link stringlengths 75 84 | letter stringclasses 5
values | answer float64 0 2,935,363,332B | problem stringlengths 14 5.33k | solution listlengths 1 13 |
|---|---|---|---|---|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_4 | A | 20 | If $a$ is $50\%$ larger than $c$ , and $b$ is $25\%$ larger than $c$ , then $a$ is what percent larger than $b$
$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \ } 50\% \qquad \mathrm{(D) \ } 100\% \qquad \mathrm{(E) \ }200\%$ | [
"Translating each sentence into an equation, $a = 1.5c$ and $b = 1.25c$\nWe want a relationship between $a$ and $b$ . Dividing the second equation into the first will cancel the $c$ , so we try that and get:\n$\\frac{a}{b} = \\frac{1.5}{1.25}$\n$\\frac{a}{b} = \\frac{150}{125}$\n$\\frac{a}{b} = \\frac{6}{5}$\n$a =... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_19 | D | 720 | If $a,b,$ and $c$ are positive real numbers such that $a(b+c) = 152, b(c+a) = 162,$ and $c(a+b) = 170$ , then $abc$ is
$\mathrm{(A)}\ 672 \qquad\mathrm{(B)}\ 688 \qquad\mathrm{(C)}\ 704 \qquad\mathrm{(D)}\ 720 \qquad\mathrm{(E)}\ 750$ | [
"Adding up the three equations gives $2(ab + bc + ca) = 152 + 162 + 170 = 484 \\Longrightarrow ab + bc + ca = 242$ . Subtracting each of the above equations from this yields, respectively, $bc = 90, ca = 80, ab = 72$ . Taking their product, $ab \\cdot bc \\cdot ca = a^2b^2c^2 = 90 \\cdot 80 \\cdot 72 = 720^2 \\Long... |
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_13 | null | 2 | If $a,b,c$ , and $d$ are non-zero numbers such that $c$ and $d$ are the solutions of $x^2+ax+b=0$ and $a$ and $b$ are
the solutions of $x^2+cx+d=0$ , then $a+b+c+d$ equals
$\textbf{(A) }0\qquad \textbf{(B) }-2\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }(-1+\sqrt{5})/2$ | [
"By Vieta's formulas, $c + d = -a$ $cd = b$ $a + b = -c$ , and $ab = d$ . From the equation $c + d = -a$ $d = -a - c$ , and from the equation $a + b = -c$ $b = -a - c$ , so $b = d$\nThen from the equation $cd = b$ $cb = b$ . Since $b$ is nonzero, we can divide both sides of the equation by $b$ to get $c = 1$ . Simi... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_18 | A | 14 | If $a,b,c$ are real numbers such that $a^2 + 2b =7$ $b^2 + 4c= -7,$ and $c^2 + 6a= -14$ , find $a^2 + b^2 + c^2.$
$\text{(A) }14 \qquad \text{(B) }21 \qquad \text{(C) }28 \qquad \text{(D) }35 \qquad \text{(E) }49$ | [
"Adding all of the equations gives $a^2 + b^2 +c^2 + 6a + 2b + 4c=-14.$ Adding 14 on both sides gives $a^2 + b^2 +c^2 + 6a + 2b + 4c+14=0.$ Notice that 14 can split into $9, 1,$ and $4,$ which coincidentally makes $a^2 +6a, b^2+2b,$ and $c^2+4c$ into perfect squares. Therefore, $(a+3)^2 + (b+1)^2 + (c+2) ^2 =0.$ An... |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_4 | null | 675 | If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube , what is the smallest possible value of $c$ | [
"Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know $b + c + d = 3c$ and $a + b + c + d + e = 5c$ . Thus, $c$ must be in the form of $3 \\cdot x^2$ based upon the first part and in the form of $5^2 \\cdot y^3$ based upon the second part, with $x$ and... |
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_20 | C | 4 | If $a=\tfrac{1}{2}$ and $(a+1)(b+1)=2$ then the radian measure of $\arctan a + \arctan b$ equals
$\textbf{(A) }\frac{\pi}{2}\qquad \textbf{(B) }\frac{\pi}{3}\qquad \textbf{(C) }\frac{\pi}{4}\qquad \textbf{(D) }\frac{\pi}{5}\qquad \textbf{(E) }\frac{\pi}{6}$ | [
"Solution by e_power_pi_times_i\nSince $a=\\frac{1}{2}$ $b=\\frac{1}{3}$ . Now we evaluate $\\arctan a$ and $\\arctan b$ . Denote $x$ and $\\theta$ such that $\\arctan x = \\theta$ . Then $\\tan(\\arctan(x)) = \\tan(\\theta)$ , and simplifying gives $x = \\tan(\\theta)$ . So $a = \\tan(\\theta_a) = \\frac{1}{2}$ an... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_24 | B | 0 | If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$
$\mathrm{(A)}\ -2 \qquad \mathrm{(B)}\ 0 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 3 \qquad \mathrm{(E)}\ 4$ | [
"Using logarithmic rules, we see that\n\\[\\log_{a}a-\\log_{a}b+\\log_{b}b-\\log_{b}a = 2-(\\log_{a}b+\\log_{b}a)\\] \\[=2-\\left(\\log_{a}b+\\frac {1}{\\log_{a}b}\\right)\\]\nSince $a$ and $b$ are both greater than $1$ , using AM-GM gives that the term in parentheses must be at least $2$ , so the largest possible ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_24 | null | 0 | If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$
$\mathrm{(A)}\ -2 \qquad \mathrm{(B)}\ 0 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 3 \qquad \mathrm{(E)}\ 4$ | [
"By the logarithmic rules, we have $2-(\\log_a{b}+\\frac{1}{\\log_a{b}})$ .\nLet $x=\\log_a{b}$ . Thus, the expression becomes $2-(x+\\frac{1}{x})$ . We want to find the maximum of the function. To do so, we will find its derivative and let it equal to 0. We get: $\\frac{d}{dx}\\big(2-(x+\\frac{1}{x})\\big)=\\frac{... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_12 | A | 4 | If $a\otimes b = \dfrac{a + b}{a - b}$ , then $(6\otimes 4)\otimes 3 =$
$\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$ | [
"$6\\otimes4=\\frac{6+4}{6-4}=5$ $5\\otimes3=\\frac{5+3}{5-3}=4, \\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_5 | E | 5 | If $b$ and $c$ are constants and $(x + 2)(x + b) = x^2 + cx + 6$ , then $c$ is
$\textbf{(A)}\ -5\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 5$ | [
"We first start out by expanding the left side of the equation, $(x+2)(x+b)=x^{2}+bx+2x+2b=x^2+(2+b)x+2b=x^2+cx+6$ . We know the constant \nterms have to be equal so we have $2b=6$ , so $b=3$ . Plugging $b=3$ back in yields $x^2+(2+3)x+6=x^2+cx+6$ . Thus, $c=5 \\implies \\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/1971_AHSME_Problems/Problem_2 | D | 2 | If $b$ men take $c$ days to lay $f$ bricks, then the number of days it will take $c$ men working at the same rate to lay $b$ bricks, is
$\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad \textbf{(E) }f/b^2$ | [
"We can use a modified version of the equation $\\text{Distance} = \\text{Rate} \\times {\\text{Time}}$ , which is $\\text{Work Done} = \\text{Rate of Work} \\times{ \\text{Time Worked}}$ . In this case, the work done is the number of bricks laid, the rate of work is the number of men working, and the time worked i... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_38 | C | 7 | If $f(a)=a-2$ and $F(a,b)=b^2+a$ , then $F(3,f(4))$ is:
$\textbf{(A)}\ a^2-4a+7 \qquad \textbf{(B)}\ 28 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 11$ | [
