id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
4,401 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_22 | 3 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | As said in previous solutions, start by factoring $72, 600,$ and $900$ . The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\] To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows su... | 15 |
4,402 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23 | 2 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | WLOG, let the largest of the three numbers drawn be $a>0$ . Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$ . The probability that their sum is greater than $a$ is $\boxed{12}$ | 12 |
4,403 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23 | 3 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | This problem is going to require some geometric probability, so let's dive right in.
Take three integers $x$ $y$ , and $z$ . Then for the triangle inequality to hold, the following $3$ inequalities must be true
\[x + y > z\] \[y + z > x\] \[x + z > y\]
Now, it would be really easy if these equations only had two variab... | 12 |
4,404 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23 | 4 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | Consider a stick of length $1$ . Cutting the stick at two random points gives a triangle from the three new segments. These two random points must be on opposite sides of the halfway mark. Thus, after the first cut is made, there is $\boxed{12}$ probability that the second cut is on the opposite side. | 12 |
4,405 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24 | 1 | There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf... | The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value.
Derive $f(x)$ so that the acceleration $f''(x)=0$ . Using the power rule, \begin{align*} f(x) &= x^3-ax^2+bx-a \\ f’(x) &= 3x^2-2ax+b \\ f’’(x) &= 6x-2a \end{align*} So $x=\frac{a}{3}$ for the inf... | 9 |
4,406 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24 | 2 | There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf... | Let the roots of the polynomial be $r, s, t$ . By Vieta's formulas we have $r+s+t=a$ $rs+st+rt=b$ , and $rst=a$ . Since both $a$ and $b$ are positive, it follows that all 3 roots $r, s, t$ are positive as well, and so we can apply AM-GM to get \[\tfrac 13 (r+s+t) \ge \sqrt[3]{rst} \quad \Rightarrow \quad a \ge 3\sqrt[3... | 9 |
4,407 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24 | 3 | There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf... | We see that with cubics, the number $3$ comes up a lot, and as $9=3\cdot3$ has the most relation to $3$ , we can assume $b=\boxed{9}$ | 9 |
4,408 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25 | 1 | Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silv... | Consider $f(2)$ . The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$ . Since $100\le (x+1)^2-x^2=2x+1$ , this first happens at $x\ge \lfloor 99/2\rfloor = 50$ . The perfect squa... | 64 |
4,409 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25 | 2 | Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silv... | We assume $n \geq 1$ for all claims.
We will let $g_k(x) = \left \lfloor \frac{x^2}{10^k}\right \rfloor$ . This is the result when the last k digits are truncated off $x^2$
Let $x_n$ = the smallest a, such that $g_{2n}(a) - g_{2n}(a-1) \geq 2$
We then have $\left \lfloor \frac{a^2}{10^{2n}} \right \rfloor - \left \lflo... | 64 |
4,410 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_1 | 1 | What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \frac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | By: Dragonfly
We find that $a^{-1}$ is the same as $2$ , since a number to the power of $-1$ is just the reciprocal of that number. We then get the equation to be
\[\frac{2\times2+\frac{2}{2}}{\frac{1}{2}}\]
We can then simplify the equation to get $\boxed{10}$ | 10 |
4,411 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_3 | 1 | Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | By: dragonfly
First of all, lets plug in all of the $x$ 's into the equation.
$\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\bigg|$ $-(-2016)$
Then we simplify to get
$\bigg|$ $|2016+2016|-2016$ $\bigg|$ $+2016$
which simplifies into
$\bigg|$ $2016$ $\bigg|$ $+2016$
and finally we get $\boxed{4032}$ | 32 |
4,412 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_3 | 2 | Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | Consider $x$ is negative.
We replace all instances of $x$ with $|x|$
$\bigg|$ $||x|+|x||-|x|$ $\bigg|$ $+|x|$
$=$ $\bigg|$ $|2x|-|x|$ $\bigg|$ $+|x|$
$=$ $|x|$ $+|x|$
$=|2x|$
$=4032 \implies \boxed{4032}$ | 32 |
4,413 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_4 | 1 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | By: dragonfly
We set up equations to find each angle. The larger angle will be represented as $x$ and the smaller angle will be represented as $y$ , in degrees. This implies that
$4x=5y$
and
$2\times(90-x)=90-y$
since the larger the original angle, the smaller the complement.
