Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: What is the smallest positive integer $n$ such that $\sqrt{n}-\sqrt{n-1}<.01$
$\textbf{(A) }2499\qquad \textbf{(B) }2500\qquad \textbf{(C) }2501\qquad \textbf{(D) }10,000\qquad  \textbf{(E) }\text{There is no such integer}$

Solution: Adding $\sqrt{n - 1}$ to both sides, we get \[\sqrt{n} < \sqrt{n - 1} + 0.01.\] Squaring both sides, we get \[n < n - 1 + 0.02 \sqrt{n - 1} + 0.0001,\] which simplifies to \[0.9999 < 0.02 \sqrt{n - 1},\] or \[\sqrt{n - 1} > 49.995.\] Squaring both sides again, we get \[n - 1 > 2499.500025,\] so $n > 2500.500025$ . The smallest positive integer $n$ that satisfies this inequality is $\boxed{2501}$

Answer:  2501
Problem: A positive integer $n$ not exceeding $100$ is chosen in such a way that if $n\le 50$ , then the probability of choosing $n$ is $p$ , and if $n > 50$ , then the probability of choosing $n$ is $3p$ . The probability that a perfect square is chosen is
$\textbf{(A) }.05\qquad \textbf{(B) }.065\qquad \textbf{(C) }.08\qquad \textbf{(D) }.09\qquad  \textbf{(E) }.1$

Solution: Let's say that we will have $3$ slips for every number not exceeding $100$ but bigger than $50.$ This is to account for the $3p$ probability part. Let's now say that we will only have one slip for each number below or equal to $50.$ The probability(or $p$ ) will then be $\frac{1}{200}.$ Now let's have all the squares under $50,$ which are $1,4,9,16,25,36,49.$ The probability for these are $\frac{7}{200}.$ The numbers above $50$ that are squares are $64,81,100.$ We then need to multiply the probability by $3$ so the probability of these are $\frac{9}{200}.$ The answer is $\frac{7}{200}+\frac{9}{200}=0.08\implies\boxed{.08}$

Answer: C .08
Problem: The following four statements, and only these are found on a card: [asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On this card exactly three statements are false.", F, SE); label("On this card exactly four statements are false.", G, SE); [/asy]
(Assume each statement is either true or false.) Among them the number of false statements is exactly
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: There can be at most one true statement on the card, eliminating $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$ . If there are $0$ true on the card, statement $4$ ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is $\boxed{3}$ , since $3$ are false and only the third statement ("On this card exactly three statements are false") is correct.

Answer: D 3
Problem: The following four statements, and only these are found on a card: [asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On this card exactly three statements are false.", F, SE); label("On this card exactly four statements are false.", G, SE); [/asy]
(Assume each statement is either true or false.) Among them the number of false statements is exactly
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: If all of them are false, that would mean that the $4$ th one is false too. Therefore, $E$ is not the correct answer. If exactly $3$ of them are false, that would mean that only $1$ statement is true. This is correct since if only $1$ statement is true, the card that is true is the one that has $3$ of these statements are false. If we have $1$ or $2$ false statements, that would mean that there is more than $1$ true statement. Therefore, our answer is $\boxed{3}$

Answer: D 3
Problem: [asy] size(100); real a=4, b=3; // import cse5; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); //Credit to chezbgone2 for the diagram [/asy]
In $\triangle ABC, AB = 10~ AC = 8$ and $BC = 6$ . Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB$ . Let $Q$ and $R$ be the points of intersection, distinct from $C$ , of circle $P$ with sides $AC$ and $BC$ , respectively. The length of segment $QR$ is
$\textbf{(A) }4.75\qquad \textbf{(B) }4.8\qquad \textbf{(C) }5\qquad \textbf{(D) }4\sqrt{2}\qquad  \textbf{(E) }3\sqrt{3}$