"We find $f(4)=(4)-2=2$ , so $F(3,f(4))=F(3,2)=(2)^2+3=\\boxed{7}$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_13 | null | 2 | If $f(x) = ax+b$ and $f^{-1}(x) = bx+a$ with $a$ and $b$ real, what is the value of $a+b$
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 1 \qquad\mathrm{(E)}\ 2$ | [
"Since $f(f^{-1}(x))=x$ , it follows that $a(bx+a)+b=x$ , which implies $abx + a^2 +b = x$ . This equation holds for all values of $x$ only if $ab=1$ and $a^2+b=0$\nThen $b = -a^2$ . Substituting into the equation $ab = 1$ , we get $-a^3 = 1$ . Then $a = -1$ , so $b = -1$ , and \\[f(x)=-x-1.\\] Likewise \\[f^{-1}(x... |
https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_14 | null | 8 | If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$ , then $f(3) =$
$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$ | [
"Substituting $x = -3$ , we get \\[f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.\\] But $f(-3) = 2$ , so $81a - 9b + 2 = 2$ , which means $81a - 9b = 0$ . Then \\[f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \\boxed{8}.\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_17 | C | 0 | If $i^2=-1$ , then $(1+i)^{20}-(1-i)^{20}$ equals
$\mathrm{(A)\ } -1024 \qquad \mathrm{(B) \ }-1024i \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 1024 \qquad \mathrm{(E) \ }1024i$ | [
"Notice that $(1+i)^2=2i$ and $(1-i)^2=-2i$ . Therefore,\n\\[(1+i)^{20}-(1-i)^{20}=(2i)^{10}-(-2i)^{10}=(2i)^{10}-(2i)^{10}=0, \\boxed{0}.\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_12 | D | 2 | If $i^2=-1$ , then $(i-i^{-1})^{-1}=$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ -2i \qquad\textbf{(C)}\ 2i \qquad\textbf{(D)}\ -\frac{i}{2} \qquad\textbf{(E)}\ \frac{i}{2}$ | [
"We simplify step by step as follows: \\begin{align*}(i-i^{-1})^{-1}&=\\frac{1}{i-i^{-1}}\\\\&=\\frac{1}{i-\\frac{1}{i}}\\\\&=\\frac{1}{\\left(\\frac{i^2-1}{i}\\right)}\\\\&=\\frac{i}{i^2-1}\\\\&=\\boxed{2}"
] |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29 | C | 35 | If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$ | [
"Working with the second part of the problem first, we know that $3n$ has $30$ divisors. We try to find the various possible prime factorizations of $3n$ by splitting $30$ into various products of $1, 2$ or $3$ integers.\n$30 \\rightarrow p^{29}$\n$2 \\cdot 15 \\rightarrow pq^{14}$\n$3\\cdot 10 \\rightarrow p^2q^9... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29 | null | 35 | If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$ | [
"Because $2n$ has $28$ factors and $3n$ has $30$ factors, we should rewrite the number $n = 2^{e_1}3^{e_2}... p_n^{e_n}$ As the formula for the number of divisors for such a number gives: $(e_1+1)(e_2+1)... (e_n+1)$ We plug in the variations we need to make for the cases $2n$ and $3n$ $2n$ has $(e_1+2)(e_2+1)(e_3+1... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_17 | B | 2 | If $n$ is an even positive integer, the $\emph{double factorial}$ notation $n!!$ represents the product of all the even integers from $2$ to $n$ . For example, $8!! = 2 \cdot 4 \cdot 6 \cdot 8$ . What is the units digit of the following sum? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]
$\textbf{(A) } 0\qquad... | [
"Notice that once $n>8,$ the units digit of $n!!$ will be $0$ because there will be a factor of $10.$ Thus, we only need to calculate the units digit of \\[2!!+4!!+6!!+8!! = 2+8+48+48\\cdot8.\\] We only care about units digits, so we have $2+8+8+8\\cdot8,$ which has the same units digit as $2+8+8+4.$ The answer is ... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_34 | A | 12 | If $n$ is any whole number, $n^2(n^2 - 1)$ is always divisible by
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ \text{any multiple of }12\qquad \textbf{(D)}\ 12-n\qquad \textbf{(E)}\ 12\text{ and }24$ | [
"Suppose $n$ is even. So, we have $n^2(n+1)(n-1).$ Out of these three numbers, at least one of them is going to be a multiple of 3. $n^2$ is also a multiple of 4. Therefore, this expression is always divisible by $\\boxed{12}.$"
] |
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_27 | E | 4 | If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$ | [
"If $p$ is a root of $x^3 - x^2 + x - 2 = 0$ , then $p^3 - p^2 + p - 2 = 0$ , or \\[p^3 = p^2 - p + 2.\\] Similarly, $q^3 = q^2 - q + 2$ , and $r^3 = r^2 - r + 2$ , so \\[p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.\\]\nBy Vieta's formulas $p + q + r = 1$ $pq + pr + qr = 1$ , and $pqr = 2$ . Squaring the ... |
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_11 | null | 2 | If $r$ is positive and the line whose equation is $x + y = r$ is tangent to the circle whose equation is $x^2 + y ^2 = r$ , then $r$ equals
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }\sqrt{2}\qquad \textbf{(E) }2\sqrt{2}$ | [
"The circle $x^2 + y^2 = r$ has center $(0,0)$ and radius $\\sqrt{r}$ . Therefore, if the line $x + y = r$ is tangent to the circle $x^2 + y^2 = r$ , then the distance between $(0,0)$ and the line $x + y = r$ is $\\sqrt{r}$\nThe distance between $(0,0)$ and the line $x + y = r$ is \\[\\frac{|0 + 0 - r|}{\\sqrt{1^2 ... |
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_15 | B | 15 | If $r$ is the remainder when each of the numbers $1059,~1417$ , and $2312$ is divided by $d$ , where $d$ is an integer greater than $1$ , then $d-r$ equals
$\textbf{(A) }1\qquad \textbf{(B) }15\qquad \textbf{(C) }179\qquad \textbf{(D) }d-15\qquad \textbf{(E) }d-1$ | [
"We are given these congruences:\nLet's make a new congruence by subtracting (i) from (ii), which results in \\[358 \\equiv 0 \\pmod{d}.\\] Subtract (ii) from (iii) to get \\[895 \\equiv 0 \\pmod{d}.\\]\nNow we know that $358$ and $895$ are both multiples of $d$ . Their prime factorizations are $358=2 \\cdot 179$ a... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_1 | E | 128 | If $x \neq 0, \frac x{2} = y^2$ and $\frac{x}{4} = 4y$ , then $x$ equals
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 128$ | [
"From $\\frac{x}{4} = 4y$ , we get $x=16y$ . Plugging in the other equation, $\\frac{16y}{2} = y^2$ , so $y^2-8y=0$ . Factoring, we get $y(y-8)=0$ , so the solutions are $0$ and $8$ . Since $x \\neq 0$ , we also have $y \\neq 0$ , so $y=8$ . Hence $x=16\\cdot 8 = \\boxed{128}$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_10 | D | 22 | If $x$ $y$ , and $z$ are positive with $xy = 24$ $xz = 48$ , and $yz = 72$ , then $x + y + z$ is
$\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24$ | [
"The first two equations in the problem are $xy=24$ and $xz=48$ . Since $xyz \\ne 0$ , we have $\\frac{xy}{xz}=\\frac{24}{48} \\implies 2y=z$ . We can substitute $z = 2y$ into the third equation $yz = 72$ to obtain $2y^2=72 \\implies y=6$ and $2y=z=12$ . We replace $y$ into the first equation to obtain $x=4$\nSince... |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_3 | D | 12 | If $x$ $y$ , and $z$ are real numbers such that
