Answer: 2.0

If $\theta$ is a constant such that $0 < \theta < \pi$ and $x + \dfrac{1}{x} = 2\cos{\theta}$ , then for each positive integer $n$ $x^n + \dfrac{1}{x^n}$ equals
$\textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta$

Solution:
Multiply both sides by $x$ and rearrange to $x^2-2x\cos(\theta)+1=0$ . Using the quadratic equation, we can solve for $x$ . After some simplifying:
\[x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}\] \[x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}\] \[x=\cos(\theta) + i\sin(\theta)\]
Substituting this expression in to the desired $x^n + \dfrac{1}{x^n}$ gives:
\[(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}\]
Using DeMoivre's Theorem:
\[=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)\]
Because $\cos$ is even and $\sin$ is odd:
\[=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)\]
$=\boxed{2},$ (Error compiling LaTeX. Unknown error_msg)


Answer: 1.0

If $\theta$ is an acute angle, and $\sin 2\theta=a$ , then $\sin\theta+\cos\theta$ equals
$\textbf{(A) }\sqrt{a+1}\qquad \textbf{(B) }(\sqrt{2}-1)a+1\qquad \textbf{(C) }\sqrt{a+1}-\sqrt{a^2-a}\qquad\\ \textbf{(D) }\sqrt{a+1}+\sqrt{a^2-a}\qquad  \textbf{(E) }\sqrt{a+1}+a^2-a$

Solution:
Let $x = \sin\theta+\cos\theta$ , so we want to find $x$ . First, square the expression to get $x^2 = \sin^2 \theta + 2\sin\theta\cos\theta + \cos^2 \theta$ . Recall that $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$ . Plugging these in, the equation simplifies to $x^2 = 1 + \sin 2\theta$ . Given that $\sin 2\theta=a$ , the equation becomes $x^2=1+a$ . Take the square root of both sides to get $x = \sqrt{1+a}$
Hence, the answer is $\boxed{1}$ jiang147369


Answer: 2.0

If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$ , then $AC+BC$ must be
$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2}$ $\textbf{(E)}\ AB^{2}$

Solution:
Because $AB$ is the diameter of the semi-circle, it follows that $\angle C = 90$ . Now we can try to eliminate all the solutions except for one by giving counterexamples.
$\textbf{(A):}$ Set point $C$ anywhere on the perimeter of the semicircle except on $AB$ . By triangle inequality, $AC+BC>AB$ , so $\textbf{(A)}$ is wrong. $\textbf{(B):}$ Set point $C$ on the perimeter of the semicircle infinitesimally close to $AB$ , and so $AC+BC$ almost equals $AB$ , therefore $\textbf{(B)}$ is wrong. $\textbf{(C):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$ , it follows that $\textbf{(C)}$ is wrong. $\textbf{(E):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$ , it follows that $\textbf{(E)}$ is wrong.
Therefore, the only possible case is $\boxed{2}$


Answer: 2.0

If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \  } 1+\sqrt{2} \qquad \mathrm{(D) \  } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \  }\text{none of these}$

Solution: Noting that $\usepackage{gensymb} 25 \degree + 20 \degree = 45 \degree$ , we apply the angle sum formula \[\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B},\] giving \begin{align*}1 &= \tan 45^{\circ} \\ &= \tan(A+B) \\ &= \frac{\tan A+\tan B}{1-\tan A\tan B},\end{align*} so \[\tan A + \tan B = 1-\tan A\tan B.\] Hence \begin{align*}(1+\tan A)(1+\tan B) &= 1+\tan A+\tan B+\tan A\tan B \\ &= 1+\left(1-\tan A\tan B\right)+\tan A\tan B \\ &= \boxed{2}

Solution:
Expanding in terms of sines and cosines, we obtain \begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= \left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right) \\ &=\frac{\left(\sin A+\cos A\right)\left(\sin B+\cos B\right)}{\cos A\cos B} \\ &= \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B}.\end{align*}
Recalling the angle sum identities \begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*} this reduces to \[\frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B}.\]
Now, using the product-to-sum formula \[\cos A\cos B = \frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right),\] we can simplify the denominator, yielding \[\left(1+\tan A\right)\left(1+\tan B\right) = \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right)}.\]
Finally, since $\usepackage{gensymb} A+B = 45 \degree$ , we have $\sin(A+B) = \cos(A+B)$ , so \begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}\left(\cos(A-B)+\sin(A+B)\right)} \\ &= \frac{1}{\left(\frac{1}{2}\right)} \\ &= \boxed{2}

Solution: As in Solution 2, we rewrite the expression as \[\frac{\cos 20^{\circ} \cos 25^{\circ} + \cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}},\] and hence as \[1 + \frac{\cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}}.\] Using the angle sum identities \begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*} we obtain \begin{align*}&\cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} = \sin\left(20^{\circ}+25^{\circ}\right) = \sin 45^{\circ} \text{ and} \\ &\cos 20^{\circ} \cos 25^{\circ} - \sin 20^{\circ} \sin 25^{\circ} = \cos\left(20^{\circ}+25^{\circ}\right) = \cos 45^{\circ}.\end{align*} Therefore the expression becomes \begin{align*}1+\frac{\sin 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}} &= 1+1 \qquad \text{(since } \sin 45^{\circ} = \cos 45^{\circ}\text{)} \\ &= \boxed{2}

Solution:
As in Solutions 2 and 3, the expression becomes \[\left(\frac{\cos A+\sin A}{\cos A}\right)\left(\frac{\cos B+\sin B}{\cos B}\right).\]
Now, using the identity $\cos^2 A + \sin^2 A = 1$ and the double-angle identity $\sin(2A) = 2\sin A\cos A$ , we observe that \begin{align*}\left(\cos A + \sin A\right)^2 &= \cos^2 A + \sin^2 A + 2\sin A \cos A \\ &= 1 + 2\sin A \cos A \\ &= 1 + \sin(2A).\end{align*}
Since $A$ and $B$ are acute, we have $\sin A,\cos A,\sin B,\cos B > 0$ , so $\cos A + \sin A > 0$ and $\cos B + \sin B > 0$ . Hence, taking the positive square root of both sides in the above identity, the expression becomes \[\left(\frac{\sqrt{1+\sin(2A)}}{\cos A}\right)\left(\frac{\sqrt{1+\sin(2B)}}{\cos B}\right).\] Recalling the further identity $\sin A = \cos\left(90^{\circ}-A\right)$ , together with the half-angle identity \[\cos\left(\frac{A}{2}\right) = \sqrt{\frac{1+\cos A}{2}} \qquad \text{for } 0^{\circ} \leq A \leq 180^{\circ},\] we finally obtain \begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= 2\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2A\right)}{2}}}{\cos A}\right)\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2B\right)}{2}}}{\cos B}\right) \\ &= 2\left(\frac{\cos\left(45^{\circ}-A\right)}{\cos A}\right)\left(\frac{\cos\left(45^{\circ}-B\right)}{\cos B}\right) \\ &= 2\left(\frac{\cos 25^{\circ} \cos 20^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}}\right) = \boxed{2}


Answer: 3.0

If $a = - 2$ , the largest number in the set $\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}$ is
$\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1$

Solution:
Since all the numbers are small, we can just evaluate the set to be \[\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}\]
The largest number is $6$ , which corresponds to $-3a$
$\boxed{3}$


Answer: 3333.0

Solution: Adding all four of the equations up, we can see that it equals \[3(a+b+c+d)\] This is equal to $3(1111) = \boxed{3333}$ ~awin


Answer: 2.0

If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1$ , then $b$ is
$\textbf{(A)}\ -2\qquad \textbf{(B)}\ -1\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2$

Solution: Using polynomial division, we find that the remainder is $(2a+b)x+(a+b+1)$ , so for the condition to hold, we need this remainder to be $0$ . This gives $2a+b=0$ and $a+b+1=0$ , so $b=-2a$ and $a-2a+1=0 \implies a=1 \implies b=-2$ , which is $\boxed{2}.$


Answer: 6.0

If $a$ and $b$ are positive real numbers and each of the equations $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots , then the smallest possible value of $a+b$ is
$\mathrm{(A) \ }2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ } 6$

Solution:
Since both of the equations have real roots, both of their discriminants are nonnegative. Therefore, we have
$a^2-4(1)(2b)=a^2-8b\geq0\implies a^2\geq8b$ from the first equation, and
$(2b)^2-4(1)(a)=4b^2-4a\geq0\implies 4b^2\geq4a\implies b^2\geq a$ from the second.
We can square the second equation to get $b^4\geq a^2$ , and combining this with the first one gives $b^4\geq a^2\geq8b$ , so $b^4\geq8b$ . We can divide both sides by $b$ , since it is positive, and take the cubed root of that to get $b\geq2$ . Therefore, we have $a^2\geq8b\geq8(2)=16$ , and since $a$ is positive, we can take the square root of this to get $a\geq4$ . Therefore, $a+b\geq 2+4=6$ , and the smallest possible value is $6, \boxed{6}$


Answer: 20.0

If $a$ is $50\%$ larger than $c$ , and $b$ is $25\%$ larger than $c$ , then $a$ is what percent larger than $b$
$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \  } 50\% \qquad \mathrm{(D) \  } 100\% \qquad \mathrm{(E) \  }200\%$

Solution:
Translating each sentence into an equation, $a = 1.5c$ and $b = 1.25c$
We want a relationship between $a$ and $b$ .  Dividing the second equation into the first will cancel the $c$ , so we try that and get:
$\frac{a}{b} = \frac{1.5}{1.25}$
$\frac{a}{b} = \frac{150}{125}$
$\frac{a}{b} = \frac{6}{5}$
$a = 1.2b$
In this case, $a$ is $1.2 - 1 = 0.2 = 20\%$ bigger than $b$ , and the answer is $\boxed{20}$

Solution:
Arbitrarily assign a value to one of the variables.  Since $c$ is the smallest variable, let $c = 100$
If $a$ is $50\%$ larger than $c$ , then $a = 150$
If $b$ is $25\%$ larger than $c$ , then $b = 125$
We see that $\frac{a}{b} = \frac{150}{125} = 1.2$ So, $a$ is $20\%$ bigger than $b$ , and the answer is $\boxed{20}$


Answer: 720.0

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
$\mathrm{(A)}\ 672 \qquad\mathrm{(B)}\ 688 \qquad\mathrm{(C)}\ 704 \qquad\mathrm{(D)}\ 720 \qquad\mathrm{(E)}\ 750$

Solution: 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, $ab \cdot bc \cdot ca = a^2b^2c^2 = 90 \cdot 80 \cdot 72 = 720^2 \Longrightarrow abc = \boxed{720}$


Answer: 2.0

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
$\textbf{(A) }0\qquad \textbf{(B) }-2\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad  \textbf{(E) }(-1+\sqrt{5})/2$

Solution:
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 $b = d$
Then from the equation $cd = b$ $cb = b$ . Since $b$ is nonzero, we can divide both sides of the equation by $b$ to get $c = 1$ . Similarly, from the equation $ab = d$ $ab = b$ , so $a = 1$ . Then $b = d = -a - c = -2$ . Therefore, $a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{2}$ . The answer is (B).