Solution: We know that triangle $RCQ$ is similar to triangle $ABC$ . We draw a line to point $D$ on hypotenuse $AB$ such that $\angle QDR$ is $90 ^\circ$ and that $RDQC$ is a rectangle. Since triangle $RCQ$ is similar to triangle $ABC$ , let $RC$ be $4x$ and $RD/CQ$ be $3x$ . Now we have line segment $AQ$ $8-3x$ , and line segment $RB$ $6-4x$ . Since $BD + DA = AB$ , we use simple algebra and Pythagorean Theorem to get $\sqrt {(3x)^2 + (6-4x)^2}$ $\sqrt {(4x)^2 + (8-3x)^2}$ $10$ . Expanding and simplifying gives us $\sqrt {25x^2-48x+36}$ $\sqrt {25x^2-48x+64}$ $10$
Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\sqrt {25x^2-48x+36}$ . Now, we can square both sides and simplify to get $0 = 72 - 20 \sqrt{25x^2-48x+36}$ . Dividing both sides by $4$ , we get $18 - 5 \sqrt {25x^2-48x+36}$ $0$ . We then add $5 \sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \sqrt {25x^2-48x+36}$ . Since this is very messy, let $25x^2 - 48x = y$ . Squaring both sides, we get $324 = 25y + 900, 25y = -576$ . Solving for $y$ , we have $y = -23.04$ . Plugging in $y$ as $25x^2-48x$ , we have $25x^2-48x+23.04 = 0$ . Using the quadratic equation, we get $\frac {48+0}{50}$ . Therefore, $x = \frac {48}{50}$
Remember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$ . Since $5x$ is the hypotenuse, our answer is $\boxed{4.8}$

Answer: B 4.8
Problem: 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: D 7
Problem: For every triple $(a,b,c)$ of non-zero real numbers, form the number $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$ . 
The set of all numbers formed is
$\textbf{(A)}\ {0} \qquad \textbf{(B)}\ \{-4,0,4\} \qquad \textbf{(C)}\ \{-4,-2,0,2,4\} \qquad \textbf{(D)}\ \{-4,-2,2,4\}\qquad \textbf{(E)}\ \text{none of these}$

Solution: Solution by e_power_pi_times_i
$\dfrac{x}{|x|} = 1$ or $-1$ depending whether $x$ is positive or negative. If $a$ $b$ , and $c$ are positive, then the entire thing amounts to $4$ . If one of the three is negative and the other two positive, the answer is $0$ . If two of the three is negative and one is positive, the answer is $0$ . If all three are negative, the answer is $-4$ . Therefore the set is $\boxed{4,0,4}$

Answer: B 4,0,4
Problem: [asy] size(120); path c = Circle((0, 0), 1); pair A = dir(20), B = dir(130), C = dir(240), D = dir(330); draw(c); pair F = 3(A-B) + B; pair G = 3(D-C) + C; pair E = intersectionpoints(B--F, C--G)[0]; draw(B--E--C--A); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E); //Credit to MSTang for the diagram [/asy]
In the adjoining figure $\measuredangle E=40^\circ$ and arc $AB$ , arc $BC$ , and arc $CD$ all have equal length. Find the measure of $\measuredangle ACD$
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }\left(\frac{45}{2}\right)^\circ\qquad \textbf{(E) }30^\circ$

Solution: Solution by e_power_pi_times_i
If arcs $AB$ $BC$ , and $CD$ are congruent, then $\measuredangle ACB = \measuredangle BDC = \measuredangle CBD = \theta$ . Because $ABCD$ is cyclic, $\measuredangle CAD = \measuredangle CBD = \theta$ , and $\measuredangle ADB = \measuredangle ACB = \theta$ . Then, $\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ$ $\theta = 55^\circ$ $\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{15}$

Answer: B 15
Problem: If $(3x-1)^7 = a_7x^7 + a_6x^6 + \cdots + a_0$ , then $a_7 + a_6 + \cdots + a_0$ equals
$\text{(A)}\ 0 \qquad  \text{(B)}\ 1 \qquad  \text{(C)}\ 64 \qquad  \text{(D)}\ -64 \qquad  \text{(E)}\ 128$