then $x + y + z =$
$\mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \ } 8 \qquad \mathrm{(D) \ } 12 \qquad \mathrm{(E) \ }50$ | [
"If the sum of three squared expressions is zero, then each expression itself must be zero, since $a^2 \\ge 0$ with the equality iff $a=0$\nIn this case, $x-3=0$ $y-4=0$ , and $z-5=0$ . Adding the three equations and moving the constant to the right gives $x + y + z = 12$ , and the answer is $\\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_25 | A | 3 | If $x$ and $y$ are non-zero real numbers such that \[|x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0,\] then the integer nearest to $x-y$ is
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5$ | [
"We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$\nSolving for $y$ in the top equation gives us $3-x$ . Plugging this in gives us: $x^3-x^2+3x=0$ . Since we're told $x$ is not zero, we can divide by $x$ , giving us: $x^2-x+3=0$\nThe discriminant ... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_3 | A | 8 | If $x$ and $y$ are positive integers for which $2^x3^y=1296$ , what is the value of $x+y$
$(\mathrm {A})\ 8 \qquad (\mathrm {B})\ 9 \qquad (\mathrm {C})\ 10 \qquad (\mathrm {D})\ 11 \qquad (\mathrm {E})\ 12$ | [
"$1296 = 2^4 3^4$ and $4+4=\\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_14 | C | 202 | If $x$ is the cube of a positive integer and $d$ is the number of positive integers that are divisors of $x$ , then $d$ could be
$\text{(A) } 200\quad\text{(B) } 201\quad\text{(C) } 202\quad\text{(D) } 203\quad\text{(E) } 204$ | [
"Solution by e_power_pi_times_i\nNotice that if $x$ is expressed in the form $a^b$ , then the number of positive divisors of $x^3$ is $3b+1$ . Checking through all the answer choices, the only one that is in the form $3b+1$ is $\\boxed{202}$",
"Solution by e_power_pi_times_i\nSince the divisors are from $x^3$ , t... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_20 | B | 1 | If $x,y,$ and $z$ are positive numbers satisfying
\[x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}\]
Then what is the value of $xyz$
$\text {(A)}\ \frac{2}{3} \qquad \text {(B)}\ 1 \qquad \text {(C)}\ \frac{4}{3} \qquad \text {(D)}\ 2 \qquad \text {(E)}\ \frac{7}{... | [
"We multiply all given expressions to get: \\[(1)xyz + x + y + z + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{xyz} = \\frac{28}{3}\\] Adding all the given expressions gives that \\[(2) x + y + z + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 4 + \\frac{7}{3} + 1 = \\frac{22}{3}\\] We subtract $(2)$ from... |
https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_3 | C | 1 | If $x=\dfrac{1-i\sqrt{3}}{2}$ where $i=\sqrt{-1}$ , then $\dfrac{1}{x^2-x}$ is equal to
$\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1+i\sqrt{3}\qquad \textbf{(D) }1\qquad \textbf{(E) }2$ | [
"Using DeMoivre's theorem, we can calculate $x^2=\\frac{1+i\\sqrt{3}}{2}$ The denominator is therefore $-1$ which makes the answer \\[\\boxed{1}.\\] ~lopkiloinm"
] |
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_1 | D | 7 | If $y = 2x$ and $z = 2y$ , then $x + y + z$ equals
$\text{(A)}\ x \qquad \text{(B)}\ 3x \qquad \text{(C)}\ 5x \qquad \text{(D)}\ 7x \qquad \text{(E)}\ 9x$ | [
"Solution by e_power_pi_times_i\n$x+y+z = x+(2x)+(4x) = \\boxed{7}$"
] |
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_18 | B | 5 | If $y=(\log_23)(\log_34)\cdots(\log_n[n+1])\cdots(\log_{31}32)$ then
$\textbf{(A) }4<y<5\qquad \textbf{(B) }y=5\qquad \textbf{(C) }5<y<6\qquad \textbf{(D) }y=6\qquad \\ \textbf{(E) }6<y<7$ | [
"Solution by e_power_pi_times_i\nNote that $\\log_{a}b = \\dfrac{\\log{b}}{\\log{a}}$ . Then $y=(\\dfrac{\\log3}{\\log2})(\\dfrac{\\log4}{\\log3})\\cdots(\\dfrac{\\log32}{\\log31}) = \\dfrac{\\log32}{\\log2} = \\log_232 = \\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_8 | E | 15 | If 1 pint of paint is needed to paint a statue 6 ft. high, then the number of pints it will take to paint (to the same thickness) 540 statues similar to the original but only 1 ft. high is
$\textbf{(A)}\ 90\qquad\textbf{(B)}\ 72\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 15$ | [
"The statues are similar, and since the height if the smaller statue is $\\tfrac{1}{6}$ of the original statue, the surface area is $\\tfrac{1}{36}$ of the original statue. Thus, $\\tfrac{1}{36}$ pints of paint is needed for one 1 ft. statue, so painting 540 of these statues requires $\\tfrac{540}{36} = \\boxed{15... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_9 | D | 250 | If 5 times a number is 2, then 100 times the reciprocal of the number is
$\text{(A)}\ 2.5 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 250 \qquad \text{(E)}\ 500$ | [
"If $5$ times a number is $2$ , then $5x = 2 \\rightarrow x = \\frac{2}{5}$ , and the number is $\\frac{2}{5}$\nIf the number is $\\frac{2}{5}$ , then $100$ times its reciprocal is $100$ times $\\frac{5}{2}$ , which is $250$ , giving an answer of $\\boxed{250}$"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_6 | C | 12 | If 554 is the base $b$ representation of the square of the number whose base $b$ representation is 24, then $b$ , when written in base 10, equals
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 16$ | [
"Write out the numbers using the definition of base numbers \\[554_b = 5b^2 + 5b + 4\\] \\[24_b = 2b+4\\] Since $554_b = (24_b)^2$ , we can write an equation. \\[5b^2 + 5b + 4 = (2b+4)^2\\] \\[5b^2 + 5b + 4 = 4b^2 + 16b + 16\\] \\[b^2 - 11b - 12 = 0\\] \\[(b-12)(b+1) = 0\\] Since base numbers must be positive, $b$ ... |
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_27 | A | 1 | If \[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\] then $N$ equals
$\textbf{(A) }1\qquad \textbf{(B) }2\sqrt{2}-1\qquad \textbf{(C) }\frac{\sqrt{5}}{2}\qquad \textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$ | [
"Let $x=\\frac{\\sqrt{\\sqrt{5}+2}+\\sqrt{\\sqrt{5}-2}}{\\sqrt{\\sqrt{5}+1}}$ and $y=\\sqrt{3-2\\sqrt{2}}.$\nNote that \\begin{align*} x^2&=\\frac{\\left(\\sqrt{\\sqrt{5}+2}+\\sqrt{\\sqrt{5}-2}\\right)^2}{\\left(\\sqrt{\\sqrt{5}+1}\\right)^2} \\\\ &=\\frac{\\left(\\sqrt{5}+2\\right)+2\\cdot\\left(\\sqrt{\\sqrt{5}+2... |
https://artofproblemsolving.com/wiki/index.php/1992_AHSME_Problems/Problem_16 | E | 2 | If \[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\] for three positive numbers $x,y$ and $z$ , all different, then $\frac{x}{y}=$
$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$ | [