Answer: 14.0

If $a,b,c$ are real numbers such that $a^2 + 2b =7$ $b^2 + 4c= -7,$ and $c^2 + 6a= -14$ , find $a^2 + b^2 + c^2.$
$\text{(A) }14 \qquad \text{(B) }21 \qquad \text{(C) }28 \qquad \text{(D) }35 \qquad \text{(E) }49$

Solution: 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 easy solution to this equation is $a=-3, b=-1,$ and $c=-2.$ Plugging in that solution, we get $a^2+b^2+c^2=-3^2+-1^2+-2^2=\boxed{14}.$


Answer: 675.0

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 $c$

Solution: 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 $y$ denoting an integers $c$ is minimized if it’s prime factorization contains only $3,5$ , and since there is a cubed term in $5^2 \cdot y^3$ $3^3$ must be a factor of $c$ $3^35^2 = \boxed{675}$ , which works as the solution.

Solution:
Let $b$ $c$ $d$ , and $e$ equal $a+1$ $a+2$ $a+3$ , and $a+4$ , respectively. Call the square and cube $k^2$ and $m^3$ , where both k and m are integers. Then:
$5a + 10 = m^3$
Now we know $m^3$ is a multiple of 125 and $m$ is a multiple of 5. The lower $m$ is, the lower the value of $c$ will be. Start from 5 and add 5 each time.
$m = 5$ gives no solution for k
$m = 10$ gives no solution for k
$m = 15$ gives a solution for k.
$10 + 5a = 15^3$
$2 + a = 675$
$c = \boxed{675}$

Solution: Let the numbers be $a,a+1,a+2,a+3,a+4.$ When then know $3a+6$ is a perfect cube and $5a+10$ is perfect cube. Since $5a+10$ is divisible by $5$ we know that $5a+10 = (5k)^3$ since otherwise we get a contradiction. This means $a = 25k^3 - 2$ in which plugging into the other expression we know $3(25k^3 - 2) + 6 = 75k^3$ is a perfect square. We know $75 = 5^2 \cdot 3$ so we let $k = 3$ to obtain the perfect square. This means that $c = a+2 = (25 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.$

Solution:
(This is literally a combination of 1 and 3)
Since $a$ $b$ $c$ $d$ , and $e$ are consecutive, $a = c-2$ $b = c-1$ $c=c$ $d = c+1$ , and $e = c+2$
Because $b+c+d = 3c$ is a perfect square, and $a+b+c+d+e = 5c$ is a perfect cube, we can express $c$ as $c = 3^{n} \cdot 5^{k}$
Now, by the problem's given information,
$k \equiv 0   \text{(mod 2)}$
$n \equiv 0   \text{(mod 3)}$
and because ALL exponents have to be cubes/squares,
$k \equiv 2   \text{(mod 3)}$
$n \equiv 1   \text{(mod 2)}$
Therefore,
$k = 2$
$n = 3$
$c = 3^3 \cdot 5^2 = \boxed{675}$


Answer: 4.0

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:
Solution by e_power_pi_times_i
Since $a=\frac{1}{2}$ $b=\frac{1}{3}$ . 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 $a = \tan(\theta_a) = \frac{1}{2}$ and $b = \tan(\theta_b) = \frac{1}{3}$ . The question asks for $\theta_a + \theta_b$ , so we try to find $\tan(\theta_a + \theta_b)$ in terms of $\tan(\theta_a)$ and $\tan(\theta_b)$ . Using the angle addition formula for $\tan(\alpha+\beta)$ , we get that $\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}$ . Plugging $\tan(\theta_a) = \frac{1}{2}$ and $\tan(\theta_b) = \frac{1}{3}$ in, we have $\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}$ . Simplifying, $\tan(\theta_a + \theta_b) = 1$ , so $\theta_a + \theta_b$ in radians is $\boxed{4}$


Answer: 0.0

If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$
$\mathrm{(A)}\ -2      \qquad \mathrm{(B)}\ 0     \qquad \mathrm{(C)}\ 2      \qquad \mathrm{(D)}\ 3      \qquad \mathrm{(E)}\ 4$

Solution:
Using logarithmic rules, we see that
\[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\] \[=2-\left(\log_{a}b+\frac {1}{\log_{a}b}\right)\]
Since $a$ 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 values is $2-2=0 \Rightarrow \boxed{0}.$

Solution:
Similar to the previous solution, we use our logarithmic rules and start off the same way. \[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\]
However, now, we use a different logarithmic rule stating that $\log_{a}b$ is simply equal to $\frac {\log_{10}b}{\log_{10}a}$ . With this, we can rewrite our previous equation to give us \[2-(\log_{a}b+\log_{b}a) = 2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right)\]
We can now cross multiply to get that \[2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right) = 2 - \left(\frac {2 \cdot \log_{10}b \cdot \log_{10}a}{\log_{10}b \cdot \log_{10}a}\right)\] Finally, we cancel to get $2-2=0 \Rightarrow \boxed{0}.$


Answer: 0.0

Solution:
By the logarithmic rules, we have $2-(\log_a{b}+\frac{1}{\log_a{b}})$ .
Let $x=\log_a{b}$ . 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: $\frac{d}{dx}\big(2-(x+\frac{1}{x})\big)=\frac{d}{dx}\big(2-x-\frac{1}{x})=-1+x^{-2}=0 \implies \frac{1}{x^2}=1, x^2=1, x=\pm1.$ Since $a\geq b>1, x=-1$ is not a solution. Thus, $x=1$ . Substituting it into the original expression $2-(x+\frac{1}{x})$ , we get $2-(1+\frac{1}{1})=2-2=\boxed{0}$


Answer: 4.0

If $a\otimes b = \dfrac{a + b}{a - b}$ , then $(6\otimes 4)\otimes 3 =$
$\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$

Solution: $6\otimes4=\frac{6+4}{6-4}=5$ $5\otimes3=\frac{5+3}{5-3}=4, \boxed{4}$


Answer: 5.0

If $b$ and $c$ are constants and $(x + 2)(x + b) = x^2 + cx + 6$ , then $c$ is
$\textbf{(A)}\ -5\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 5$

Solution:
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
terms 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, $c=5 \implies \boxed{5}$


Answer: 2.0

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
$\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad  \textbf{(E) }f/b^2$

Solution: 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 is the number of days. With these definitions, we see from the first equation that $f = b \cdot c$ . If we let $X$ be the number of days it will take $c$ men working at the same rate to $b$ bricks, then we have the equation $b = c \cdot x$ . So, $x = \frac{b}{c}$ . The first equation says that $\frac{f}{c^2} = \frac{b}{c}$ , which leads to $x = \frac{f}{c^2}$ . This doesn't match any of our answer choices though, so we have to fiddle around a bit before we realize that $c = \frac{f}{b}$ , a substitution we can make to see that $x = \frac{b^2}{f}$ . Thus, the answer is $\boxed{2}$


Answer: 7.0

If $f(a)=a-2$ and $F(a,b)=b^2+a$ , then $F(3,f(4))$ is:
$\textbf{(A)}\ a^2-4a+7 \qquad \textbf{(B)}\ 28 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 11$

Solution: We find $f(4)=(4)-2=2$ , so $F(3,f(4))=F(3,2)=(2)^2+3=\boxed{7}$


Answer: 2.0

If $f(x) = ax+b$ and $f^{-1}(x) = bx+a$ with $a$ and $b$ real, what is the value of $a+b$
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 1 \qquad\mathrm{(E)}\ 2$

Solution:
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 $a^2+b=0$
Then $b = -a^2$ . Substituting into the equation $ab = 1$ , we get $-a^3 = 1$ . Then $a = -1$ , so $b = -1$ , and \[f(x)=-x-1.\] Likewise \[f^{-1}(x)=-x-1.\] These are inverses to one another since \[f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.\] \[f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.\] Therefore $a+b=\boxed{2}$


Answer: 8.0

If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$ , then $f(3) =$
$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$

Solution: 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 \[f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.\]


Answer: 0.0

Solution:
Notice that $(1+i)^2=2i$ and $(1-i)^2=-2i$ . Therefore,
\[(1+i)^{20}-(1-i)^{20}=(2i)^{10}-(-2i)^{10}=(2i)^{10}-(2i)^{10}=0, \boxed{0}.\]


Answer: 2.0

Solution: 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}


Answer: 35.0

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?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$