Solution: Solution by e_power_pi_times_i
Notice that if $x=1$ , then $a_7x^7 + a_6x^6 + \cdots + a_0 =  a_7 + a_6 + \cdots + a_0$ . Therefore the answer is $(3(1)-1)^7) = \boxed{128}$

Answer: D 128
Problem: Al's age is $16$ more than the sum of Bob's age and Carl's age, and the square of Al's age is $1632$ more than the square of the sum of 
Bob's age and Carl's age. What is the sum of the ages of Al, Bob, and Carl?
$\text{(A)}\ 64 \qquad  \text{(B)}\ 94 \qquad  \text{(C)}\ 96 \qquad  \text{(D)}\ 102 \qquad  \text{(E)}\ 140$

Solution: Solution by e_power_pi_times_i
Denote Al's age, Bob's age, and Carl's age by $a$ $b$ , and $c$ , respectively. Then, $a = 16 + b + c$ and $a^2 = 1632 + b^2 + c^2$ . Substituting the first equation into the second, $(16 + b + c)^2 = b^2 + c^2 + 2bc + 32b + 32c + 256 = b^2 + c^2 + 1632$ . Thus, $bc + 16b + 16c = 688$ , and $(b+16)(c+16) = 944$ . Since $944 = 2^4\cdot59$ $(b,c) = (0,43)$ or $(43,0)$ . Then $a + b + c = 2b + 2c + 16 = \boxed{102}$

Answer: D 102
Problem: How many pairs $(m,n)$ of integers satisfy the equation $m+n=mn$
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad  \textbf{(E) }\text{more than }4$

Solution: Solution by e_power_pi_times_i
If $m+n=mn$ $mn-m-n = (m-1)(n-1)-1 = 0$ . Then $(m-1)(n-1) = 1$ , and $(m,n) = (2,2) or (0,0)$ . The answer is $\boxed{2}$

Answer: B 2
Problem: 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: B 5
Problem: Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n$
$\textbf{(A) }102\qquad \textbf{(B) }112\qquad \textbf{(C) }249\qquad \textbf{(D) }502\qquad \textbf{(E) }\text{none of the above}\qquad$

Solution: We first observe that since there will be more 2s than 5s in $1005!$ , we are looking for the largest $n$ such that $5^n$ divides $1005!$ . We will use the fact that:
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor \cdots\]
(This is an application of Legendre's formula).
From $k=5$ and onwards, $\left \lfloor {\frac{1005}{5^k}}\right \rfloor = 0$ . Thus, our calculation becomes
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor + \left \lfloor {\frac{1005}{5^4}}\right \rfloor\]
\[n = \left \lfloor {\frac{1005}{5}}\right \rfloor + \left \lfloor {\frac{1005}{25}}\right \rfloor + \left \lfloor {\frac{1005}{125}}\right \rfloor + \left \lfloor {\frac{1005}{625}}\right \rfloor\]
\[n = 201 + 40 + 8 + 1 = \boxed{250}\]

Answer:  250
Problem: Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$

Solution: Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$ . Then $r(x)$ is the unique polynomial such that \[g(x^{12}) - r(x) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 - r(x)\] is divisible by $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$ , and $\deg r(x) < 5$
Note that $(x - 1)(x^5 + x^4 + x^3 + x^2 + 1) = x^6 - 1$ is a multiple of $g(x)$ . Also, \[g(x^{12}) - 6 = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} - 5 \\ = (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).\] Each term is a multiple of $x^6 - 1$ . For example, \[x^{60} - 1 = (x^6 - 1)(x^{54} + x^{48} + \dots + x^6 + 1).\] Hence, $g(x^{12}) - 6$ is a multiple of $x^6 - 1$ , which means that $g(x^{12}) - 6$ is a multiple of $g(x)$ . Therefore, the remainder is $\boxed{6}$ . The answer is (A).