"Since \\[\\frac{a_1}{b_1} = \\frac{a_2}{b_2} = ... = \\frac{a_n}{b_n} = \\frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n},\\] we can say that \\[\\frac{y}{x-z} = \\frac{x+y}{z} = \\frac{x}{y} = \\frac {y+(x+y)+x}{(x-z)+z+y} = \\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_27 | C | 51.005 | If \[x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20,\] then \[x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=\]
$\textbf{(A) } 5.05 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 51.005 \qquad \textbf{(D) } 61.25 \qquad \textbf{(E) } 400$ | [
"We rationalize the denominator in the given equation, then solve for $x:$ \\begin{align*} x+\\sqrt{x^2-1}+\\frac{x+\\sqrt{x^2-1}}{\\left(x-\\sqrt{x^2-1}\\right)\\left(x+\\sqrt{x^2-1}\\right)} &= 20 \\\\ x+\\sqrt{x^2-1}+x+\\sqrt{x^2-1} &= 20 \\\\ x+\\sqrt{x^2-1} &= 10 \\\\ \\sqrt{x^2-1} &= 10-x \\\\ x^2-1 &= 100-20... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_27 | C | 4 | If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ \text{more than }6$ | [
"Let the angle be $\\theta$ and the sides around it be $a$ and $b$ . \nThe area of the triangle can be written as \\[A =\\frac{a \\cdot b \\cdot \\sin(\\theta)}{2}\\] The doubled sides have length $2a$ and $2b$ , while the angle is still $\\theta$ . Thus, the area is \\[\\frac{2a \\cdot 2b \\cdot \\sin(\\theta)}{2}... |
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_14 | null | 80 | If an integer $n > 8$ is a solution of the equation $x^2 - ax+b=0$ and the representation of $a$ in the base- $n$ number system is $18$ ,
then the base-n representation of $b$ is
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 80 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 280$ | [
"Assuming the solutions to the equation are n and m, by Vieta's formulas, $n_n + m_n = 18_n$\n$n_n = 10_n$ , so $10_n + m_n = 18_n$\n\\[m_n = 8_n\\]\nAlso by Vieta's formulas, $n_n \\cdot m_n = b_n$ \\[10_n \\cdot 8_n = \\boxed{80}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_44 | C | 11 | If an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by
$\textbf{(A) \ } 9-k \qquad \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1$ | [
"Let $n = 10a+b$ . The problem states that $10a+b=k(a+b)$ . We want to find $x$ , where $10b+a=x(a+b)$ . Adding these two equations gives $11(a+b) = (k+x)(a+b)$ . Because $a+b \\neq 0$ , we have $11 = k + x$ , or $x = \\boxed{11}$"
] |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_36 | C | 2 | If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3}$ | [
"The equality implies $x-8$ and $x-10$ are both powers of two; since they differ by two, it must be the case that $(x-8,x-10) = (4,2)$ or $(x-8,x-10) = (-2,-4)$ . (Note that $(1,-1)$ is not allowed because then the product is negative.) These yield $(x,y) = (12,3)$ or $(x,y) = (6,3)$ , for a total of $\\boxed{2}$ ... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_24 | D | 27 | If circular arcs $AC$ and $BC$ have centers at $B$ and $A$ , respectively, then there exists a circle tangent to both $\overarc {AC}$ and $\overarc{BC}$ , and to $\overline{AB}$ . If the length of $\overarc{BC}$ is $12$ , then the circumference of the circle is
[asy] label("A", (0,0), W); label("B", (64,0), E); label("... | [
"First, note the triangle $ABC$ is equilateral. Next, notice that since the arc $BC$ has length 12, it follows that we can find the radius of the sector centered at $A$ $\\frac {1}{6}({2}{\\pi})AB=12 \\implies AB=36/{\\pi}$ . Next, connect the center of the circle to side $AB$ , and call this length $r$ , and call ... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_12 | E | 19 | If each of the three operation signs, $+$ $\text{--}$ $\times$ , is used exactly ONCE in one of the blanks in the expression
\[5\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}4\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}6\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}3\]
then the value of the result coul... | [
"There are a reasonable number of ways to place the operation signs, so guess and check to find that $5-4+6 \\times 3 = \\boxed{19}$"
] |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_5 | A | 648 | If five geometric means are inserted between $8$ and $5832$ , the fifth term in the geometric series:
$\textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these}$ | [
"We can let the common ratio of the geometric sequence be $r$ $5832$ is given to be the seventh term in the geometric sequence as there are five terms between it and $a_1$ if we consider $a_1=8$ .\nBy the formula for each term in a geometric sequence, we find that $a_n=a_1r^{n-1}$ or $(5382)=(8)r^6$ We divide by ei... |
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_2 | C | 1 | If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is
$\textbf{(A) }\frac{1}{\pi^2}\qquad \textbf{(B) }\frac{1}{\pi}\qquad \textbf{(C) }1\qquad \textbf{(D) }\pi\qquad \textbf{(E) }\pi^2$ | [
"Creating equations, we get $4\\cdot\\frac{1}{2\\pi r} = 2r$ . Simplifying, we get $\\frac{1}{\\pi r} = r$ . Multiplying each side by $r$ , we get $\\frac{1}{\\pi} = r^2$ . Because the formula of the area of a circle is $\\pi r^2$ , we multiply each side by $\\pi$ to get $1 = \\pi r^2$ .\nTherefore, our answer is $... |
https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_25 | D | 16 | If it is known that $\log_2(a)+\log_2(b) \ge 6$ , then the least value that can be taken on by $a+b$ is:
$\text{(A) } 2\sqrt{6}\quad \text{(B) } 6\quad \text{(C) } 8\sqrt{2}\quad \text{(D) } 16\quad \text{(E) none of these}$ | [
"We use the logarithm property of addition: \\begin{align*} \\log_2(a)+\\log_2(b) \\ge 6 &= \\log_2(ab) \\ge 6\\\\ &\\Rightarrow 2^{log_2(ab)} \\ge 2^6\\\\ &= ab \\ge 64 \\end{align*} Due to the Quadratic Optimization or the AM-GM Inequality , the least value is obtained when $a = b$ .\nTherefore, $a = b = 8 \\Righ... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_34 | C | 24 | If one side of a triangle is $12$ inches and the opposite angle is $30^{\circ}$ , then the diameter of the circumscribed circle is:
$\textbf{(A)}\ 18\text{ inches} \qquad \textbf{(B)}\ 30\text{ inches} \qquad \textbf{(C)}\ 24\text{ inches} \qquad \textbf{(D)}\ 20\text{ inches}\\ \textbf{(E)}\ \text{none of these}$ | [
"By the Extended Law of Sines, the diameter, or twice the circumradius $R$ , is given by \\[2R=\\frac{12\\text{ inches}}{\\sin30^\\circ}=\\boxed{24}.\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_13 | D | 75 | If rose bushes are spaced about $1$ foot apart, approximately how many bushes are needed to surround a circular patio whose radius is $12$ feet?