Solution:
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.
$30 \rightarrow p^{29}$
$2 \cdot 15 \rightarrow pq^{14}$
$3\cdot 10 \rightarrow p^2q^9$
$5\cdot 6 \rightarrow p^4q^5$
$2\cdot 3\cdot 5 \rightarrow pq^2r^4$
The variables $p, q, r$ are different prime factors, and one of them must be $3$ .  We now try to count the factors of $2n$ , to see which prime factorization is correct and has $28$ factors.
In the first case, $p=3$ is the only possibility.  This gives $2n = 2\cdot p^{28}$ , which has $2\cdot {29}$ factors, which is way too many.
In the second case, $p=3$ gives $2n = 2q^{14}$ .  If $q=2$ , then there are $16$ factors, while if $q\neq 2$ , there are $2\cdot 15 = 30$ factors.
In the second case, $q=3$ gives $2n = 2p3^{13}$ .  If $p=2$ , then there are $3\cdot 13$ factors, while if $p\neq 2$ , there are $2\cdot 2 \cdot 13$ factors.
In the third case, $p=3$ gives $2n = 2\cdot 3\cdot q^9$ .  If $q=2$ , then there are $11\cdot 2 = 22$ factors, while if $q \neq 2$ , there are $2\cdot 2\cdot 10$ factors.
In the third case, $q=3$ gives $2n = 2\cdot p^2\cdot 3^8$ .  If $p=2$ , then there are $4\cdot 9$ factors, while if $p \neq 2$ , there are $2\cdot 3\cdot 9$ factors.
In the fourth case, $p=3$ gives $2n = 2\cdot 3^3\cdot q^5$ .  If $q=2$ , then there are $7\cdot 4= 28$ factors.  This is the factorization we want.
Thus, $3n = 3^4 \cdot 2^5$ , which has $5\cdot 6 = 30$ factors, and $2n = 3^3 \cdot 2^6$ , which has $4\cdot 7 = 28$ factors.
In this case, $6n = 3^4\cdot 2^6$ , which has $5\cdot 7 = 35$ factors, and the answer is $\boxed{35}$


Answer: 35.0

Solution:
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 $(e_1+2)(e_2+1)(e_3+1)... (e_n+1) = 28$ $3n$ has $(e_1+1)(e_2+2)(e_3+1)...(e_n+1) = 30$
If we take the  top and divide by the bottom, we get the following equation: $\frac{(e_1+2)(e_2+1)}{(e_1+1)(e_2+2)} = \frac{14}{15}$ . Letting $e_1=x$ and $e_2 = y$ for convenience and expanding this out gives us: $xy-13x+16y+2=0$
We can use Simon's Favorite Factoring Trick (SFFT) to turn this back into: $(x+16)(y-13) +2 + 208 = 0$ or $(x+16)(y-13) = - 210$
As we want to be dealing with rather reasonable numbers for $x$ and $y$ , we try to make the $x+16$ term the slightly larger term and the $y-13$ term the slightly smaller term. This effect is achieved when $x+16 = 21$ and $y-13 = -10$ . Therefore, $x = 5, y = 3$ . We get that this already satisfies the requirements for the number we are looking for, and we take $(5+2)(3+2) = \boxed{35}$

Solution: Let $\, 2^{e_1} 3^{e_2} 5^{e_3} \cdots \,$ be the prime factorization of $\, n$ . Then the number of positive divisors of $\, n \,$ is $\, (e_1 + 1)(e_2 + 1)(e_3 + 1) \cdots \;$ . In view of the given information, we have \[28 = (e_1 + 2)(e_2 + 1)P\] and \[30 = (e_1 + 1)(e_2 + 2)P,\] where $\, P = (e_3 + 1)(e_4 + 1) \cdots \;$ . Subtracting the first equation from the second, we obtain $\, 2 = (e_1 - e_2)P, \,$ so either $\, e_1 - e_2 = 1 \,$ and $\, P = 2, \,$ or $\, e_1 - e_2 = 2 \,$ and $\, P = 1$ . The first case yields $\, 14 = (e_1 + 2)e_1 \,$ and $\, (e_1 + 1)^2 = 15$ ; since $\, e_1 \,$ is a nonnegative integer, this is impossible. In the second case, $\, e_2 = e_1 - 2 \,$ and $\, 30 = (e_1 + 1)e_1, \,$ from which we find $\, e_1 = 5 \,$ and $\, e_2 = 3$ . Thus $\, n = 2^5 3^3, \,$ so $\, 6n = 2^6 3^4 \,$ has $\, (6+1)(4+1) = \boxed{35} \,$ positive divisors.


Answer: 2.0

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? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]
$\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Solution: 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 $\boxed{2}.$


Answer: 12.0

If $n$ is any whole number, $n^2(n^2 - 1)$ is always divisible by
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ \text{any multiple of }12\qquad \textbf{(D)}\ 12-n\qquad \textbf{(E)}\ 12\text{ and }24$

Solution: 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 $\boxed{12}.$


Answer: 4.0

If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$

Solution:
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 \[p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.\]
By Vieta's formulas $p + q + r = 1$ $pq + pr + qr = 1$ , and $pqr = 2$ . Squaring the equation $p + q + r = 1$ , we get \[p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.\] Subtracting $2pq + 2pr + 2qr = 2$ , we get \[p^2 + q^2 + r^2 = -1.\]
Therefore, $p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6 = (-1) - 1  + 6 = \boxed{4}$

Solution:
We know that $p^3+q^3+r^3=(p+q+r)(p^2+q^2+r^2-pq-qr-pr)+3pqr$ . By Vieta's formulas, $p+q+r=1$ $pqr=2$ , and $pq+qr+pr=1$ .
So if we can find $p^2+q^2+r^2$ , we are done. Notice that $(p+q+r)^2=p^2+q^2+r^2+2pq+2qr+2pr$ , so $p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=1^2-2\cdot1=-1$ , which means that $p^3+q^3+r^3=1\cdot-2+3\cdot2=\boxed{4}$

Solution:
Use Vieta's formulas to get $p+q+r=1$ $pq+qr+pr=1$ , and $pqr=2$
Square $p+q+r=1$ , and get $p^2+q^2+r^2+2pq+2pr+2qr=1$
Substitute $pq+qr+pr=1$ and simplify to get $-1=p^2+q^2+r^2$
After that, multiply both sides by $1=p+q+r$ , to get $-1=p^3+q^3+r^3+p^2q+q^2r+p^2r+q^2r+r^2p+r^2q$
Then, factor out $pq$ $qr$ , and $pr$ $-1=p^3+q^3+r^3+pq(p+q)+qr(q+r)+pr(p+r)$
Then, substitute the first equation into $p+q$ $q+r$ , and $p+r$ $-1=p^3+q^3+r^3+pq(1-r)+qr(1-p)+pr(1-q)$
Then, multiply it out: $-1=p^3+q^3+r^3+pq+qr+pr-3pqr$
After that, substitute the equations $pq+qr+pr=1$ and $pqr=2$ $-1=p^3+q^3+r^3+1-6$
Solving that, you get $p^3+q^3+r^3=\boxed{4}$ ~EZ PZ Ms.Lemon SQUEEZY


Answer: 2.0

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
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }\sqrt{2}\qquad  \textbf{(E) }2\sqrt{2}$

Solution:
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 $\sqrt{r}$
The distance between $(0,0)$ and the line $x + y = r$ is \[\frac{|0 + 0 - r|}{\sqrt{1^2 + 1^2}} = \frac{r}{\sqrt{2}}.\] Hence, \[\frac{r}{\sqrt{2}} = \sqrt{r}.\] Then $r = \sqrt{r} \cdot \sqrt{2}$ , so $\sqrt{r} = \sqrt{2}$ , which means $r = \boxed{2}$ or (B), $2$


Answer: 15.0

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
$\textbf{(A) }1\qquad \textbf{(B) }15\qquad \textbf{(C) }179\qquad \textbf{(D) }d-15\qquad  \textbf{(E) }d-1$

Solution:
We are given these congruences:
Let's make a new congruence by subtracting (i) from (ii), which results in \[358 \equiv 0 \pmod{d}.\] Subtract (ii) from (iii) to get \[895 \equiv 0 \pmod{d}.\]
Now we know that $358$ and $895$ are both multiples of $d$ . Their prime factorizations are $358=2 \cdot 179$ and $895=5 \cdot 179$ , so their common factor is $179$ , which means $d=179$
Plug $d=179$ back into any of the original congruences to get $r=164$ . Then, $d-r=179-164= \boxed{15}$ . ~ jiang147369


Answer: 128.0

If $x \neq 0, \frac x{2} = y^2$ and $\frac{x}{4} = 4y$ , then $x$ equals
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 128$

Solution: 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 $x=16\cdot 8 = \boxed{128}$


Answer: 22.0

If $x$ $y$ , and $z$ are positive with $xy = 24$ $xz = 48$ , and $yz = 72$ , then $x + y + z$ is
$\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24$

Solution:
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 $x=4$
Since we know every variable's value, we can substitute them in to find $x+y+z = 4+6+12 = \boxed{22}$

Solution: These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives $(xyz)^2 = (xy)(yz)(xz) = (24)(48)(72) = (24 \times 12)^2 \implies xyz = 288$ . We divide $xyz = 288$ by each of the given equations, which yields $x = 4$ $y = 6$ , and $z = 12$ . The desired sum is $4+6+12 = 22$ , so the answer is $\boxed{22}$


Answer: 12.0

If $x$ $y$ , and $z$ are real numbers such that
then $x + y + z =$
$\mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \  } 8 \qquad \mathrm{(D) \  } 12 \qquad \mathrm{(E) \  }50$

Solution:
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 $a=0$
In 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 $\boxed{12}$


Answer: 3.0

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
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5$

Solution:
We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$
Solving 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: $x^2-x+3=0$
The discriminant of this is $(-1)^2-4(1)(3)=-11$ , which means the equation has no real solutions.
We conclude that $x$ is negative. In this case $-x+y=3$ and $-xy+x^3=0$ .  Negating the top equation gives us $x-y=-3$ . We seek $x-y$ , so the answer is $\boxed{3}$


Answer: 8.0

If $x$ and $y$ are positive integers for which $2^x3^y=1296$ , what is the value of $x+y$
$(\mathrm {A})\ 8 \qquad (\mathrm {B})\ 9 \qquad (\mathrm {C})\ 10 \qquad (\mathrm {D})\ 11 \qquad (\mathrm {E})\ 12$

Solution: $1296 = 2^4 3^4$ and $4+4=\boxed{8}$


Answer: 202.0

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:
Solution by e_power_pi_times_i
Notice 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:
Solution by e_power_pi_times_i
Since the divisors are from $x^3$ , then the answer must be something in (mod $3$ ). Since $200$ and $203$ are the same (mod $3$ ), as well as $201$ and $204$ $\boxed{202}$ is the only answer left.