Answer:  6
Problem: Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$

Solution: We express the quotient and remainder as follows. \[g(x^{12}) = Q(x) g(x) + R(x)\] Note that the solutions to $g(x)$ correspond to the 6th roots of unity, excluding $1$ . Hence, we have $x^6 = 1$ , allowing us to set: \[g(x^{12}) = 6\] \[g(x) = 0\] We have $5$ values of $x$ that return $R(x) = 6$ . However, $g(x)$ is quintic, implying the remainder is of degree at most $4$ . Since there are $5$ solutions, the only possibility is that the remainder is a constant $\boxed{6}$

Answer:  6
Problem: Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$

Solution: We can use the Chinese remainder theorem over $\mathbb{Q}[x].$ Since $g(x)=(x+1)(x^2-x+1)(x^2+x+1)$ $\mathbb{Q}[x]/g\cong \mathbb{Q}[x]/(x+1)\times \mathbb{Q}[x]/(x^2-x+1)\times \mathbb{Q}[x]/(x^2+x+1).$ This means that if we can find the remainder of $g(x^{12})$ modulo $x+1,x^2-x+1,x^2+x+1$ , we can reconstruct the remainder modulo $g.$ We can further use that each factor is irreducible and that if $p(x)$ is an irreducible polynomial over $\mathbb{Q}$ with root $\alpha$ $\mathbb{Q}[x]/p\cong \mathbb{Q}(\alpha)$ so to evaluate the remainders of $g(x^{12})$ , we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root $-1$ , the second has roots the primitive sixth roots of unity, and the third as roots the primitive cube roots of unity (this is easily seen as $g(x)(x-1)=x^6-1$ ). Evaluating $g(x^{12})$ on each of these values yields $g(1)=6$ so the remainder is $6$ on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo $g$ must be $\boxed{6}$ as well.

Answer:  6
Problem: A supermarket has $128$ crates of apples. Each crate contains at least $120$ apples and at most $144$ apples. 
What is the largest integer $n$ such that there must be at least $n$ crates containing the same number of apples?
$\textbf{(A) }4\qquad \textbf{(B) }5\qquad \textbf{(C) }6\qquad \textbf{(D) }24\qquad  \textbf{(E) }25$

Solution: To find the largest number of "repeated" crates necessary, we must account for all the possibilities of the number of apples in each crate. Since each crate contains a minimum of $120$ apples and a maximum of $144$ apples, there are $144 - 120 + 1 = 25$ different amounts possible for the number of apples per crate.
Now, we have to count for the worst case scenario: the $25$ amounts are repeated as many times as possible.
$25$ can go into $128$ exactly $5$ times because $5 \cdot 25 = 125$ , which is less than $128$ . This leaves a remainder of $3$ crates.
The worst case scenario would be that these $3$ crates have a different number of apples each. It doesn't actually matter how many apples are in these $3$ crates because any of the $25$ values would be repeated again anyway. So, the answer is $5 + 1 = \boxed{6}$ jiang147369

Answer: C 6
Problem: The measures of the interior angles of a convex polygon are in arithmetic progression. 
If the smallest angle is $100^\circ$ , and the largest is $140^\circ$ , then the number of sides the polygon has is
$\textbf{(A) }6\qquad \textbf{(B) }8\qquad \textbf{(C) }10\qquad \textbf{(D) }11\qquad  \textbf{(E) }12$

Solution: Let $n$ equal the number of sides the polygon has. The sum of all the interior angles of a polygon is: $180(n-2)$
The formula for an arithmetic series is $\frac{n(a_1 + a_n)}{2}$ . Set this equal to $180(n-2)$ and solve. In this case, $a_1=100$ and $a_n=140$
Our equation becomes $\frac{n(100+140)}{2} = 180(n-2) \Rightarrow 240n = 360(n-2) \Rightarrow 120n = 720$
Simplifying, we get $n = \boxed{6}$ jiang147369

Answer: A 6
Problem: 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: B 15
Problem: 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: A 1
Problem: What is the smallest positive odd integer $n$ such that the product $2^{1/7}2^{3/7}\cdots2^{(2n+1)/7}$ is greater than $1000$ ? 
(In the product the denominators of the exponents are all sevens, and the numerators are the successive odd integers from $1$ to $2n+1$ .)
$\textbf{(A) }7\qquad \textbf{(B) }9\qquad \textbf{(C) }11\qquad \textbf{(D) }17\qquad  \textbf{(E) }19$