$\text{(A)}\ 12 \qquad \text{(B)}\ 38 \qquad \text{(C)}\ 48 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 450$ | [
"The circumference of the patio is $2\\cdot 12\\cdot \\pi =24\\pi \\approx 75.398$ . Since the bushes are spaced $1$ foot apart, about $75\\rightarrow \\boxed{75}$ are needed."
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6 | D | 72 | If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$ | [
"The sum of the ratios is $10$ . Since the sum of the angles of a triangle is $180^{\\circ}$ , the ratio can be scaled up to $54:54:72$ $(3\\cdot 18:3\\cdot 18:4\\cdot 18).$ The numbers in the ratio $54:54:72$ represent the angles of the triangle. The question asks for the largest, so the answer is $\\boxed{72}$"... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_38 | B | 2 | If the expression $\begin{pmatrix}a & c\\ d & b\end{pmatrix}$ has the value $ab-cd$ for all values of $a, b, c$ and $d$ , then the equation $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$
$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \... | [
"By $\\begin{pmatrix}a & c\\\\ d & b\\end{pmatrix}=ab-cd$ , we have $2x^2-x=3$ . Subtracting $3$ from both sides, giving $2x^2-x-3=0$ . This factors to $(2x-3)(x+1)=0$ . Thus, $x=\\dfrac{3}{2},-1$ , so the equation is $\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_7 | null | 925 | If the integer $k$ is added to each of the numbers $36$ $300$ , and $596$ , one obtains the squares of three consecutive terms of an arithmetic series. Find $k$ | [
"Call the terms of the arithmetic progression $a,\\ a + d,\\ a + 2d$ , making their squares $a^2,\\ a^2 + 2ad + d^2,\\ a^2 + 4ad + 4d^2$\nWe know that $a^2 = 36 + k$ and $(a + d)^2 = 300 + k$ , and subtracting these two we get $264 = 2ad + d^2$ (1). Similarly, using $(a + d)^2 = 300 + k$ and $(a + 2d)^2 = 596 + k$ ... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_48 | D | 35 | If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude,
then the ratio of the smaller base to the larger base is:
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{2}{5... | [
"\nLet $a$ be the length of the smaller base of isosceles trapezoid $ABCD$ , and $1$ be the length of the larger base of the trapezoid. The ratio of the smaller base to the larger base is $\\frac a1=a$ . Let point $E$ be the foot of the altitude from $C$ to $\\overline{AD}$\nSince the larger base of the isosceles t... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_19 | B | 10 | If the length and width of a rectangle are each increased by $10\%$ , then the perimeter of the rectangle is increased by
$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$ | [
"Let the width be $w$ and the length be $l$ . Then, the original perimeter is $2(w+l)$\nAfter the increase, the new width and new length are $1.1w$ and $1.1l$ , so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$\nTherefore, the percent change is \\begin{align*} \\frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \\frac{.2(w+l)}{2(w+l)... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_21 | D | 80 | If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%$ , then the area is increased by
$\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\%$ | [
"If a rectangle had dimensions $10 \\times 10$ and area $100$ , then its new dimensions would be $12 \\times 15$ and area $180$ . The area is increased by $180-100=80$ or $80/100 = \\boxed{80}$"
] |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_21 | null | 80 | If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%$ , then the area is increased by
$\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\%$ | [
"Let the dimensions of the rectangle be $x \\times y$ . This rectangle has area $xy$ . The new dimensions would be $1.2x \\times 1.5y$ , so the area is $=1.8xy$ , which is $\\boxed{80}$ more than the original area."
