Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Call a fraction $\frac{a}{b}$ , not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\  10 \qquad\textbf{(C)}\  11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$

Solution: We split this up into two cases:
Case 1: integer + integer
The whole numbers we have are $\frac{10}{5}$ (or $2$ ), $\frac{12}{3}$ (or $4$ ), and $\frac{14}{1}$ (or $14$ ). There are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Case 2: fraction + fraction
The fractions we have are $\frac{5}{10}$ (or $\frac{1}{2}$ ), $\frac{9}{6}$ (or $\frac{3}{2}$ ), and $\frac{13}{2}$ . Similarly, there are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Thus, $6+6=12$
So now you would just go ahead and innocently choose $\textbf{(D) }12$ , right? No! We overcounted $8$ , as $\frac{13}{2}+\frac96=\frac{12}{3}+\frac{12}{3}=8$ . Therefore, the correct answer is actually $12-1=\boxed{11}$

Answer: C 11
Problem: The greatest prime number that is a divisor of $16384$ is $2$ because $16384 = 2^{14}$ . What is the sum of the digits of the greatest prime number that is a divisor of $16383$
$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$

Solution: We have \begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*}
Since $129$ is composite, $127$ is the largest prime which can divide $16383$ . The sum of $127$ 's digits is $1+2+7=\boxed{10}$

Answer: C 10
Problem: Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$

Solution: Let Alice have the number A, Bob B.
When Alice says that she can't tell who has the larger number, it means that $A$ cannot equal $1$ . Therefore, it makes sense that Bob has $2$ because he now knows that Alice has the larger number. $2$ is also prime. The last statement means that $200+A$ is a perfect square. The three squares in the range $200-300$ are $225$ $256$ , and $289$ . So, $A$ could equal $25$ $56$ , or $89$ , so $A+B$ is $27$ $58$ , or $91$ , of only $\boxed{27}$ is an answer choice.

Answer:  27
Problem: Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$

Solution: Denote by $A$ and $B$ the numbers drawn by Alice and Bob, respectively.
Alice's sentence “I can't tell who has the larger number.” implies $A \in \left\{ 2 , \cdots , 39 \right\}$
Bob's sentence “I know who has the larger number.” implies $B \in \left\{ 1 , 2 , 39, 40 \right\}$
Their subsequent conversation that $B$ is prime implies $B = 2$
Then, Alice's next sentence  “In that case, if I multiply your number by 100 and add my number, the result is a perfect square.” implies $200 + A$ is a perfect square.
Hence, $A = 25$
Therefore, the answer is $\boxed{27}$

Answer: A 27
Problem: Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$

Solution: We see that $225$ is one such square that works. Bob gets $2$ and Alice gets $25$ which is valid. Thus, $2 + 25 = 27.$ So $\boxed{27}$ is our answer.

Answer: A 27
Problem: In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$

Solution: Note that $\triangle APB \cong \triangle BQC.$ Then, it follows that $\overline{PB} \cong \overline{QC}.$ Thus, $QC = PB = PR + RB = 7 + 6 = 13.$ Define $x$ to be the length of side $CR,$ then $RQ = 13-x.$ Because $\overline{BR}$ is the altitude of the triangle, we can use the property that $QR \cdot RC = BR^2.$ Substituting the given lengths, we have \[(13-x) \cdot x = 36.\] Solving, gives $RQ = 4$ and $RC = 9.$ We eliminate the possibility of $x=4$ because $RC > QR.$ Thus, the side length of the square, by Pythagorean Theorem, is \[\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.\] Thus, the area of the square is $(\sqrt{117})^2 = 117,$ so the answer is $\boxed{117}.$