We then find that $x=75$ and $y=60$ , and t... | 135 |
4,414 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_4 | 2 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | We can visualize the problem like so: [asy] path b = brace((0,10),(90,10),5); draw(b); label("$90^\circ$",b,N); draw("$5x$",(0,0)--(75,0),N); draw((75,2.5)--(75,-2.5)); draw("$1y$",(75,0)--(90,0),N); draw("$4x$",(0,-10)--(60,-10),S); draw((60,-7.5)--(60,-12.5)); draw("$2y$",(60,-10)--(90,-10),S); draw((0,5)--(0,-15)... | 135 |
4,415 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_6 | 1 | All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$ , with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64$ . What is the length of $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$ | By: Albert471
Plotting points $B$ and $C$ on the graph shows that they are at $\left( -x,x^2\right)$ and $\left( x,x^2\right)$ , which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}*2x*x^2 = 64=x^3$ Making x=4, and the length of $BC$ is $2x$ , so the answer is $\boxed{8}$ | 8 |
4,416 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_7 | 1 | Josh writes the numbers $1,2,3,\dots,99,100$ . He marks out $1$ , skips the next number $(2)$ , marks out $3$ , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number $(2)$ , skips the next number $(4)$ , marks out ... | Following the pattern, you are crossing out...
Time 1: Every non-multiple of $2$
Time 2: Every non-multiple of $4$
Time 3: Every non-multiple of $8$
Following this pattern, you are left with every multiple of $64$ which is only $\boxed{64}$ | 64 |
4,417 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_9 | 1 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the sho... | By Albert471
To start, use algebra to determine the number of posts on each side. You have (the long sides count for $2$ because there are twice as many) $6x = 20 + 4$ (each corner is double counted so you must add $4$ ) Making the shorter end have $4$ , and the longer end have $8$ $((8-1)*4)*((4-1)*4) = 28*12 = 336$ .... | 336 |
4,418 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_10 | 1 | A quadrilateral has vertices $P(a,b)$ $Q(b,a)$ $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$ | Note that the slope of $PQ$ is $\frac{a-b}{b-a}=-1$ and the slope of $PS$ is $\frac{b+a}{a+b}=1$ . Hence, $PQ\perp PS$ and we can similarly prove that the other angles are right angles. This means that $PQRS$ is a rectangle. By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$ . Simplifying we get $(a-b)(a+b... | 4 |
4,419 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_10 | 2 | A quadrilateral has vertices $P(a,b)$ $Q(b,a)$ $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$ | Solution by e_power_pi_times_i
By the Shoelace Theorem, the area of the quadrilateral is $2a^2 - 2b^2$ , so $a^2 - b^2 = 8$ . Since $a$ and $b$ are integers, $a = 3$ and $b = 1$ , so $a + b = \boxed{4}$ | 4 |
4,420 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_11 | 1 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}... | Solution by e_power_pi_times_i
Revised by Kinglogic and RJ5303707
[asy] Label l; l.p=fontsize(8); xaxis(-1,8,Ticks(l, 1.0)); yaxis(-1,16,Ticks(l, 1.0)); real f(real x) { return x * pi; } D(graph(f,-1/pi,5.1)); D((5.1,-1)--(5.1,16)); D((-1,-0.1)--(8,-0.1)); for(int x = 0; x < 5.1; ++x) { for(int y = 0; y < 16; ++... | 50 |
4,421 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14 | 1 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | The sum of the geometric sequence is $\frac{a}{1 - r}$ where $a$ is the first term and $r$ is the common ratio. We know the second term, $ar,$ is equal to $1.$ Thus $ar = 1 \Rightarrow a = \frac{1}{r}.$ This means, \[S = \frac{a}{1 - r} = \frac{1/r}{1 - r} = \frac{1}{r(1 - r)}.\] In order to minimize $S,$ we maximize t... | 4 |
4,422 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14 | 2 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | A geometric sequence always looks like
\[a,ar,ar^2,ar^3,\dots\]
and they say that the second term $ar=1$ . You should know that the sum of an infinite geometric series (denoted by $S$ here) is $\frac{a}{1-r}$ . We now have a system of equations which allows us to find $S$ in one variable.
\begin{align*} ar&=1 \\ S&=\fr... | 4 |
4,423 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14 | 3 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | \[\textbf{Completing the Square and Quadratics}\] Let $r$ be the common ratio. If the second term is $1$ , the first must be $\frac{1}{r}$ . By the infinite geometric series formula, the sum must be \[S=\frac{\frac{1}{r}}{1-r}\] This equals $\frac{1}{r(1-r)}$ . To find the minimum value of S, we must find the maximum v... | 4 |
4,424 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15 | 1 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of th... | First assign each face the letters $a,b,c,d,e,f$ . The sum of the product of the faces is $abc+acd+ade+aeb+fbc+fcd+fde+feb$ . We can factor this into $(a+f)(b+c)(d+e)$ which is the product of the sum of each pair of opposite faces. In order to maximize $(a+f)(b+c)(d+e)$ we use the numbers $(7+2)(6+3)(5+4)$ or $\boxed{7... | 729 |
4,425 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15 | 2 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of th... | We'll proceed from the factoring process above.