Answer: 1.0

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 $xyz$
$\text {(A)}\ \frac{2}{3} \qquad \text {(B)}\ 1 \qquad \text {(C)}\ \frac{4}{3} \qquad \text {(D)}\ 2 \qquad \text {(E)}\ \frac{7}{3}$

Solution: 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 $(1)$ to get that $xyz + \frac{1}{xyz} = 2$ . Hence, by inspection, $\boxed{1}$ \[\] ~AopsUser101


Answer: 1.0

If $x=\dfrac{1-i\sqrt{3}}{2}$ where $i=\sqrt{-1}$ , then $\dfrac{1}{x^2-x}$ is equal to
$\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1+i\sqrt{3}\qquad \textbf{(D) }1\qquad  \textbf{(E) }2$

Solution: 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


Answer: 7.0

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:
Solution by e_power_pi_times_i
$x+y+z = x+(2x)+(4x) = \boxed{7}$


Answer: 5.0

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:
Solution by e_power_pi_times_i
Note that $\log_{a}b = \dfrac{\log{b}}{\log{a}}$ . Then $y=(\dfrac{\log3}{\log2})(\dfrac{\log4}{\log3})\cdots(\dfrac{\log32}{\log31}) = \dfrac{\log32}{\log2} = \log_232 = \boxed{5}$


Answer: 15.0

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
$\textbf{(A)}\ 90\qquad\textbf{(B)}\ 72\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 15$

Solution: 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 $\tfrac{540}{36} = \boxed{15}$ pints of paint.


Answer: 250.0

If 5 times a number is 2, then 100 times the reciprocal of the number is
$\text{(A)}\ 2.5 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 250 \qquad \text{(E)}\ 500$

Solution:
If $5$ times a number is $2$ , then $5x = 2 \rightarrow x = \frac{2}{5}$ , and the number is $\frac{2}{5}$
If the number is $\frac{2}{5}$ , then $100$ times its reciprocal is $100$ times $\frac{5}{2}$ , which is $250$ , giving an answer of $\boxed{250}$


Answer: 12.0

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
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 16$

Solution: 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, $b$ in base 10 equals $\boxed{12}$


Answer: 1.0

If \[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\] then $N$ equals
$\textbf{(A) }1\qquad \textbf{(B) }2\sqrt{2}-1\qquad \textbf{(C) }\frac{\sqrt{5}}{2}\qquad \textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$

Solution:
Let $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$ and $y=\sqrt{3-2\sqrt{2}}.$
Note 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}\cdot\sqrt{\sqrt{5}-2}\right)+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ &=\frac{\left(\sqrt{5}+2\right)+2+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ &=\frac{2\sqrt{5}+2}{\sqrt{5}+1} \\ &=2. \end{align*} Since $x>0,$ we have $x=\sqrt{2}.$
On the other hand, note that \begin{align*} y^2&=3-2\sqrt{2} \\ &=2-2\sqrt{2}+1 \\ &=\left(\sqrt{2}-1\right)^2. \end{align*} Since $y>0,$ we have $y=\sqrt{2}-1.$
Finally, the answer is \[N=x-y=\boxed{1}.\]

Solution:
Let $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$ and $y=\sqrt{3-2\sqrt{2}}.$
Note that \[x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}. \hspace{15mm} (\bigstar)\] We rewrite each term in the numerator separately:
Substituting these results into $(\bigstar),$ we have \[x=\frac{\left(\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}\right)+\left(\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}\right)}{2}=\sqrt2.\] On the other hand, we have \[y=\sqrt2-1\] by the argument of either Solution 1 or Solution 2.
Finally, the answer is \[N=x-y=\boxed{1}.\]


Answer: 2.0

If \[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\] for three positive numbers $x,y$ and $z$ , all different, then $\frac{x}{y}=$
$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$

Solution: 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}$


Answer: 51.005

Solution: 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 &= 100-20x+x^2 \\ 20x &= 101 \\ x &= 5.05. \end{align*} We rationalize the denominator in the requested expression, then simplify the result: \begin{align*} x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}} &= x^2+\sqrt{x^4-1}+\frac{x^2-\sqrt{x^4-1}}{\left(x^2+\sqrt{x^4-1}\right)\left(x^2-\sqrt{x^4-1}\right)} \\ &= x^2+\sqrt{x^4-1}+x^2-\sqrt{x^4-1} \\ &= 2x^2 \\ &= \boxed{51.005}


Answer: 4.0

If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ \text{more than }6$

Solution:
Let the angle be $\theta$ and the sides around it be $a$ and $b$ .
The 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 \[\frac{2a \cdot 2b \cdot \sin(\theta)}{2}\] \[\Rrightarrow  \frac{4ab \sin \theta}{2} = 4A\] \[\boxed{4}\]


Answer: 80.0

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
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 80 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 280$

Solution:
Assuming the solutions to the equation are n and m, by Vieta's formulas, $n_n + m_n = 18_n$
$n_n = 10_n$ , so $10_n + m_n = 18_n$
\[m_n = 8_n\]
Also by Vieta's formulas, $n_n \cdot m_n = b_n$ \[10_n \cdot 8_n = \boxed{80}\]


Answer: 11.0

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
$\textbf{(A) \ } 9-k  \qquad \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1$

Solution: 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 $x = \boxed{11}$


Answer: 2.0

If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3}$

Solution: 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 $\boxed{2}$ solutions.


Answer: 27.0

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("C", (32, 32*sqrt(3)), N); draw(arc((0,0),64,0,60)); draw(arc((64,0),64,120,180)); draw((0,0)--(64,0)); draw(circle((32, 24), 24)); [/asy]
$\textbf {(A)}\ 24 \qquad \textbf {(B)}\ 25 \qquad \textbf {(C)}\ 26 \qquad \textbf {(D)}\ 27 \qquad \textbf {(E)}\ 28$

Solution: 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 the foot $M$ . Since $ABC$ is equilateral, it follows that $MB=18/{\pi}$ , and $OA$ (where O is the center of the circle) is $36/{\pi}-r$ .  By the Pythagorean Theorem, you get $r^2+(18/{\pi})^2=(36/{\pi}-r)^2 \implies r=27/2{\pi}$ . Finally, we see that the circumference is $2{\pi}\cdot 27/2{\pi}=\boxed{27}$


Answer: 19.0

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 could equal
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 19$

Solution: There are a reasonable number of ways to place the operation signs, so guess and check to find that $5-4+6 \times 3 = \boxed{19}$


Answer: 648.0

If five geometric means are inserted between $8$ and $5832$ , the fifth term in the geometric series:
$\textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these}$

Solution:
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 $a_1=8$ .
By 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 eight to find: $r^6=729$ $r=\pm 3$
Because $a_2$ will not be between $8$ and $5832$ if $r=-3$ we can discard it as an extraneous solution. We find $r=3$ and $a_5=a_1r^4=(8)(3)^4=\boxed{648}$


Answer: 1.0

If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is
$\textbf{(A) }\frac{1}{\pi^2}\qquad \textbf{(B) }\frac{1}{\pi}\qquad \textbf{(C) }1\qquad \textbf{(D) }\pi\qquad \textbf{(E) }\pi^2$

Solution:
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 $1 = \pi r^2$ .
Therefore, our answer is $\boxed{1}$


Answer: 16.0

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:
$\text{(A) } 2\sqrt{6}\quad \text{(B) } 6\quad \text{(C) } 8\sqrt{2}\quad \text{(D) } 16\quad \text{(E) none of these}$

Solution:
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 $a = b$ .
Therefore, $a = b = 8 \Rightarrow a + b = \boxed{16}$


Answer: 24.0

If one side of a triangle is $12$ inches and the opposite angle is $30^{\circ}$ , then the diameter of the circumscribed circle is:
$\textbf{(A)}\ 18\text{ inches} \qquad \textbf{(B)}\ 30\text{ inches} \qquad \textbf{(C)}\ 24\text{ inches} \qquad \textbf{(D)}\ 20\text{ inches}\\ \textbf{(E)}\ \text{none of these}$

Solution: By the Extended Law of Sines, the diameter, or twice the circumradius $R$ , is given by \[2R=\frac{12\text{ inches}}{\sin30^\circ}=\boxed{24}.\]


Answer: 75.0

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?
$\text{(A)}\ 12 \qquad \text{(B)}\ 38 \qquad \text{(C)}\ 48 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 450$

Solution: 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.


Answer: 72.0

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?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$

Solution: 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 $\boxed{72}$

Solution: We can denote the angles of the triangle as $3x$ $3x$ $4x$ . Due to the sum of the angles in a triangle, $3x+3x+4x=180^{\circ}\implies x=18^{\circ}$ . The greatest angle is $4x$ and after substitution we get $\boxed{72}$

Solution: We know the longest side must be denoted by the 4 in the ratio. Since the ratio is 3:3:4, we know that the longest side must be $\frac{4}{3+3+4}$ of the degree total (which for all triangles is 180). Thus, \[\frac{4}{3+3+4} \cdot 180 = \frac{4}{10} \cdot 180 = \boxed{72}\]

Solution: Since we see the ratio is $3:3:4$ , we can rule out the answer of ${\textbf{(E) }90}$ because the numbers in the ratio are too big to have $90^\circ$ . Also, we are trying to find the largest angle and all the other angles except for 72 are too small to be the largest angle. Using all this, our answer is $\boxed{72}$


Answer: 2.0

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 $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$
$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \textbf{(C)}\ \text{Is satisified for no values of }x\qquad\\ \textbf{(D)}\ \text{Is satisfied for an infinite number of values of }x\qquad\\ \textbf{(E)}\ \text{None of these.}$

Solution: 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 $\boxed{2}$


Answer: 925.0

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 $k$

Solution:
Call the terms of the arithmetic progression $a,\ a + d,\ a + 2d$ , making their squares $a^2,\ a^2 + 2ad + d^2,\ a^2 + 4ad + 4d^2$
We 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 $(a + 2d)^2 = 596 + k$ , subtraction yields $296 = 2ad + 3d^2$ (2).
Subtracting the first equation from the second, we get $2d^2 = 32$ , so $d = 4$ . Substituting backwards yields that $a = 31$ and $k = \boxed{925}$

Solution: Since terms in an arithmetic progression have constant differences, \[\sqrt{300+k}-\sqrt{36+k}=\sqrt{596+k}-\sqrt{300+k}\] \[\implies 2\sqrt{300+k} = \sqrt{596+k}+\sqrt{36+k}\] \[\implies 4k+1200=596+k+36+k+2\sqrt{(596+k)(36+k)}\] \[\implies 2k+568=2\sqrt{(596+k)(36+k)}\] \[\implies k+284=\sqrt{(596+k)(36+k)}\] \[\implies k^2+568k+80656=k^2+632k+21456\] \[\implies 568k+80656=632k+21456\] \[\implies 64k = 59200\] \[\implies k = \boxed{925}\]