Answer: D 117
Problem: In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$

Solution: As above, note that $\bigtriangleup BPA \cong \bigtriangleup CQB$ , which means that $QC =  13$ . In addition, note that $BR$ is the altitude of a right triangle to its hypotenuse, so $\bigtriangleup BQR \sim \bigtriangleup CBR \sim \bigtriangleup CQB$ . Let the side length of the square be $x$ ; using similarity side ratios of $\bigtriangleup BQR$ to $\bigtriangleup CQB$ , we get \[\frac{6}{x} = \frac{QB}{13} \implies QB \cdot x =  78\] Note that $QB^2 + x^2 = 13^2 = 169$ by the Pythagorean theorem, so we can use the expansion $(a+b)^2 = a^2+2ab+b^2$ to produce two equations and two variables;
\[(QB + x)^2 = QB^2 + 2QB\cdot x + x^2 \implies (QB+x)^2 = 169 + 2 \cdot 78 \implies QB+x = \sqrt{13(13)+13(12)} = \sqrt{13 \cdot 25} = 5\sqrt{13}\] \[(QB-x)^2 = QB^2 - 2QB\cdot x + x^2 \implies (QB - x)^2 = 169 - 2\cdot 78 \implies \pm(QB-x) = \sqrt{13(13) - 13(12)}\]
Since $QB-x$ is negative, it doesn't make sense in the context of this problem, so we go with \[x-QB = \sqrt{13(13) - 13(12)} = \sqrt{13 \cdot 1} = \sqrt{13}\]
We want $x^2$ , so we want to find $x$ . Adding the first equation to the second, we get \[2x = 6\sqrt{13} \implies x = 3\sqrt{13}\]
Then $x^2$ $(3\sqrt{13}^2) = 9 \cdot 13 = 117 = \boxed{117}$

Answer: D 117
Problem: In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$

Solution: We have that $\triangle CRB \sim \triangle BAP.$ Thus, $\frac{\overline{CB}}{\overline{CR}} = \frac{\overline{PB}}{\overline{AB}}$ . Now, let the side length of the square be $s.$ Then, by the Pythagorean theorem, $CR = \sqrt{x^2-36}.$ Plugging all of this information in, we get \[\frac{s}{\sqrt{s^2-36}} = \frac{13}{s}.\] Simplifying gives \[s^2=13\sqrt{s^2-36},\] Squaring both sides gives \[s^4 = 169s^2- 169\cdot 36 \implies s^4-169s^2 + 169\cdot 36 = 0.\] We now set $s^2=t,$ and get the equation $t^2-169t + 169\cdot 36 = 0.$ From here, notice we want to solve for $t$ , as it is precisely $s^2,$ or the area of the square. So we use the Quadratic formula , and though it may seem bashy, we hope for a nice cancellation of terms. \[t = \frac{169\pm\sqrt{169^2-4\cdot 36 \cdot 169}}{2}.\] It seems scary, but factoring $169$ from the square root gives us \[t = \frac{169\pm \sqrt{169 \cdot (169-144)}}{2} = \frac{169 \pm \sqrt{169 \cdot 25}}{2} = \frac{169 \pm 13\cdot 5}{2} = \frac{169\pm 65}{2},\] giving us the solutions \[t=52, 117.\] We instantly see that $t=52$ is way too small to be an area of this square ( $52$ isn't even an answer choice, so you can skip this step if out of time) because then the side length would be $2\sqrt{13}$ and then, even the largest line you can draw inside the square (the diagonal)  is $2\sqrt{26},$ which is less than $13$ (line $PB$ ) And thus, $t$ must be $117$ , and our answer is $\boxed{117}.$ $\blacksquare$

Answer: D 117
Problem: In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$

Solution: Denote $a = RC$ . Now tilt your head to the right and view $R, \overrightarrow{RB}$ and $\overrightarrow{RC}$ as the origin, $x$ -axis and $y$ -axis, respectively. In particular, we have points $B(6,0), C(0,a), P(-7,0)$ . Note that side length of the square $ABCD$ is $BC = \sqrt{a^2 + 36}$ . Also equation of line $BC$ is \[\underbrace{\frac{x}{6} + \frac{y}{a} = 1}_{\text{intercepts form}}     \quad \implies \quad     ax + 6y - 6a = 0.\] Because the distance from $P(-7,0)$ to line $\color[rgb]{0,0.4,0.65}BC: ax + 6y - 6a = 0$ is also the side length $\sqrt{a^2 + 36}$ , we can apply the point-line distance formula to get \[\frac{|a\cdot(-7) + 6 \cdot 0 - 6a|}{\sqrt{a^2 + 36}} = {\sqrt{a^2 + 36}}\] which reduces to $|13a| = a^2 + 36$ . Since $a$ is positive, the last equations factors as $a^2 - 13a + 36 = (a-4)(a-9) = 0$ . Now judging from the figure, we learn that $a > RB = 6$ . So $a = 9$ .
Therefore, the area of the square $ABCD$ is $BC^2 = RC^2 + RB^2 = a^2 + 6^2 = \boxed{117}$