By the AM-GM inequality,
\[\frac{a_1+a_2+a_3}{3}\geq\sqrt[3]{a_1a_2a_3}\]
Cubing both sides,
\[\left(\frac{a_1+a_2+a_3}{3}\right)^3\geq{a_1a_2a_3}\]
Let $a_1=(a+f)$ $a_2=(b+c)$ , and $a_3=(d+e)$ . Let's substitute in these values.
\[\left(\frac{a+b+c+d+e+f}{3}\right)^3\ge... | 729 |
4,426 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15 | 4 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of th... | It is obvious to put $5$ $6$ , and $7$ on the faces that share the same vertex. As $4$ is the next biggest number, the face with $4$ has to be next to the faces with $6$ and $7$ . As $4$ is the next biggest number, the face with $3$ has to be next to the faces with $5$ and $7$ . making the face with $2$ next to the fac... | 729 |
4,427 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | 1 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.
For the first case, we can cleverly choose the convenient form of our sequence to be \[a-n,\cdots, a-1, a, a+1, \cdots, a+n\]
because then our sum will just... | 7 |
4,428 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | 2 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | The sum from $a$ to $b$ where $a$ and $b$ are integers and $a>b$ is
$S=\dfrac{(a-b+1)(a+b)}{2}$
$345=\dfrac{(a-b+1)(a+b)}{2}$
$2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)$
Let $c=a-b+1$ and $d=a+b$
$2\cdot 3\cdot 5\cdot 23=c\cdot d$
If we factor $690$ into all of its factor groups $(\text{exg}~ (10,69) ~\text{or} ~(15,46))$ w... | 7 |
4,429 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | 3 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | The consecutive sums can be written as $345=kn+\sum_{i=1}^{k-1}{i}$ where $k$ is the number of terms in a sequence, and $n$ is the first term. Then, $\{k,n\}\in \mathbb{N}$ and $k\geq2$ . Evaluating the sum and rearranging yields $n=\frac{345}{k}-\frac{k-1}{2}$
The prime factorization of $345$ is $1\cdot3\cdot5\cdot23$... | 7 |
4,430 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | 5 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | By the sum of an arithmetic sequence... this ultimately comes to $n+n+1+n+2....+n+p=345=(2n+p)(p+1)=690=23\cdot3\cdot5\cdot2$
Quick testing (would take you roughly a minute)
We see that the first 7 values of $p$ that work are
$p=1,2,4,5,9,14,22$
We see that each one of them works.
Hence, the answer is $\boxed{7}$ | 7 |
4,431 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | 6 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | The sum of all integers from $x$ to $y$ inclusive is equal to
\[\frac{y(y+1)}{2}-\left(\frac{x(x+1)}{2}\right)+x.\]
In words, $x$ added to sum of the first $x$ positive inegers subtracted from the sum of the first positive $y$ integers.
Setting this equal to $345$ and multiplying by $2$ to clear fractions, we can see t... | 7 |
4,432 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_18 | 1 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2} \qquad\textbf{(B)}\ \pi+2 \qquad\textbf{(C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | Consider the case when $x \geq 0$ $y \geq 0$ \[x^2+y^2=x+y\] \[(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}\] Notice the circle intersect the axes at points $(0, 1)$ and $(1, 0)$ . Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of $\frac{\sqrt{2}}{2}$ and a tria... | 2 |
4,433 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_20 | 1 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 6... | We use complementary counting. First, because each team played $20$ other teams, there are $21$ teams total. All sets that do not have $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A$ have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other t... | 385 |
4,434 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_20 | 2 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 6... | As above, note that there are 21 teams, and call them A, B, C, ... T, U. WLOG, assume that A beat teams B-L and lost to teams M-U. We will count the number of sets satisfying the “cycle-win” condition—e.g. here, A beats a team in X which beats a team in Y which beats A. The first and third part of the condition are alr... | 385 |
4,435 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_20 | 3 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 6... | Note that there are $21$ teams total and $\binom{21}{3}=1330$ ways to pick ${A,B,C}.$ The possible arrangements are one team beats the other two or they each win/lose equally (we want the second case). Approximately $\frac{1}{4}$ of all the arrangements satisfy the second case, and $\frac{1330}{4}=332.5,$ which is by f... | 385 |
4,436 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_25 | 1 | The sequence $(a_n)$ is defined recursively by $a_0=1$ $a_1=\sqrt[19]{2}$ , and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$ . What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ ... | Let $b_i=19\text{log}_2a_i$ . Then $b_0=0, b_1=1,$ and $b_n=b_{n-1}+2b_{n-2}$ for all $n\geq 2$ . The characteristic polynomial of this linear recurrence is $x^2-x-2=0$ , which has roots $2$ and $-1$
Therefore, $b_n=k_12^{n}+k_2(-1)^n$ for constants to be determined $k_1, k_2$ . Using the fact that $b_0=0, b_1=1,$ we ... | 17 |
4,437 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_25 | 3 | The sequence $(a_n)$ is defined recursively by $a_0=1$ $a_1=\sqrt[19]{2}$ , and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$ . What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ ... | Like in Solution 2 , calculate the first few terms of the sequence, but also keep a running sum $c_n$ of the logarithms (not modulo $19$ here): \[0,1,2,5,10,21,42,\dots\] Notice that $c_n=2c_{n-1}+1$ for odd $n$ and $c_n=2c_{n-1}$ for even $n$ . Since $2$ is relatively prime to $19$ , we can ignore even $n$ and calcula... | 17 |