] |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_6 | C | 12 | If the markings on the number line are equally spaced, what is the number $\text{y}$
[asy] draw((-4,0)--(26,0),Arrows); for(int a=0; a<6; ++a) { draw((4a,-1)--(4a,1)); } label("0",(0,-1),S); label("20",(20,-1),S); label("y",(12,-1),S); [/asy]
$\text{(A)}\ 3 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(... | [
"Five steps are taken to get from $0$ to $20$ . Each step is of equal size, so each step is $4$ . Three steps are taken from $0$ to $y$ , so $y=3\\times 4=12\\rightarrow \\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_19 | C | 0 | If the parabola $y = ax^2 + bx + c$ passes through the points $( - 1, 12)$ $(0, 5)$ , and $(2, - 3)$ , the value of $a + b + c$ is:
$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$ | [
"Substituting in the $(x, y)$ pairs gives the following system of equations: \\[a-b+c=12\\] \\[c=5\\] \\[4a+2b+c=-3\\] We know $c=5$ , so plugging this in reduces the system to two variables: \\[a-b=7\\] \\[4a+2b=-8\\] Dividing the second equation by 2 gives $2a+b=-4$ , which can be added to the first equation to g... |
https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_11 | B | 50 | If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $\overline{AC}$ , in inches, is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \sqrt{50}\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ \sqrt{200}\qquad \textbf{(E)}\ \text{none of these}$ | [
"For rectangle $ABCD$ with perimeter 20, the diagonal $AC$ is given by: \\[AC = \\sqrt{l^2 + w^2}\\] To minimize $AC$ $l$ and $w$ should be equal (i.e., the rectangle is a square). Thus, $l = w = 5$ .\nSo, the minimum $AC$ is: \\[AC = \\sqrt{5^2 + 5^2} = \\boxed{50}\\] ~ proloto"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_23 | C | 15 | If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ b... | [
"Consider positive integers $a, b$ with a difference of $20$ . Suppose $b = a-20$ . Then, we have $(a)(a-20) = n$ . If there is another pair of two integers that multiply to $n$ but have a difference of 23, one integer must be greater than $a$ , and one must be smaller than $a-20$ . We can create two cases and set ... |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_19 | D | 35 | If the product $\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9$ , what is the sum of $a$ and $b$
$\text{(A)}\ 11 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 37$ | [
"Notice that the numerator of the first fraction cancels out the denominator of the second fraction, and the numerator of the second fraction cancels out the denominator of the third fraction, and so on.\nThe only numbers left will be $a$ in the numerator from the last fraction and $2$ in the denominator from the f... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_8 | C | 300 | If the radius of a circle is increased $100\%$ , the area is increased:
$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$ | [
"Increasing by $100\\%$ is the same as doubling the radius. If we let $r$ be the radius of the old circle, then the radius of the new circle is $2r.$\nSince the area of the circle is given by the formula $\\pi r^2,$ the area of the new circle is $\\pi (2r)^2 = 4\\pi r^2.$ The area is quadrupled, or increased by $\\... |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_12 | B | 9,902 | If the sequence $\{a_n\}$ is defined by
$a_1=2$
$a_{n+1}=a_n+2n$
where $n\geq1$
Then $a_{100}$ equals
$\mathrm{(A) \ }9900 \qquad \mathrm{(B) \ }9902 \qquad \mathrm{(C) \ } 9904 \qquad \mathrm{(D) \ }10100 \qquad \mathrm{(E) \ } 10102$ | [
"We begin to evaluate the first couple of terms of the sequence, hoping to find a pattern: $2, 4, 8, 14, 22, ....$ . We notice that the difference between succesive terms of the sequence are $2, 4, 6, 8, ....$ , a clear pattern. We can see that this pattern continues infinitely because of the recursive definition: ... |
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_4 | A | 2 | If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \sqrt2 \qquad \textbf{(C)}\ 1/2 \qquad \textbf{(D)}\ 2\sqrt2 \qquad \textbf{(E)}\ 4$ | [
"Solution by e_power_pi_times_i\nDenote the side of one square as $s$ . Then the diagonal of the second square is $s$ , so the side of the second square is $\\dfrac{s\\sqrt{2}}{2}$ . The area of the second square is $\\dfrac{1}{2}s^2$ , so the ratio of the areas is $\\dfrac{s^2}{\\dfrac{1}{2}s^2} = \\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_16 | B | 15 | If the sum of all the angles except one of a convex polygon is $2190^{\circ}$ , then the number of sides of the polygon must be
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 19 \qquad \textbf{(E)}\ 21$ | [
"Let $n$ be the number of sides in the polygon. The number of interior angles in the polygon is $180(n-2)$ . We know that the sum of all but one of them is $2190^{\\circ}$ , so the sum of all the angles is more than that. \\[180(n-2) > 2190\\] \\[n-2 > 12 \\tfrac{1}{6}\\] \\[n > 14 \\tfrac{1}{6}\\]\nThe sum of th... |
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_11 | D | 110 | If the sum of the first $10$ terms and the sum of the first $100$ terms of a given arithmetic progression are $100$ and $10$ ,
respectively, then the sum of first $110$ terms is:
$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$ | [
"Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.\nSum of the first 10 terms: $\\frac{10}{2}(2a+9d)=100 \\Longleftrightarrow 2a+9d=20$ Sum of the first 100 terms: $\\frac{100}{2}(2a+99d)=10 \\Longleftrightarrow 2a+99d=\\frac{1}{5}$\nSolving the system, we get $d=-\\fra... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_7 | D | 35 | If the value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes, then $n=$
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45$ | [
"We have \\begin{align*} 20(25)+10(10) &= 10(25)+n(10) \\\\ 600 &= 250+10n \\\\ 35 &= n \\implies \\boxed{35}"
] |
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30 | C | 16 | If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$ | [
"The two poles formula says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is $\\frac1{\\frac1{20}+\\frac1{80}}$ , or $\\frac1{\\frac1{16}}=\\boxed{16}$",
"The two lines can be represented as $y=\\frac{-x}{5}+20$ and $y=\\frac{4... |
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_45 | A | 17 | If you are given $\log 8\approx .9031$ and $\log 9\approx .9542$ , then the only logarithm that cannot be found without the use of tables is:
$\textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4$ | [
"While $\\log 17 = \\log(8 + 9)$ , we cannot easily deal with the logarithm of a sum. Furthermore, $17$ is prime, so none of the logarithm rules involving products or differences works. It therefore cannot be found without the use of a table (note: in 1951, calculators were very rare). The correct answer is therefo... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_17 | D | 91 | If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$ , then your score on the seventh test was
$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$ | [
"If the average score of the first six is $84$ , then the sum of those six scores is $6\\times 84=504$\nThe average score of the first seven is $85$ , so the sum of the seven is $7\\times 85=595$\nTaking the difference leaves us with just the seventh score, which is $595-504=91$ , so the answer is $\\boxed{91}$",
... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_1 | D | 9 | Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,
and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how
many items will they buy?
$\textb... | [
"We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$ . Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas,... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_9 | D | 2 | In $2005$ Tycoon Tammy invested $100$ dollars for two years. During the first year
her investment suffered a $15\%$ loss, but during the second year the remaining
investment showed a $20\%$ gain. Over the two-year period, what was the change
in Tammy's investment?