Answer: 35.0

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:
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{2}{5}$

Solution:
Let $a$ 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 $\overline{AD}$
Since the larger base of the isosceles trapezoid equals a diagonal, $AC=AD=1$ . Since the smaller base equals the altitude, $BC=CE=a$ . Since the trapezoid is isosceles, $DE=\frac{1-a}{2}$ , so $AE = 1-\frac{1-a}{2} = \frac{a+1}{2}$ . Using the Pythagorean Theorem on right triangle $ACE$ , we have \[a^2+\left(\frac{a+1}{2}\right)^2=1\] Multiplying both sides by $4$ gives \[4a^2+(a+1)^2=4\] Expanding the squared binomial and rearranging gives \[5a^2+2a-3=0\] This can be factored into $(5a-3)(a+1)=0$ . Since a must be positive, $5a+3=0$ , so $a=\frac 35$ . The ratio of the smaller base to the larger base is $\boxed{35}$


Answer: 10.0

If the length and width of a rectangle are each increased by $10\%$ , then the perimeter of the rectangle is increased by
$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$

Solution:
Let the width be $w$ and the length be $l$ . Then, the original perimeter is $2(w+l)$
After the increase, the new width and new length are $1.1w$ and $1.1l$ , so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$
Therefore, the percent change is \begin{align*} \frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \frac{.2(w+l)}{2(w+l)} \\ &= \frac{.2}{2} \\ &= 10\% \\ \end{align*}
$\boxed{10}$

Solution: Assume WLOG that the rectangle is a square with length $10$ and width $10$ .  Thus, the square has a perimeter of $40$ .  Increasing the length and width by $10\%$ will increase the dimensions of the square to 11x11.  Thus, the new square's perimeter is $44$ , and because $44$ is $110\%$ of $40$ , our answer is $\boxed{10}$


Answer: 80.0

If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%$ , then the area is increased by
$\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\%$

Solution: 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 $80/100 = \boxed{80}$


Answer: 80.0

Solution: 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.


Answer: 12.0

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]
$\text{(A)}\ 3 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 16$

Solution: 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 $y=3\times 4=12\rightarrow \boxed{12}$


Answer: 0.0

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:
$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$

Solution: 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 get $3a=3$ , or $a=1$ . So the solution set is $(1, -6, 5)$ , and the sum is $\boxed{0}$


Answer: 50.0

If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $\overline{AC}$ , in inches, is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \sqrt{50}\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ \sqrt{200}\qquad \textbf{(E)}\ \text{none of these}$

Solution:
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, $l = w = 5$ .
So, the minimum $AC$ is: \[AC = \sqrt{5^2 + 5^2} = \boxed{50}\] ~ proloto


Answer: 15.0

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 by $23$ . What is the sum of the digits of $N$
$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution:
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 both equal. We have $(a)(a-20) = (a+1)(a-22)$ , and $(a)(a-20) = (a+2)(a-21)$ . Starting with the first case, we have $a^2-20a = a^2-21a-22$ ,or $0=-a-22$ , which gives $a=-22$ , which is not possible. The other case is $a^2-20a = a^2-19a-42$ , so $a=42$ . Thus, our product is $(42)(22) = (44)(21)$ , so $c = 924$ . Adding the digits, we have $9+2+4 = \boxed{15}$ .
-Sepehr2010
-minor edits by aidanb
-extremely minor edits by cc21
-no edits in the actual solution by get-rickrolled

Solution:
We have 4 integers in our problem. Let's call the smallest of them $a$ $a(a+23) =$ either $(a+1)(a+21)$ or $(a+2)(a+22)$ . So, we have the following:
$a^2 + 23a = a^2 + 22a +21$ or
$a^2+23a = a^2 + 24a +44$
The second equation has negative solutions, so we discard it. The first equation has $a = 21$ , and so $a + 23 = 44$ . If we check $(a+1)(a+21)$ we get $22 \cdot 42 = 21 \cdot 44$ $44$ is $2$ times $22$ , and $42$ is $2$ times $21$ , so our solution checks out. Multiplying $21$ by $44$ , we get $924$ => $9 + 2 + 4 = \boxed{15}$

Solution:
From the problems, it follows that
\begin{align*} x(x+20)&=y(y+23) = N\\ x^2+20x&=y^2+23y\\ 4x^2+4\cdot20x &= 4y^2+4\cdot23y\\ 4x^2+4\cdot20x+20^2-20^2 &= 4y^2+4\cdot23y+23^2-23^2\\ (2x+20)^2-20^2 &= (2y+23)^2-23^2\\ 23^2-20^2 &= (2y+23)^2-(2x+20)^2\\ (23+20)(23-20) &= (2y+23+2x+20)(2y+23-2x-20)\\ 43\cdot 3 &= (2y+2x+43)(2y-2x+3)\\ 129\cdot 1 &= (2y+2x+43)(2y-2x+3)\\ \end{align*} Since both $(2y+2x+43)$ and $(2y-2x+3)$ must be integer, we get two equations. \begin{align} 129 or 43 &= (2y+2x+43)\\ 1 or 3 &= 2y-2x+3\\ \end{align} 43 & 1 yields (0,0) which is not what we want.
129 & 1 yields (22,21) which is more interesting.
Simplifying the equations, we get: \begin{align*} x+y &= 43\\ x-y &= 1\\ x=22&, y=21\\ N &= (22)(22+20) = 924. \end{align*}
So, the answer is $\boxed{15}$

Solution: Say one factorization is $n(n+23).$ The two cases for the other factorization are $(n+1)(n+21)$ and $(n+2)(n+22).$ We know it must be the first because of AM-GM intuition: lesser factors of a number are closer together than larger factors of a number. (We can also try both and see which works.) Thus, $n(n+23)=(n+1)(n+21)$ and we find that $n=21,N=924$ meaning the answer is $\boxed{15}.$

Solution:
Since we are given that some pairs of divisors differ by 20 and 23 and we can let the pair be $(x-10)$ and $(x+10)$ as well as $(y-\frac{23}{2})$ and $(y+\frac{23}{2})$ . We also know the product of both the complementary divisors give the same number so $(x-10)(x+10)=(y-\frac{23}{2})(y+\frac{23}{2})$ .
Now we let $y=\frac{a}{2}$ . Then we substitute and get $x^2-100=\frac{(a^2-529)}{4}$ . Finally we multiply by 4 and get $4x^2-a^2=-129, a^2-4x^2=129$ .
Then we use differences of squares and get $a$ $2x$ =129, $a$ $2x$ =1. We finish by getting $a=$ 65 and $x=32$ . So $(42)(22) = 924$ Adding the digits, we have $9+2+4 = \boxed{15}$

Solution:
$N$ can be written $N = \left( a - 10 \right) \left( a + 10 \right)$ with a positive integer $a > 10$ and $N = \left( \frac{2b + 1}{2} - \frac{23}{2} \right) \left( \frac{2b + 1}{2} + \frac{23}{2} \right)$ with a positive integer $b > 11$
The above equations can be reorganized as \[ \left( 2b + 1 + 2 a \right) \left( 2 b + 1 - 2 a \right) = 43 \cdot 3 . \]
The only solution is $2b + 1 + 2a = 129$ and $2b + 1 - 2a = 1$ .
Thus, $a = b = 32$ .
Therefore, $N = 924$ .
So the sum of the digits of $N$ is $9 + 2 + 4 = \boxed{15}$

Solution:
We can write $N$ as $a(a+20)$ or $b(b+23)$ where $a$ and $b$ are divisors of $N.$ Since $a(a+20) = b(b+23),$ we know that $a^2 + 20a - b^2 - 23b = 0$ , and we can view this as a quadratic in $a.$
Since the solution for $a$ must be an integer, the discriminant for this quadratic must be a perfect square and therefore $20^2 - 4(-b^2 - 23b) = (2c)^2 = 400 + 4b^2 + 92b$ so $b^2 + 23b -c^2 + 100 = 0.$
Since the discriminant of this quadratic in $b$ must also be a perfect square we know that $23^2 - 4(-c^2+100) = d^2$ which we can simplify as $d^2 - 4c^2 = (d-2c)(d+2c) = 129.$ Since they are both positive integers $d - 2c$ and $d + 2c$ are factors of $129 = 3 \cdot 43$ so $d - 2c = 1$ and $d + 2c = 129$ or $d - 2c = 3$ and $d - 2c = 43.$
These systems of equations give us $(c,d) = (32,65)$ and $(c,d) = (10,23)$ respectively, if we plug our values for $c$ into the equation for $b$ we get $b^2 + 23b - 924 = 0$ and $b^2 + 23b = 0$ respectively. The first equation gives us $b = 21$ or $b = -44$ and the second gives us $b = 0$ or $b = -23$ , since $b$ is positive we know that $b = 21$ and $N = (21)(21 + 23) = 924$ , therefore the sum of the digits of $N$ is $9 + 2 + 4 = \boxed{15}.$

Solution: Consider the numbers of the form $a(a+20)$ . Since $b(b+23)$ is always even, $a$ is even. Thus, for $a \ge 2$ , we calculate $a(a+20)$ for even values of $a$ . Then, we check if it can also be represented as a product of numbers that differ by $23$ . Checking, we see that $22 \cdot 42 = 21 \cdot 44 = 924$ works. Thus, the answer is $9 + 2 + 4 = \boxed{15}$


Answer: 35.0

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 $b$
$\text{(A)}\ 11 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 37$

Solution:
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.
The only numbers left will be $a$ in the numerator from the last fraction and $2$ in the denominator from the first fraction.  (The $b$ will cancel with the numerator of the preceeding number.)  Thus, $\frac{a}{2} = 9$ , and $a=18$
Since the numerator is always one more than the denominator, $b = a-1 = 17$ , and $a+ b = 18 + 17 = 35$ , giving an answer of $\boxed{35}$

