) = 1</math> since the only possible valid sequence is <math>010</math>. <math>f(4)=1</math> since the only possible valid sequence is <math>0110</math>. <math>f(5)=1</math> since the only possible valid sequence is <math>01010</math>. <br />
<br />
The recursive sequence is then as follows:<br />
<br />
<cmath>f(3)=1</cmath><br />
<cmath>f(4)=1</cmath><br />
<cmath>f(5) = 1</cmath><br />
<cmath>f(6) = 1 + 1 = 2</cmath> <br />
<cmath>f(7) = 1 + 1 = 2</cmath><br />
<cmath>f(8) = 1 + 2 = 3</cmath> <br />
<cmath>f(9) = 2 + 2 = 4</cmath><br />
<cmath>f(10) = 2 + 3 = 5</cmath><br />
<cmath>f(11) = 3 + 4 = 7</cmath> <br />
<cmath>f(12) = 4 + 5 = 9</cmath> <br />
<cmath>f(13) = 5 + 7 = 12</cmath><br />
<cmath>f(14) = 7 + 9 = 16</cmath> <br />
<cmath>f(15) = 9 + 12 = 21</cmath><br />
<cmath>f(16) = 12 + 16 = 28</cmath><br />
<cmath>f(17) = 16 + 21 = 37</cmath> <br />
<cmath>f(18) = 21 + 28 = 49</cmath><br />
<cmath>f(19) = 28 + 37 = 65</cmath><br />
<br />
So, our answer is <math>\boxed{\text{\bf{(C)} } 65}</math>.<br />
<br />
<br />
'''Contributors:'''<br />
<br />
~Original Author<br />
<br />
~solasky<br />
<br />
~BakedPotato66<br />
<br />
==Solution 2 (casework)==<br />
After any particular <math>0</math>, the next <math>0</math> in the sequence must appear exactly <math>2</math> or <math>3</math> positions down the line. In this case, we start at position <math>1</math> and end at position <math>19</math>, i.e. we move a total of <math>18</math> positions down the line. Therefore, we must add a series of <math>2</math>s and <math>3</math>s to get <math>18</math>. There are a number of ways to do this:<br />
<br />
'''Case 1''': nine <math>2</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.<br />
<br />
'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose4} = 35</math> ways to arrange them.<br />
<br />
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.<br />
<br />
==Solution 3 (casework and blocks)==<br />
We can simplify the original problem into a problem where there are <math>2^{17}</math> binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of <math>0</math>s, <math>1</math>s, and <math>11</math>s. Now, we use casework: <br />
<br />
'''Case 1''': Alternating 1s and 0s. There is simply 1 way to do this: <math>0101010101010101010</math>. <br />
Now, we note that there cannot be only one block of <math>11</math> in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of <math>11</math>s this cannot be satisfied. This is true for all odd numbers of <math>11</math> blocks. <br />
<br />
'''Case 2''': There are 2 <math>11</math> blocks. Using the zeroes in the sequence as dividers, we have a sample as <math>0110110101010101010</math>. We know there are 8 places for <math>11</math>s, which will be filled by <math>1</math>s if the <math>11</math>s don't fill them. This is <math>{8\choose2} = 28</math> ways. <br />
<br />
'''Case 3''': Four <math>11</math> blocks arranged. Using the same logic as Case 2, we have <math>{7\choose4} = 35</math> ways to arrange four <math>11</math> blocks. <br />
<br />
'''Case 4''': No single <math>1</math> blocks, only <math>11</math> blocks. There is simply one case for this, which is <math>0110110110110110110</math>. <br />
<br />
Adding these four cases, we have <math>1+28+35+1=\boxed{\textbf{(C) }65}</math> as our final answer. <br />
<br />
~Equinox8<br />
<br />
==Solution 4 (similar to #3)==<br />
Any valid sequence must start with a <math>0</math>. We can then think of constructing a sequence as adding groups of terms to this <math>0</math>, each ending in <math>0</math>. (This is always possible because every valid string ends in <math>0</math>.) For example, we can represent the string <math>01011010110110</math> as: <math>0-10-110-10-110-110</math>.<br />
