Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
A sample transformation is shown in the figure below. [asy]         import geometry;         unitsize(0.6cm);          void ds(pair x) {             filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible);         }          ds((1,1));         ds((2,1));         ds((3,1));         ds((1,3));          for (int i = 0; i <= 5; ++i) {             draw((0,i)--(5,i));             draw((i,0)--(i,5));         }          label("Initial", (2.5,-1));         draw((6,2.5)--(8,2.5),Arrow);          ds((10,2));         ds((11,1));         ds((11,0));          for (int i = 0; i <= 5; ++i) {             draw((9,i)--(14,i));             draw((i+9,0)--(i+9,5));         }          label("Transformed", (11.5,-1)); [/asy] Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.) [asy]         import geometry;         unitsize(0.6cm);          void ds(pair x) {             filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible);         }          for (int i = 1; i < 4; ++ i) {             for (int j = 1; j < 4; ++j) {                 label("?",(i + 0.5, j + 0.5));             }         }          for (int i = 0; i <= 5; ++i) {             draw((0,i)--(5,i));             draw((i,0)--(i,5));         }          label("Initial", (2.5,-1));         draw((6,2.5)--(8,2.5),Arrow);          ds((11,2));          for (int i = 0; i <= 5; ++i) {             draw((9,i)--(14,i));             draw((i+9,0)--(i+9,5));         }          label("Transformed", (11.5,-1)); [/asy] $\textbf{(A)}\ 14 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 22 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 30$

Solution: There are two cases for the initial configuration:
Together, the answer is $2+20=\boxed{22}.$

Answer: C 22
Problem: Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$

Solution: Without loss of generality, we assume the length of each side of $ABCD$ is $2$ .
Because $E$ is the midpoint of $CD$ $CE = 1$
Because $ABCD$ is a rhombus, $\angle BCE = 180^\circ - \angle D$
In $\triangle BCE$ , following from the law of sines, \[ \frac{CE}{\sin \angle FBC} = \frac{BC}{\sin \angle BEC} . \]
We have $\angle BEC = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC$
Hence, \[ \frac{1}{\sin \angle FBC} = \frac{2}{\sin \left( 46^\circ - \angle FBC \right)} . \]
By solving this equation, we get $\tan \angle FBC = \frac{\sin 46^\circ}{2 + \cos 46^\circ}$
Because $AF \perp BF$ \begin{align*} BF & = AB \cos \angle ABF \\ & = 2 \cos \left( 46^\circ - \angle FBC \right) . \end{align*}
In $\triangle BFC$ , following from the law of sines, \[ \frac{BF}{\sin \angle BCF} = \frac{BC}{\sin \angle BFC} . \]
Because $\angle BCF = 180^\circ - \angle BFC - \angle FBC$ , the equation above can be converted as \[ \frac{BF}{\sin \left( \angle BFC + \angle FBC \right)} = \frac{BC}{\sin \angle BFC} . \]
Therefore, \begin{align*} \tan \angle BFC & = \frac{\sin \angle FBC}{\cos \left( 46^\circ - \angle FBC \right) - \cos \angle FBC} \\ & = \frac{1}{\sin 46^\circ - \left( 1 - \cos 46^\circ \right) \cot \angle FBC} \\ & = \frac{\sin 46^\circ}{\cos 46^\circ - 1} \\ & = - \frac{\sin 134^\circ}{1 + \cos 134^\circ} \\ & = - \tan \frac{134^\circ}{2} \\ & = - \tan 67^\circ \\ & = \tan \left( 180^\circ - 67^\circ \right) \\ & = \tan 113^\circ . \end{align*}
Therefore, $\angle BFC = \boxed{113}$

Answer: D 113
Problem: Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$

Solution: Extend segments $\overline{AD}$ and $\overline{BE}$ until they meet at point $G$
Because $\overline{AB} \parallel \overline{ED}$ , we have $\angle ABG = \angle DEG$ and $\angle GDE = \angle GAB$ , so $\triangle ABG \sim \triangle DEG$ by AA.
Because $ABCD$ is a rhombus, $AB = CD = 2DE$ , so $AG = 2GD$ , meaning that $D$ is a midpoint of segment $\overline{AG}$
Now, $\overline{AF} \perp \overline{BE}$ , so $\triangle GFA$ is right and median $FD = AD$
So now, because $ABCD$ is a rhombus, $FD = AD = CD$ . This means that there exists a circle from $D$ with radius $AD$ that passes through $F$ $A$ , and $C$
AG is a diameter of this circle because $\angle AFG=90^\circ$ . This means that $\angle GFC = \angle GAC = \frac{1}{2} \angle GDC$ , so $\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ$ , which means that $\angle BFC = \boxed{113}$