Solution:
Find a pattern.  If $a=4$ , then the expression is just the first two terms, which is $2$
If $a=5$ , then the expression is the first three terms, giving $2.5$
If $a=6$ , the expression is the first four terms, giving $3$
If $a=7$ , the expression will be the first five terms, giving $3.5$
Conjecture that the expression is always going to equal $\frac{a}{2}$ , and thus when $a=18$ , the expression will be $9$ , as desired.
As above, when $a=18$ $b=17$ , and the sum is $35$ , or $\boxed{35}$


Answer: 300.0

If the radius of a circle is increased $100\%$ , the area is increased:
$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$

Solution:
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 $2r.$
Since 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 $\boxed{300}$


Answer: 9902.0

If the sequence $\{a_n\}$ is defined by
$a_1=2$
$a_{n+1}=a_n+2n$
where $n\geq1$
Then $a_{100}$ equals
$\mathrm{(A) \ }9900 \qquad \mathrm{(B) \ }9902 \qquad \mathrm{(C) \ } 9904 \qquad \mathrm{(D) \ }10100 \qquad \mathrm{(E) \ } 10102$

Solution:
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: each term is the previous term plus the next even number. Therefore, since the differences of consecutive terms form an arithmetic sequence, then the terms satisfy a quadratic , specifically, the one that contains the points $(1, 2), (2, 4),$ , and $(3, 8)$ . Let the quadratic be $f(x)=ax^2+bx+c$ , so:
$a+b+c=2$ (1)
$4a+2b+c=4$ (2)
$9a+3b+c=8$ (3)
Subtracting (1) from (2) and (2) from (3) yields the two-variable system of equations
$3a+b=2$ (4)
$5a+b=4$ (5)
We can subtract (4) from (5) to find that $2a=2$ , so $a=1$ . Substituting this back in yields $b=-1$ , and substituting these back into one of the original equations yields $c=2$ , so the closed form for the terms is
$f(x)=x^2-x+2$ , or
$a_n=n^2-n+2$
Substituting in $n=100$ yields $a_{100}=100^2-100+2=9902, \boxed{9902}$


Answer: 2.0

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:
Solution by e_power_pi_times_i
Denote 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 $\dfrac{s^2}{\dfrac{1}{2}s^2} = \boxed{2}$


Answer: 15.0

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
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 19 \qquad \textbf{(E)}\ 21$

Solution:
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. \[180(n-2) > 2190\] \[n-2 > 12 \tfrac{1}{6}\] \[n > 14 \tfrac{1}{6}\]
The sum of the angles in a 15-sided polygon is $2340^{\circ}$ , making the remaining angle $150^{\circ}$ .  The angles of a convex polygon are all less than $180^{\circ}$ , and since adding one more side means adding $180^{\circ}$ to the measure of the remaining angle, we can confirm that there are $\boxed{15}$ sides in the polygon.


Answer: 110.0

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:
$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$

Solution:
Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.
Sum of the first 10 terms: $\frac{10}{2}(2a+9d)=100 \Longleftrightarrow 2a+9d=20$ Sum of the first 100 terms: $\frac{100}{2}(2a+99d)=10 \Longleftrightarrow 2a+99d=\frac{1}{5}$
Solving the system, we get $d=-\frac{11}{50}$ $a=\frac{1099}{100}$ . The sum of the first 110 terms is $\frac{110}{2}(2a+109d)=55(-2)=-110$
Therefore, $\boxed{110}$


Answer: 35.0

If the value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes, then $n=$
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45$

Solution: We have \begin{align*} 20(25)+10(10) &= 10(25)+n(10) \\ 600 &= 250+10n \\ 35 &= n \implies \boxed{35}


Answer: 16.0

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:
$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$

Solution: 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 $\frac1{\frac1{16}}=\boxed{16}$

Solution:
The two lines can be represented as $y=\frac{-x}{5}+20$ and $y=\frac{4x}{5}$ .
Solving the system,
$\frac{-x}{5}+20=\frac{4x}{5}$
$20=x.$
So the lines meet at an $x$ -coordinate of 20.
Solving for the height they meet,
\[y=\frac{4\cdot 20}{5}\] \[y=\boxed{16}.\]


Answer: 17.0

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:
$\textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4$

Solution: 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 therefore $\boxed{17}$


Answer: 91.0

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
$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$

Solution:
If the average score of the first six is $84$ , then the sum of those six scores is $6\times 84=504$
The average score of the first seven is $85$ , so the sum of the seven is $7\times 85=595$
Taking the difference leaves us with just the seventh score, which is $595-504=91$ , so the answer is $\boxed{91}$

Solution:
Let's remove the condition that the average of the first seven tests is $85$ , and say the 7th test score was a $0$ . Then, the average of the first seven tests would be \[\frac{6\times 84}{7}=72\]
If we increase the seventh test score by $7$ , the average will increase by $\frac{7}{7}=1$ . We need the average to increase by $85-72=13$ , so the seventh test score is $7\times 13=91$ more than $0$ , which is clearly $91$ .  This is choice $\boxed{91}$


Answer: 9.0

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?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

Solution: 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, they bought a total of $9$ items. Therefore, the answer is $\boxed{9}$

Solution:
Let $s$ be the number of sandwiches and $d$ be the number of sodas. We have to satisfy the equation of \[4.50s+d=30\] In the question, it states that Ike and Mike buys as many sandwiches as possible.
So, we drop the number of sodas for a while.
We have: \begin{align*} 4.50s&=30 \\ s&=\frac{30}{4.5} \\ s&=6R3 \end{align*} We don't want a remainder so the maximum number of sandwiches is $6$ .
The total money spent is $6\cdot 4.50=27$ .
The number of dollar left to spent on sodas is $30-27=3$ dollars. $3$ dollars can buy $3$ sodas leading us to a total of $6+3=9$ items.
Hence, the answer is $\boxed{9}$


Answer: 2.0

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?
$\textbf{(A)}\  5\%\text{ loss}\qquad \textbf{(B)}\ 2\%\text{ loss}\qquad \textbf{(C)}\ 1\%\text{ gain}\qquad \textbf{(D)}\ 2\% \text{ gain} \qquad \textbf{(E)}\  5\%\text{ gain}$

Solution: 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 $\boxed{2}$


Answer: 61.0

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 $BC$
$\textbf{(A)} \ 11 \qquad  \textbf{(B)} \ 28 \qquad  \textbf{(C)} \ 33 \qquad  \textbf{(D)} \ 61 \qquad  \textbf{(E)} \ 72$

Solution:
Let 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
\[CX \cdot CB = CD \cdot CE\] \[CX(CX+XB) = (97-86)(97+86)\] \[CX(CX+XB) = 3 \cdot 11 \cdot 61.\]
Since lengths cannot be negative, we must have $CX+XB \ge CX.$ This generates the four solution pairs for $(CX,CX+XB)$ \[(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).\]
However, by the Triangle Inequality on $\triangle ACX,$ we see that $CX>13.$ This implies that we must have $CX+XB= \boxed{61}.$

Solution: Let $BX = q$ $CX = p$ , and $AC$ meet the circle at $Y$ and $Z$ , with $Y$ on $AC$ .  Then $AZ = AY = 86$ .  Using the Power of a Point, we get that $p(p+q) = 11(183) = 11 * 3 * 61$ .  We know that $p+q>p$ , and that $p>13$ by the triangle inequality on $\triangle ACX$ .  Thus, we get that $BC = p+q = \boxed{61}$

Solution:
Let $x$ represent $CX$ , and let $y$ represent $BX$ . Since the circle goes through $B$ and $X$ $AB = AX = 86$ .
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$ , dividing both sides of the equation by $y$ is allowed.)
$x(x+y) = (97+86)(97-86)$
$x(x+y) = 2013$
The prime factors of $2013$ are $3$ $11$ , and $61$ . Obviously, $x < x+y$ . In addition, by the Triangle Inequality, $BC < AB + AC$ , so $x+y < 183$ . Therefore, $x$ must equal $33$ , and $x+y$ must equal $\boxed{61}$


Answer: 420.0

In $\bigtriangleup ABC$ $AB=BC=29$ , and $AC=42$ . What is the area of $\bigtriangleup ABC$
$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$

Solution: 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 $\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\boxed{420}$

Solution: Splitting the isosceles triangle in half, we get a right triangle with hypotenuse $29$ and leg $21$ . Using the Pythagorean Theorem , we know the height is $\sqrt{29^2-21^2}=20$ . Now that we know the height, the area is $\dfrac{(20)(42)}{2} = \boxed{420}$


Answer: 41.0

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 integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$

Solution:
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 $DF=EF=\frac{5x}{\sqrt{2}}$
Note that $\triangle DGE \sim \triangle ABC$ by SAS similarity, so $\angle BAC = \angle GDE$ and $\angle ACB = \angle DEG$ . Since $DEFG$ is a cyclic quadrilateral, $\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE$ and $\angle ACB = \angle DEG = \angle GFD$ , implying that $\triangle AFE$ and $\triangle CDF$ are isosceles. As a result, $AE=CD=\frac{5x}{\sqrt{2}}$ , so $BE=3-\frac{5x}{\sqrt{2}}$ and $BD =4-\frac{5x}{\sqrt{2}}$
Finally, using the Pythagorean Theorem on $\triangle BDE$ \[\left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2\] Solving for $x$ , we get that $x=\frac{5\sqrt{2}}{14}$ , so $DE= 5x =\frac{25 \sqrt{2}}{14}$ . Thus, the answer is $25+2+14=\boxed{041}$

Solution:
First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $\overline{DE}$ the same as the diameter of $\omega$ . We also note that $\triangle DGE \sim \triangle ABC$ since $\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$
From congruent arc intersections, we know that $\angle GED \cong \angle GBC$ , and that from similar triangles $\angle GED$ is also congruent to $\angle GCB$ . Thus, $\triangle BGC$ is an isosceles triangle with $BG = GC$ , so $G$ is the midpoint of $\overline{AC}$ and $AG = GC = 5/2$ . Similarly, we can find from angle chasing that $\angle ABF = \angle EDF = \frac{\pi}4$ . Therefore, $\overline{BF}$ is the angle bisector of $\angle B$ . From the angle bisector theorem, we have $\frac{AF}{AB} = \frac{CF}{CB}$ , so $AF = 15/7$ and $CF = 20/7$
Lastly, we apply power of a point from points $A$ and $C$ with respect to $\omega$ and have $AE \times AB=AF \times AG$ and $CD \times CB=CG \times CF$ , so we can compute that $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$ . From the Pythagorean Theorem, we result in $DE = \frac{25 \sqrt{2}}{14}$ , so $a+b+c=25+2+14= \boxed{041}$