To not have any consecutive 0s, we must have at least one <math>1</math> before the next <math>0</math>. However, we cannot have three or more <math>1</math>s before the next <math>0</math> because we cannot have three consecutive <math>1</math>s. Consequently, we can only have one or two <math>1</math>s. <br />
<br />
So we can have the groups: <math>10</math> and <math>110</math>.<br />
<br />
After the initial <math>0</math>, we have <math>18</math> digits left to fill in the string. Let the number of <math>10</math> blocks be <math>x</math>, and <math>110</math> be <math>y</math>. Then <math>x</math> and <math>y</math> must satisfy <math>2x+3y=18</math>. We recognize this as a Diophantine equation. Taking <math>\pmod{2}</math> yields <math>y=0 \pmod{2}</math>. Since <math>x</math> and <math>y</math> must both be nonnegative, we get the solutions <math>(9, 0)</math>, <math>(6, 2)</math>, <math>(3, 4)</math>, and <math>(0, 6)</math>. We now handle each of these cases separately.<br />
<br />
<math>(9, 0)</math>: Only one arrangement, namely all <math>10</math>s.<br />
<br />
<math>(6, 2)</math>: We have 6 groups of <math>11</math>, and <math>2</math> groups of <math>110</math>. This has <math>\binom{6+2}{2}=28</math> cases.<br />
<br />
<math>(3, 4)</math>: This means we have 3 groups of <math>10</math>, and 4 groups of <math>110</math>. This has <math>\binom{3+4}{3}=35</math> cases.<br />
<br />
<math>(0, 6)</math>: Only one arrangement, namely all <math>110</math>.<br />
<br />
Adding these, we have <math>1+28+35+1=65 \longrightarrow \boxed{(C)}</math>.<br />
~Math4Life2020<br />
<br />
~edited by alpha_2 for spelling and and typos (<math>001</math> instead of <math>110</math>)<br />
<br />
==Solution 5 (Constructive Counting)==<br />
Suppose the number of <math>0</math>s is <math>n</math>. We can construct the sequence in two steps:<br />
<br />
Step 1: put <math>n-1</math> of <math>1</math>s between the <math>0</math>s;<br />
<br />
Step 2: put the rest <math>19-n-(n-1)=20-2n</math> of <math>1</math>s in the <math>n-1</math> spots where there is a <math>1</math>. There are <math>\binom{n-1}{20-2n}</math> ways of doing this.<br />
<br />
Now we find the possible values of <math>n</math>:<br />
<br />
First of all <math>n+(n-1) \leq 19 \Rightarrow n\leq 10</math> (otherwise there will be two consecutive <math>0</math>s); <br />
<br />
And secondly <math>20-2n \leq n-1\Rightarrow n\geq 7</math> (otherwise there will be three consecutive <math>1</math>s). <br />
<br />
Therefore the answer is<br />
<cmath><br />
\sum_{n=7}^{10} \binom{n-1}{20-2n} = \binom{6}{6} + \binom{7}{4} + \binom{8}{2} + \binom{9}{0} = \boxed{\textbf{(C) }65}.<br />
</cmath><br />
<br />
~ asops<br />
<br />
==Video Solution==<br />
For those who want a video solution: https://youtu.be/VamT49PjmdI<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_18&diff=1623002016 AMC 10B Problems/Problem 182021-09-15T20:00:15Z<p>Asops: </p>
<hr />
<div>==Problem==<br />
<br />
In how many ways can <math>345</math> be written as the sum of an increasing sequence of two or more consecutive positive integers?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math><br />
<br />
==Solution 1==<br />
Factor <math>345=3\cdot 5\cdot 23</math>.<br />
<br />
Suppose we take an odd number <math>k</math> of consecutive integers, with the median as <math>m</math>. Then <math>mk=345</math> with <math>\tfrac12k<m</math>.<br />
Looking at the factors of <math>345</math>, the possible values of <math>k</math> are <math>3,5,15,23</math> with medians as <math>115,69,23,15</math> respectively.<br />
<br />
Suppose instead we take an even number <math>2k</math> of consecutive integers, with median being the average of <math>m</math> and <math>m+1</math>. Then <math>k(2m+1)=345</math> with <math>k\le m</math>.<br />
Looking again at the factors of <math>345</math>, the possible values of <math>k</math> are <math>1,3,5</math> with medians <math>(172,173),(57,58),(34,35)</math> respectively.<br />