$\textbf{(A)}\ 5\%\text{ loss}\qquad \textbf{(B)}\ 2\%... | [
"After the $15 \\%$ loss, Tammy has $100 \\cdot 0.85 = 85$ dollars. After the $20 \\%$ gain, she has $85 \\cdot 1.2 = 102$ dollars. This is an increase in $2$ dollars from her original $100$ dollars, a $\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_19 | D | 61 | In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61... | [
"\nLet circle $A$ intersect $AC$ at $D$ and $E$ as shown. We apply Power of a Point on point $C$ with respect to circle $A.$ This yields the diophantine equation\n\\[CX \\cdot CB = CD \\cdot CE\\] \\[CX(CX+XB) = (97-86)(97+86)\\] \\[CX(CX+XB) = 3 \\cdot 11 \\cdot 61.\\]\nSince lengths cannot be negative, we must ha... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_6 | B | 420 | In $\bigtriangleup ABC$ $AB=BC=29$ , and $AC=42$ . What is the area of $\bigtriangleup ABC$
$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$ | [
"We know the semi-perimeter of $\\triangle ABC$ is $\\frac{29+29+42}{2}=50$ . Next, we use Heron's Formula to find that the area of the triangle is just $\\sqrt{50(50-29)^2(50-42)}=\\sqrt{50 \\cdot 21^2 \\cdot 8}=\\boxed{420}$",
"Splitting the isosceles triangle in half, we get a right triangle with hypotenuse $2... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_15 | null | 41 | In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive i... | [
"Since $\\angle DBE = 90^\\circ$ $DE$ is the diameter of $\\omega$ . Then $\\angle DFE=\\angle DGE=90^\\circ$ . But $DF=FE$ , so $\\triangle DEF$ is a 45-45-90 triangle. Letting $DG=3x$ , we have that $EG=4x$ $DE=5x$ , and $DF=EF=\\frac{5x}{\\sqrt{2}}$\nNote that $\\triangle DGE \\sim \\triangle ABC$ by SAS similar... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | D | 61 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)... | [
"Let $BX = q$ $CX = p$ , and $AC$ meets the circle at $Y$ and $Z$ , with $Y$ on $AC$ . Then $AZ = AY = 86$ . Using the Power of a Point (Secant-Secant Power Theorem), we get that $p(p+q) = 11(183) = 11 * 3 * 61$ . We know that $p+q>p$ , so $p$ is either $3$ $11$ , or $33$ . We also know that $p>11$ by the trian... |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_9 | null | 306 | In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$ | [
"Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar $\\triangle ABC \\sim \\triangle DPD' \\sim \\triangle PEE' \\sim \\triangle F'PF$ ). The remaining three se... |
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_31 | A | 110 | In $\triangle ABC$ $AB=AC$ $\angle A=40^\circ$ . Point $O$ is within the triangle with $\angle OBC \cong \angle OCA$ .
The number of degrees in $\angle BOC$ is:
$\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}$ | [
"Since $\\triangle ABC$ is an isosceles triangle, $\\angle ABC = \\angle ACB = 70^{\\circ}$ . Let $\\angle OBC = \\angle OCA = x$ . Since $\\angle ACB = 70$ $\\angle OCB = 70 - x$ . The angle of $\\triangle OBC$ add up to $180$ , so $\\angle BOC = 180 - (x + 70 - x) = \\boxed{110}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_9 | C | 56 | In $\triangle ABC$ $AB=AC=28$ and $BC=20$ . Points $D,E,$ and $F$ are on sides $\overline{AB}$ $\overline{BC}$ , and $\overline{AC}$ , respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$ , respectively. What is the perimeter of parallelogram $ADEF$
[asy] siz... | [
"Note that because $\\overline{DE}$ and $\\overline{EF}$ are parallel to the sides of $\\triangle ABC$ , the internal triangles $\\triangle BDE$ and $\\triangle EFC$ are similar to $\\triangle ABC$ , and are therefore also isosceles triangles.\nIt follows that $BD = DE$ . Thus, $AD + DE = AD + DB = AB = 28$\nSince ... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | D | 75 | In $\triangle ABC$ $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$ | [
"$\\angle ADB = 120^\\circ$ $\\angle ADC = 60^\\circ$ $\\angle DAB = 15^\\circ$ , let $\\angle ACB = \\theta$ $\\angle DAC = 120^\\circ - \\theta$\nBy the Law of Sines , we have $\\frac{CD}{\\sin(120^\\circ - \\theta)} = \\frac{AD}{\\sin \\theta}$\n$\\space$ $\\frac{BD}{\\sin15^\\circ} = \\frac{AD}{\\sin45^\\circ}$... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | null | 75 | In $\triangle ABC$ $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$ | [
"Draw a good diagram! Now, let's call $BD=t$ , so $DC=2t$ . Given the rather nice angles of $\\angle ABD = 45^\\circ$ and $\\angle ADC = 60^\\circ$ as you can see, let's do trig. Drop an altitude from $A$ to $BC$ ; call this point $H$ . We realize that there is no specific factor of $t$ we can call this just yet, s... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_34 | B | 120 | In $\triangle ABC$ , side $a = \sqrt{3}$ , side $b = \sqrt{3}$ , and side $c > 3$ . Let $x$ be the largest number such that the magnitude,
in degrees, of the angle opposite side $c$ exceeds $x$ . Then $x$ equals:
$\textbf{(A)}\ 150^{\circ} \qquad \textbf{(B)}\ 120^{\circ}\qquad \textbf{(C)}\ 105^{\circ} \qquad \textbf... | [
"Using the Law of Cosines \\[\\sqrt{3 + 3 - 2\\cdot 3 \\cdot \\cos{x^\\circ}}>3\\]\nBoth sides are positive, so squaring both sides will not affect the inequality.\n\\[6 - 6 \\cos{x^\\circ}>9\\] \\[\\cos{x^\\circ} < -\\frac{1}{2}\\]\nNote that $\\cos{120^\\circ} = -\\frac{1}{2}$ . As $x$ gets closer to $180^{\\cir... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_11 | null | 20 | In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose th... | [
"Let the tangent circle be $\\omega$ . Some notation first: let $BC=a$ $AB=b$ $s$ be the semiperimeter, $\\theta=\\angle ABC$ , and $r$ be the inradius. Intuition tells us that the radius of $\\omega$ is $r+\\frac{2rs}{s-a}$ (using the exradius formula). However, the sum of the radius of $\\omega$ and $\\frac{rs}{s... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_23 | C | 2 | In $\triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$
$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$ | [
"There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have $\\triangle{ABC}$ , and you draw a median from point $A$ to side $BC$ (label this as $M$ ), then: $(AM)^2 = \\dfrac{2(b^2) + 2(c^2) - (a^2)}{4}$ . Note that $b$ is the length of side $\\overline{AC}$ $c$ is the length of si... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_23 | null | 2 | In $\triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$
$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$ | [
"From Stewart's Theorem , we have $(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.$ Simplifying, we get $(5/4)a^3 = (5/2)a \\implies (5/4)a^2 = 5/2 \\implies a^2 = 2 \\implies a = \\boxed{2}.$ - awu2014"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_10 | A | 3 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | [
"Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$ . By the Pythagorean Theorem $CH=\\sqrt{48}$ . Since $CD=8$ $HD=\\sqrt{8^2-48}=\\sqrt{16}=4$ , and $BD=HD-1$ , so $BD=\\boxed{3}$",