Solution:
Call $DE=x$ and as a result $DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}$ . Since $EFGD$ is cyclic we just need to get $DG$ and using LoS(for more detail see the $2$ nd paragraph of Solution $2$ ) we get $AG=\frac{5}{2}$ and using a similar argument(use LoS again) and subtracting you get $FG=\frac{5}{14}$ so you can use Ptolemy to get $x=\frac{25\sqrt{2}}{14} \implies \boxed{041}$ .
~First

Solution: See inside the $\triangle DEF$ , we can find that $AG>AF$ since if $AG<AF$ , we can see that Ptolemy Theorem inside cyclic quadrilateral $EFGD$ doesn't work. Now let's see when $AG>AF$ , since $\frac{DG}{EG} = \frac{3}{4}$ , we can assume that $EG=4x;GD=3x;ED=5x$ , since we know $EF=FD$ so $\triangle EFD$ is isosceles right triangle. We can denote $DF=EF=\frac{5x\sqrt{2}}{2}$ .Applying Ptolemy Theorem inside the cyclic quadrilateral $EFGD$ we can get the length of $FG$ can be represented as $\frac{x\sqrt{2}}{2}$ . After observing, we can see $\angle AFE=\angle EDG$ , whereas $\angle A=\angle EDG$ so we can see $\triangle AEF$ is isosceles triangle. Since $\triangle ABC$ is a $3-4-5$ triangle so we can directly know that the length of AF can be written in the form of $3x\sqrt{2}$ . Denoting a point $J$ on side $AC$ with that $DJ$ is perpendicular to side $AC$ . Now with the same reason, we can see that $\triangle DJG$ is a isosceles right triangle, so we can get $GJ=\frac{3x\sqrt{2}}{2}$ while the segment $CJ$ is $2x\sqrt{2}$ since its 3-4-5 again. Now adding all those segments together we can find that $AC=5=7x\sqrt{2}$ and $x=\frac{5\sqrt{2}}{14}$ and the desired $ED=5x=\frac{25\sqrt{2}}{14}$ which our answer is $\boxed{041}$ ~bluesoul

Solution:
The main element of the solution is the proof that $BF$ is bisector of $\angle B.$
Let $O$ be the midpoint of $DE.$ $\angle EBF = 90^\circ \implies$
$O$ is the center of the circle $BDGFE.$ $\angle EOF = 90^\circ \implies \overset{\Large\frown} {EF} = 90^\circ \implies \angle EBF = 45^\circ \implies$ BF is bisector of $\angle ABC\implies BF = \frac {2AB \cdot BC}{AB+BC} \cos 45^\circ =\frac {12 \cdot \sqrt{2}}{7}.$ \[\angle EGD = 90^\circ, \frac {EG}{GD}=\frac{4}{3} \implies\] \[\angle GED = \angle GCD =\gamma \implies  \overset{\Large\frown} {DG} = 2\gamma.\] \[2\angle ACB =  \overset{\Large\frown} {BEF} -  \overset{\Large\frown} {DG} \implies  \overset{\Large\frown} {BEF} = 4 \gamma \implies\] \[\angle BOF = 4 \gamma \implies \angle OBF = \angle OFB = 90^\circ – 2 \gamma.\] Let $BO = EO = DO = r \implies BF = 2 r \cos(90^\circ – 2\gamma) =$ \[=2 r \sin 2\gamma = 4r \sin \gamma \cdot \cos \gamma = 4 r\cdot \frac {3}{5} \cdot \frac {4}{5} = \frac {48}{25} = \frac {12 \cdot \sqrt{2}}{7}\implies\] \[r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = 2r =  \frac {25 \cdot \sqrt{2}}{14}\implies  \boxed{041}.\] vladimir.shelomovskii@gmail.com, vvsss

Solution:
The main element of the solution is the proof that $G$ is midpoint of $AC.$
As in Solution 5 we get $\angle GED = \angle DBG =\gamma \implies$
$\triangle BCG$ is isosceles triangle with $BG=CG.$
Similarly $BG = AG \implies AG = CG = BG = \frac {AC}{2} =\frac {5}{2}.$
\[\overset{\Large\frown} {FG} = 90^\circ – \overset{\Large\frown} {GD} =  90^\circ – 2\gamma \implies\] \[\overset{\Large\frown} {BFG} = 4\gamma + 90^\circ – 2\gamma = 90^\circ + 2\gamma \implies\] \[\angle BOG = 90^\circ + 2\gamma \implies \angle BGO = \angle GBO = 45^\circ - \gamma.\] Let $\hspace{10mm} BO = EO = DO = r \implies$ \[BG = 2 r \cos(45^\circ – \gamma) = 2 r (\sin \gamma + \cos \gamma)\frac {\sqrt {2}}{2} =\] \[r \biggl(\frac {3}{5} + \frac {4}{5}\biggr) \sqrt {2} = r \frac {7 \sqrt{2}}{5} = \frac {5}{2}\implies\] \[r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = \frac {25 \cdot \sqrt{2}}{14}\implies  \boxed{041}.\] vladimir.shelomovskii@gmail.com, vvsss

Solution:
Let $(BEFGD) = \omega$ .
By Incenter-Excenter(Fact $5$ ), $F$ is the angle bisector of $\angle B$ .
Then by Ratio Lemma we have \[\frac{AG}{CG} = \frac{\sin(ABG)}{\sin(CBG)} \cdot \frac{AB}{BC} = \frac{\sin(GDE)}{\sin(DEG)} \cdot \frac{3}{4} = 1\] Thus, $G$ is the midpoint of $AC$
We can calculate $AF$ and $CF$ to be $\frac{15}{7}$ and $\frac{20}{7}$ respectively.
And then by Power of a Point, we have $\newline$ \[\operatorname{Pow}_{\omega}(A) = AE \cdot AB = AF \cdot AG \implies AE = \frac{25}{14}\] And then similarly, we have $CD = AE = \frac{25}{14}$ $\newline$
Then $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$ and by Pythagorean we have $DE = \frac{25\sqrt{2}}{14}$ , so our answer is $\boxed{041}.$


Answer: 61.0

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 $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$

Solution: 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 triangle inequality on $\triangle ACX$ .  Thus, $p$ is $33$ so we get that $BC = p+q = \boxed{61}$

Solution:
Let $x$ represent $CX$ , and let $y$ represent $BX$ . Since the circle goes through $B$ and $X$ $AB = AX = 86$ .
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 y + xy^2 + 86^2 y = 97^2 y$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$ , dividing both sides of the equation by $y$ is allowed.)
$x(x+y) = (97+86)(97-86)$
$x(x+y) = 2013$
The prime factors of $2013$ are $3$ $11$ , and $61$ . Obviously, $x < x+y$ . In addition, by the Triangle Inequality, $BC < AB + AC$ , so $x+y < 183$ . Therefore, $x$ must equal $33$ , and $x+y$ must equal $\boxed{61}$

Solution:
Let $CX=x, BX=y$ . Let the circle intersect $AC$ at $D$ and the diameter including $AD$ intersect the circle again at $E$ .
Use power of a point on point C to the circle centered at A.
So $CX \cdot CB=CD \cdot CE \Rightarrow x(x+y)=(97-86)(97+86) \Rightarrow x(x+y)=3*11*61$
Obviously $x+y>x$ so we have three solution pairs for $(x,x+y)=(1,2013),(3,671),(11,183),(33,61)$ .
By the Triangle Inequality, only $x+y=61$ yields a possible length of $BX+CX=BC$
Therefore, the answer is $\boxed{61}$

Solution:
We first draw the height of isosceles triangle $ABD$ and get two equations by the Pythagorean Theorem .
First, $h^2 + k^2 = 86^2$ . Second, $h^2 + (k + m)^2 = 97^2$ .
Subtracting these two equations, we get $2km + m^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 2013$ .
We then add $k^2$ to both sides to get $k^2 + 2km + m^2 = 2013 + k^2$ .
We then complete the square to get $(k + m)^2 = 2013 + k^2$ . Because $k$ and $m$ are both integers, we get that $2013 + k^2$ is a square number. Simple guess and check reveals that $k = 14$ .
Because $k$ equals $14$ , therefore $m = 33$ . We want $\overline{BC} = 2k + m$ , so we get that $\overline{BC} = \boxed{61}$

Solution: Let $E$ be the foot of the altitude from $A$ to $BX.$ Since $\triangle ABX$ is isosceles $AX=AB=86,EB=EX,$ and the answer is $EC+EB=EC+EX.$ $(EC+EX)(EC-EX)=EC^2-EX^2=(97^2-AE^2)-(86^2-AE^2)=97^2-86^2=2013$ by the Pythagorean Theorem. Only $EC+EX=\boxed{61}.$


Answer: 306.0

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 $d$

Solution:
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 sections are parallelograms
By similar triangles, $BE'=\frac{d}{510}\cdot450=\frac{15}{17}d$ and $EC=\frac{d}{425}\cdot450=\frac{18}{17}d$ . Since $FD'=BC-EE'$ , we have $900-\frac{33}{17}d=d$ , so $d=\boxed{306}$

Solution:
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 sections are parallelograms
Since $PDAF'$ is a parallelogram, we find $PF' = AD$ , and similarly $PE = BD'$ . So $d = PF' + PE = AD + BD' = 425 - DD'$ . Thus $DD' = 425 - d$ . By the same logic, $EE' = 450 - d$
Since $\triangle DPD' \sim \triangle ABC$ , we have the proportion
Doing the same with $\triangle PEE'$ , we find that $PE' =510 - \frac{17}{15}d$ . Now, $d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}$