<br />
Thus the answer is <math>\boxed{\textbf{(E) }7}</math>.<br />
<br />
==Solution 2==<br />
<br />
We need to find consecutive numbers (an arithmetic sequence that increases by <math>1</math>) that sums to <math>345</math>. This calls for the sum of an arithmetic sequence given that the first term is <math>k</math>, the last term is <math>g</math> and with <math>n</math> elements, which is: <math>\frac {n \cdot (k+g)}{2}</math>.<br />
<br />
We look for sequences of <math>n</math> consecutive numbers starting at <math>k</math> and ending at <math>k+n-1</math>. We can now substitute <math>g</math> with <math>k+n-1</math>. Now we substiute our new value of <math>g</math> into <math>\frac {n \cdot (k+g)}{2}</math> to get that the sum is <math>\frac {n \cdot (k+k+n-1)}{2} = 345</math>.<br />
<br />
This simplifies to <math>\frac {n \cdot (2k+n-1)}{2} = 345</math>. This gives a nice equation. We multiply out the 2 to get that <math>n \cdot (2k+n-1)=690</math>. This leaves us with 2 integers that multiply to <math>690</math> which leads us to think of factors of <math>690</math>. We know the factors of <math>690</math> are: <math>1,2,3,5,6,10,15,23,30,46,69,115,138,230,345,690</math>. So through inspection (checking), we see that only <math>2,3,5,6,10,15</math> and <math>23</math> work. This gives us the answer of <math>\boxed{\textbf{(E) }7}</math> ways.<br />
<br />
~~jk23541<br />
<br />
<br />
An alternate way to finish.<br />
<br />
Let <br />
<cmath><br />
\begin{align*}<br />
2k+n-1 &=\frac{690}{k} \\<br />
n &= k \\<br />
\end{align*}<br />
</cmath><br />
where <math>k</math> is a factor of <math>690.</math><br />
We find <math>2k = 1+\frac{690}{k}-k</math> so we need <math>\frac{690}{k} - k</math> to be positive and odd. Fortunately, regardless of the parity of <math>k</math> we see that <math>\frac{690}{k} - k</math> is odd. Furthermore, we need <math>\frac{690}{k} >k</math> which eliminates exact half of the factors. Now, since we need more than <math>1</math> integer to sum up we need <math>k \ge 2</math> which eliminates one more case. There were <math>16</math> cases to begin with, so our answer is <math>\frac{16}{2}-1 = \boxed{\textbf{(E) }7}</math> ways.<br />
<br />
==Solution 2.1==<br />
At the very end of Solution 2, where we find the factors of 690, instead of inspection, notice that all numbers will work until you get to <math>30</math>, and that is because <math>\frac{345}{30}=11.5</math>, which means <math>11</math> and <math>12</math> must be the middle 2 numbers; however, a sequence of length <math>30</math> with middle numbers <math>11</math> and <math>12</math> that consists only of integers would go into the negatives, so any number from 30 onwards wouldn't work, and since <math>1</math> is a trivial, non-counted solution, we get <math>\boxed{\textbf{(E) }7}</math> -ColtsFan10<br />
<br />
==Solution 3 (Fast And Clean)==<br />
The median of the sequence <math>m</math> is either an integer or a half integer. Let <math>m=\frac{i}{2}, i \in N</math>, then <math>P=i\cdot n=2\cdot 3 \cdot 5 \cdot 23</math>. <br />
<br />
On the other hand we have two constraints: <br />
<br />
1) <math>m \geq \frac{n+1}{2} \iff i>n</math> because the integers in the sequence are all positive, and <math>n>1</math>;<br />
<br />
2) If <math>n</math> is odd then <math>m</math> is an integer, <math>i</math> is even; if <math>n</math> is even then <math>m</math> is a half integer, <math>i</math> is odd. Therefore, <math>n</math> and <math>i</math> have opposite parity.<br />
<br />
Now <math>P</math> has <math>16</math> factors and it is not a perfect square. There are <math>8-1=7</math> choices for <math>1 < n < \sqrt{P}</math>. Also since <math>2|P, 4\nmid P</math>, we know <math>n</math> and <math>\frac{P}{n}</math> must have opposite parity. Therefore the answer is <math>\boxed{\textbf{(E) }7}</math>.<br />