Answer: D 117
Problem: In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$

Solution: Denote $\angle PBA = \alpha$ .
Because $\angle QRB = \angle QBC = 90^\circ$ $\angle BCQ = \alpha$
Hence, $AB = BP \cos \angle PBA = 13 \cos \alpha$ $BC = \frac{BR}{\sin \angle BCQ} = \frac{6}{\sin \alpha}$
Because $ABCD$ is a square, $AB = BC$ .
Hence, $13 \cos \alpha = \frac{6}{\sin \alpha}$
Therefore, \begin{align*} \sin 2 \alpha & = 2 \sin \alpha \cos \alpha \\ & = \frac{12}{13} . \end{align*}
Thus, $\cos 2 \alpha = \pm \frac{5}{13}$
$\textbf{Case 1}$ $\cos 2 \alpha = \frac{5}{13}$
Thus, $\cos \alpha = \sqrt{\frac{1 + \cos 2 \alpha}{2}} = \frac{3}{\sqrt{13}}$
Hence, $AB = 13 \cos \alpha = 3 \sqrt{13}$
Therefore, ${\rm Area} \ ABCD = AB^2 = 117$
$\textbf{Case 2}$ $\cos 2 \alpha = - \frac{5}{13}$
Thus, $\cos \alpha = \sqrt{\frac{1 + \cos 2 \alpha}{2}} = \frac{2}{\sqrt{13}}$
Hence, $AB = 13 \cos \alpha = 2 \sqrt{13}$
However, we observe $BQ = \frac{BR}{\cos \alpha} = 3 \sqrt{13} > AB$ .
Therefore, in this case, point $Q$ is not on the segment $AB$
Therefore, this case is infeasible.
Putting all cases together, the answer is $\boxed{117}$

Answer: D 117
Problem: In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$

Solution: Note that if we connect points $P$ and $C$ , we get a triangle with height $RC$ and length $13$ . This triangle has an area of $\frac {1}{2}$ the square. We can now use answer choices to our advantage!
Answer choice A: If $BC$ was $\sqrt {85}$ $RC$ would be $7$ . The triangle would therefore have an area of $\frac {91}{2}$ which is not  half of the area of the square. Therefore, A is wrong.
Answer choice B: If $BC$ was $\sqrt {93}$ $RC$ would be $\sqrt {57}$ . This is obviously wrong.
Answer choice C: If $BC$ was $10$ , we would have that $RC$ is $8$ . The area of the triangle would be $52$ , which is not half the area of the square. Therefore, C is wrong.
Answer choice D: If $BC$ was $\sqrt {117}$ , that would mean that $RC$ is $9$ . The area of the triangle would therefore be $\frac {117}{2}$ which IS half the area of the square. Therefore, our answer is $\boxed{117}$

Answer: D 117
Problem: Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpositions?
$(\textbf{A})\: 1.6\qquad(\textbf{B}) \: 1.8\qquad(\textbf{C}) \: 2.0\qquad(\textbf{D}) \: 2.2\qquad(\textbf{E}) \: 2.4$

Solution: After the first swap, we do casework on the next swap.
Case 1: Silva swaps the two balls that were just swapped
There is only one way for Silva to do this, and it leaves 5 balls occupying their original position.
Case 2: Silva swaps one ball that has just been swapped with one that hasn't been swapped
There are two ways for Silva to do this, and it leaves 2 balls occupying their original positions.
Case 3: Silva swaps two balls that have not been swapped
There are two ways for Silva to do this, and it leaves 1 ball occupying their original position.
Our answer is the average of all 5 possible swaps, so we get \[\frac{5+2\cdot2+2\cdot1}{5}=\frac{11}5=\boxed{2.2}.\]

Answer: D 2.2
Problem: Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpositions?
$(\textbf{A})\: 1.6\qquad(\textbf{B}) \: 1.8\qquad(\textbf{C}) \: 2.0\qquad(\textbf{D}) \: 2.2\qquad(\textbf{E}) \: 2.4$

Solution: The "expected value" in the question tips us off to this technique. Consider any ball. The probability it returns to the same position is the probability of being swapped twice plus the probability of never being swapped: \[\frac{2}{5} \cdot \frac{1}{5} + \left(\frac{3}{5}\right)^2 = \frac{11}{25}.\] Multiply by 5 for 5 balls to get $\boxed{2.2}.$