Solution: Combine the terms in the product to get $2^{\frac{1+3+5+ \dots +(2n-1)+(2n+1)}{7}}$
The exponent can be simplified to \[\frac{1+3+5+ \dots +(2n-1)+(2n+1)}{7} \Rightarrow \frac{\frac{n(1+(2n+1))}{2}}{7} \Rightarrow \frac{n^2}{7}.\]
We want this inequality to be true with the smallest positive odd integer value of $n$ \[2^{\frac{n^2}{7}} > 1000.\]
Now, let's test the answer choices. For $n=7$ , we have $2^{49/7}=2^{7}<1000$ . For $n=9$ , we have $2^{81/7}>2^{10}>1000$
So our answer is $\boxed{9}$ . ~ jiang147369

Answer: B 9
Problem: In the adjoining figure, circle $K$ has diameter $AB$ ; circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K$ ; and circle $M$ tangent to circle $K$ , to circle $L$ and $AB$ . The ratio of the area of circle $K$ to the area of circle $M$ is [asy] /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ size(150); pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25); draw(circle(K,1)^^A--B); draw(circle(L,0.5)^^circle(M,.25)); label("$A$", A, W); label("$K$", K, S); label("$B$", B, E); label("$L$", L); label("$M$", M); [/asy] $\textbf{(A) }12\qquad \textbf{(B) }14\qquad \textbf{(C) }16\qquad \textbf{(D) }18\qquad  \textbf{(E) }\text{not an integer}$

Solution: Let $R$ and $r$ be the radius of $\odot K$ and the radius of $\odot M,$ respectively. It follows that the radius of $\odot L$ is $\frac{R}{2}.$
Suppose $P$ is the foot of the perpendicular from $M$ to $\overline{KL}.$ We construct the auxiliary lines, as shown below: [asy] /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ size(200); pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25),I=(2*sqrt(2)/3,1/3),E=(sqrt(2)/3,1/3),P=(0,0.25); draw(circle(K,1)^^A--B); draw(circle(L,0.5)^^circle(M,.25)); draw(L--K,red); draw(L--M,red); draw(K--I,red); draw(P--M,red); label("$A$", A, (-5/4,0)); label("$K$", K, (0,-5/4)); label("$B$", B, (5/4,0)); label("$L$", L, (0,5/4)); label("$M$", M, (0,5/4)); label("$P$", P, (-5/4,0)); dot(K,linewidth(4)); dot(L,linewidth(4)); dot(M,linewidth(4)); dot(I,linewidth(4)); dot(E,linewidth(4)); dot(P,linewidth(4)); [/asy] In right $\triangle KPM,$ we have $KP=r$ and $KM=R-r.$ By the Pythagorean Theorem, we get $PM^2=(R-r)^2-r^2.$
In right $\triangle LPM,$ we have $LP=\frac{R}{2}-r$ and $LM=\frac{R}{2}+r.$ By the Pythagorean Theorem, we get $PM^2=\left(\frac{R}{2}+r\right)^2-\left(\frac{R}{2}-r\right)^2.$
We equate the expressions for $PM^2,$ then simplify: \begin{align*} (R-r)^2-r^2&=\left(\frac{R}{2}+r\right)^2-\left(\frac{R}{2}-r\right)^2 \\ \left(R^2-2Rr+r^2\right)-r^2&=\left(\frac{R^2}{4}+Rr+r^2\right)-\left(\frac{R^2}{4}-Rr+r^2\right) \\ R^2-2Rr&=2Rr \\ R^2&=4Rr \\ R&=4r. \end{align*} Therefore, the ratio of the area of $\odot K$ to the area of $\odot M$ is $\frac{\pi R^2}{\pi r^2}=\left(\frac{R}{r}\right)^2=\boxed{16}.$

Answer: C 16
Problem: 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 \[N=x-y=\boxed{1}.\]

Answer: A 1
Problem: 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 \[x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}. \hspace{15mm} (\bigstar)\] 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: A 1