<br />
~ asops<br />
==Video Solution 1==<br />
https://youtu.be/dwEm_PcmaYg<br />
<br />
~savannahsolver<br />
<br />
== Video Solution 2==<br />
https://youtu.be/ZhAZ1oPe5Ds?t=950<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=17|num-a=19}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_5&diff=1622891986 AIME Problems/Problem 52021-09-15T14:48:13Z<p>Asops: </p>
<hr />
<div>== Problem ==<br />
What is that largest [[positive integer]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>?<br />
<br />
== Video Solution ==<br />
https://youtu.be/zfChnbMGLVQ?t=1458<br />
<br />
~ pi_is_3.14<br />
<br />
https://youtu.be/TA_Ug1-qBCU<br />
<br />
~ momeme<br />
<br />
== Solution 1 ==<br />
If <math>n+10 \mid n^3+100</math>, <math>\gcd(n^3+100,n+10)=n+10</math>. Using the [[Euclidean algorithm]], we have <math>\gcd(n^3+100,n+10)= \gcd(-10n^2+100,n+10)</math> <math>= \gcd(100n+100,n+10)</math> <math>= \gcd(-900,n+10)</math>, so <math>n+10</math> must divide <math>900</math>. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides <math>900</math> is <math>\boxed{890}</math>; we can double-check manually and we find that indeed <math>900\mid 890^3+100</math>.<br />
<br />
== Solution 2 (Simple) ==<br />
Let <math>n+10=k</math>, then <math>n=k-10</math>. Then <math>n^3+100 = k^3-30k^2+300k-900</math> Therefore, <math>900</math> must be divisible by <math>k</math>, which is largest when <math>k=900</math> and <math>n=\boxed{890}</math><br />
<br />
== Solution 3 ==<br />
In a similar manner, we can apply synthetic division. We are looking for <math>\frac{n^3 + 100}{n + 10} = n^2 - 10n + 100 - \frac{900}{n + 10}</math>. Again, <math>n + 10</math> must be a factor of <math>900 \Longrightarrow n = \boxed{890}</math>.<br />
<br />
==Solution 4==<br />
The key to this problem is to realize that <math>n+10 \mid n^3 +1000</math> for all <math>n</math>. Since we are asked to find the maximum possible <math>n</math> such that <math>n+10 \mid n^3 +100</math>, we have: <math>n+10 \mid ((n^3 +1000) - (n^3 +100) \longrightarrow n+10 \mid 900</math>. This is because of the property that states that if <math>a \mid b</math> and <math>a \mid c</math>, then <math>a \mid b \pm c</math>. Since, the largest factor of 900 is itself we have: <math>n+10=900 \Longrightarrow \boxed{n = 890}</math><br />
<br />
~qwertysri987<br />
<br />
==Solution 5 (Easy Modular Arithmetic)==<br />
Notice that <math>n\equiv -10 \pmod{n+10}</math>. Therefore <br />
<cmath><br />
0 \equiv n^3+100\equiv(-10)^3+100=-900 \pmod{n+10} \Rightarrow n+10 | 900 \Rightarrow \max_{n+10 | n^3+100} {n} = \boxed{890}.<br />
</cmath><br />
<br />
~asops<br />
<br />
== See also ==<br />
{{AIME box|year=1986|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_25&diff=1622362020 AMC 10B Problems/Problem 252021-09-14T19:20:01Z<p>Asops: </p>
<hr />
<div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #25]] and [[2020 AMC 12B Problems|2020 AMC 12B #24]]}}<br />
<br />
==Problem==<br />
<br />
Let <math>D(n)</math> denote the number of ways of writing the positive integer <math>n</math> as a product<cmath>n = f_1\cdot f_2\cdots f_k,</cmath>where <math>k\ge1</math>, the <math>f_i</math> are integers strictly greater than <math>1</math>, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number <math>6</math> can be written as <math>6</math>, <math>2\cdot 3</math>, and <math>3\cdot2</math>, so <math>D(6) = 3</math>. What is <math>D(96)</math>?<br />
<br />
<math>\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184</math><br />
<br />
==Solution 1==<br />
<br />
Note that <math>96 = 2^5 \cdot 3</math>. Since there are at most six not necessarily distinct factors <math>>1</math> multiplying to <math>96</math>, we have six cases: <math>k=1, 2, ..., 6.</math> Now we look at each of the six cases.<br />