Answer: D 113
Problem: Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$

Solution: Let $\overline{AC}$ meet $\overline{BD}$ at $O$ , then $AOFB$ is cyclic and $\angle FBO = \angle FAO$ . Also, $AC \cdot BO = [ABCD] = 2 \cdot [ABE] = AF \cdot BE$ , so $\frac{AF}{BO} = \frac{AC}{BE}$ , thus $\triangle AFC \sim \triangle BOE$ by SAS, and $\angle OEB = \angle ACF$ , then $\angle CFE = \angle EOC = \angle DAC = 67^\circ$ , and $\angle BFC = \boxed{113}$

Answer: D 113
Problem: Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$

Solution: Observe that all answer choices are close to $112.5 = 90+\frac{45}{2}$ . A quick solve shows that having $\angle D = 90^\circ$ yields $\angle BFC = 135^\circ = 90 + \frac{90}{2}$ , meaning that $\angle BFC$ increases with $\angle D$ . 
Substituting, $\angle BFC = 90 + \frac{46}{2} = \boxed{113}$

Answer: D 113
Problem: Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$

Solution: This solution refers to the Diagram section.
We extend $AD$ and $BE$ to point $G$ , as shown below: [asy] /* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NW,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); [/asy] We know that $AB=AD=2$ and $CE=DE=1$
By AA Similarity, $\triangle ABG \sim \triangle DEG$ with a ratio of $2:1$ . This implies that $2AD=AG$ and $AD \cong DG$ , so $AG=2AD=2\cdot2=4$ . That is, $D$ is the midpoint of $AG$
Now, let's redraw our previous diagram, but construct a circle with radius $AD$ or $2$ centered at $D$ and by extending $CD$ to point $H$ , which is on the circle, as shown below: [asy] /* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NE,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); draw(Circle(D,6),dashed); [/asy] Notice how $F$ and $C$ are on the circle and that $\angle CFE$ intercepts with $\overset{\Large\frown} {CG}$
Let's call $\angle CFE = \theta$
Note that $\angle CDG$ also intercepts $\overset{\Large\frown} {CG}$ , So $\angle CDG = 2\angle CFE$
Let $\angle CDG = 2\theta$ . Notice how $\angle CDG$ and $\angle ADC$ are supplementary to each other. We conclude that \begin{align*} 2\theta &= 180-\angle ADC \\ 2\theta &= 180-46 \\ 2\theta &= 134 \\ \theta &= 67.  \end{align*} Since $\angle BFC=180-\theta$ , we have $\angle BFC=180-67=\boxed{113}$

Answer: D 113
Problem: Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$

Solution: If angle $ADC$ was a right angle, it would be much easier. Thus, first pretend that $ADC$ is a right angle. $ABCD$ is now a square. WLOG, let each of the side lengths be 1. We can use the Pythagorean Theorem to find the length of line $AE$ , which is $\sqrt{5}/2$ . We want the measure of angle $BFC$ , so to work closer to it, we should try finding the length of line $BF$ . Angle $FAB$ and angle $ABF$ are complementary. Angle $ABF$ and angle $FBC$ are also complementary. Thus, $\sin FAB=\cos ABF=\sin FBC$ $\sin FAB=\sin FBC=(1/2)/(\sqrt{5}/2)=1/\sqrt{5}$ . Since $\sin FAB=1\sqrt{5}$ ,and $AB=1$ $FB=\sin FAB$ . It follows now that $FE=3*\sqrt{5}/10$
Now, zoom in on triangle $BEC$ . To use the Law of Cosines on triangle $FBC$ , we need the length of $FC$ . Use the Law of Cosines on triangle $EFC$ . Cos $E=1/\sqrt{5}$ . Thus, after using the Law of Cosines, $FC=\sqrt{2/5}$
Since we now have SSS on $BEC$ , we can get use the Law of Cosines. $\cos BFC=1/-\sqrt{2}$ $\arccos 1/-\sqrt{2}$ is 45, but if the cosine is negative that means that the angle is the supplement of the positive cosine value. $180-45=135$ . Angle $BFC$ is $135^\circ$
Realize that, around point F, there will always be 3 right angles, regardless of what angle $ADC$ is. There are only two angles that change when $ADC$ changes. Break up angle $BFC$ into angle $BFB'$ , which is always 90 degrees, and angle $B'FC$ , which we have discovered to to be half of $ADC$ . Thus, when angle $ADC$ is 46 degrees, then $B'FC$ will be 23. $23+90=113$ . Angle $BFC$ is $\boxed{113}$ degrees.