4,438 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_2 | 1 | Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
$\textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 67\qquad\textbf{(E)}\ 72$ | Letting $x$ be the third side, then by the triangle inequality, $20-15 < x < 20+15$ , or $5 < x < 35$ . Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so $\boxed{72}$ is our answer. | 72 |
4,439 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_3 | 1 | Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textb... | If the average of the first $14$ peoples' scores was $80$ , then the sum of all of their tests is $14 \cdot 80 = 1120$ . When Payton's score was added, the sum of all of the scores became $15 \cdot 81 = 1215$ . So, Payton's score must be $1215-1120 = \boxed{95}$ | 95 |
4,440 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_3 | 2 | Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textb... | The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first $14$ students each scored $80$ . If Payton also scored an $80$ , the average would still be $80$ . In order to increase the overall average to $81$ , we need to add one more point to all of... | 95 |
4,441 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_4 | 1 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | Let $a$ be the bigger number and $b$ be the smaller.
$a + b = 5(a - b)$
Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives
$\frac{a}{b} = \frac32$ , so the answer is $\boxed{32}$ | 32 |
4,442 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_4 | 2 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{32}$ | 32 |
4,443 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_6 | 1 | Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ $1$
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$ | This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.
The first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the n... | 4 |
4,444 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_7 | 1 | Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?
$\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is ... | Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \[r_2=\frac{11r_1}{10}\] \[\pi r_1^2h_1=\pi r_2^2h_2\] Substituting the first equation into the second and divid... | 21 |
4,445 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_10 | 1 | Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$ . What is $x$
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26$ | Use SFFT to get $(x+1)(y+1)=81$ . The terms $(x+1)$ and $(y+1)$ must be factors of $81$ , which include $1, 3, 9, 27, 81$ . Because $x > y$ $x+1$ is equal to $27$ or $81$ . But if $x+1=81$ , then $y=0$ and so $x=\boxed{26}$ | 26 |
4,446 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_11 | 1 | On a sheet of paper, Isabella draws a circle of radius $2$ , a circle of radius $3$ , and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly $k \ge 0$ lines. How many different values of $k$ are possible?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \... | Isabella can get $0$ lines if the circles are concentric, $1$ if internally tangent, $2$ if overlapping, $3$ if externally tangent, and $4$ if non-overlapping and not externally tangent. There are $\boxed{5}$ values of $k$ | 5 |
4,447 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_15 | 1 | What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$ | We can rewrite the fraction as $\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}$ . Since the last digit of the numerator is odd, a $5$ is added to the right if the numerator is divided by $2$ , and this will continuously happen because $5$ , itself, is odd. Indeed, this happens twenty-two times since we... | 26 |
4,448 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_15 | 2 | What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$ | Multiply the numerator and denominator of the fraction by $5^{22}$ (which is the same as multiplying by 1) to give $\frac{5^{22} \cdot 123456789}{10^{26}}$ . Now, instead of thinking about this as a fraction, think of it as the division calculation $(5^{22} \cdot 123456789) \div 10^{26}$ . The dividend is a huge number... | 26 |
4,449 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_15 | 3 | What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$ | The denominator is $10^4 \cdot 2^{22}$ . Each $10$ adds one digit to the right of the decimal, and each additional $2$ adds another digit. The answer is $4 + 22 = \boxed{26}$ | 26 |
4,450 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | 1 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.
The quadratic formula gives the roots of the quadratic equation: $x=\frac{a\pm\sqrt{a^2-8a}}{2}$
As long as the numerator is an even integer, the roots are both integers. But first of all, the... | 16 |
4,451 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | 2 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | By the quadratic formula, the roots $r$ can be represented by \[r=\frac{a\pm\sqrt{a^2-8a}}{2}\] For $r\in\mathbb{Z}$ $a\in\mathbb{Z}$ , since $\frac{\sqrt{a^2-8a}}{2}$ and $\frac{a}{2}$ will have different mantissas (mantissae?).
Now observe the discriminant $\sqrt{a^2-8a}=\sqrt{a(a-8)}$ and have two cases.