<br />
<br />
<math>k=1</math>: We see that there is <math>1</math> way, merely <math>96</math>.<br />
<br />
<math>k=2</math>: This way, we have the <math>3</math> in one slot and <math>2</math> in another, and symmetry. The four other <math>2</math>'s leave us with <math>5</math> ways and symmetry doubles us so we have <math>10</math>.<br />
<br />
<math>k=3</math>: We have <math>3, 2, 2</math> as our baseline. We need to multiply by <math>2</math> in <math>3</math> places, and see that we can split the remaining three powers of <math>2</math> in a manner that is <math>3-0-0</math>, <math>2-1-0</math> or <math>1-1-1</math>. A <math>3-0-0</math> split has <math>6 + 3 = 9</math> ways of happening (<math>24-2-2</math> and symmetry; <math>2-3-16</math> and symmetry), a <math>2-1-0</math> split has <math>6 \cdot 3 = 18</math> ways of happening (due to all being distinct) and a <math>1-1-1</math> split has <math>3</math> ways of happening (<math>6-4-4</math> and symmetry) so in this case we have <math>9+18+3=30</math> ways.<br />
<br />
<math>k=4</math>: We have <math>3, 2, 2, 2</math> as our baseline, and for the two other <math>2</math>'s, we have a <math>2-0-0-0</math> or <math>1-1-0-0</math> split. The former grants us <math>4+12=16</math> ways (<math>12-2-2-2</math> and symmetry and <math>3-8-2-2</math> and symmetry) and the latter grants us also <math>12+12=24</math> ways (<math>6-4-2-2</math> and symmetry and <math>3-4-4-2</math> and symmetry) for a total of <math>16+24=40</math> ways.<br />
<br />
<math>k=5</math>: We have <math>3, 2, 2, 2, 2</math> as our baseline and one place to put the last two: on another two or on the three. On the three gives us <math>5</math> ways due to symmetry and on another two gives us <math>5 \cdot 4 = 20</math> ways due to symmetry. Thus, we have <math>5+20=25</math> ways.<br />
<br />
<math>k=6</math>: We have <math>3, 2, 2, 2, 2, 2</math> and symmetry and no more twos to multiply, so by symmetry, we have <math>6</math> ways.<br />
<br />
<br />
Thus, adding, we have <math>1+10+30+40+25+6=\boxed{\textbf{(A) } 112}</math>.<br />
<br />
~kevinmathz<br />
<br />
==Solution 2==<br />
<br />
As before, note that <math>96=2^5\cdot3</math>, and we need to consider 6 different cases, one for each possible value of <math>k</math>, the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with <math>k</math> factors. First, the factorization needs to contain one factor that is itself a multiple of <math>3</math>, and there are <math>k</math> to choose from, and that leaves <math>k-1</math> slots left to fill, each of which must contain at least one factor of <math>2</math>. Once we have filled in a two to each of the remainder slots, we're left with <math>5-(k-1)=6-k</math> twos. <br />
<br />
Consider the remaining <math>6-k</math> factors of <math>2</math> left to assign to the <math>k</math> factors. Using stars and bars, the number of ways to do this is:<br />
<cmath>{{(6-k)+(k-1)}\choose{6-k}}={5\choose{6-k}}</cmath> This makes <math>k{5\choose{6-k}}</math> possibilities for each k.<br />
<br />
To obtain the total number of factorizations, add all possible values for k: <cmath>\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{\textbf{(A) } \text{112}}.</cmath><br />
<br />
==Solution 3==<br />
<br />
Begin by examining <math>f_1</math>. <math>f_1</math> can take on any value that is a factor of <math>96</math> except <math>1</math>. For each choice of <math>f_1</math>, the resulting <math>f_2...f_k</math> must have a product of <math>96/f_1</math>. This means the number of ways the rest <math>f_a</math>, <math>1<a<=k</math> can be written by the scheme stated in the problem for each <math>f_1</math> is equal to <math>D(96/f_1)</math>, since the product of <math>f_2 \cdot f_3... \cdot f_k=x</math> is counted as one valid product if and only if <math>f_1 \cdot x=96</math>, the product <math>x</math> has the properties that factors are greater than <math>1</math>, and differently ordered products are counted separately.<br />