Answer: D 113
Problem: Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$

Solution: Given that all the answer choices and coefficients are integers, we hope that $P(x)$ has positive integer coefficients.
Throughout this solution, we will express all polynomials in base $x$ . E.g. $x^2 + x + 1 = 111_{x}$
We are given: \[111a + 12 = 101b + 21 = P(x).\] We add $111$ and $101$ to each side and balance respectively: \[111(a - 1) + 123 = 101(b - 1) + 122 = P(x).\] We make the unit's digits equal: \[111(a - 1) + 123 = 101(b - 2) + 223 = P(x).\] We now notice that: \[111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).\] Therefore $a = 11_{x} = x + 1$ $b = 12_{x} = x + 2$ , and $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$ $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$ ‘s digit in $101b + 21$ when $b$ is an integer. The desired sum is $1^2 + 2^2 +3^2+ 3^2 = \boxed{23}$

Answer: E 23
Problem: Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$

Solution: Let $P(x) = Q(x)(x^2+x+1) + x + 2$ , then $P(x) = Q(x)(x^2+1) + xQ(x) + x + 2$ , therefore $xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}$ , or $xQ(x) \equiv x-1 \pmod{x^2+1}$ . Clearly the minimum is when $Q(x) = x+1$ , and expanding gives $P(x) = x^3+2x^2+3x+3$ . Summing the squares of coefficients gives $\boxed{23}$

Answer: E 23
Problem: Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$

Solution: Let $P(x) = (x^2+x+1)Q_1(x) + x + 2$ , 
then $P(x) = (x^2+1)Q_1(x) + xQ_1(x) + x + 2$
Also $P(x) = (x^2+1)Q_2(x) + 2x + 1$
We infer that $Q_1(x)$ and $Q_2(x)$ have same degree, we can assume $Q_1(x) = x + a$ , and $Q_2(x) = x + b$ , since $P(x)$ has least degree. If this cannot work, we will try quadratic, etc.
Then we get: $(x^2+1)(Q_1(x) - Q_2(x)) + xQ_1(x) - x + 1 = 0$
The constant term gives us: $(Q_1(x) - Q_2(x)) + 1 = 0$
So $Q_1(x) - Q_2(x) = -1$
Substituting this in gives: $-(x^2+1) + xQ_1(x) - x + 1 = 0$
Solving this equation, we get $Q_1(x) = x + 1$
Plugging this into our original equation we get $P(x) = x^3 + 2x^2 + 3x + 3$
Verify this works with $P(x) = (x^2+1)Q_2(x) + 2x + 1$
Therefore the answer is $1^2 + 2^2 + 3^2 + 3^2 = \boxed{23}$

Answer: E 23
Problem: Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$

Solution: Notice that we cannot have the quotients equal to some constants, since the same constant will yield different constant terms for $P(x)$ (which is bad) and different constants will yield different first coefficients (also bad). Thus, we try setting the quotients equal to linear terms (for minimizing degree).
Let $P(x)=(x^2+x+1)(ax+b)+(x+2)$ and $P(x)=(x^2+1)(ax+c)+(2x+1)$ . The quotients have the same $x$ coefficient, since $P(x)$ must have the same $x^3$ coefficient in both cases. Expanding, we get \[P(x)=ax^3+(a+b)x^2+(a+b+1)x+(b+2)\] and \[P(x)=ax^3+cx^2+(a+2)x+(c+1).\]
Equating coefficients, we get $b+2=c+1$ $a+b+1=a+2$ , and $a+b=c$ . From the second equation, we get $b=1$ , then substituting into         the first, $c=2$ . Finally, from $a+b=c$ , we have $a=1$ . Now, $P(x)=(x^2+x+1)(ax+b)+(x+2)=(x^2+x+1)(x+1)+(x+2)=x^3+2x^2+3x+3$ and our answer is \[1^2+2^2+3^2+3^2=\boxed{23}.\]

Answer: E 23
Problem: Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$

Solution: We construct the following equations in terms of $P(x)$ and the information given by the problem: \[\textbf{(1) } P(x)=(x^2+x+1)\cdot Q(x)+x+2\] \[\textbf{(2) } P(x)=(x^2+1)\cdot R(x)+2x+1\] Upon inspection, $Q(x)$ and $R(x)$ cannot be constant, so the smallest possible degree of $P(x)$ is $3,$ and both $Q(x)$ and $R(x)$ are linear.
Let $Q(x)=x-q$ and $R(x)=x-r.$ We know there will be values for $q$ and $r$ that make the below equation hold, so we can assume that $P(x)$ has a leading coefficient of $1$
Substituting these values in, and setting $\textbf{(1)}$ and $\textbf{(2)}$ equal to each  other, \[(x^2+x+1)(x-q)+x+2=(x^2+1)(x-r)+2x+1.\] We plug in $x=0$ , yielding $r+1=q.$ Substituting this value into the above equation, \[(x^2+x+1)(x-r-1)+x+2=(x^2+1)(x-r)+2x+1.\] Letting $x=1,$ we conclude that $r=-2,$ so $R(x)=x+2.$ Therefore, \[P(x)=(x^2+1)(x+2)+2x+1 = x^3+2x^3+3x+3.\] The requested sum is \[1^2+2^2+3^2+3^2=\boxed{23}\]