Solution: Refer to the diagram in solution 2; let $a^2=[E'EP]$ $b^2=[D'DP]$ , and $c^2=[F'FP]$ . Now, note that $[E'BD]$ $[D'DP]$ , and $[E'EP]$ are similar, so through some similarities we find that $\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2$ . Similarly, we find that $[D'AF]=(b+c)^2$ and $[F'CE]=(c+a)^2$ , so $[ABC]=(a+b+c)^2$ . Now, again from similarity, it follows that $\frac{d}{510}=\frac{a+b}{a+b+c}$ $\frac{d}{450}=\frac{b+c}{a+b+c}$ , and $\frac{d}{425}=\frac{c+a}{a+b+c}$ , so adding these together, simplifying, and solving gives $d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}$ $=\frac{10}{\frac{10}{306}}=\boxed{306}$

Solution:
Refer to the diagram above. Notice that because $CE'PF$ $AF'PD$ , and $BD'PE$ are parallelograms, $\overline{DD'} = 425-d$ $\overline{EE'} = 450-d$ , and $\overline{FF'} = 510-d$
Let $F'P = x$ . Then, because $\triangle ABC \sim \triangle F'PF$ $\frac{AB}{AC}=\frac{F'P}{F'F}$ , so $\frac{425}{510}=\frac{x}{510-d}$ . Simplifying the LHS and cross-multiplying, we have $6x=2550-5d$ . From the same triangles, we can find that $FP=\frac{18}{17}x$
$\triangle PEE'$ is also similar to $\triangle F'PF$ . Since $EF'=d$ $EP=d-x$ . We now have $\frac{PE}{EE'}=\frac{F'P}{FP}$ , and $\frac{d-x}{450-d}=\frac{17}{18}$ . Cross multiplying, we have $18d-18x=450 \cdot 17-17d$ . Using the previous equation to substitute for $x$ , we have: \[18d-3\cdot2550+15d=450\cdot17-17d\] This is a linear equation in one variable, and we can solve to get $d=\boxed{306}$


Answer: 110.0

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:
$\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}$

Solution: 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 $\angle BOC = 180 - (x + 70 - x) = \boxed{110}$


Answer: 56.0

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] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,S); label("$F$",F,dir(0)); [/asy]
$\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }60\qquad \textbf{(E) }72\qquad$

Solution:
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.
It follows that $BD = DE$ . Thus, $AD + DE = AD + DB = AB = 28$
Since opposite sides of parallelograms are equal, the perimeter is $2 * (AD + DE) =   \boxed{56}$

Solution:
We can set point $F$ to be on point $C$ , and point $D$ to be on point $A$
This makes a degenerate parallelogram with sides of length $28$ and $0$ , so it has a perimeter of $28 + 28 =  \boxed{56}$


Answer: 75.0

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 $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$

Solution:
$\angle ADB = 120^\circ$ $\angle ADC = 60^\circ$ $\angle DAB = 15^\circ$ , let $\angle ACB = \theta$ $\angle DAC = 120^\circ - \theta$
By the Law of Sines , we have $\frac{CD}{\sin(120^\circ - \theta)} = \frac{AD}{\sin \theta}$
$\space$ $\frac{BD}{\sin15^\circ} = \frac{AD}{\sin45^\circ}$
$\frac{BD}{CD} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}$
$\frac{1}{2} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}$
By the Triple-angle Identities $\sin 45^\circ = 3\sin15^\circ -4\sin^3 15^\circ$
$\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{3\ -4\sin^2 15^\circ}$
$\sin^2 15^\circ = \frac{1 - \cos 30^\circ}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}$
$\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{3 - 4 \cdot \frac{2 - \sqrt{3}}{4}} = \frac{\sin \theta}{1+\sqrt{3}}$
$\frac{\sin \theta}{\sin(120^\circ - \theta)} = \frac{1+\sqrt{3}}{2}$
$\sin(120^\circ - \theta) = \sin120^\circ \cos\theta - \sin\theta \cos120^\circ = \frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{1}{2} \cdot \sin\theta$ ，so
$\frac{\sin \theta}{\frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{1}{2} \cdot \sin\theta} = \frac{1+\sqrt{3}}{2}$
$\frac{\sqrt{3} \cdot \cos\theta + \sin\theta}{2 \sin \theta} = \frac{2}{1+\sqrt{3}}$
$\frac{\sqrt{3}}{2} \cdot \cot \theta = \frac{2}{1+\sqrt{3}} - \frac{1}{2} = \frac{3 - \sqrt{3}}{2 + 2 \sqrt{3}}$
$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
Suppose $\cos \theta = k(\sqrt{3} - 1)$ , and $\sin \theta = k(\sqrt{3} + 1)$
$\sin^2 \theta + \cos^2 \theta = 1$ $k^2(\sqrt{3} + 1)^2 + k^2(\sqrt{3} - 1)^2 = 8k^2 = 1$ $k = \frac{1}{2 \sqrt{2}}$
$\sin \theta = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$ $\cos \theta = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$
$\sin 2 \theta = 2 \cdot \frac{\sqrt{3} + 1}{2 \sqrt{2}} \cdot \frac{\sqrt{3} - 1}{2 \sqrt{2}} = \frac{1}{2}$
Two possible values of $2 \theta$ are $150^\circ$ and $30^\circ$ . However we can rule out $30^\circ$ because $\cos 15^\circ$ is positive, while $\cos \theta$ is negative.
Therefore $2 \theta = 150^\circ$ $\angle ACB = \boxed{75}$


Answer: 75.0

Solution:
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, so let $AH=kt$ . Notice that in $\triangle{ABH}$ we get $BH=kt$ . Using the 60-degree angle in $\triangle{ADH}$ , we obtain $DH=\frac{\sqrt{3}}{3}kt$ . The comparable ratio is that $BH-DH=t$ . If we involve our $k$ , we get:
$kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t$ . Eliminating $t$ and removing radicals from the denominator, we get $k=\frac{3+\sqrt{3}}{2}$ . From there, one can easily obtain $HC=3t-kt=\frac{3-\sqrt{3}}{2}t$ . Now we finally have a desired ratio. Since $\tan\angle ACH = 2+\sqrt{3}$ upon calculation, we know that $\angle ACH$ can be simplified. Indeed, if you know that $\tan(75)=2+\sqrt{3}$ or even take a minute or two to work out the sine and cosine using $\sin(x)^2+\cos(x)^2=1$ , and perhaps the half- or double-angle formulas, you get $\boxed{75}$

Solution:
Without loss of generality, we can assume that $BD = 1$ and $CD = 2$ . As above, we are able to find that $\angle ADC = 60^\circ$ and $\angle ADB = 120^\circ$
Using Law of Sines on triangle $ADB$ , we find that \[\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}.\] Since we know that \[\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4},\] \[\sin 45^\circ = \frac{\sqrt{2}}{2},\] \[\sin 120^\circ = \frac{\sqrt{3}}{2},\] we can compute $AD$ to equal $1+\sqrt{3}$ and $AB$ to be $\frac{3\sqrt{2}+\sqrt{6}}{2}$
Next, we apply Law of Cosines to triangle $ADC$ to see that \[AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ).\] Simplifying the right side, we get $AC^2 = 6$ , so $AC = \sqrt{6}$
Now, we apply Law of Sines to triangle $ABC$ to see that \[\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}.\] After rearranging and noting that $\sin 45^\circ = \frac{\sqrt{2}}{2}$ , we get \[\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}.\]
Dividing the right side by $\sqrt{3}$ , we see that \[\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4},\] so $\angle ACB$ is either $75^\circ$ or $105^\circ$ . Since $105^\circ$ is not a choice, we know $\angle ACB = \boxed{75}$

Solution:
For starters, we have $\angle ABD=120^\circ.$ Dropping perpendiculars $\overline{DX}$ and $\overline{CY}$ from $D$ and $C$ to $\overline{AB}$ gives $\angle ADX=120^\circ-45^\circ=75^\circ,$ since $\angle BDX=45^\circ.$
Without loss of generality, let $BD=1$ and $CD=2.$ This tells us that $BX=DX=\sqrt{2}/2.$ Using trigonometric identities, we find that
\[\tan \angle ADX=\tan 75^\circ=\dfrac{\sqrt 6+\sqrt 2}{\sqrt 6-\sqrt 2}.\] Thus, $AX/DX=\dfrac{\sqrt 6+\sqrt 2}{\sqrt 6-\sqrt 2},$ which gives $AX=\dfrac{2\sqrt 2+\sqrt 6}{2}.$ Thus, $AB=AX+BX=\dfrac{3\sqrt 2+\sqrt 6}{2}.$
Now, note that $BCY$ is a $45-45-90$ triangle, so $CY=BY=\dfrac{3\sqrt 2}{2}.$ Thus, we have
\[[ABC]=\dfrac{1}{2}\cdot AB\cdot CY=\dfrac{1}{2}\cdot \dfrac{3\sqrt 2+\sqrt 6}{2}\cdot \dfrac{3\sqrt 2}{2}=\dfrac{9+3\sqrt 3}{4}.\] Additionally, note that $AY=AB-BY=\sqrt{6}/2.$ Applying the Pythagorean Theorem to triangle $AYC$ then tells us that $AC=\sqrt{6}.$ By the trigonometric formula for area,
\[[ABC]=\dfrac{1}{2}\cdot BC\cdot AC\cdot \sin \angle ACB=\dfrac{3\sqrt 6}{2}\sin \angle ACB.\] Setting this equal to our other area and solving gives $\sin \angle ACB=\dfrac{\sqrt 6+\sqrt 2}{4},$ so $\angle ACB=\boxed{75}.$ ~vaporwave


Answer: 120.0

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:
$\textbf{(A)}\ 150^{\circ} \qquad \textbf{(B)}\ 120^{\circ}\qquad \textbf{(C)}\ 105^{\circ} \qquad \textbf{(D)}\ 90^{\circ} \qquad \textbf{(E)}\ 60^{\circ}$

Solution:
Using the Law of Cosines \[\sqrt{3 + 3 - 2\cdot 3 \cdot \cos{x^\circ}}>3\]
Both sides are positive, so squaring both sides will not affect the inequality.
\[6 - 6 \cos{x^\circ}>9\] \[\cos{x^\circ} < -\frac{1}{2}\]
Note that $\cos{120^\circ} = -\frac{1}{2}$ .  As $x$ gets closer to $180^{\circ}$ $\cos{x}$ decreases towards $-1$ .  Thus, $x > 120$ , so the answer is $\boxed{120}$