<br />
For example, say the first factor is <math>2</math>. Then, the remaining numbers must multiply to <math>48</math>, so the number of ways the product can be written beginning with <math>2</math> is <math>D(48)</math>. To add up all the number of solutions for every possible starting factor, <math>D(96/f_1)</math> must be calculated and summed for all possible <math>f_1</math>, except <math>96</math> and <math>1</math>, since a single <math>1</math> is not counted according to the problem statement. The <math>96</math> however, is counted, but only results in <math>1</math> possibility, the first and only factor being <math>96</math>. This means <br />
<br />
<math>D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1</math>. <br />
<br />
Instead of calculating D for the larger factors first, reduce <math>D(48)</math>, <math>D(32)</math>, and <math>D(24)</math> into sums of <math>D(m)</math> where <math>m<=16</math> to ease calculation. Following the recursive definition <math>D(n)=(</math>sums of <math>D(c))+1</math> where c takes on every divisor of n except for 1 and itself, the sum simplifies to <br />
<br />
<math>D(96)=(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)</math><br />
+<math>(D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1.</math><br />
<br />
<math>D(24)=D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1</math>, so the sum further simplifies to <br />
<br />
<math>D(96)=3D(16)+4D(12)+5D(8)+4D(6)+5D(4)+4D(3)+5D(2)+5</math>, after combining terms. From quick casework,<br />
<br />
<math>D(16)=8, D(12)=8, D(8)=4, D(6)=3, D(4)=2, D(3)=1</math> and <math>D(2)=1</math>. Substituting these values into the expression above, <br />
<br />
<math>D(96)=3 \cdot 8+4 \cdot 8+5 \cdot 4+4 \cdot 3+5 \cdot 2+4 \cdot 1+5 \cdot 1+5=\boxed{\textbf{(A) } 112}</math>. <br />
<br />
~monmath a.k.a Fmirza<br />
<br />
==Solution 4==<br />
Note that <math>96 = 3 \cdot 2^5</math>, and that <math>D</math> of a perfect power of a prime is relatively easy to calculate. Also note that you can find <math>D(96)</math> from <math>D(32)</math> by simply totaling the number of ways there are to insert a <math>3</math> into a set of numbers that multiply to <math>32</math>.<br />
<br />
First, calculate <math>D(32)</math>. Since <math>32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2</math>, all you have to do was find the number of ways to divide the <math>2</math>'s into groups, such that each group has at least one <math>2</math>. By stars and bars, this results in <math>1</math> way with five terms, <math>4</math> ways with four terms, <math>6</math> ways with three terms, <math>4</math> ways with two terms, and <math>1</math> way with one term. (The total, <math>16</math>, is not needed for the remaining calculations.)<br />
<br />
Then, to get <math>D(96)</math>, in each possible <math>D(32)</math> sequence, insert a <math>3</math> somewhere in it, either by placing it somewhere next to the original numbers (in one of <math>n+1</math> ways, where <math>n</math> is the number of terms in the <math>D(32)</math> sequence), or by multiplying one of the numbers by <math>3</math> (in one of <math>n</math> ways). There are <math>2+1=3</math> ways to do this with one term, <math>3+2=5</math> with two, <math>7</math> with three, <math>9</math> with four, and <math>11</math> with five.<br />
<br />
The resulting number of possible sequences is <math>3 \cdot 1 + 5 \cdot 4 + 7 \cdot 6 + 9 \cdot 4 + 11 \cdot 1 = \boxed{\textbf{(A) }112}</math>. ~[[User:emerald_block|emerald_block]]<br />
<br />
==Solution 5 (Minimal Casework)==<br />