Positive $a... | 16 |
4,452 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | 3 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | Let $m$ and $n$ be the roots of $x^2-ax+2a$
By Vieta's Formulas, $n+m=a$ and $mn=2a$
Substituting gets us $n+m=\frac{mn}{2}$
$2n-mn+2m=0$
Using Simon's Favorite Factoring Trick:
$n(2-m)+2m=0$
$-n(2-m)-2m=0$
$-n(2-m)-2m+4=4$
$(2-n)(2-m)=4$
This means that the values for $(m,n)$ are $(0,0),(4,4),(3,6),(1,-2)$ giving us $... | 16 |
4,453 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | 4 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | The quadratic formula gives \[x = \frac{a \pm \sqrt{a(a-8)}}{2}\] . For $x$ to be an integer, it is necessary (and sufficient!) that $a(a-8)$ to be a perfect square. So we have $a(a-8) = b^2$ ; this is a quadratic in itself and the quadratic formula gives \[a = 4 \pm \sqrt{16 + b^2}\]
We want $16 + b^2$ to be a perfect... | 16 |
4,454 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | 5 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | First of all, we know that $a$ is the sum of the quadratic's two roots, by Vieta's formulas. Thus, $a$ must be an integer. Then, we notice that the discriminant $a^2-8a$ must be equal to a perfect square so that the roots are integers. Thus, $a(a-8)=b^2$ where $b$ is an integer.
We can complete the square and rearrange... | 16 |
4,455 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_19 | 1 | For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }6... | Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that \[x^2 + (y - 2)^2 = y^2.\] Simplifying yields $x^2 - 4y + 4 = 0$ , so $x^2 = 4(y - 1)$ . Thus, $y$ is one more than a perfect square.
The perimeter $p = 2 + x + 2y = 2y + 2\sqrt{y... | 31 |
4,456 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_19 | 2 | For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }6... | Let $BC = x$ and $CD = AD = z$ be positive integers. Drop a perpendicular from $A$ to $CD$ . Denote the intersection point of the perpendicular and $CD$ as $E$
$AE$ 's length is $x$ , as well.
Call $ED$ $y$ .
By the Pythagorean Theorem, $x^2 + y^2 = (y + 2)^2$ .
And so: $x^2 = 4y + 4$ , or $y = (x^2-4)/4$
Writing this ... | 31 |
4,457 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | 1 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf... | The area of $T$ is $\dfrac{1}{2} \cdot 8 \cdot 3 = 12$ and the perimeter is 18.
The area of $T'$ is $\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ and the perimeter is $2a + b$
Thus $2a + b = 18$ , so $2a = 18 - b$
Thus $12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ , so $48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b... | 3 |
4,458 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | 2 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf... | The area is $12$ , the semiperimeter is $9$ , and $a = 9 - \frac12b$ . Using Heron's formula, $\sqrt{9\left(\frac{b}{2}\right)\left(\frac{b}{2}\right)(9-b)} = 12$ . Squaring both sides and simplifying, we have $-b^3+9b^2-64=0$ . Since we know $b = 8$ is a solution, we divide by $b - 8$ to get the other solution. Thus, ... | 3 |
4,459 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | 3 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf... | Triangle $T$ , being isosceles, has an area of $\frac{1}{2}(8)\sqrt{5^2-4^2}=12$ and a perimeter of $5+5+8=18$ .
Triangle $T'$ similarly has an area of $\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12$ and $2a+b=18$
Now we apply our computational fortitude.
\[\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12\... | 3 |
4,460 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | 4 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf... | Triangle T has perimeter $5 + 5 + 8 = 18$ so $18 = 2a + b$
Using Heron's, we get $\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})\left(\frac{b}{2}\right)^2(\frac{2a-b}{2})}$
We know that $2a + b = 18$ from above so we plug that in, and we also know that then $2a - b = 18 - 2b$
$12 = \frac{3b}{2}\sqrt{9-b}$
$64 = 9b^2 - b^3... | 3 |
4,461 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | 5 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf... | For this new triangle, say its legs have length $d$ and the base length $2c$ . To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that $c+d=9$ and $c*\sqrt{d^2-c^2}=12$ . It's better to let a side be some variable so we avoid having to add non-square ... | 3 |
4,462 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | 6 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf... | Since triangles $T'$ and $T$ have the same area and the same perimeter, $2a+b=18$ and $9*(9-a)^2(9-b) = 9*4^2*1$ By trying each answer choice, it is clear that the answer is $\boxed{3}$ | 3 |
4,463 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_22 | 2 | For each positive integer $n$ , let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$ , with no more than three $A$ s in a row and no more than three $B$ s in a row. What is the remainder when $S(2015)$ is divided by $12$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}... | We can start off by finding patterns in $S(n)$ . When we calculate a few values we realize either from performing the calculation or because the calculation was performed in the exact same way that $S(n) = 2^n - 2((n_4)- (n_5) \dots (n_n))$ . Rearranging the expression we realize that the terms aside from $2^{2015}$ ar... | 8 |
4,464 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_23 | 1 | Let $S$ be a square of side length 1. Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\frac12$ is $\frac{a-b\pi}{c}$ , where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a+b+c$
$\textbf{(A)... | WLOG, let the first point be on the bottom side of the square. The points where the second point could exist are outside a circle of radius 0.5 centered on the first point. The parts of the square that lie in this circle are the distance from the point to the closest side of the square $n$ , the distance from the point... | 59 |
4,465 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_3 | 1 | Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?