Answer: E 23
Problem: Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$

Solution: By remainder theorem, the polynomial can be written as follows.
\[P(x) = (x^2+x+1)Q_{1}(x)+x+2 = (x^2+1)Q_{2}(x)+2x+1\] This is a timed exam, we can use the information given by answer choices. The answer choices tell us this is the polynomial with integer coefficients, and we need to find the polynomial with the least degree so we can assume both $Q_{1}(x)$ and $Q_{2}(x)$ are linear (the coefficient of x should be same).
Then we can write $P(x)$ as a cubic polynomial.
\[P(x) = (x^2+x+1)(ax+b)+x+2 = (x^2+1)(ax+c)+2x+1\] Substituting $x=0,1,-1$ to determine the value of $a$ and $b$
We have: \[P(0) = b+2 = c+1\] \[P(1) = 3a+3b+3 = 2a+2c+3\] \[P(-1) = -a+b+1 = -2a+2c-1\]
We can solve the simultaneous equations: $a=1,b=1,c=2$
Hence, $P(x)=(x^2+x+1)(x+1)+x+2=x^3+2x^2+3x+3$ . The answer is $1^2+2^2+3^2+3^2=\boxed{23}$

Answer: E 23
Problem: Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$ $x^{2}+y^{2}=64$ , and $(x-5)^{2}+y^{2}=3$ . What is the sum of the areas of all circles in $S$
$\textbf{(A)}~48\pi\qquad\textbf{(B)}~68\pi\qquad\textbf{(C)}~96\pi\qquad\textbf{(D)}~102\pi\qquad\textbf{(E)}~136\pi\qquad$

Solution: We denote by $C_1$ the circle that has the equation $x^2 + y^2 = 4$ .
We denote by $C_2$ the circle that has the equation $x^2 + y^2 = 64$ .
We denote by $C_3$ the circle that has the equation $(x-5)^2 + y^2 = 3$
We denote by $C_0$ a circle that is tangent to $C_1$ $C_2$ and $C_3$ .
We denote by $\left( u, v \right)$ the coordinates of circle $C_0$ , and $r$ the radius of this circle.
From the graphs of circles $C_1$ $C_2$ $C_3$ , we observe that if $C_0$ is tangent to all of them, then $C_0$ must be internally tangent to $C_2$ .
We have \[ u^2 + v^2 = \left( 8 - r \right)^2 . \hspace{1cm} (1) \]
We do the following casework analysis in terms of the whether $C_0$ is externally tangent to $C_1$ and $C_3$
Case 1: $C_0$ is externally tangent to $C_1$ and $C_3$
We have \[ u^2 + v^2 = \left( r + 2 \right)^2   \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 2: $C_1$ is internally tangent to $C_0$ and $C_3$ is externally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r - 2 \right)^2  \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 3: $C_1$ is externally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r + 2 \right)^2  \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 4: $C_1$ is internally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r - 2 \right)^2  \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Because the graph is symmetric with the $x$ -axis, and for each case above, the solution of $v$ is not 0. Hence, in each case, there are two congruent circles whose centers are symmetric through the $x$ -axis.
Therefore, the sum of the areas of all the circles in $S$ is $2\left( 3^2 \pi +5^2 \pi +3^2 \pi +5^2 \pi \right) = \boxed{136}$

Answer: E 136
Problem: Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$

Solution: We have \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| &&(\bigstar) \\ &\leq \frac12\left|\frac12|800-400|\right| \\ &= 100, \end{align*} from which we eliminate answer choices $\textbf{(D)}$ and $\textbf{(E)}.$
Note that \begin{alignat*}{8} |f(800)-f(300)| &\leq \frac12|800-300| &&= 250, \\ |f(800)-f(900)| &\leq \frac12|800-900| &&= 50, \\ |f(400)-f(300)| &\leq \frac12|400-300| &&= 50, \\ |f(400)-f(900)| &\leq \frac12|400-900| &&= 250. \\ \end{alignat*} Let $a=f(300)=f(900).$ Together, it follows that \begin{align*} |f(800)-a|&\leq 50, \\ |f(400)-a|&\leq 50. \\ \end{align*} We rewrite $(\bigstar)$ as \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\ &= \frac12|(f(800)-a)-(f(400)-a)| \\ &\leq \frac12|50-(-50)| \\ &=\boxed{50} ~MRENTHUSIASM

Answer: B 50