Consider the arrangement of the prime factors of 96 in a line <math>(2,2, 2, 2, 2, 3)</math>. An arrangement of factors can be created by placing "dividers" to group primes. For example, <math>(2, 2, |, 2, 2, 2, |, 3)</math> is equivalent to the arrangement <math>4 \cdot 8 \cdot 3</math>. Because there are <math>6</math> ways to order the prime factors, and <math>2^5</math> ways to place dividers, this gives us an initial <math>6 \cdot 2^5</math> ways to arrange divisors. <br />
<br />
However, through this method, we overcount cases where <math>3</math> is combined with another factor. For example, the arrangement <math>4 \cdot 6 \cdot 4</math> can be written as <math>(2, 2, |, 2, 3, |, 2, 2)</math> or <math>(2, 2, |, 3, 2, |, 2, 2)</math>. Precisely, we double count any case with <math>6</math> as a factor, triple count any case with <math>12</math>, quadruple count any case with <math>24</math>, etc. <br />
<br />
Now, consider all cases where <math>3</math> must be grouped with at least one <math>2</math>. This can be expressed in the same "line" format as <math>(2, 2, 2, 2, 6)</math>, where dividers can again be placed to group divisors. In this case, there are <math>5</math> ways to order divisors, and <math>2^4</math> ways to place dividers, so we have an <math>5 \cdot 2^4</math> possible sequences for this case. Notice that in this format, we double count cases where <math>12</math> is a factor, we triple count cases where <math>24</math> is a factor, etc. Precisely, for any case counted <math>n</math> times in the first step, it is counted <math>n - 1</math> times in this step. Thus, if we subtract, we count each case exactly once. <br />
<br />
So, we get:<br />
<br />
<math>6 \cdot 2^5 - 5 \cdot 2^4 = \boxed{\textbf{(A) }112}</math>. ~[[User:hdai1122|hdai1122]]<br />
<br />
==Solution 6 (Another Fast Way)==<br />
<br />
First we factor <math>32</math> into <math>m</math> numbers <math>g_1, \cdots, g_m</math> where <math>g_i>1,i=1,\ldots,m</math>. By applying stars and bars there are <math>\binom{5-1}{m-1}</math> ways. Then we can either insert <math>3</math> into each of the <math>m+1</math> spaces between (or beyond) <math>g_i</math>'s, or multiply it to one of the <math>g_i</math>'s, a total of <math>2m+1</math> ways. Hence the answer to the problem is<br />
<br />
<cmath><br />
\sum_{m=1}^5 (2m+1)\binom{5-1}{m-1} = \sum_{n=0}^4 (2n+3) \binom{4}{n} = 8\sum_{n=0}^4 \frac{n}{4} \binom{4}{n} + 3 \sum_{n=0}^4 \binom{4}{n} = 8 \sum_{n=0}^4 \binom{3}{n-1} + 3\sum_{n=0}^4 \binom{4}{n} = 8 \cdot 2^3 + 3\cdot 2^4 = \boxed{\textbf{(A) }112}.<br />
</cmath><br />
<br />
~ asops<br />
<br />
==Video Solutions==<br />
=== Video Solution 1 ===<br />
https://youtu.be/8WrdYLw9_ns?t=1444 - Quicker Way<br />
<br />
~ pi_is_3.14<br />
<br />
===Video Solution 2===<br />
https://www.youtube.com/watch?v=977F9lBb37E&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=11&t=0s ~ MathEx<br />
<br />
==See Also==<br />
{{AMC10 box|year=2020|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2020|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Asopshttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_22&diff=1476472021 AMC 10A Problems/Problem 222021-02-21T13:52:06Z<p>Asops: </p>
<hr />
<div>==Problem==<br />
Hiram's algebra notes are <math>50</math> pages long and are printed on <math>25</math> sheets of paper; the first sheet contains pages <math>1</math> and <math>2</math>, the second sheet contains pages <math>3</math> and <math>4</math>, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly <math>19</math>. How many sheets were borrowed?<br />
<br />
<math>\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20</math><br />
<br />
==Solution==<br />