Answer: D 2.2
Problem: Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

Solution: First note the useful fact that if $R$ is the circumradius of a dodecagon ( $12$ -gon) the area of the figure is $3R^2.$ If we connect the vertices of the $3$ squares we get a dodecagon. The radius of circumcircle of the dodecagon is simply half the diagonal of the square, which is $3\sqrt{2}.$ Thus the area of the dodecagon is $3 \cdot (3\sqrt{2})^2 = 3 \cdot 18 = 54.$ But, the problem asked for the area of the combined figure which was made from the rotated squares. This area is the area of the dodecagon, which was found, subtracting the $12$ isosceles triangles, which are formed when connecting the vertices of the squares to created the dodecagon. To find this area, we need to know the base of the isosceles triangle, call this $x.$ Then, we can use the Law of Cosines on the triangle that is formed from the two vertices of the square and the center of the square. After computing, we get that $x = 3\sqrt{3} -3.$ Realize that the $12$ isosceles are congruent with an angle measure of $120^{\circ},$ this means that we can create $4$ congruent equilateral triangles with side length of $3\sqrt3 - 3.$ The area of the equilateral triangle is $\frac{\sqrt{3}}{4} \cdot (3\sqrt{3} -3)^2 =  \frac{\sqrt{3}}{4} \cdot (36 - 18\sqrt{3}) = \frac{36\sqrt{3} - 54}{4}.$ Thus, the area of all the twelve small equilateral traingles are $4 \cdot \frac{36\sqrt{3} - 54}{4} = 36\sqrt{3} - 54$ . Thus, the requested area is $54 - (36\sqrt{3} - 54) = 108 - 36\sqrt{3}.$ Thus, $a+b+c = 108 + 36 + 3 = 147,$ so the answer is $\boxed{147}.$

Answer: E 147
Problem: Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

Solution: As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png , all 12 vertices of three squares form a regular dodecagon (12-gon).
Denote by $O$ the center of this dodecagon.
Hence, $\angle AOB = \frac{360^\circ}{12} = 30^\circ$
Because the length of a side of a square is 6, $AO = 3 \sqrt{2}$
Hence, $AB = 2 AO \sin \frac{\angle AOB}{2} = 3 \left( \sqrt{3} - 1 \right)$
We notice that $\angle MAB = \angle MBA = 30^\circ$ .
Hence, $AM = \frac{AB}{2\cos \angle MAB} = 3 - \sqrt{3}$
Therefore, the area of the region that three squares cover is \begin{align*} &  \ [ABCDEFGHIJKL] - 12[MAB] \\ & = 12 [OAB] - 12 [MAB] \\ & = 12 \cdot \frac{1}{2} OA \cdot OB \sin \angle AOB - 12 \cdot \frac{1}{2} MA \cdot MB \sin \angle AMB \\ & = 6 OA^2 \sin \angle AOB - 6 MA^2 \sin \angle AMB \\ & = 108 - 36\sqrt{3} \end{align*}
Therefore, the answer is $\boxed{147}$

Answer: E 147
Problem: Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

Solution: Let $O$ be the center of the polygon, $A$ be the bottom right corner of the first square, $C$ be the next vertex to the left of $A$ , and $M$ be the midpoint between $A$ and $B$ , where $B$ is the bottom left corner of the first square. Note that because there are three $90^{\circ}$ squares separated by $\frac{90^{\circ}}{3} = 30^{\circ}$ , each side of the 24-sided polygon is equal in length, meaning to calculate the area of the whole polygon, we find the area of $\bigtriangleup OAC$ and multiply by 24.
To find $[\bigtriangleup OAC]$ , we already know the height $\overline{OM}$ is the sidelength of the square over $2$ , or $\frac{6}{2}=3$ , so we just need the length of the base $\overline{AC}$ . Notice that $\bigtriangleup OCM$ is a $30-60-90$ triangle since $\angle COM = \frac{360^{\circ}}{12} = 30^{\circ}$ , so $\overline{CM} = \frac{\overline{OM}}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ . Then $\overline{AC} = \overline{AM} - \overline{CM} = \frac{6}{2} - \sqrt{3} = 3 - \sqrt{3}$ , so \begin{align*} & [\bigtriangleup AOC] = \frac{1}{2} \cdot \overline{OM} \cdot \overline{AC} \\ & = \frac{1}{2} (3)(3 - \sqrt{3}) \\ & = \frac{9 - 3\sqrt{3}}{2} \end{align*}
Then the whole area of the polygon is $\frac{9 - 3\sqrt{3}}{2} \cdot 24 = 108 - 36\sqrt{3}$ . The desired solution is then $108 + 36 + 3 = 147$ , so the answer is $\boxed{147}$

Answer: E 147