$\textbf{(A)}\; 8 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 14 \qquad\textbf{(D)}\; 15 \qquad\textbf{(E)}\; 18$ | Let $a$ be the number written two times, and $b$ the number written three times. Then $2a + 3b = 100$ . Plugging in $a = 28$ doesn't yield an integer for $b$ , so it must be that $b = 28$ , and we get $2a + 84 = 100$ . Solving for $a$ , we obtain $a = \boxed{8}$ | 8 |
4,466 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_7 | 1 | A regular 15-gon has $L$ lines of symmetry, and the smallest positive angle for which it has rotational symmetry is $R$ degrees. What is $L+R$
$\textbf{(A)}\; 24 \qquad\textbf{(B)}\; 27 \qquad\textbf{(C)}\; 32 \qquad\textbf{(D)}\; 39 \qquad\textbf{(E)}\; 54$ | From consideration of a smaller regular polygon with an odd number of sides (e.g. a pentagon), we see that the lines of symmetry go through a vertex of the polygon and bisect the opposite side. Hence $L=15$ , the number of sides / vertices. The smallest angle for a rotational symmetry transforms one side into an adjace... | 39 |
4,467 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | 1 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | $(625^{\log_5 2015})^\frac{1}{4} = ((5^4)^{\log_5 2015})^\frac{1}{4} = (5^{4 \cdot \log_5 2015})^\frac{1}{4} = (5^{\log_5 2015 \cdot 4})^\frac{1}{4} = ((5^{\log_5 2015})^4)^\frac{1}{4} = (2015^4)^\frac{1}{4} = \boxed{2015}$ | 15 |
4,468 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | 2 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | We can rewrite $\log_5 2015$ as as $5^x = 2015$ . Thus, $625^{x \cdot \frac{1}{4}} = 5^x = \boxed{2015}.$ | 15 |
4,469 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | 3 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | $(625^{\log_5 2015})^{\frac{1}{4}} = (625^{\frac{1}{4}})^{\log_5 2015} = 5^{\log_5 2015} = \boxed{2015}$ | 15 |
4,470 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | 4 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | We note that the year number is just $2015$ , so just guess $\boxed{2015}$ | 15 |
4,471 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_10 | 1 | How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?
$\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7$ | Since we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths $(a,b,c)$ with $a<b<c$ . Furthermore, "positive area" tells us that $c < a + b$ and the perimeter constraints means $a+b+c < 15$
There are no triangles when $a = 1$ because then $c$ must be less than $b+1$ , ... | 5 |
4,472 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_13 | 1 | Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$ . What is $AC$
$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$ | $\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$ , hence $\angle ACB = \angle ADB = 40^\circ$ . By considering $\triangle ABC$ , it follows that $\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ$ . Hence $\triangle ABC$ is isosceles, and $AC = BC = \boxed{6}.$ | 6 |
4,473 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_18 | 1 | For every composite positive integer $n$ , define $r(n)$ to be the sum of the factors in the prime factorization of $n$ . For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$ , and $2 + 5 + 5 = 12$ . What is the range of the function $r$ $\{r(n): n \text{ is a composite positive integer... | This problem becomes simple once we recognize that the domain of the function is $\{4, 6, 8, 9, 10, 12, 14, 15, \dots\}$ . By evaluating $r(4)$ to be $4$ , we can see that $\textbf{(E)}$ is incorrect. Evaluating $r(6)$ to be $5$ , we see that both $\textbf{(B)}$ and $\textbf{(C)}$ are incorrect. Since our domain consis... | 3 |
4,474 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_18 | 2 | For every composite positive integer $n$ , define $r(n)$ to be the sum of the factors in the prime factorization of $n$ . For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$ , and $2 + 5 + 5 = 12$ . What is the range of the function $r$ $\{r(n): n \text{ is a composite positive integer... | Think backwards. The range is the same as the numbers $y$ that can be expressed as the sum of two or more prime positive integers.