Suppose the roommate took pages <math>a</math> through <math>b</math>, or equivalently, page numbers <math>2a-1</math> through <math>2b</math>. Because there are <math>(2b-2a+2)</math> numbers taken, <cmath>\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50*51}{2} \implies (2a+2b-39)(b-a+1)=\frac{50*13}{2}=25*13.</cmath> The first possible solution that comes to mind is if <math>2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12</math>, which indeed works, giving <math>b=22</math> and <math>a=10</math>. The answer is <math>22-10+1=\boxed{(\textbf{B})13}</math><br />
<br />
~Lcz<br />
<br />
==Solution 2 (Different Variable Choice, Similar Logic)==<br />
Suppose the smallest page number removed is <math>k,</math> and <math>n</math> pages are removed. It follows that the largest page number removed is <math>k+n-1.</math><br />
<br />
<b>Remarks:</b><br />
<br />
1. <math>n</math> pages are removed means that <math>\frac{n}{2}</math> sheets are removed, from which <math>n</math> must be even.<br />
<br />
2. <math>k</math> must be odd, as the smallest page number removed is on the right side (odd-numbered).<br />
<br />
3. <math>1+2+3+\cdots+50=\frac{51(50)}{2}=1275.</math><br />
<br />
4. The sum of the page numbers removed is <math>\frac{(2k+n-1)n}{2}.</math><br />
<br />
Together, we have <cmath>\begin{align*}<br />
\frac{1275-\frac{(2k+n-1)n}{2}}{50-n}&=19 \\<br />
1275-\frac{(2k+n-1)n}{2}&=19(50-n) \\<br />
2550-(2k+n-1)n&=38(50-n) \\<br />
2550-(2k+n-1)n&=1900-38n \\<br />
650&=(2k+n-39)n.<br />
\end{align*}</cmath><br />
The factors of <math>650</math> are <cmath>1,2,5,10,13,25,26,50,65,130,325,650.</cmath> Since <math>n</math> is even, we only have a few cases to consider:<br />
<br />
<cmath>\begin{array}{ c c c }<br />
\boldsymbol{n} & \boldsymbol{2k+n-39} & \boldsymbol{k} \\ <br />
\hline<br />
2 & 325 & 181 \\ <br />
10 & 65 & 47 \\<br />
26 & 25 & 19 \\<br />
50 & 13 & 1 \\<br />
130 & 5 & \text{negative} \\<br />
650 & 1 & \text{negative} \\<br />
\end{array}</cmath><br />
<br />
Since <math>1\leq k \leq 50,</math> only <math>k=47,19,1</math> are possible:<br />
<br />
If <math>k=47,</math> then the notebook will run out if we take <math>10</math> pages starting from page <math>47.</math><br />
<br />
If <math>k=1,</math> then the average page number of the remaining pages will be undefined, as there is no page remaining (after taking <math>50</math> pages starting from page <math>1</math>).<br />
<br />
So, the only possibility is <math>k=19,</math> from which <math>n=26</math> pages are taken out, which is <math>\frac n2=\boxed{\textbf{(B)} ~13}</math> sheets.<br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3== <br />
<br />
Let <math>n</math> be the number of sheets borrowed, with an average page number <math>k+25.5</math>. The remaining <math>25-n</math> sheets have an average page number of <math>19</math> which is less than <math>25.5</math>, the average page number of all <math>50</math> pages, therefore <math>k>0</math>. Since the borrowed sheets start with an odd page number and end with an even page number we have <math>k \in \mathbb N</math>. We notice that <math>n < 25</math> and <math>k \le (49+50)/2-25.5=24<25</math>.<br />
<br />
The weighted increase of average page number from <math>25.5</math> to <math>k+25.5</math> should be equal to the weighted decrease of average page number from <math>25.5</math> to <math>19</math>, where the weights are the page number in each group (borrowed vs. remained), therefore <br />
<br />
<cmath>2nk=2(25-n)(25.5-19)=13(25-n) \implies 13 | n \text{ or } 13 | k</cmath><br />
<br />
Since <math>n, k < 25</math> we have either <math>n=13</math> or <math>k=13</math>. If <math>n=13</math> then <math>k=6</math>. If <math>k=13</math> then <math>2n=25-n</math> which is impossible. Therefore the answer should be <math>n=\boxed{\textbf{(B)} ~13}</math><br />
<br />
~asops<br />
<br />
== Video Solution by OmegaLearn (Arithmetic Sequences and System of Equations) ==<br />
https://youtu.be/dWOLIdTxwa4<br />
<br />
~ pi_is_3.14<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Asops