"After drawing out a diagram, let $\\angle{ABC}=\\theta$ . By the Law of Cosines, $7^2=2^2+7^2-2(7... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_6 | A | 3 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | [
"Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$ . By the Pythagorean Theorem $CH=\\sqrt{48}$ . Since $CD=8$ $HD=\\sqrt{8^2-48}=\\sqrt{16}=4$ , and $BD=HD-1$ , so $BD=\\boxed{3}$",
"After drawing out a diagram, let $\\angle{ABC}=\\theta$ . By the Law of Cosines, $7^2=2^2+7^2-2(7... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_28 | B | 35 | In $\triangle ABC$ , we have $\angle C = 3\angle A$ $a = 27$ and $c = 48$ . What is $b$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C--... | [
"Let $\\angle A = x^{\\circ}$ , so $\\angle C = 3x^{\\circ}$ , and thus $\\angle B = \\left(180-4x\\right)^{\\circ}$ . Now let $D$ be a point on side $AB$ such that $\\angle ACD = x^{\\circ}$ , so $\\angle BCD = 3x^{\\circ}-x^{\\circ} = 2x^{\\circ}$ , which gives \\[\\angle CDB = 180^{\\circ}-2x^{\\circ}-\\left(180... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | null | 13 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relativ... | [
"Suppose we label the angles as shown below. As $\\angle BCD$ and $\\angle BAD$ intercept the same arc, we know that $\\angle BAD=\\gamma$ . Similarly, $\\angle ABD=\\gamma$ . Also, using $\\triangle ICA$ , we find $\\angle CIA=180-\\alpha-\\gamma$ . Therefore, $\\angle AID=\\alpha+\\gamma$ . Therefore, $\\angle D... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_1 | null | 547 | In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n... | [
"\nIf we set $\\angle{BAC}$ to $x$ , we can find all other angles through these two properties:\n1. Angles in a triangle sum to $180^{\\circ}$ .\n2. The base angles of an isosceles triangle are congruent.\nNow we angle chase. $\\angle{ADE}=\\angle{EAD}=x$ $\\angle{AED} = 180-2x$ $\\angle{BED}=\\angle{EBD}=2x$ $\\an... |
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_10 | D | 10.8 | In $\triangle ABC$ with $\overline{AB}=\overline{AC}=3.6$ , a point $D$ is taken on $AB$ at a distance $1.2$ from $A$ . Point $D$ is joined to $E$ in the prolongation of $AC$ so that $\triangle AED$ is equal in area to $ABC$ . Then $\overline{AE}$ is: $\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qq... | [
"Note that $\\frac{1}{2}AB * AC *\\sin\\angle BAC = \\frac{1}{2}AD * AE *\\sin\\angle DAE$ . Since $\\angle BAC = \\angle DAE$ , we have $AB*AC = AD*AE$ , so that $3.6*3.6 = 1.2*AE$ . Therefore, $AE = \\frac{3.6^2}{1.2} = 10.8$ . Thusly, our answer is $\\boxed{10.8}$ , and we are done."
] |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_19 | A | 9 | In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$ | [
"Notice that by the Law of Sines, $a:b:c = \\sin{A}:\\sin{B}:\\sin{C}$ , so let's flip all the cosines using $\\sin^{2}{x} + \\cos^{2}{x} = 1$ $\\sin{x}$ is positive for $0^{\\circ} < x < 180^{\\circ}$ , so we're good there).\n$\\sin A=\\frac{3\\sqrt{15}}{16}, \\qquad \\sin B= \\frac{\\sqrt{15}}{8}, \\qquad \\text{... |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_9 | E | 4 | In $\triangle ABC$ with right angle at $C$ , altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$ , then the area of $\triangle ABC$ is
$\textbf{(A)}\ 6K\qquad\textbf{(B)}\ 4\sqrt3\ K\qquad\textbf{(C)}\ 3\sqrt3\ K\qquad\textbf{(D)}\ 3K\qquad\textbf{(E)}\ 4K$ | [
"\nDraw diagram as shown (note that $A$ and $B$ can be interchanged, but it doesn’t change the solution).\nNote that because $CM$ is a median, $AM = BM$ . Also, by ASA Congruency, $\\triangle CHA = \\triangle CHM$ , so $AH = HM$ . That means $HM = \\tfrac{1}{4} \\cdot AB$ , and since $\\triangle CHM$ and $\\trian... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | null | 247 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$... | [
"Because $M$ is the midpoint of $BC$ , following from the Stewart's theorem, $AM = 2 \\sqrt{37}$\nBecause $A$ $B$ $C$ , and $P$ are concyclic, $\\angle BPA = \\angle C$ $\\angle CPA = \\angle B$\nDenote $\\theta = \\angle PBQ$\nIn $\\triangle BPQ$ , following from the law of sines, \\[ \\frac{BQ}{\\sin \\angle BPA}... |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_11 | B | 6.5 | In $\triangle ABC, AB = 13, BC = 14$ and $CA = 15$ . Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC$ .
The length of $HM$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A); draw(B--C--A--B^^M--H--A^^rightanglemark(... | [
"Warning: this solution is very intensive in calculation. Please do NOT try this on the test!\nLet's start by finding $AH$ . By Heron's Formula, $s=\\frac{13+14+15}{2}=21, [ABC]=\\sqrt{21*(21-13)(21-14)(21-15)}=84$ . Using the area formula $A=0.5bh$ $AH=12$ . Now using the Pythagorean Theorem, $BH=5, HC=9$\nNow $AM... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_15 | null | 289 | In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overl... | [
"By the Angle Bisector Theorem, we know that $[CBD]=\\frac{169}{289}[ABC]$ . Therefore, by finding the area of triangle $CBD$ , we see that \\[\\frac{507\\cdot BD}{2}\\sin\\frac{B}{2}=\\frac{169}{289}[ABC].\\] Solving for $BD$ yields \\[BD=\\frac{2[ABC]}{3\\cdot289\\sin\\frac{B}{2}}.\\] Furthermore, $\\cos\\frac{B}... |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_16 | C | 9 | In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$ is similar to $\triangle PCA$ . The length of $PC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("A... | [
"Since we are given that $\\triangle{PAB}\\sim\\triangle{PCA}$ , we have $\\frac{PC}{PA}=\\frac{6}{8}=\\frac{PA}{PC+7}$\nSolving for $PA$ in $\\frac{PC}{PA}=\\frac{6}{8}=\\frac{3}{4}$ gives us $PA=\\frac{4PC}{3}$\nWe also have $\\frac{PA}{PC+7}=\\frac{3}{4}$ . Substituting $PA$ in for our expression yields $\\frac{... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | null | 289 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | [
"By the Law of Cosines on $\\triangle ABC$ , we have: \\[\\cos(A) = \\frac{10^2+10^2-12^2}{2*10*10} = \\frac{7}{25}\\] By the Law of Cosines on $\\triangle ADE$ , then \\[\\frac{7}{25} = \\frac{10-x}{2x} \\iff x =\\frac{250}{39}\\] So, our answer is $250+39=\\boxed{289}$",
"We draw the altitude from $B$ to $\\ov... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20 | D | 14 | In $\triangle BAC$ $\angle BAC=40^\circ$ $AB=10$ , and $AC=6$ . Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sq... | [
"Let $C_1$ be the reflection of $C$ across $\\overline{AB}$ , and let $C_2$ be the reflection of $C_1$ across $\\overline{AC}$ . Then it is well-known that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$ . (Proving this is a simple application of the triangle inequality; for an example of a simple... |
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