The lowest number we can get is $y = 2+2 = 4$ . For any number greater than 4, we can get to it by adding some amount of 2's and then possibly a 3 if that number is odd. For example, 23 can... | 3 |
4,475 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_20 | 1 | For every positive integer $n$ , let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:
\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text\\ f(i-1,1) &... | Simply take some time to draw a table of values of $f(i,j)$ for the first few values of $i$
\[\begin{array}{|c || c | c | c | c | c |} \hline i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline 0 & 1 & 2 & 3 & 4 & 0\\ \hline 1 & 2 & 3 & 4 & 0 & 1\\ \hline 2 & 3 & 0 & 2 & 4 & 1\\ \hline 3 & 0 & 3 & 4 & 1 & ... | 1 |
4,476 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_20 | 2 | For every positive integer $n$ , let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:
\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text\\ f(i-1,1) &... | We are given that \[f(0,n) \equiv n+1\pmod{5} .\] Then, $f(1,n) = f(0, f(1, n-1)) \equiv f(1,n-1) + 1\pmod{5}$ . Thus $f(1,n) \equiv n+f(1,0)\pmod{5}$ . Since $f(1,0)=f(0,1)=2$ , we get \[f(1,n) \equiv n+2\pmod{5} .\] Then, $f(2,n) = f(1, f(2, n-1)) \equiv f(2,n-1) + 2\pmod{5}$ . Thus $f(2,n) \equiv 2n+f(2,0)\pmod{5}$ ... | 1 |
4,477 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 1 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessar... | We can translate this wordy problem into this simple equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil\]
We will proceed to solve this equation via casework.
Case 1: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}$
Our equation becomes $\frac{s}{2} - 19 = \frac{s}{5} + \fr... | 13 |
4,478 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 2 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessar... | It can easily be seen that the problem can be expressed by the equation: \[\left\lceil \frac{s}{2} \right\rceil - \left\lceil \frac{s}{5} \right\rceil = 19\]
However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:
\[\frac{s+a}{2} - \frac{s+b}{5} = 19\] Where $a \in \{0,... | 13 |
4,479 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 3 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessar... | As before, we write the equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil.\]
To get a ballpark estimate of where $s$ might lie, we remove the ceiling functions to find:
\[\frac{s}{2} - 19 = \frac{s}{5}.\]
This gives $\frac{3s}{10} = 19$ , and thus values for $s$ will be around... | 13 |
4,480 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 4 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessar... | Notice that the possible number of steps in the staircase is around 60 to 70. By testing all of the values between 60 and 70, we see that 63, 64 and 66 work. Adding those up gives 193, so the answer is $1+9+3=\boxed{13}.$ | 13 |
4,481 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 5 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessar... | We represent C's steps with $2c + a = n$ and D's steps with $5d + b = n$ , where $a \in \{1,2\}$ and $b \in \{1,2,3,4,5\}$ , where $n$ is the number of steps, $c$ is the number of jumps C takes bar the last one, and $d$ is the number of jumps D takes bar the last.
The reason for starting at 1 and ending at 5 instead of... | 13 |
4,482 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22 | 1 | Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\;... | Consider shifting every person over three seats left after each person has gotten up and sat back down again. Now, instead of each person being seated not in the same chair and not in an adjacent chair, each person will be seated either in the same chair or an adjacent chair. The problem now becomes the number of ways ... | 20 |
4,483 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22 | 2 | Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\;... | Label the people sitting at the table $A, B, C, D, E, F,$ and assume that they are initially seated in the order $ABCDEF$ . The possible new positions for $A, B, C, D, E,$ and $F$ are respectively (a dash indicates a non-allowed position):
\[\begin{tabular}{| c | c | c | c | c | c |} \hline - & - & A & A & A & - \\ \hl... | 20 |
4,484 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22 | 3 | Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\;... | We can represent each rearrangement as a permutation of the six elements $\{1,2,3,4,5,6\}$ in cycle notation. Note that any such permutation cannot have a 1-cycle, so the only possible types of permutations are 2,2,2-cycles, 4,2-cycles, 3,3-cycles, and 6-cycles. We deal with each case separately.
For 2,2,2-cycles, supp... | 20 |
4,485 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_23 | 1 | A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; ... | We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $a\le b, ac \le bc$ , from the first equation we get $abc \le 6bc$ . Thus $a\le 6$ . From the second equation we see that $a > 2$ . Thus $a\in \{3, 4, 5, 6\}$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions. | 10 |
4,486 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_23 | 2 | A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; ... | The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields
\[2(ab+bc+ca)=abc.\]
Divide both sides by $2abc$ to obtain \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\]
First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we ... | 10 |
4,487 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | 1 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and... | First, note that $PQ$ lies on the radical axis of any of the pairs of circles. Suppose that $O_1$ and $O_2$ are the centers of two circles $C_1$ and $C_2$ that intersect exactly at $P$ and $Q$ , with $O_1$ and $O_2$ lying on the same side of $PQ$ , and $O_1 O_2=39$ . Let $x=O_1 R$ $y=O_2 R$ , and suppose that the radiu... | 192 |
4,488 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | 2 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and... | Note that if a circle passes through a pair of points, the center of the circle is on the perpendicular bisector of the line segment between the pair of points. This means that $A$ $B$ $C$ , and $D$ are all on the perpendicular bisector of $PQ$ . Let us say the distance from $A$ to the line $PQ$ is some $a$ . Therefore... | 192 |
4,489 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | 3 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and... | Let's start by drawing $PQ$ . Because all circles contain $P$ and $Q$ , all the centers lie on the perpendicular bisector of $PQ$ , and point $R$ is on this bisector.
For all the circle radii to be different (there can't be two congruent circles), two centers are on the same side of $PQ$ , and two are on the opposite s... | 192 |
4,490 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | 4 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and... | Since the radical axis $PQ$ is perpendicular to the line connecting the center of the circles, we have that $A$ $B$ $C$ $D$ , and $R$ are collinear. WLOG, assume that $A$ and $B$ are on the same side of $R$ and let $AR=y$ and let $BP=x$ so that $AP=\frac{5}{8}x$
Then, using the Pythagorean Theorem on right triangles $P... | 192 |
4,491 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | 5 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and... | Note that the four circles are coaxial, meaning $A,B,C,D,R$ are all collinear. Let $AR=x$ . By Pythagorean Theorem , the radius of circle $A$ squared would be $r_A^2 = x^2+24^2$ and the radius of circle $B$ squared would be $r_B^2 = (x+39)^2+24^2.$ Since $r_A^2 = \dfrac{25}{64}r_B^2$ \[x^2+24^2 = \dfrac{25}{64}((x+39)^... | 192 |
4,492 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | 1 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away fr... | Let $x=e^{i\pi/6}$ , a $30^\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:
\[1+2x+3x^2+\cdots+(k+1)x^k\]
We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetico-geometric series
\begin{align*} S &=1+2x+3x^2+\cdots+2015x^{2014} \\ xS &=... | 24 |
4,493 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | 2 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away fr... | Here is an alternate solution that does not use complex numbers:
The distance from $P_{2015}$ to $P_0$ can be evaluated using the Pythagorean theorem . Assuming $P_0$ lies at the origin, we can calculate the distance the bee traveled to $P_{2015}$ by evaluating the distance the bee traveled in the x-direction and the y... | 24 |
4,494 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | 3 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away fr... | We first notice that if the bee is turning 30 degrees each turn, it will take 12 turns to be looking in the same direction when the bee initially left. This means we simply need to answer the question; how far will the bee be when the bee is facing in the same direction?
First we use the fact that after 3 turns, the be... | 24 |
4,495 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | 4 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away fr... | Suppose that the bee makes a move of distance $i$ . After $6$ turns it will be facing the opposite direction and move $i+6$ units. Combining these opposite movements gives a total movement of $-6$ units in the original direction. This means that every $12$ moves, the bee will move $-6$ units in each direction of $0^\ci... | 24 |
4,496 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | 5 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away fr... | Let $P_0$ be the origin. East would be the real axis in the positive direction. Then we can assign each $P_n$ a complex value. The displacement would then be the magnitude of the complex number.
Notice that after the $n$ th move the value of $P_n$ is $P_{n-1}+ne^{\frac{i(n-1)\pi}{6}}$ . Also notice that after six moves... | 24 |
4,497 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | 6 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away fr... | After each 12 moves, the bee will be facing the same direction as it started. Let $P_0$ be the origin and let $P_n$ (with $n$ divisible by $12$ ) be $(x,y)$ . We now notice that each of the move pairs with lengths $n+1$ $n+7$ $n+2$ $n+8$ $n+3$ $n+9$ $n+4$ $n+10$ $n+5$ $n+11$ $n+6$ $n+12$ will move the bee 6 units in th... | 24 |
4,498 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | 4 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | Let's use casework on the yellow house. The yellow house $(\text{Y})$ is either the $3^\text{rd}$ house or the last house.
Case 1: $\text{Y}$ is the $3^\text{rd}$ house.
The only possible arrangement is $\text{B}-\text{O}-\text{Y}-\text{R}$
Case 2: $\text{Y}$ is the last house.
There are two possible arrangements:
$\te... | 3 |
4,499 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | 5 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | There are $4!=24$ arrangements without restrictions. There are $3!\cdot2!=12$ arrangements such that the blue house neighboring the yellow house (calculating the arrangments of [ $\text{BY}$ ], $\text{O}$ , and $\text{R}$ ). Hence, there are $24-12=12$ arrangements with the blue and yellow houses non-adjacent.
By symme... | 3 |
4,500 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | 6 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | To start with, the blue house is either the first or second house.
If the blue house is the first, then the orange must follow, leading to $2$ cases: $\text{B-O-R-Y}$ and $\text{B-O-Y-R}$
If the blue house is second, then the orange house must be first and the yellow house last, leading to $1$ case: $\text{O-B-R-Y}$ .
... | 3 |
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